zBEZZMWo5uM-019|So we'll have 3 moles of gas in the reagents and 3 moles of gas after the reaction goes to completion in the products.
zBEZZMWo5uM-020|So 3 moles of gas on one side, 3 moles of gas on the other side under the same conditions of volume and temperature give us the same pressure.
zBEZZMWo5uM-021|So the total pressure is the same after that reaction occurs.
zBEZZMWo5uM-022|The correct answer here, B.
xVv8aSBCCi4-000|In a galvanic cell, electrons flow from high potential to low potential.
xVv8aSBCCi4-002|And at the copper electrode, copper ions are reduced to copper metal.
xVv8aSBCCi4-003|We can make that reaction go in the opposite direction.
xVv8aSBCCi4-004|If we take a battery or another half cell or set of half cells that have a higher natural potential than the copper zinc.
xVv8aSBCCi4-008|So I'm just going to put everything together in one beaker.
xVv8aSBCCi4-009|A copper electrode, a zinc electrode, connected by an external voltage.
xVv8aSBCCi4-013|So I still have my same designation of the electrodes in electrolysis that I did when I had the galvanic reaction.
xVv8aSBCCi4-014|What I need to get the galvanic reaction to be overcome is a voltage greater than the standard galvanic potential.
xVv8aSBCCi4-018|It's based on the configuration of the cells, and the shape of the electrodes.
xVv8aSBCCi4-020|The current flow, when that occurs, is measured in amps.
xVv8aSBCCi4-021|An amps, one ampere, symbol A, is one Coulomb of charge per second.
xVv8aSBCCi4-022|So I can measure the current flow in amps in an electrolytic cell that forces a galvanic cell to go in reverse.
6Bnu5c_-Bpo-001|I can plot chemical reactions on a reaction profile.
6Bnu5c_-Bpo-002|And when I do, this reaction coordinate tells me the progress of the chemical reaction.
6Bnu5c_-Bpo-003|But this is not a time coordinate.
6Bnu5c_-Bpo-004|This reaction coordinate is the progress of the reaction, which could go in a few seconds or 1,000 years.
6Bnu5c_-Bpo-007|Here, a product building up over time.
6Bnu5c_-Bpo-008|So this part of the plot, as the concentrations are changing, and I'm approaching equilibrium, is the domain of kinetics.
6Bnu5c_-Bpo-014|Here's a reaction proceeding more rapidly.
6Bnu5c_-Bpo-015|The concentration's building up more rapidly, as a function of time, and achieving equilibrium at an earlier time.
9GuHD1DtCPA-000|Let's do a calculation where we estimate the enthalpy of a chemical reaction using bond enthalpies.
9GuHD1DtCPA-001|So the reaction of the formation of benzene from three moles of acetylene.
9GuHD1DtCPA-002|Now, when we do these we're going to use bond enthalpies, so we'll think of all the bonds that are broken and all the bonds that are formed.
9GuHD1DtCPA-003|So the bonds that are broken in the chemical reaction would be all the bonds in the acetylene.
9GuHD1DtCPA-004|And all the bonds formed would be all the bonds in the benzene.
9GuHD1DtCPA-008|I have to break 6 moles of carbon hydrogen bonds, because there's 2 carbon hydrogen bonds and each acetylene.
9GuHD1DtCPA-012|And this is one of the strengths of using bond enthalpies to estimate chemical reactions.
9GuHD1DtCPA-013|Very frequently in chemical reactions it's only one or two bonds that are really changing.
9GuHD1DtCPA-014|So if you know a few bond energies, you can calculate the enthalpy for a lot of reactions just because a simple chemical reaction occurs.
9GuHD1DtCPA-015|In this case, I have alternating double and single bonds.
9GuHD1DtCPA-016|And depending on which table you look at, you might find a delocalized carbon carbon bond.
9GuHD1DtCPA-017|But that's not very common.
9GuHD1DtCPA-018|I actually just remember the double and single bonds.
9GuHD1DtCPA-022|I have to break all these bonds.
9GuHD1DtCPA-023|That's an endothermic step.
9GuHD1DtCPA-024|I put energy in.
9GuHD1DtCPA-025|The amount of energy I have to put in?
9GuHD1DtCPA-026|2,511 kilojoules.
9GuHD1DtCPA-031|So just by knowing a few bond enthalpies, I can calculate enthalpies for a wide variety of chemical reactions.
V5mlJRNZ0m4-000|Let's look at the standard state-free energy difference for a chemical reaction, the atomization of H2.
V5mlJRNZ0m4-001|I can write the chemical reaction H2 molecules breaking down into H2 atoms.
V5mlJRNZ0m4-005|What is the enthalpy difference for this chemical reaction?
V5mlJRNZ0m4-006|Well, I don't really have to go to a table or look anything up because all that is happening here is I'm breaking the hydrogen-hydrogen bond and making hydrogen atoms.
V5mlJRNZ0m4-007|And if you break a bond, that requires energy.
V5mlJRNZ0m4-010|What about delta S?
V5mlJRNZ0m4-011|I can also get delta S without looking at a table or doing a calculation because all I need is the sign of delta S, not the magnitude, just like delta H.
V5mlJRNZ0m4-012|So delta AS is positive for this.
V5mlJRNZ0m4-013|I'm going from one molecule to two atoms.
V5mlJRNZ0m4-014|I'm increasing the number of particles.
V5mlJRNZ0m4-015|When I increase the number of particles, I increase the number of microstates.
V5mlJRNZ0m4-019|And delta S is positive, so T is always positive.
V5mlJRNZ0m4-020|So this negative sign means the slope will be negative.
V5mlJRNZ0m4-022|So this is just a sketch of what the plot of delta G standard versus T looks like for the atomization of hydrogen gas.
ULdcQkZeRlA-000|The reaction of breaking a bond is an interesting one to look at.
ULdcQkZeRlA-002|In order to break it, you have to overcome that, put energy in.
ULdcQkZeRlA-003|It's an endothermic, or up-hill, energetic reaction.
ULdcQkZeRlA-007|So bonded molecules, atoms bonded together in molecules, are downhill from free atoms.
ULdcQkZeRlA-008|And that fits together with our picture of the universe.
ULdcQkZeRlA-009|When we look around us, we find atoms more commonly in molecules bonded together.
ULdcQkZeRlA-010|So that's the down hill, or lower, energy state of molecules is the bonded together state.
ULdcQkZeRlA-012|So this one we can take to the bank.
ULdcQkZeRlA-013|It's always true.
ULdcQkZeRlA-014|If you break a bond, you always have to put energy in.
ULdcQkZeRlA-015|It requires energy, and forming bonds always releases energy.
dQtNrUa5Az0-000|Hi.
dQtNrUa5Az0-001|Today, we're going to talk about atomic structure.
dQtNrUa5Az0-002|Now an atom, as you may know, is the tiniest particle that retains the properties of an element.
dQtNrUa5Az0-003|I have an element here, pure carbon.
dQtNrUa5Az0-004|Now, how can we get down to atoms.
dQtNrUa5Az0-005|Well, you could do a thought experiment or a Gedanken-experiment, as Einstein used to say.
dQtNrUa5Az0-006|That's an experiment that you can't actually do, but you can perform it in your mind to help you think about a concept.
dQtNrUa5Az0-007|So here's a Gedanken-experiment.
dQtNrUa5Az0-008|Take this piece of carbon, cut it in half.
dQtNrUa5Az0-009|And then take that half and cut it in half again.
dQtNrUa5Az0-010|And take that half and cut it in half again.
dQtNrUa5Az0-011|If you keep cutting it in half, eventually you'll get down to a tiny piece of carbon that retains the properties of carbon.
dQtNrUa5Az0-012|If I cut it in half again, I don't have carbon anymore.
dQtNrUa5Az0-013|Now, that tiniest piece that retains the properties of carbon, that would be an atom of carbon.
dQtNrUa5Az0-014|What's an atom composed of?
dQtNrUa5Az0-016|It's positively charged, because it contains protons.
dQtNrUa5Az0-017|Protons contain a positive one electrical charge.
dQtNrUa5Az0-018|Now a proton is very important for the nucleus, because the proton is the single defining factor that determines the identity of the element.
dQtNrUa5Az0-019|If you have one proton in your nucleus, you are a hydrogen atom.
dQtNrUa5Az0-020|If you have six protons, you're a carbon atom.
dQtNrUa5Az0-021|I don't care what else is in the Nucleus, six protons you are carbon.
dQtNrUa5Az0-022|So what else is in the nucleus then?
dQtNrUa5Az0-023|Well, nuclei often contain neutrons.
dQtNrUa5Az0-024|Now neutrons are not charged as their name implies.
dQtNrUa5Az0-025|They're neutral particles.
dQtNrUa5Az0-026|They simply add to the mass of the atom, but they don't change the identity of the element, and that's very interesting.
dQtNrUa5Az0-027|You can have carbon that has different masses.
dQtNrUa5Az0-028|It's just the protons that determine the element carbon, but with different numbers of neutrons, you'll have different masses, all of them carbon.
dQtNrUa5Az0-030|And in the neutral atom, there's an electron for every proton in the nucleus.
dQtNrUa5Az0-031|That is the charges balance out, so you have neutral atoms with an equal number of protons and electrons.
dQtNrUa5Az0-034|We can give it the atomic mass, and we can give it the atomic number.
dQtNrUa5Az0-035|The atomic number is the number of protons in the nucleus.
dQtNrUa5Az0-036|That's what determines the identity of the element.
dQtNrUa5Az0-038|The mass is determined by neutrons plus protons.
dQtNrUa5Az0-039|The electrons don't really contribute to the mass, because electrons are so tiny.
dQtNrUa5Az0-040|They're about one 2000th the mass of a proton or a neutron.
dQtNrUa5Az0-041|Protons and neutrons have about the same mass.
dQtNrUa5Az0-042|So atomic mass given by the sun, atomic number determining the element of the nuclei-- of the element.
dQtNrUa5Az0-043|The atomic number, we actually often don't write down with a symbol, because the symbol and the atomic number are redundant information.
dQtNrUa5Az0-044|If you know the symbol, you know the number of protons in the nucleus.
dQtNrUa5Az0-045|If you know the number of protons in the nucleus, you know the symbol or the nature of that element.
dQtNrUa5Az0-046|So you'll often see an element written as its symbol with the atomic mass.
dQtNrUa5Az0-047|The atomic mass is not unique, because the number of neutrons doesn't have to be unique.
dQtNrUa5Az0-049|All of those are hydrogen, but they're different isotopes of hydrogen. They have the same number of protons with a different number of neutrons.
dQtNrUa5Az0-050|And we can look at the isotopes of various elements on the periodic table.
aSpt0ct0aKc-000|We define the pH scale to help us keep track of the concentration of acids and bases in solution.
aSpt0ct0aKc-001|So pX stands for minus log base 10 of the concentration of X.
aSpt0ct0aKc-004|So if you have a number like 10 to the minus 10, minus log of 10 to the minus 10 is 10.
aSpt0ct0aKc-005|So you can just-- instead of looking at 10 to the minus 10, you can say, what's the magnitude, the exponent?
aSpt0ct0aKc-009|So the pH of pure water minus log of 10 to the minus 7th-- 7.
aSpt0ct0aKc-012|So if H3O+ goes up, OH- must go down.
aSpt0ct0aKc-013|They're locked in synchrony with each other by the equilibrium of the autodissociation of water.
aSpt0ct0aKc-020|So we have a scale based essentially on the exponent of the H3O+ concentration.
aSpt0ct0aKc-022|That's the pH scale.
qZxrftuYRRQ-000|Let's look at a chemical reaction here, the dimerization of NO2 to form N2O4.
qZxrftuYRRQ-001|And if I tell you the reaction is spontaneous at 25 degrees C, what can you predict about the enthalpy change for that chemical reaction.
qZxrftuYRRQ-010|We're looking at the dimerization of NO2 to form N2O4.
qZxrftuYRRQ-011|And we're told it's spontaneous.
qZxrftuYRRQ-012|That is, delta G is negative at 25 degrees C.
qZxrftuYRRQ-013|When we look at this reaction, we can predict a couple of things about the entropy and the enthalpy.
qZxrftuYRRQ-014|The entropy, for instance, in this case, decreases.
qZxrftuYRRQ-015|And how do I know that?
qZxrftuYRRQ-016|Well, the number of particles decrease.
qZxrftuYRRQ-017|And if the number of particles decrease, the number of accessible microstates has to decrease as well.
qZxrftuYRRQ-021|So that means this is a positive contribution.
qZxrftuYRRQ-022|Minus T delta S is a positive contribution to delta G.
qZxrftuYRRQ-023|So this doesn't favor.
qZxrftuYRRQ-024|Entropically, this is not favored.
qZxrftuYRRQ-025|The entropic contribution tends to make delta G positive.
qZxrftuYRRQ-026|So we have to have an enthalpic contribution that's negative.
qZxrftuYRRQ-031|Now, I should point out that you probably could have predicted that anyway.
qZxrftuYRRQ-032|Because NO2 going to N2O4, the only thing that's happening here is the formation of a bond.
qZxrftuYRRQ-033|And NO2 monomers become N2O4 dimers.
qZxrftuYRRQ-034|A single bond is formed.
qZxrftuYRRQ-035|And if all that's happened is a bond formation, then that must release energy.
qZxrftuYRRQ-036|Forming bonds always releases energy.
qZxrftuYRRQ-037|Breaking bonds always requires energy.
qZxrftuYRRQ-038|So you could have approached this from two directions.
qZxrftuYRRQ-039|In either case, you get an exothermic chemical reaction.
IHPOtneICaY-000|Let's look at a situation where we take a strong acid, HCl add pH 3 and dilute that by a factor of 10.
IHPOtneICaY-001|What is the new pH?
IHPOtneICaY-002|So an HCl solution, I add water, dilute it by a factor of 10.
IHPOtneICaY-011|We're talking about taking HCl, a strong acid, add pH 3, and diluting it by a factor of 10.
IHPOtneICaY-012|Now, strong acids are particularly easy to work with and make calculations for because they completely dissociate.
IHPOtneICaY-015|Now, you add water and dilute by a factor of 10, that means the H3O plus concentration goes from 10 to the minus 3 to 10 to the minus 4.
IHPOtneICaY-016|A simple tenfold dilution.
IHPOtneICaY-017|Calculate the pH now, well, the pH is 4.
IHPOtneICaY-018|Minus log of 10 to the minus 4.
IHPOtneICaY-019|The correct answer here is the pH changes from 3 to 4.
wrgnaV-cVrs-000|If we look at ionic bonding, we have to transfer an electron from one element to another element.
wrgnaV-cVrs-001|Sodium chloride, for instance, when that's formed, sodium gives up an electron and gives it to chlorine.
wrgnaV-cVrs-003|So let's look at that.
wrgnaV-cVrs-004|Here's the ionization energy for sodium, 496 kilojoules per mole.
wrgnaV-cVrs-005|I have to put in 496 kilojoules per mole to pull a mole of electrons off a mole of sodium atoms.
wrgnaV-cVrs-006|When I add a mole of electrons to a mole of chlorine atoms, what I get is 349 kilojoules released.
wrgnaV-cVrs-008|So why does sodium chloride form?
wrgnaV-cVrs-009|Pulling off the electron costs more than adding it to the chlorine.
wrgnaV-cVrs-010|Well, obviously there's another component.
wrgnaV-cVrs-011|There's several components involved.
wrgnaV-cVrs-013|And that's a tremendously stable, strong interaction.
wrgnaV-cVrs-014|And the formation of that lattice releases a tremendous amount of energy.
wrgnaV-cVrs-015|So that extra release of energy, the formation of the ionic bond is what gives you enough energy to overcome the ionization energy of the sodium.
wrgnaV-cVrs-016|In fact, the formation of sodium chloride is a spectacular reaction, and we'll look at it in the demonstration lab.
wrgnaV-cVrs-017|We'll see metallic sodium and gaseous chlorine dramatically form sodium chloride.
wrgnaV-cVrs-019|Sodium and chlorine form an ionic bond due mainly to the fact that the coulombic interaction is so strong.
jHbuu287uxc-000|The natural direction of processes in the universe is determined by entropy changes.
jHbuu287uxc-001|The natural direction is to increase the entropy of the universe.
jHbuu287uxc-002|That is, to disperse energy over as many microstates as possible.
jHbuu287uxc-003|But practically, it's difficult to measure the number of microstates.
jHbuu287uxc-004|So what we need is a thermodynamic parameter, entropy, but related to a parameter that we can measure.
jHbuu287uxc-005|And it turns out heat is that parameter that we can measure.
jHbuu287uxc-006|And when you think about heat and work, you would probably associate entropy with heat over work, and why is that?
jHbuu287uxc-007|Well, work is a concerted action of particles, all moving in the same direction to compress something.
jHbuu287uxc-008|There's relatively few microstates involved with all the particles moving in the same direction.
jHbuu287uxc-009|Where heat is the random motion of particles, and there's many microstates involved in the random motion.
jHbuu287uxc-010|So heat is the more naturally associated parameter for entropy.
jHbuu287uxc-011|And you can think of that-- here's a ball bouncing.
jHbuu287uxc-014|Slight raise of temperature, more random motion of the molecules in the floor.
jHbuu287uxc-018|So balls don't spontaneously bounce because that concerted motion is a low entropy situation.
jHbuu287uxc-019|Now, how do we measure this heat and entropy?
jHbuu287uxc-020|Well, it's very closely correlated.
jHbuu287uxc-021|The entropy change is given by the heat involved in a system over the temperature.
jHbuu287uxc-023|Heat always goes from a hotter system to a cooler system.
jHbuu287uxc-024|In fact, some people call that the second law of thermodynamics-- heat always moves from hot to cool.
jHbuu287uxc-025|Now, you can never get two systems that are at the same temperature to have all of the heat spontaneously move to one side.
jHbuu287uxc-026|That would be like all the particles moving to one side of a container and leaving a vacuum on the other side.
jHbuu287uxc-027|That's not the spontaneous direction, not the favored direction of the universe.
jHbuu287uxc-031|You can see already-- this entropy change is smaller than this entropy change.
jHbuu287uxc-032|This has a high temperature.
jHbuu287uxc-033|So this negative entropy, this decrease in entropy is small.
jHbuu287uxc-034|This increase in entropy, because the temperature is lower, is large.
jHbuu287uxc-035|So the heat that moves has a bigger effect on the entropy of the cold system than the hot system.
jHbuu287uxc-036|And there's kind of an analogy.
jHbuu287uxc-037|Think about it-- if you were to sneeze in a crowded room, a hot, noisy room, that sneeze wouldn't have much effect.
jHbuu287uxc-038|But think about that same sneeze in an absolutely quiet, cool room.
jHbuu287uxc-039|Then there'd be a big effect of that sneeze.
jHbuu287uxc-040|So you're seeing that same thing.
jHbuu287uxc-041|The same amount of heat transferring have a bigger effect on the entropy on the cold side than the hot side.
jHbuu287uxc-042|So if you add those two, you find that the overall entropy increases.
jHbuu287uxc-043|Now, what if the temperature changes?
jHbuu287uxc-049|So the temperature change and the heat capacity determine the entropy change if the temperature changes.
jHbuu287uxc-050|In fact, it's very strongly related to the heat capacity of the system.
jHbuu287uxc-051|It determines how the entropy changes in the system.
jHbuu287uxc-053|Delta-S, q-reversible over T gives us a way to measure heat and calculate entropy.
jHbuu287uxc-054|Much easier than taking microstates and trying to count them.
jHbuu287uxc-055|The statistical and thermodynamic versions of entropy, turns out the thermodynamic one-- much easier to measure.
6EOSsIa9wDg-000|In oxidation reduction chemistry, we tabulate standard reduction potentials for half cells.
6EOSsIa9wDg-001|And we tabulate them relative to the standard hydrogen electrode.
6EOSsIa9wDg-002|We arbitrarily set that electrode to zero, and we measure all our potentials relative to that one.
6EOSsIa9wDg-006|We can look at the top one here is permanganate being reduced.
6EOSsIa9wDg-007|Now, when permanganate is reduced, that has a high reduction potential.
6EOSsIa9wDg-010|So as it is being reduced, as this reduction occurs, the permanganate is oxidizing something else.
6EOSsIa9wDg-011|That is, when you pull electrons off and become reduced, you had to oxidize something.
6EOSsIa9wDg-012|The process of pulling electrons off something is an oxidation.
6EOSsIa9wDg-013|So high reduction potential means you have a strong ability to oxidize.
6EOSsIa9wDg-014|So the permanganate ion can oxidize any of the products here listed below.
6EOSsIa9wDg-015|And in fact, all these products and the products in the lower half of our table as well.
6EOSsIa9wDg-020|Let's look at the lower half of the table.
6EOSsIa9wDg-021|We can also look at those reduction potentials lower than the standard hydrogen electrode.
6EOSsIa9wDg-024|In contrast, it means calcium metal is a strong reducing agent.
6EOSsIa9wDg-025|That is the preferred direction is essentially the reverse direction, to give up electrons and form calcium ions.
6EOSsIa9wDg-026|If you give up your electrons, something else must accept them, and that thing that accepts them becomes reduced.
6EOSsIa9wDg-027|Its oxidation number is reduced by accepting negatively charged electrons.
3ofEcJm3U7Y-000|Now let's see how good your memory is.
3ofEcJm3U7Y-001|We're going to look at those three atoms and see if we can choose which will have this ionization profile.
3ofEcJm3U7Y-002|Remember, when you ionize an electron, if you give excess photon energy, that will go into kinetic energy of the ejected electron.
3ofEcJm3U7Y-011|So we're testing your memory a little bit.
3ofEcJm3U7Y-013|You can have some transitions in the visible, but if you're going to ionize, that's going to be in the ultraviolet.
3ofEcJm3U7Y-014|And among these three, we said hydrogen will be the most difficult to ionize.
3ofEcJm3U7Y-015|They'll all behave exactly like hydrogen because we saw that their effective nuclear charges are about 1.
3ofEcJm3U7Y-019|So they would come here at lower frequency thresholds than the ultraviolet.
yMvYRVCIEfo-000|One situation where the thermodynamic definition of entropy that is the heat evolved at a constant temperature is particularly appealing our phase changes.
yMvYRVCIEfo-001|That's because phase changes occur at a fixed temperature, and there's a specific heat involved, the enthalpy of either vaporization or melting.
yMvYRVCIEfo-003|And you might expect that regardless of the liquid that this entropy of vaporization is going to be about the same.
yMvYRVCIEfo-005|So you're going to expect an increase in entropy, and it will be fairly independent of the nature of the particle.
yMvYRVCIEfo-006|And that's what you find.
yMvYRVCIEfo-007|If you have a higher enthalpy of vaporization you boil at a higher temperature.
UNmXHX7WfWs-000|Now let's talk about electrons grouping around a nucleus.
UNmXHX7WfWs-001|So they have to occupy many of the orbitals, and we've already seen when they do that.
UNmXHX7WfWs-002|They occupy the orbitals in regular ways.
UNmXHX7WfWs-003|There's rules that say, no two electrons can have the same four quantum numbers.
UNmXHX7WfWs-004|N l m sub l and m sub s, at least one must be different for each electron.
UNmXHX7WfWs-005|And we've seen they always go in spin parallel to degenerate orbitals.
UNmXHX7WfWs-006|But another effect occurs, and that is low energy electrons will shield the higher energy electrons.
UNmXHX7WfWs-008|You can't solve it exactly like you can the two body problem, one nucleus and one electron.
UNmXHX7WfWs-009|So what we do is we take the one nucleus one electron problem, and we say, let that one electron define all the orbitals.
UNmXHX7WfWs-010|And then when we start putting more electrons in, they'll serve as just a slight perturbation to that one electron system.
UNmXHX7WfWs-011|It's easier than solving all the mathematics over again.
UNmXHX7WfWs-013|Now, s electrons are very good shielders.
UNmXHX7WfWs-014|That is, they can shield some of the nuclear charge from outer electrons.
UNmXHX7WfWs-015|So here's a big nuclear positive charge.
UNmXHX7WfWs-016|An s electron has no node at the nucleus.
UNmXHX7WfWs-017|That is, the orbitals are aligned such that the electron has access to the nucleus.
UNmXHX7WfWs-018|If you're a p electron or a d electron, if l is one or two, then you have an angular node at the nucleus.
UNmXHX7WfWs-019|You're forbidden mathematically to exist at the nucleus.
UNmXHX7WfWs-022|Now, what do we mean by shielding?
UNmXHX7WfWs-023|Shielding means I remove some of the effective nuclear charge.
UNmXHX7WfWs-025|And it'll seem more like maybe just a plus one-ish effective nuclear charge.
UNmXHX7WfWs-026|So if it sees a lower nuclear charge, it's easier to ionize.
UNmXHX7WfWs-027|Its energy state is raised, it's closer to that zero ionized state.
UNmXHX7WfWs-028|Let's look at that in more detail.
UNmXHX7WfWs-029|Here I've plotted the one s, two s, three s, four s energy levels.
UNmXHX7WfWs-030|L equals zero.
UNmXHX7WfWs-031|Very good shielding electrons, because l equals 0 means s, no node at the nucleus.
UNmXHX7WfWs-033|These levels are raised slightly because of the shielding effect of the s electron.
UNmXHX7WfWs-034|And that effect continues.
UNmXHX7WfWs-036|And the effect is pretty dramatic.
UNmXHX7WfWs-039|S electrons that have this inner state don't shield each other very well.
UNmXHX7WfWs-040|All s electrons have equal access to the nucleus.
UNmXHX7WfWs-042|But s electrons are great shielders of p and d electrons.
UNmXHX7WfWs-043|So you get two p energies that are higher than 2 s, and you get d energies that are even higher than four s.
idWS5CF3e8U-000|We measure heat transfer using a technique called "calorimetry." Calorimetry revolves around the conservation of energy.
idWS5CF3e8U-001|That is, every joule of heat lost by a system is absorbed by the surroundings.
idWS5CF3e8U-002|Or, every joule of heat absorbed by the system comes from the surroundings.
idWS5CF3e8U-007|Now, what causes heat capacities?
idWS5CF3e8U-008|Well, if you add heat to a system and raise its temperature, what you're doing is changing the kinetic energies of the particles.
idWS5CF3e8U-009|You can always associate higher temperatures with more molecular motion.
idWS5CF3e8U-010|So, higher temperature, and I've absorbed some heat, means that heat went into increasing the kinetic energy of those particles.
idWS5CF3e8U-015|More molecular motion is higher temperature.
idWS5CF3e8U-018|That is, I'm getting more motion for every joule of heat that I add.
idWS5CF3e8U-019|Well, what if I don't get more relative motion?
idWS5CF3e8U-023|I completely overcome the hydrogen bonding interactions in liquid water and add energy to do that transition.
idWS5CF3e8U-024|Now, there I'm adding energy, and I'm getting a phase transition, rather than an increase in kinetic energy.
idWS5CF3e8U-025|That's a potential energy contribution to the heat capacity.
idWS5CF3e8U-026|Or I could have intramolecular bonds breaking or forming.
W_ngfrFO7f0-000|Let's look at a sequential series of reactions, A going to B going to C.
W_ngfrFO7f0-008|We're talking about a 2-step reaction.
W_ngfrFO7f0-009|A goes to B and then B goes to C.
W_ngfrFO7f0-010|The enthalpy for the first step, delta H1, the enthalpy for the second step, delta H2.
W_ngfrFO7f0-013|That is, I put in energy here to break those bonds but then I get all of that back because I form the same bonds.
W_ngfrFO7f0-014|So the sequential reaction A going to C, delta H1 plus delta H2 would be the enthalpy for that reaction.
W_ngfrFO7f0-015|Those are equal and opposite.
W_ngfrFO7f0-016|So they add to zero.
W_ngfrFO7f0-017|So this reaction has an enthalpy zero.
RCWYbfXHxXw-000|Let's look at a sample of sulfur atoms in some form.
RCWYbfXHxXw-001|And we're going to trap them in 22 liters, about a quarter atmosphere of pressure, and about 546 kelvin.
RCWYbfXHxXw-002|The question I have for you is, what's the form of sulfur?
RCWYbfXHxXw-010|We're talking about a mole of sulfur particles.
RCWYbfXHxXw-013|But I didn't have 1 atmosphere of pressure in my sample.
RCWYbfXHxXw-014|I only had a quarter atmosphere of pressure.
RCWYbfXHxXw-015|Well, how can you have that mole of particles give you a quarter atmosphere of pressure?
RCWYbfXHxXw-017|You have to have fewer particles.
RCWYbfXHxXw-019|That would give me 1/4 the particles and 1/4 the pressure.
RCWYbfXHxXw-020|Now, in this sample, the temperature was also twice the standard temperature.
RCWYbfXHxXw-021|How do I get a factor of two in temperature and stay at a quarter of the pressure in 22 liters?
RCWYbfXHxXw-025|So those particles must be in groups of eight.
RCWYbfXHxXw-026|The correct answer here, S8.
4kxMdkzCJGA-001|Those energy levels will be one to a wave function-- energy level 1, wave function number 1, energy level 2, wave function number 2.
4kxMdkzCJGA-002|They might be the same energy, that is, wave function 3 may have the same energy as wave function number 2.
4kxMdkzCJGA-003|But they'll all have different energies associated with different sets of quantum numbers.
4kxMdkzCJGA-004|So let's look at those quantum numbers and the energy levels for an atom.
4kxMdkzCJGA-005|The principle quantum number n again runs just as the integers-- 1, 2, 3.
4kxMdkzCJGA-006|And it gives you the overall energy, as we said, just like in a particle, a box, of the hydrogen wave functions.
4kxMdkzCJGA-007|So the wave functions energies that actually look quite simple-- Z squared-- that's the charge on the nucleus.
4kxMdkzCJGA-008|So for a hydrogen atom, that's one proton, so that charge is 1; n, the quantum number; and a constant.
4kxMdkzCJGA-009|This is Rydberg's constant.
4kxMdkzCJGA-010|It has a value that we can calculate and a value that we tabulate.
4kxMdkzCJGA-011|That value is 2.18 times 10 to the minus 18th joules.
4kxMdkzCJGA-013|So how does this look?
4kxMdkzCJGA-021|Why do we do this negative energy level?
4kxMdkzCJGA-022|Well, whenever you have energy, you have to give your energy relative to something.
4kxMdkzCJGA-023|And the relative to something, we say, for atoms is the electron and the nucleus infinitely far apart and at rest.
4kxMdkzCJGA-024|So that's the 0 in energy.
4kxMdkzCJGA-025|When you bring the nucleus and the electron closer together, they become more stable.
4kxMdkzCJGA-026|They go down in energy.
4kxMdkzCJGA-027|So when you're at 0, the only way to go down in energy is to go to negative energy values.
4kxMdkzCJGA-028|So we're going to use a negative system to describe this.
4kxMdkzCJGA-029|If we bring in n equal 2, we still have 1, the charge of the nucleus.
4kxMdkzCJGA-030|We're still talking about a hydrogen atom.
4kxMdkzCJGA-031|The charge of the nucleus is still 1.
4kxMdkzCJGA-032|n is now 2.
4kxMdkzCJGA-038|And of course, you don't even have to go to infinite to get ionized.
4kxMdkzCJGA-039|But the high levels of n are systems where the energy is about 0.
4kxMdkzCJGA-040|And since it's 1 over n squared, you don't have to go to a very big n for this energy to get very, very small and close to 0 very rapidly.
4kxMdkzCJGA-041|The ionized state is the 0 energy state for the hydrogen atom.
Li26UtxovcM-000|Let's look at what happens when you burn a hydrocarbon in oxygen.
Li26UtxovcM-001|A hydrocarbon is a compound composed of hydrogen and carbon only.
Li26UtxovcM-002|Now I've written it here generically as HC reacting with O2, oxygen molecules.
Li26UtxovcM-003|When that occurs, you can take the mass spectrum of the product.
Li26UtxovcM-005|And they'll have distinct mass spectrum signatures.
Li26UtxovcM-006|The water, mass 18, and the carbon dioxide at mass 44.
Li26UtxovcM-007|So the combustion of hydrocarbons gives a very distinctive mass spectrum of the products.
UdINHPWJRYc-000|Let's apply the first law of thermodynamics to an actual process.
UdINHPWJRYc-001|Here's an ideal gas expanding isothermally-- that is no temperature change-- against a vacuum.
UdINHPWJRYc-002|What's the heat flow?
UdINHPWJRYc-003|Is it less than 0, equal to 0, or greater than 0?
UdINHPWJRYc-013|We're talking about an ideal gas expanding against a vacuum isothermally.
UdINHPWJRYc-014|And whenever you hear isothermal for an ideal gas, you have to be excited, because if it's isothermal, the temperature didn't change.
UdINHPWJRYc-015|Isothermal, delta T is 0.
UdINHPWJRYc-016|If the temperature doesn't change, the energy of the gas could not have changed.
UdINHPWJRYc-021|One will be a negative sign.
UdINHPWJRYc-025|So no heat flows in this process and no work is done.
UdINHPWJRYc-026|The correct answer here, B, the heat flow is 0.
Mbq2QNVY10k-000|For higher steric numbers, we hybridize with some of the d orbitals.
Mbq2QNVY10k-001|So which energy diagram for phosphorous atomic orbitals in PCl5 is true after hybridization?
Mbq2QNVY10k-002|Here are the standard atomic orbitals for phosphorus.
Mbq2QNVY10k-011|We're talking about the hybridization in PCl5.
Mbq2QNVY10k-012|In PCl5 the phosphorus has Steric Number 5 as to accommodate five chlorines.
Mbq2QNVY10k-013|So we need five equivalent atomic orbitals.
Mbq2QNVY10k-014|The five equivalent atomic orbitals are going to come from hybridizing the standard atomic orbitals.
Mbq2QNVY10k-018|So the best answer here is A-- four unchanged d orbitals and five equivalent dsp3's.
8_ZRFjqe6F4-000|When you talk about ideal gases, we can take a volume and a temperature and a number of moles and calculate a pressure.
8_ZRFjqe6F4-002|Now, it's interesting to note that the ideal gas law doesn't say, well, what kind of particle is it?
8_ZRFjqe6F4-003|If it's a hydrogen particle H2, it will exert the same pressure as a bromine particles, so BR2.
8_ZRFjqe6F4-005|And that's true of gases that behave ideally.
8_ZRFjqe6F4-006|The pressure is independent of the nature of the particle.
8_ZRFjqe6F4-007|So why is that?
8_ZRFjqe6F4-008|Well, we can talk about how pressure originates from first principles.
8_ZRFjqe6F4-010|A transfer of momentum is a force applied to the walls of the container, and force per unit area is pressure.
8_ZRFjqe6F4-012|The bromine does it by coming in with a heavy mass and striking the particle, striking the walls of the container.
8_ZRFjqe6F4-013|The hydrogen particle, though much lighter, just moves very much faster.
8_ZRFjqe6F4-014|So even though it has a lower mass, it has a higher velocity.
8_ZRFjqe6F4-015|And it can transfer that same momentum in tiny little hits.
8_ZRFjqe6F4-016|So two different particles can exert the same pressure on the wall by momentum transfer.
8_ZRFjqe6F4-018|Either way, I can exert the same pressure on the walls of the container.
Hy9dsigzTiE-000|Let's look at two gases dissolved in a liquid, helium and nitrogen, aqueous solution for both of them.
Hy9dsigzTiE-001|Let's say they're both at equal concentrations in the solution.
Hy9dsigzTiE-002|What does that mean about the partial pressure above the solution when that is at equilibrium?
Hy9dsigzTiE-010|We're talking about helium and nitrogen dissolved in solution.
Hy9dsigzTiE-011|Their aqueous concentrations are the same.
Hy9dsigzTiE-012|So what does that mean about their partial pressures above the solution?
Hy9dsigzTiE-013|Well, to solve this, we need to look at the equilibrium constants, the Henry's law constants for these gases.
Hy9dsigzTiE-015|Here's helium, 0.4, and nitrogen 0.7.
Hy9dsigzTiE-020|The correct answer here is helium partial pressure is greater than nitrogen.
Bl-zUmeYj74-000|Let's talk more about the properties of electrons.
Bl-zUmeYj74-001|So far we've talked about one electron systems, a hydrogen atom with only one electron.
Bl-zUmeYj74-002|We need to add more electrons to get to the rest of the elements on the periodic table.
Bl-zUmeYj74-003|So we need to talk about groups of electrons.
Bl-zUmeYj74-004|And electrons have special properties.
Bl-zUmeYj74-005|So let's look at some of them.
Bl-zUmeYj74-006|Electrons have a property called spin.
Bl-zUmeYj74-007|Now, it's called spin for historical reasons because it behaves like a spinning charge.
Bl-zUmeYj74-008|Electrons are charged, and if they spun around their axis, they would create a tiny magnetic field.
Bl-zUmeYj74-009|Now, electrons undoubtedly don't spin about the axis.
Bl-zUmeYj74-010|They might not even have an axis.
Bl-zUmeYj74-011|No one has ever seen an electron.
Bl-zUmeYj74-012|It's one of the interesting parts of science.
Bl-zUmeYj74-013|No one has ever seen one of those things.
Bl-zUmeYj74-014|We don't know what it looks like.
Bl-zUmeYj74-015|But the electron has a property that behaves like it's spinning around an axis.
Bl-zUmeYj74-016|Remember when we had photons, and they behaved like they had momentum?
Bl-zUmeYj74-017|The photons don't have mass, so they can't really have an MV momentum, but they have a property that behaves just like momentum, that photon energy.
Bl-zUmeYj74-018|Same thing with electrons.
Bl-zUmeYj74-019|They probably don't spin about an axis, but they have a property that behaves just like an electric charge spinning about its axis.
Bl-zUmeYj74-020|So if a electric charge spins about an axis, it acts like a little magnet.
Bl-zUmeYj74-021|So here I have a little magnet.
Bl-zUmeYj74-022|Let's look at that.
Bl-zUmeYj74-024|So here's a little bar magnet.
Bl-zUmeYj74-025|If it's in a magnetic field or not in a magnetic field, depending on when I bring this magnet in, that will determine the properties of this little bar magnet.
Bl-zUmeYj74-026|Let's take the magnet away right now.
Bl-zUmeYj74-027|It's in no magnetic field.
Bl-zUmeYj74-028|So it can point in any direction in space.
Bl-zUmeYj74-030|Just like an electron, it has a preferred direction in a magnetic field.
Bl-zUmeYj74-031|Here's a low-energy direction aligned with the magnetic field.
Bl-zUmeYj74-032|If I push it away from that low-energy direction, I have to put energy in.
Bl-zUmeYj74-033|And if I release my finger, it'll go back to the low-energy direction.
Bl-zUmeYj74-034|This little bar magnet can point in any direction with respect to the field.
Bl-zUmeYj74-035|I can give it any amount of energy and have it spin back to its preferred direction in the magnetic field.
Bl-zUmeYj74-036|For electrons, that direction in the magnetic field is quantized.
Bl-zUmeYj74-037|It can only be spin up with the magnetic field or spin down against the magnetic field, quantized energy levels, two of them.
Bl-zUmeYj74-038|And we're used to quantization now for tiny particles.
Bl-zUmeYj74-039|So we understand electron magnetic field can't point in any direction.
Bl-zUmeYj74-040|It can point spin up or spin down.
Bl-zUmeYj74-041|Let's look at an experiment that manifests that.
Bl-zUmeYj74-042|If you take electrons and you send them through an inhomogeneous magnetic field, they will separate based on their spin.
Bl-zUmeYj74-043|Out here, where there's no magnetic field, they can either be spin up or spin down.
Bl-zUmeYj74-044|As they pass through the magnetic field, that will sort them based on whether they're spin up or spin down.
Bl-zUmeYj74-045|Now, if they could point anywhere with the magnetic field, you'd expect them to go through the magnetic field and just end up any old place.
Bl-zUmeYj74-046|The ones that were pointing slightly up would go high.
Bl-zUmeYj74-047|The ones that were pointing slightly down would go low.
Bl-zUmeYj74-048|The ones that were pointing in the middle would go straight.
Bl-zUmeYj74-049|But we know that the electron spin is quantized.
Bl-zUmeYj74-050|It can only be spin up or spin down.
Bl-zUmeYj74-051|So when you do this experiment, you get a group of electrons, the ones that were spin up and give you a bright spot here.
Bl-zUmeYj74-052|The ones that were spin down give you a bright spot here.
Bl-zUmeYj74-053|Two possible orientations of the magnetic field for electrons.
Bl-zUmeYj74-054|So we're going to call this property of the electrons.
Bl-zUmeYj74-058|Plus 1/2 and minus 1/2 will be the spin values for the electron spin.
Bl-zUmeYj74-059|So that's another quantum number.
Bl-zUmeYj74-060|So we have n, l, m sub l that defined the orbital.
Bl-zUmeYj74-061|And now we have m sub s, which defines the spin of the electron.
Bl-zUmeYj74-062|So four possible quantum numbers now can describe an electron about an atom.
Bl-zUmeYj74-063|N, l, m sub l, and m sub s.
Bl-zUmeYj74-064|Now, it's interesting.
Bl-zUmeYj74-065|This quantum number has a half integer value.
Bl-zUmeYj74-066|All the other quantum numbers have this integer value of 0, 1, 2, minus 1, minus 2, etc.
Bl-zUmeYj74-067|Now we have a half integer value.
Bl-zUmeYj74-068|And you may wonder why that is.
Bl-zUmeYj74-069|Well, it has to do with the strange properties of electrons.
Bl-zUmeYj74-070|Electrons are crazy little particles.
Bl-zUmeYj74-071|Like I said, no one has ever seen one, so we don't know what they look like.
Bl-zUmeYj74-072|But they have strange properties that don't behave like normal particles.
Bl-zUmeYj74-074|And that's kind of odd right?
Bl-zUmeYj74-075|Usually expect something-- if you say, I'll take this and I'll go around in a closed loop, it should look the way it looked when it started.
Bl-zUmeYj74-076|Electrons don't behave that way.
Bl-zUmeYj74-077|They have a twist in their space relative to our space.
Bl-zUmeYj74-078|And they act kind of like this device here.
Bl-zUmeYj74-079|This is a Mobius strip.
Bl-zUmeYj74-080|This has a twist in its space relative to our space.
Bl-zUmeYj74-082|I have to go around this loop twice to get back to the same state that I started in.
Bl-zUmeYj74-083|That's how electrons behave in our space because their space is twisted relative to ours.
Bl-zUmeYj74-084|They're very, very crazy little particles.
Bl-zUmeYj74-085|We're going to look at now taking those electrons and placing them into orbitals more than one at a time.
Bl-zUmeYj74-086|So we're going to move beyond the hydrogen atom to the rest of the elements of the periodic table and look at multi electron atoms.
1qiwlqrG6VY-000|I can form molecular orbitals from atomic orbitals by taking linear combinations of atomic orbitals.
1qiwlqrG6VY-001|I can add them and subtract them with constant coefficients.
1qiwlqrG6VY-002|So my atomic orbitals, when I'm talking about the p orbitals can be added and subtracted, I have to choose an internuclear axis though.
1qiwlqrG6VY-003|So I'm going to choose the internuclear axis, the axis that contains both nuclei, as the positive z-axis.
1qiwlqrG6VY-004|If I choose the positive z-axis, then I can add the pz orbitals together, pz atomic orbitals to form a molecular orbital.
1qiwlqrG6VY-005|Now notice I'm going to add these two orbitals, but look how the signs overlap.
1qiwlqrG6VY-006|Remember, red and green represent opposite signs.
1qiwlqrG6VY-007|Green can be positive, red can be negative.
1qiwlqrG6VY-009|So I'm going to bring those two together and make that linear combination.
1qiwlqrG6VY-010|I choose that linear combination because that gives me a higher amplitude, a higher probability of finding electrons, between the two nuclei.
1qiwlqrG6VY-011|Constructive interference between the two nuclei leads to a bonding orbital.
1qiwlqrG6VY-014|I can show you that with these little models.
1qiwlqrG6VY-015|Here comes one atom with a pz orbital and another atom with a pz orbital.
1qiwlqrG6VY-016|Multiply this one by negative 1, so that I get constructive interference.
1qiwlqrG6VY-017|And I add to form constructive interference between the two nuclei.
1qiwlqrG6VY-018|And I call this a sigma bonding interaction because the electron density is along the internuclear axis.
1qiwlqrG6VY-019|That wil be my definition of sigma orbitals.
1qiwlqrG6VY-023|So here, I can add together px and px and get a bonding orbital.
1qiwlqrG6VY-024|In this case, I'll call it a pi bonding orbital.
1qiwlqrG6VY-025|Pi bonding orbitals and pi antibonding orbitals have their electron density above and below the internuclear axis.
1qiwlqrG6VY-026|So I can get a pi bonding from the x, but I can also get a pi bonding from adding py and py.
1qiwlqrG6VY-027|So 2 pi bonding orbitals.
yHIDWAYbYBs-000|I have a question for you.
yHIDWAYbYBs-009|We're talking about the reaction of hydrogen in balloons.
yHIDWAYbYBs-011|Now, that's actually a slow step, the mixing of the hydrogen and the oxygen in the atmosphere.
yHIDWAYbYBs-012|So this reaction will occur, and there'll be an explosion.
yHIDWAYbYBs-013|But it'll actually be more of a [INAUDIBLE] explosion.
yHIDWAYbYBs-014|The mixing step slows it down a bit.
yHIDWAYbYBs-015|This second balloon has only oxygen.
yHIDWAYbYBs-016|Now, many people think that oxygen is a very dangerous gas, very explosive.
yHIDWAYbYBs-017|But it turns out oxygen is an explosive gas in the presence of other molecules to react with.
yHIDWAYbYBs-018|Oxygen by itself is not an explosive gas.
yHIDWAYbYBs-019|It will react rapidly with other molecules.
yHIDWAYbYBs-020|In this case, we haven't given it any other molecules, except maybe the material in the balloon.
yHIDWAYbYBs-021|But in general, this is going to give almost no explosion.
yHIDWAYbYBs-022|The third case, I have hydrogen and oxygen intimately mixed together.
yHIDWAYbYBs-023|Now in this case when the flame comes through the balloon, hydrogen and oxygen are primed to react.
yHIDWAYbYBs-024|They're very close to each other.
yHIDWAYbYBs-025|A hydrogen doesn't have to go very far to find an oxygen.
yHIDWAYbYBs-028|Now Chemical Quizzes that we're going to do-- ChemQuizzes-- throughout this series are going to require you to think outside the box or outside the balloon.
yHIDWAYbYBs-029|This one had a lot of chemical principles, especially for early on in the course, but I wanted to get you a feel for what you have to think about to do ChemQuizzes.
yHIDWAYbYBs-030|You might not have got this one because of the complicated nature of the quick reaction versus the slow reaction, but keep trying.
yHIDWAYbYBs-031|We're going to give you more principles and more tools to be able to perform on these ChemQuizzes.
eqHY6DH8QRI-000|For an ideal gas, the product of the pressure and the volume is a constant for a constant temperature.
eqHY6DH8QRI-001|If you change the temperature, the value of the constant changes, but the product of pressure and volume is still a constant.
eqHY6DH8QRI-002|Let's look at how changing the temperature affects our curves.
eqHY6DH8QRI-007|Now the product of pressure and volume being a constant, means pressure and volume are inversely proportional.
eqHY6DH8QRI-008|As I increase pressure, volume decreases.
eqHY6DH8QRI-009|Decrease pressure volume increases.
eqHY6DH8QRI-010|So their product is always a constant.
eqHY6DH8QRI-011|It turns out, that's how your lungs work.
eqHY6DH8QRI-012|In order for your lungs to draw air in, your chest cavity expands higher value, that means lower pressure.
eqHY6DH8QRI-013|And high pressure air from the atmosphere can rush into your lungs.
eqHY6DH8QRI-016|There's a model chest cavity.
eqHY6DH8QRI-017|The model chest cavity has that muscle in your chest, the diaphragm.
eqHY6DH8QRI-018|And when that contracts, the volume of your chest cavity increases.
eqHY6DH8QRI-019|Higher volume, lower pressure, air rushes in.
eqHY6DH8QRI-022|Decrease volume, higher pressure, air rushes out.
eqHY6DH8QRI-025|Taking advantage of the relationship p times v is the constant.
9mTP53W2_RY-000|Glucose is an energy source for your body.
9mTP53W2_RY-001|That is glucose is oxidized to release energy.
9mTP53W2_RY-002|Now, that oxidation you could do on the benchtop.
9mTP53W2_RY-003|You could burn glucose in oxygen and it would release energy.
9mTP53W2_RY-004|In your body, it's done via enzymes, and the reaction occurs at body temperatures.
9mTP53W2_RY-006|So if the initial and final states are the same, we know that free energy and enthalpy and entropy changes are the same.
9mTP53W2_RY-007|They're a state function.
9mTP53W2_RY-008|So the same amount of energy that you release on the benchtop is available to your body when you oxidize glucose in your body.
9mTP53W2_RY-013|So rather than drawing it as a ring, I just bent this bond all the way around to show that this carbon is attached to that carbon.
9mTP53W2_RY-014|Here's oxygen, carbon dioxide, and water.
9mTP53W2_RY-015|And we know how to assign oxidation numbers from Lewis structures.
9mTP53W2_RY-019|So oxygen gets these two.
9mTP53W2_RY-020|Carbon and carbon here, that's the same elements.
9mTP53W2_RY-021|So I'll share these electrons and assign one to this carbon.
9mTP53W2_RY-022|And carbon is slightly more electronegative than hydrogen, so I'll assign both of them to the carbon here.
9mTP53W2_RY-023|So I'm assigning one, two, three, four electrons to the carbon based on relative electronegativities.
9mTP53W2_RY-024|And carbon normally has four electrons around it in its valence shell.
9mTP53W2_RY-025|So it has oxidation number 0.
9mTP53W2_RY-027|Then I see that it has a plus 4 oxidation number here.
9mTP53W2_RY-028|So it goes from 0 to plus 4, that's an increase in oxidation number-- formerly, an oxidation.
9mTP53W2_RY-029|So something has to be reduced.
9mTP53W2_RY-030|If there's an oxidation, there must be a reduction as well.
9mTP53W2_RY-031|Here's elemental oxygen, where the oxidation number is 0.
9mTP53W2_RY-032|And what about the oxygen over here?
9mTP53W2_RY-033|The oxygen in water, I assign all of the electrons to the oxygen. It's the more electronegative element.
9mTP53W2_RY-034|So it has 1, 2, 3, 4, 5, 6, 7, 8 electrons around it.
9mTP53W2_RY-035|The atomic oxygen would have 6, so it has two more than it normally has as an atom.
9mTP53W2_RY-036|So that's an oxidation number of minus 2.
9mTP53W2_RY-037|Indeed, that's a reduction.
9mTP53W2_RY-038|I go from 0 to minus 2.
9mTP53W2_RY-039|A reduction in oxidation number is a reduction.
g0nvNTN4XHc-000|Let's look at the valence electrons of some atoms and see how they participate in bonding.
g0nvNTN4XHc-001|Now, we've talked about several kinds of bonding-- ionic bonding and covalent bonding.
g0nvNTN4XHc-002|Let's look at ionic bonding first.
g0nvNTN4XHc-003|If you look at lithium and fluorine, sodium and chlorine and their valence electrons.
g0nvNTN4XHc-004|You'll notice that fluorine and chlorine has seven valence electrons.
g0nvNTN4XHc-005|If they could pick up one more, they'd get to eight and that stable octet.
g0nvNTN4XHc-006|Sodium and lithium, should they lose one, then they'll have a closed shell underneath that primary shell.
g0nvNTN4XHc-007|So lithium and sodium are kind of primed to give up an electron and go to the closed shell underneath.
g0nvNTN4XHc-008|Fluorine and chlorine are primed to accept electrons and close their existing shells at that stable octet.
g0nvNTN4XHc-009|So one way that you can describe the bonding is an electron transfer.
g0nvNTN4XHc-010|One element gives up an electron to another element.
g0nvNTN4XHc-011|You form a positive ion and a negative ion.
g0nvNTN4XHc-012|And then Coulombic attraction, the attraction between plus and minus charges, holds those elements together in a bond.
g0nvNTN4XHc-013|So for lithium and fluorine, an electron goes from the lithium to the fluorine.
g0nvNTN4XHc-014|You get a Li plus and F minus holding each other together.
g0nvNTN4XHc-016|And this can happen with multiple atoms as well.
g0nvNTN4XHc-018|It can give up one to each of two chlorines.
g0nvNTN4XHc-019|It becomes plus 2 charge.
g0nvNTN4XHc-020|Each chlorine is negative 1 charge.
g0nvNTN4XHc-021|And all three of those are held together by Coulombic interaction.
g0nvNTN4XHc-022|So ionic bonding is the transfer of electrons from one species to another species.
g0nvNTN4XHc-023|And then the plus-minus interaction holds those two together in a bond.
-WxXRDp-IuU-000|Let's look at the sublimation of two different solids, iodine and carbon dioxide.
-WxXRDp-IuU-002|Would it look like A, B, or C?
-WxXRDp-IuU-003|We can start these two subliming.
-WxXRDp-IuU-012|We're talking about the sublimation of iodine and carbon dioxide.
-WxXRDp-IuU-015|So with that information, can we determine how the plot of lnK versus 1/T look relative to each other?
-WxXRDp-IuU-016|Well, lnK versus 1/T, the slope is determined by the enthalpy, the intercept by the entropy.
-WxXRDp-IuU-017|So the entropy of sublimation probably about the same.
-WxXRDp-IuU-019|So we expect the entropy of sublimation is determined mainly by the fact that it's a phase change from solid to gas.
-WxXRDp-IuU-020|Entropy of sublimation about the same means the intercept of these two plots should be about the same.
-WxXRDp-IuU-023|That sublimes at a very low temperature.
-WxXRDp-IuU-024|And the bonds of carbon dioxide, not held together in the solid as strongly as iodine.
-WxXRDp-IuU-025|Because we see the iodine at room temperature and we don't see the solid carbon dioxide at room temperature.
-WxXRDp-IuU-026|So we expect the enthalpy of sublimation to be lower.
-WxXRDp-IuU-027|The slope determines the enthalpy.
-WxXRDp-IuU-028|Enthalpy is determined by the slope for lnK versus 1/T.
-WxXRDp-IuU-029|So we expect a less dramatic slope for the carbon dioxide, a lower enthalpy of sublimation, so a less negative slope.
-WxXRDp-IuU-030|In this case, the answer is C, a less negative slope, but the same intercept for a plot of lnK versus 1/T.
oYHyZarcCQQ-000|Let's talk about the entropy of vaporization of water.
oYHyZarcCQQ-002|Here I have two diagrams showing how water is slightly oriented, even in the liquid phase, due to the strength of the hydrogen bonding.
oYHyZarcCQQ-003|My question for you is, how will that affect the entropy of vaporization for water, versus other liquids?
oYHyZarcCQQ-004|Most liquids, it's about 90 kilojoules per mole.
oYHyZarcCQQ-012|We're talking about the entropy of vaporization of water.
oYHyZarcCQQ-016|So the entropy of vaporization will be slightly larger.
oYHyZarcCQQ-017|The entropy of the gas is about the same.
oYHyZarcCQQ-018|All the gases are particles spread out and not interacting.
oYHyZarcCQQ-019|In the liquid phase, the interaction of the particles and the hydrogen bonding lowered the entropy of the liquid for water.
oYHyZarcCQQ-020|So, slightly lower number here means a slightly larger number for the enthalpy of vaporization.
oYHyZarcCQQ-021|And that is true.
oYHyZarcCQQ-022|The entropy of vaporization for water is more than 90 joules per Kelvin mole.
oYHyZarcCQQ-023|Now, you may have been able to guess this for another reason.
oYHyZarcCQQ-025|And you know water boils at around 400-ish Kelvin.
oYHyZarcCQQ-027|So, two ways to approach this problem.
oYHyZarcCQQ-028|In each case, you wind up with the same conclusion, that the entropy of vaporization for water, slightly higher than for other liquids.
qRkR-Gg29A0-001|That's what I'd like to look at now.
qRkR-Gg29A0-002|Chemical kinetics.
qRkR-Gg29A0-003|Here's a chemical reaction, and we'll define the rate as the change in concentration over time.
qRkR-Gg29A0-012|But in general, what we do is we write down common rate laws.
qRkR-Gg29A0-013|The rates as we measure them are proportional to the concentrations in some sense.
qRkR-Gg29A0-015|And I'm usually talking about the instantaneous rate.
qRkR-Gg29A0-017|So I can write rates as rate is k, a proportionality constant, times the concentration.
qRkR-Gg29A0-018|Now that proportionality constant k is called the rate constant.
qRkR-Gg29A0-019|And the rate constant will have units that are appropriate for the system.
qRkR-Gg29A0-024|This rate law-- both of these, whether the laws are first order or second order-- they're determined by experiment.
qRkR-Gg29A0-025|That is this power of 2 here, and this power of 1 here, are not related necessarily to the stoichiometric coefficients.
qRkR-Gg29A0-026|The rate laws must be determined by experiment.
qRkR-Gg29A0-028|But in kinetics, these powers are not related to stoichiometric coefficients.
qRkR-Gg29A0-029|You have to determine by experiment, you have to measure concentrations versus time, you have to plot them and see if they follow one of these rate laws.
qRkR-Gg29A0-030|So here's a second order rate law.
qRkR-Gg29A0-031|So we're saying that rates are often proportional to concentrations in some power.
qRkR-Gg29A0-032|And we can see that.
qRkR-Gg29A0-033|Here's an example.
qRkR-Gg29A0-036|So, let's just do that.
qRkR-Gg29A0-037|Here it is, 6 molar and 1 molar HCL being added to zinc metal.
qRkR-Gg29A0-038|And you can see 6 molar reacting rapidly and the 1 molar reacting more slowly.
qRkR-Gg29A0-040|And we could determine a rate law, and that's how chemical kinetics is done.
I3JPXKeFojo-000|When we talk about molecules in chemistry, it's often under two different circumstances.
I3JPXKeFojo-001|We'll talk about the individual molecule, and we'll look very closely at the molecule, the number of electrons, the orientation of bonds, the overall structure of the molecule.
I3JPXKeFojo-002|But we also need to be concerned with collections of molecules.
I3JPXKeFojo-005|The reason is in the gas phase, the molecules don't interact with each other very often.
I3JPXKeFojo-006|In fact, we'll take the ideal situation as our starting point.
I3JPXKeFojo-007|An ideal gas is where the particles fly around the container.
I3JPXKeFojo-008|And when they do interact briefly, they'll just bounce off each other.
I3JPXKeFojo-009|That will be a perfectly elastic collision.
I3JPXKeFojo-010|When they bounce off the walls, that will be an elastic collision.
I3JPXKeFojo-011|So, no interactions between the particles and the particles fly around freely in the gas phase.
I3JPXKeFojo-012|Can we describe how that macroscopic properties of pressure, volume, and temperature relate to those particles?
I3JPXKeFojo-013|Well, it turns out we can.
I3JPXKeFojo-014|Let's start with the pressure and the volume.
I3JPXKeFojo-016|That's a very interesting property.
I3JPXKeFojo-017|If the pressure goes up, the volume must go down, because their product has to be a constant.
I3JPXKeFojo-018|So, here's a sample of gas trapped in a balloon.
I3JPXKeFojo-019|And if I took and squeezed this down, increased the pressure, I would decrease the volume.
I3JPXKeFojo-020|The pressure goes up, volume goes down.
I3JPXKeFojo-021|That inverse relationship is very important and it's a classic example of what happens in gases.
I3JPXKeFojo-027|Now, that means we could also plot the pressure versus 1 over the volume.
I3JPXKeFojo-028|And we could do that P equals a constant over a V to get a linear relationship.
I3JPXKeFojo-029|It's often good in science to plot something that's linear, because then you can predict very easily trends.
I3JPXKeFojo-030|So for a fixed temperature, pressure versus one over volume, is a straight line.
I3JPXKeFojo-031|So the product of pressure and volume equals a constant is a fundamental property of gases, and that's one we'll use very frequently in this course.
J6txSOQgVLo-000|Let's look at a calculation involving empirical formula.
J6txSOQgVLo-001|Remember, empirical formula is the simplest ratio of the atoms involved.
J6txSOQgVLo-002|So methane, CH4-- the empirical formula and the molecular formula are the same.
J6txSOQgVLo-003|Ethylene, though, the ratio of carbon to hydrogen atoms is 1 to 2.
J6txSOQgVLo-004|But the actual molecular formula is C2H4.
J6txSOQgVLo-005|So there's a difference between the empirical and the molecular formula.
J6txSOQgVLo-006|Let's look at a calculation involving these empirical formulas.
J6txSOQgVLo-007|It's going to involve skunk spray.
J6txSOQgVLo-008|We're going to take skunk spray-- a two milligram sample-- and we're going to break it down into its components.
J6txSOQgVLo-009|Carbon will contain one milligram.
J6txSOQgVLo-011|Those are the mass ratios.
J6txSOQgVLo-012|What we need to do is convert those to mole ratios to get to the empirical formula.
J6txSOQgVLo-013|So let's do that.
J6txSOQgVLo-019|Now we know the mole ratios.
J6txSOQgVLo-020|They just aren't very neat.
J6txSOQgVLo-022|That is, see the ratios in a simpler way.
J6txSOQgVLo-027|So the molar ratios are 1 to 8 to 4 in this compound.
J6txSOQgVLo-028|The empirical formula-- 4 carbon atoms to every 8 hydrogen atoms to every 1 sulfur atom.
J6txSOQgVLo-029|That's the empirical formula.
J6txSOQgVLo-031|So the molar mass, if you did the mass spectrum of this compound, turned out to be somewhere around 88.
J6txSOQgVLo-034|So this empirical formula has the same mass as the molecular formula.
J6txSOQgVLo-035|So in this case, it's like methane.
J6txSOQgVLo-036|The empirical and molecular formula are the same.
ZuwQf_7yXrg-000|In the titration of a weak acid by a strong base, there's a region where the pH changes slowly as you add the base.
ZuwQf_7yXrg-001|That's called the buffer region.
ZuwQf_7yXrg-005|That's where you've converted half of the acid into the conjugate base, and those concentrations are equal.
ZuwQf_7yXrg-006|Now at this point, the titration can continue, but the pH will change slowly.
ZuwQf_7yXrg-007|And the pH will change slowly prior to that point.
ZuwQf_7yXrg-008|And why is that?
ZuwQf_7yXrg-009|Well, at this point and in this region, you have acid and conjugate base.
ZuwQf_7yXrg-010|They are both ready to accept strong acid, or strong base that's added.
ZuwQf_7yXrg-014|The K for both of these reactions is much greater than 1.
ZuwQf_7yXrg-015|And how do I know that?
ZuwQf_7yXrg-016|Well, look at the reverse reactions.
ZuwQf_7yXrg-017|For instance, the reverse reaction here, HA plus H2O goes to H3O plus and A minus.
ZuwQf_7yXrg-018|By now, that should jump out at you as the acid dissociation reaction for a weak acid.
ZuwQf_7yXrg-019|For instance, acetic acid, if that were our acid, this reaction going in this direction would have a K of around 10 to the minus 5.
ZuwQf_7yXrg-020|That means going back in this direction, we'd have 1 over that, something like K to the plus 5.
ZuwQf_7yXrg-021|Same thing here.
ZuwQf_7yXrg-022|This is the reaction of the weak base with water going in reverse.
ZuwQf_7yXrg-023|So for acetic acid, this has something like 10 to the minus 10.
ZuwQf_7yXrg-024|So going in this direction, 10 to the plus 10.
ZuwQf_7yXrg-025|So both of these equilibria lie strongly towards products.
ZuwQf_7yXrg-026|Both of them consume acid and base.
ZuwQf_7yXrg-027|Strong acid and base consumed, converted into weak acid and base.
ZuwQf_7yXrg-028|So whether I add a little acid or a little base, the pH doesn't change very much.
ZuwQf_7yXrg-029|And we can watch that happen.
ZuwQf_7yXrg-031|I'm going to add acid to all three of these, and I'm going to do it in a unique way.
ZuwQf_7yXrg-032|I'm going to add solid carbon dioxide.
ZuwQf_7yXrg-033|When I add solid carbon dioxide, that dissolves and forms carbonic acid.
ZuwQf_7yXrg-034|So that will acidify the solutions.
ZuwQf_7yXrg-035|I'll acidify the solutions, and let's watch how fast the pH changes.
ZuwQf_7yXrg-036|So here to the buffered solution, and here to the unbuffered solution.
ZuwQf_7yXrg-038|As the solutions are acidified, the unbuffered solution, pH changes rapidly.
ZuwQf_7yXrg-039|And you can see that by the color change in the indicator.
ZuwQf_7yXrg-040|In the buffered solution, a slow change in pH.
ZuwQf_7yXrg-041|This buffered solution resists change in pH.
ZuwQf_7yXrg-042|As acid is added, it's converted to a weak base, and the pH changes slowly.
ZuwQf_7yXrg-043|That's the effect of a buffer in solution.
Zghfd6Nz5Rk-000|Let's do some calculations with the ideal gas law.
Zghfd6Nz5Rk-001|I'm going to take a flask, one liter, at room temperature, put a mole of a metal in it.
Zghfd6Nz5Rk-002|Then I'm going to evacuate the flask, so there's no gas particles.
Zghfd6Nz5Rk-003|And add in chlorine gas, Cl2 gas, until the pressure comes up to 48.6 atmospheres.
Zghfd6Nz5Rk-004|Now that 48.6 atmospheres refers to a specific number of particles of chlorine gas.
Zghfd6Nz5Rk-005|I'm going to let the reaction occur.
Zghfd6Nz5Rk-006|After the reaction occurs, the pressure of chlorine gas has gone down to 24.3 atmospheres.
Zghfd6Nz5Rk-007|So some of the chlorine has been consumed.
Zghfd6Nz5Rk-011|And it's the ratio of the metal to the chlorine, this y to x ratio, that'll allow us to predict what metal that is.
Zghfd6Nz5Rk-012|So let's see if we can figure that out.
Zghfd6Nz5Rk-013|We know a in this equation, the coefficient of the metal is 1.
Zghfd6Nz5Rk-014|B is the number of moles of chlorine gas, which I can figure out, because I know the pressure went from 48.6 to 24.3.
Zghfd6Nz5Rk-015|The pressure dropped by a factor of two.
Zghfd6Nz5Rk-016|That means half of the original chlorine molecules were used up in that chemical reaction.
Zghfd6Nz5Rk-026|And I know the pressure, volume and the temperature.
Zghfd6Nz5Rk-027|So I know how many atmospheres of chlorine were consumed, the volume and their temperature.
Zghfd6Nz5Rk-030|Excuse me, one mole.
Zghfd6Nz5Rk-031|So one mole of chlorine is consumed.
Zghfd6Nz5Rk-033|So the actual relationship is C, 1.
Zghfd6Nz5Rk-034|That is, this coefficient is 1.
Zghfd6Nz5Rk-035|I have 1 mole of metal with 2 moles of chlorine atoms.
Zghfd6Nz5Rk-036|So y is 1, x is 2.
Zghfd6Nz5Rk-039|Well, it's all the metals in row two of the periodic table.
Zghfd6Nz5Rk-040|Like magnesium and calcium, for instance.
Zghfd6Nz5Rk-041|So we can tell about which metal that's going to be or predict which metal that is based on how it reacts with chlorine.
Zghfd6Nz5Rk-042|And the number of moles of chlorine it reacts with, we figure from the ideal gas law.
Zghfd6Nz5Rk-043|How the pressure changed in a chemical reaction.
_JgKgx7Jn-g-000|The reaction of alkali metals with water is a really good way to demonstrate the trend in ionization energies in the alkali metals.
_JgKgx7Jn-g-006|And this reaction, the formation of the hydroxide ion, releases quite a bit of energy.
_JgKgx7Jn-g-007|But the reaction of ionization we know absorbs energy.
_JgKgx7Jn-g-008|You have to put energy in to ionize, but then you get some back out when you form the hydroxide.
_JgKgx7Jn-g-010|So if you have a lower ionization energy, you'll have a more vigorous reaction overall because more energy is allowed to be released.
_JgKgx7Jn-g-012|We have metal being ionized, water forming hydroxide ion, more vigorous reaction for lower ionization energies.
XBua9WZkTeU-000|Let's look at a galvanic cell where we use concentration difference instead of two different cell components.
XBua9WZkTeU-001|So on each side we have a copper ion, copper metal half cell.
XBua9WZkTeU-002|The only difference is the concentration on each side.
XBua9WZkTeU-012|We're looking at a cell where the difference is the concentration of the ions on either side.
XBua9WZkTeU-013|So the overall cell reaction is just high concentration copper going to low concentration copper.
XBua9WZkTeU-014|And we know intuitively if we were to mix these two, that's what would happen, the concentration would equalize.
XBua9WZkTeU-015|So intuitively, we know the concentration here of copper ions will decrease and the concentration here will increase.
XBua9WZkTeU-016|But how does that correspond to a cell potential?
XBua9WZkTeU-017|Well, let's look at that.
XBua9WZkTeU-019|That would be 0 potential.
XBua9WZkTeU-020|We don't have one molar on both sides.
XBua9WZkTeU-021|We have 0.1 molar and one molar.
XBua9WZkTeU-022|So we have a situation where we have reactants over products, the Q inserted here.
XBua9WZkTeU-023|And that Q is less than 1.
XBua9WZkTeU-024|That means this natural log term is negative.
XBua9WZkTeU-025|So the overall potential difference is going to be positive.
XBua9WZkTeU-026|So a positive cell potential difference in our convention means current flows from left to right, and that we understand.
XBua9WZkTeU-027|Current flowing from left to right would have an oxidation occurring here.
m6SHjX_QcwE-000|Let's look at chemical reactions and we'll compare them by plotting the initial rate versus the initial concentration of a reactant.
m6SHjX_QcwE-001|Now reaction I plotted here is first order.
m6SHjX_QcwE-002|The question is, what is the order of reaction II?
m6SHjX_QcwE-003|Is that first order, second order, or is it indeterminant?
m6SHjX_QcwE-010|We're looking at initial rates versus concentration for two reactions.
m6SHjX_QcwE-011|Now, reaction I, we've plotted it here.
m6SHjX_QcwE-012|And we've measured initial rates versus initial concentrations.
m6SHjX_QcwE-013|So how would you do that experiment?
m6SHjX_QcwE-014|You'd set up the reaction, you'd measure the rate.
m6SHjX_QcwE-015|Then you'd set up the reaction again at a different initial concentration, and you'd measure the initial rate.
m6SHjX_QcwE-016|Then, you'd set up the reaction again at a different initial concentration, measure the rate, and you would plot those out.
m6SHjX_QcwE-017|And the plots you get for reaction I is linear, and that makes sense for a first order rate law.
m6SHjX_QcwE-018|We said it was first order.
m6SHjX_QcwE-019|The rate is proportional to the concentration.
m6SHjX_QcwE-020|The proportionality constant is k, so the slope of this line would be the rate constant k.
m6SHjX_QcwE-021|If it were a second order reaction, a second order reaction has overall powers add 2, in this case, there's just a single power of 2.
m6SHjX_QcwE-024|So it can't be second order.
m6SHjX_QcwE-025|It must be first orders well the quadratic relationship doesn't appear.
m6SHjX_QcwE-027|The correct answer here is A, first order.
dVdQsZ8kmWg-000|Let's compare three carbo-bases.
dVdQsZ8kmWg-001|Which is the strongest base, acetate, ethanoate, or carbonate?
dVdQsZ8kmWg-002|And here I've written their structures, acetate, ethanoate, carbonate.
dVdQsZ8kmWg-010|We're talking about three bases, ethanoate, acetate, and carbonate.
dVdQsZ8kmWg-011|We want to know which is the strongest.
dVdQsZ8kmWg-015|Well, what's going on here?
dVdQsZ8kmWg-016|This is a sequence where the oxidation number, the oxidation state of the carbon is changing.
dVdQsZ8kmWg-017|I'm adding more electron withdrawing oxygens.
dVdQsZ8kmWg-018|So the acid strength should change.
dVdQsZ8kmWg-019|And indeed, you can calculate the oxidation number of this carbon, and that's a good exercise.
dVdQsZ8kmWg-020|You can draw those Lewis dot structures, and remember how to calculate oxidation numbers from Lewis dot structures.
dVdQsZ8kmWg-023|So the strongest acid here, the weakest acid here, and the weakest acid will have the strongest conjugate base.
65q2f8Naf8M-001|Now, remember, standard state with this little degree sign, that means all gases are present at one atmosphere of pressure.
65q2f8Naf8M-002|If there's a concentration, it's one molar.
65q2f8Naf8M-003|Liquids and solids are present in their pure state.
65q2f8Naf8M-004|So with that in mind, which of these is a plot of standard state free energy versus temperature for H2O liquid going to H2O gas?
65q2f8Naf8M-012|We're talking about the liquid-to-gas phase change for water, trying to plot the free energy in the standard state versus temperature.
65q2f8Naf8M-017|If you look at this in terms of a line, this looks like y equals mx plus b.
65q2f8Naf8M-019|So if delta S is positive, the slope of this system must be negative because there's a negative sign there.
65q2f8Naf8M-020|So this slope must be negative.
65q2f8Naf8M-021|So let's see.
65q2f8Naf8M-022|We've got negative slope here, negative slope here.
65q2f8Naf8M-026|In order for liquid water to go to gaseous water, we have to absorb energy.
65q2f8Naf8M-027|Absorbing energy, that's a positive delta H.
65q2f8Naf8M-028|That says the intercept of this line must be positive.
65q2f8Naf8M-029|So now, between C and A, here we have a negative intercept.
65q2f8Naf8M-033|You could have said, well, water, liquid water going to gaseous water, that's favorable at high temperatures.
65q2f8Naf8M-035|The higher temperature you go, the more gas is favored.
65q2f8Naf8M-036|So one atmosphere, because we're talking about standard state, one atmosphere of gas is favored at higher temperatures.
65q2f8Naf8M-037|So delta G should be negative at higher temperatures.
65q2f8Naf8M-038|And the lower the temperature, the liquid should be favored.
65q2f8Naf8M-039|So delta G should be positive.
65q2f8Naf8M-040|And that's what you have here-- low temperatures, positive delta G, high temperatures, negative delta G.
65q2f8Naf8M-041|And that's the only situation where that's true.
65q2f8Naf8M-042|So two ways to arrive at answer C.
9HIhEu8J79A-000|When electromagnetic radiation interacts with matter, that radiation can be changed, or it can be emitted, or it can be absorbed in some way.
9HIhEu8J79A-001|This is a really important phenomenon in chemistry.
9HIhEu8J79A-002|Because we can't see atoms and molecules with our eyes.
9HIhEu8J79A-003|In fact, when you see on TV, a scientist, they'll always have a lab coat, he'll have some safety glasses.
9HIhEu8J79A-004|And regardless of what kind of scientist is being portrayed, they'll always have a microscope.
9HIhEu8J79A-005|And when they want to tell you something about an atom or a molecule, they look through their microscope, and they say, oh, yeah, well, that molecule is blue.
9HIhEu8J79A-006|That is just crazy.
9HIhEu8J79A-007|We can't see-- there's no optical microscope that can resolve atoms and molecules.
9HIhEu8J79A-008|What we do is we have radiation of various wavelengths interact with atoms and molecules.
9HIhEu8J79A-009|And we deduce things about the molecules based on how those atoms and molecules interact with the radiation.
9HIhEu8J79A-010|So when radiation hits a molecule or atom or any kind of matter, many things can happen.
9HIhEu8J79A-011|It can be absorbed.
9HIhEu8J79A-012|Radiation can be emitted by excited atoms.
9HIhEu8J79A-013|The radiation can change from high frequency to low frequency radiation.
9HIhEu8J79A-014|There can be a reflection process-- all kinds of different things that help us understand the matter.
9HIhEu8J79A-015|So let's talk about absorption and emission.
9HIhEu8J79A-016|It can happen in many different ways.
9HIhEu8J79A-017|You can have a continuous absorption.
9HIhEu8J79A-018|So you can have a continuous absorption of many wavelengths, so a band of wavelengths, different colors hitting you all at once.
9HIhEu8J79A-019|So for instance, when we see white light, that's all the colors mix together coming at us.
9HIhEu8J79A-022|And this is actually why you perceive color.
9HIhEu8J79A-023|When white light, a combination of all the colors, hits an object, some of those colors can be absorbed.
9HIhEu8J79A-024|The colors that aren't absorbed pass through or are reflected back and hit your eyeballs.
9HIhEu8J79A-025|And the wavelengths that hit your eyeballs can be either red or blue or green.
9HIhEu8J79A-033|So we can actually look at absorption and emission from atoms.
9HIhEu8J79A-034|And I can show you a continuous emission spectrum.
9HIhEu8J79A-035|It's actually several lines being emitted at once from atoms.
9HIhEu8J79A-036|So let's look at that.
jp_X4YHAygg-001|So let's keep track of the energetics of chemical reactions, and we'll do that in a subject called thermodynamics.
jp_X4YHAygg-002|Thermodynamics involves heat transfer and tracking of the energy of systems and surroundings.
jp_X4YHAygg-003|When we work in thermodynamics, we'll define a system and say OK, everything that happens in this beaker will be the system.
jp_X4YHAygg-004|Everything else will be the surroundings.
jp_X4YHAygg-005|And we'll track how heat moves between the system and the surroundings.
jp_X4YHAygg-006|Now, energy-- and I'll use the symbol E-- is going to change in these systems.
jp_X4YHAygg-011|Heat and work are mechanisms by which energy is transferred.
jp_X4YHAygg-015|So we give heat that leaves the system the negative sign.
jp_X4YHAygg-016|We call that exothermic-- heat that leaves the system.
jp_X4YHAygg-019|And we give that heat a positive sign.
jp_X4YHAygg-024|That's kind of a natural thing that you might already understand.
jp_X4YHAygg-025|If some of your energy is used to do work, then that should lower your internal energy, and we'll give that work a negative sign.
jp_X4YHAygg-029|By conserved, we mean energy isn't lost or created.
jp_X4YHAygg-030|If it goes from the system to the surroundings, it goes joule for joule.
jp_X4YHAygg-033|And I do that with heat and work.
jp_X4YHAygg-034|Energy is a state function, and that means it depends only on the initial and final states of the system, not on the path to get there.
jp_X4YHAygg-035|That is, if I'm at the top of a hill and I jump down to the bottom of the hill, my gravitational potential energy changes.
jp_X4YHAygg-036|I have more potential energy here than I do here.
jp_X4YHAygg-037|And that doesn't matter if I jump off the hill directly, or if I run down the hill in circles, or if I fly up in the air and then come down to the bottom of the hill.
jp_X4YHAygg-038|If I start here and I end here, the energy change, that difference is always the same.
jp_X4YHAygg-039|But the work I do to get from here to here might be different.
jp_X4YHAygg-040|So work and heat are not state functions.
jp_X4YHAygg-041|They depend on the path you take.
jp_X4YHAygg-042|But the energy change is a state function.
jp_X4YHAygg-043|Energy is a state functions, it's conserved, and when energy is moved from a system to the surroundings, it's moved joule for joule.
jp_X4YHAygg-044|If I do a joule of work on the surroundings, I lose a joule of work.
jp_X4YHAygg-045|If a joule of heat is absorbed by me from the surroundings, that's joule for joule.
jp_X4YHAygg-046|The surroundings lose a joule of heat, I gain a joule of heat.
jp_X4YHAygg-047|That's the essence of the first law of thermodynamics.
gAWuFYOUPSg-000|Now let's talk about the atoms.
gAWuFYOUPSg-001|Here's hydrogen atoms, slightly excited state, helium atoms, excited state, lithium atoms.
gAWuFYOUPSg-002|So no charge on the species.
gAWuFYOUPSg-003|They're all neutral atoms but slightly excited.
gAWuFYOUPSg-004|Which atom has the lowest ionization energy?
gAWuFYOUPSg-010|We're talking about ionizing three atoms.
gAWuFYOUPSg-011|Each of the atoms is in a slightly excited state.
gAWuFYOUPSg-012|So we have hydrogen in a 2p state, helium in the 3p state, and lithium in a 4p state.
gAWuFYOUPSg-013|So we've talked about this already.
gAWuFYOUPSg-014|When there's s electrons shielding p electrons, the shielding is very effective.
gAWuFYOUPSg-015|So helium with its two plus charges has a 1s electron shielding and outer 3p.
gAWuFYOUPSg-016|That one s shields almost one full nuclear charge.
gAWuFYOUPSg-017|The minus one of the electron shields almost plus one on the nucleus.
gAWuFYOUPSg-019|Same thing for the lithium.
gAWuFYOUPSg-020|It has three protons in its nucleus, but it has two s electrons shielding that 4p electron.
gAWuFYOUPSg-021|So those two s electrons do a very effective job of shielding nearly two full positive charges.
gAWuFYOUPSg-022|So the outer 4p sees like one positive charge.
gAWuFYOUPSg-023|So the effective charge on all of them is about one.
gAWuFYOUPSg-024|For hydrogen, it's exactly one.
gAWuFYOUPSg-025|And we can calculate these ionization energies.
gAWuFYOUPSg-026|Hydrogen in the 2p state, its ionization energy still a quarter of a Rydberg.
gAWuFYOUPSg-027|Helium in the 3p state, effective nuclear charge about 1, gives you about a ninth of a Rydberg.
gAWuFYOUPSg-028|And then lithium in the 4p state, effective nuclear charge about one, gives you 1/16 of a Rydberg.
D8JCG-Opfwk-001|The way you do that is you follow this recipe.
D8JCG-Opfwk-002|You take the electrons, and you share them equally about each atom.
D8JCG-Opfwk-004|And the difference will be formal charge.
D8JCG-Opfwk-005|So let's do that.
D8JCG-Opfwk-006|Nitrogen, in this case, has 1, 2, 3, 4 electrons in a bond.
D8JCG-Opfwk-008|What about the oxygens?
D8JCG-Opfwk-011|And it has to share these two with nitrogen, so one more.
D8JCG-Opfwk-012|6 and 1 is 7.
D8JCG-Opfwk-013|So seven electrons around oxygen in this molecule.
D8JCG-Opfwk-016|It has the same number of electrons in the molecule and the neutral atoms.
D8JCG-Opfwk-017|And it'll be minus one for this oxygen.
D8JCG-Opfwk-018|It has one more electron in the molecule than it does in the neutral atom.
D8JCG-Opfwk-019|Formal charge of zero and minus one will help us determine the quality of Lewis electron dot structures.
D8JCG-Opfwk-020|And we'll look at that next.
fOVAF8nTATI-000|Let's look at the ionization of some diatomic molecules.
fOVAF8nTATI-001|I have H2 the molecule, He2 the molecule, and Li2 the molecule.
fOVAF8nTATI-011|We're talking about the bond order of several ions.
fOVAF8nTATI-012|So I'm going to ionize H2, He2, and Li2, and see what the bond strength is.
fOVAF8nTATI-013|We'll coordinate the bond strength with the bond order.
fOVAF8nTATI-014|So here are the molecular orbital diagrams for H2, He2, and Li2 with the appropriate electrons.
fOVAF8nTATI-017|So 1, 2 bonding electrons minus 0 divided by 2 gives me 1.
fOVAF8nTATI-024|The highest energy electron is a sigma bonding electron from H2.
fOVAF8nTATI-025|And you already can imagine, if I'm removing a bonding electron, that can't make the bond more stable.
fOVAF8nTATI-026|In fact, removing an electron leaves one electron in the sigma bonding orbital.
fOVAF8nTATI-027|So if I calculate the bond order here, it's 1 minus 0 divided by 2.
fOVAF8nTATI-028|The bond order has gone down to 1/2.
fOVAF8nTATI-029|Here for helium, when I ionize, I remove an antibonding electron.
fOVAF8nTATI-032|So the bond orders go to 1/2.
fOVAF8nTATI-033|But the one that became more stable was helium.
fOVAF8nTATI-034|Unstable as He2 the molecule, but as He2+ the ion has a higher bond order.
fOVAF8nTATI-035|So in this case, the answer is He2, more stable upon ionization.
Q62PPzLln8E-000|Let's look at that helium ion in a little more detail.
Q62PPzLln8E-001|The helium plus ion, which transition has a wavelength that's equal to the 2-1 transition in the hydrogen atom?
Q62PPzLln8E-002|So we're going to line up transitions between helium plus and hydrogen, which transition in helium is the same as 2-1 in hydrogen?
Q62PPzLln8E-011|We're comparing hydrogen atoms to helium plus ions, both one electron systems.
Q62PPzLln8E-016|The n equal 2 and n equal one, n equal 4 and n equal 2 energy levels line up.
qm1SjMuodYI-000|Let's describe the bonding and the shape in SeF4.
qm1SjMuodYI-001|In order to do that, you need to start with a good Lewis electron dot structure.
qm1SjMuodYI-002|So you have to be very good at drawing your Lewis electron dot structure.
qm1SjMuodYI-003|Here's SeF4.
qm1SjMuodYI-005|And what we really need is the steric number around selenium.
qm1SjMuodYI-006|So that steric number is 1, 2, 3, 4, 5.
qm1SjMuodYI-007|I have to accommodate four fluorines and a lone pair for steric number five.
qm1SjMuodYI-008|Once I have these steric number, I can get to the geometry.
qm1SjMuodYI-009|It also gives me the hybridization.
qm1SjMuodYI-010|So steric number five means I'm going to use five equivalent hybrid orbitals.
qm1SjMuodYI-014|So our trigonal bipyramidal arrangement, but in this case, when we go to the molecular shape, we ignore electron pairs.
qm1SjMuodYI-015|So we always name our shapes without the electron pair.
qm1SjMuodYI-016|So removing that electron pair gives me a shape that looks like this, a little seesaw.
qm1SjMuodYI-017|So there's an electron pair here, but the molecular shape is named just by the atoms.
pB0yR5kXQjI-000|Let's look at some examples of molecular shapes.
pB0yR5kXQjI-001|We'll start with the VSEPR, the Valence Shell Electron Pair Repulsion of the groups around a central atom.
pB0yR5kXQjI-002|We'll count up the number of lone pairs and the number of bonded atoms around a central atom to determine the overall shape.
pB0yR5kXQjI-003|Let's look at some examples.
pB0yR5kXQjI-004|Here's carbon dioxide.
pB0yR5kXQjI-005|Carbon dioxide has steric number 2.
pB0yR5kXQjI-006|And I determine that from the Lewis dot structure.
pB0yR5kXQjI-007|Now it's absolutely critical to get the Lewis dot structure correct, because that will tell you where the lone pairs and where the bonded atoms are, and it'll determine the steric number.
pB0yR5kXQjI-008|Here, I've drawn the 16 electron system of carbon dioxide.
pB0yR5kXQjI-009|Haven't drawn the full Lewis electron dot structure, there's electrons, as you know, around these oxygens, but it's the central atom administered in here.
pB0yR5kXQjI-013|How do I fit two things as far away from each other as possible?
pB0yR5kXQjI-014|180 degrees.
pB0yR5kXQjI-015|Another example, boron dichloride with a positive charge.
pB0yR5kXQjI-016|Three things-- steric number 2 around the central atom, linear arrangement.
pB0yR5kXQjI-017|What about three things?
pB0yR5kXQjI-022|The carbons in ethylene each have steric number 3.
pB0yR5kXQjI-023|This carbon has to accommodate one, two, three things.
pB0yR5kXQjI-024|This carbon also accommodates one, two, three things.
pB0yR5kXQjI-025|All the bond angles in this molecule are 120 degrees.
pB0yR5kXQjI-026|Interestingly, this molecule is also planar.
pB0yR5kXQjI-027|All the atoms lie in one plane with bond angles of 120 degrees.
pB0yR5kXQjI-028|We can keep going.
pB0yR5kXQjI-030|Steric number 4, if you recall, is a tetrahedral arrangement.
pB0yR5kXQjI-031|Not a square planar, but a tetrahedral arrangement in space.
pB0yR5kXQjI-032|Tetrahedral angle is 109.5.
pB0yR5kXQjI-033|That's how to put four things as far away from each other in space.
pB0yR5kXQjI-034|So ammonia, the nitrogen has steric number 4, this ammonium ion with 1, 2, 3, 4 hydrogens around it.
pB0yR5kXQjI-035|The oxygen in water, steric number 4-- 1, 2 hydrogens, and 1, 2 lone pairs.
pB0yR5kXQjI-036|Now, steric number 4 for both of these, but ammonia we call a tetrahedral molecule.
pB0yR5kXQjI-037|Water, we actually don't call tetrahedral, we just call it a bent molecule.
pB0yR5kXQjI-038|The reason is, we only talk about the atomic centers when we're naming the shape of the molecule.
pB0yR5kXQjI-039|This molecule is bent, these lone pairs don't factor in to our structure of the molecule.
pB0yR5kXQjI-040|They influence it, but when we name this structure, we just call that bent, referring only to the atoms.
pB0yR5kXQjI-041|Now this bent molecule, this bond angle is actually a little less than the 109.5 of true tetrahedral.
pB0yR5kXQjI-043|In fact, this bond angle is closer to 105 degrees.
pB0yR5kXQjI-044|104.5 is usually the bond angle reported for water.
pB0yR5kXQjI-045|We can go to steric number 5.
pB0yR5kXQjI-047|Two different positions-- the axial positions-- 180 degrees from each other-- and equatorial positions-- 120 degrees from each other.
pB0yR5kXQjI-049|Phosphorus has steric number 5 in PCl5, it's a trigonal bipyramidal molecule.
pB0yR5kXQjI-050|That's quite a mouthful, but you can practice saying that-- trigonal bipyramidal.
pB0yR5kXQjI-052|In this case, two of the things are lone pairs.
pB0yR5kXQjI-054|Knowing electron pairs, that molecule looks T-shaped.
pB0yR5kXQjI-056|Now, it's interesting, you might say, well why didn't I take those two lone pairs and put them in the axial positions rather than the equatorial positions?
pB0yR5kXQjI-058|But that gives me many more 90 degree lone pair-bond pair interactions.
pB0yR5kXQjI-059|And those are very sterically bad.
pB0yR5kXQjI-060|The 90-degree lone pair-bond pair arrangements we want to minimize in our structures.
pB0yR5kXQjI-061|So this is a better arrangement of five things-- two lone pairs, three atoms.
pB0yR5kXQjI-062|We can go to steric number 6.
pB0yR5kXQjI-063|Steric number 6 will have an octahedron configuration.
pB0yR5kXQjI-064|Now remember, it's octahedral because this shape is an octahedron, which has eight sides but six vertices.
pB0yR5kXQjI-065|So six vertices, each position identical, 90-degree bond angles between all of them.
pB0yR5kXQjI-066|Some examples-- SF6.
pB0yR5kXQjI-067|Sulfur hexafluoride, that's steric number 6 around the fluorine.
pB0yR5kXQjI-068|Each fluorine identical, steric number 6 around the sulfur, each fluorine identical.
pB0yR5kXQjI-069|Or xenon tetrafluoride.
pB0yR5kXQjI-070|Again, steric number 6, but this case, two of the items are lone pairs.
pB0yR5kXQjI-071|And in this case, they are arranged axially, and they give an overall shape to xenon tetrafluoride of square planar.
pB0yR5kXQjI-074|So we have bonded atoms that determine the overall shape.
pB0yR5kXQjI-075|When we have bonded atoms and lone pairs, we ignore the lone pairs and name the shape based on bonded atoms alone.
pB0yR5kXQjI-076|That's how we do shapes of molecules and molecular geometry.
81v3NxDRdLk-000|Let's look at a few stepwise equilibria and look at manipulations with the equilibrium constants.
81v3NxDRdLk-001|So here I have nickel hydrate, a complex ion, reacting with ammonia to form a nickel ammonia complex ion.
81v3NxDRdLk-006|If I take my green solution of the hydrate and add ethylene diamine I'll get a purple ethylene diamine complex.
81v3NxDRdLk-008|Well, let's look at these equilibrium constants more carefully and see if we can make a determination.
81v3NxDRdLk-011|And here the nickel ammonia complex with the ethylene diamine as a reactant, displacing that.
81v3NxDRdLk-012|This is the K we want to determine.
81v3NxDRdLk-015|And when you add reactions, you multiply equilibrium constants.
81v3NxDRdLk-016|And we can prove that.
81v3NxDRdLk-017|Let's do this even a little more explicitly.
81v3NxDRdLk-018|We'll show K2 is K3 over K1.
81v3NxDRdLk-019|Or K3 is K1 times K2.
81v3NxDRdLk-020|We'll write out.
81v3NxDRdLk-036|So now we're in a position to say, well, what is the reaction, K2?
81v3NxDRdLk-037|How big is K2?
81v3NxDRdLk-038|We know it's K3 over K1.
81v3NxDRdLk-039|We can just do the math here.
81v3NxDRdLk-040|So if K3 is larger than K1, then K2 will be bigger than 1.
81v3NxDRdLk-041|And I'll go towards the ethylene diamine.
81v3NxDRdLk-042|If K3 is less than K1, then this K will be less than 1.
81v3NxDRdLk-045|So in order to figure this out, we're just going to have to do the experiment.
81v3NxDRdLk-046|So let's go ahead and do that.
81v3NxDRdLk-047|I have these down here.
81v3NxDRdLk-048|And I have the nickel hydrate, two vials.
81v3NxDRdLk-049|And I have some ammonia here.
81v3NxDRdLk-052|And indeed, I produced my blue nickel ammonia complex.
81v3NxDRdLk-053|Now here's the nickel hydrate.
81v3NxDRdLk-058|Which of these Ks is larger, K3 or K1?
81v3NxDRdLk-060|So here's ammonia.
81v3NxDRdLk-064|It looks like this K is going to be bigger than 1.
81v3NxDRdLk-065|It favors the ethylene diamine Let's prove that.
81v3NxDRdLk-066|Here is their exact reaction.
81v3NxDRdLk-070|It will be this violet color if K3 is bigger.
81v3NxDRdLk-071|It would be a blue color if K3 was smaller.
81v3NxDRdLk-072|But what we've shown is K3 is bigger.
81v3NxDRdLk-073|The ethylene diamine is the stronger binding ligand.
81v3NxDRdLk-074|And what we have is an analysis of equilibrium constants in multiple equilibria.
6BZb96mqmbg-000|Let's look at some second order chemical kinetics.
6BZb96mqmbg-004|Or the rate is the product of two concentrations.
6BZb96mqmbg-005|So the sum of the powers 1 plus 1 gives you second order kinetics.
6BZb96mqmbg-007|So a plot of 1 over concentration versus time is linear.
6BZb96mqmbg-008|And that's how you determine if you have a second order chemical reaction.
6BZb96mqmbg-012|Now, we can look at the half life for second order reactions as well.
6BZb96mqmbg-014|So the initial amount-- the initial concentration matters in terms of the half life.
6BZb96mqmbg-015|The amount of time it takes to go from initial concentration to molar down to initial concentration one molar depends on the fact that it's two molar.
6BZb96mqmbg-016|That is it's going to take longer for four molar to go to two molar than one molar down to half molar.
6BZb96mqmbg-017|And that tells you something about the kinetics.
6BZb96mqmbg-018|A second order reaction means there's some collision somewhere in there.
6BZb96mqmbg-020|Those have to come together.
6BZb96mqmbg-021|If the concentration of iodine increases, the rate increases, because the number of collisions increase.
6BZb96mqmbg-022|More iodine there, more collisions.
6BZb96mqmbg-023|So the frequency of collisions is reflected in this half life.
6BZb96mqmbg-024|If you have a high initial concentration, you'll have high initial concentrations and high initial collisions.
6BZb96mqmbg-025|And that will give you a low half life.
6BZb96mqmbg-026|But if you're a dilute solution, low initial concentration, then you'll have a longer half life.
6BZb96mqmbg-027|Collisions aren't as frequent.
fiJ6UDSt8hU-000|Let's talk about the p orbitals in terms of a ChemQuiz.
fiJ6UDSt8hU-001|Remember, an orbital is a group of three quantum numbers that define a wave function.
fiJ6UDSt8hU-002|So we need a value of n, a value of l, and a value of m sub l.
fiJ6UDSt8hU-003|If the value of n is 2, the value of l is 1, and the value of m sub l is 0, that's a p orbital.
fiJ6UDSt8hU-004|So for p orbitals, I've written a plot of psi and psi squared versus phi, that angle from the positive x-axis.
fiJ6UDSt8hU-009|We're talking about the 2p orbitals and we're plotting the wave function psi versus the angle phi from the positive x-axis.
fiJ6UDSt8hU-010|Now, there's three possible 2p orbitals.
fiJ6UDSt8hU-014|I've chosen to plot the 2py orbital here.
fiJ6UDSt8hU-015|And I know it's the 2py, because the maximum occurs along the y-axis.
fiJ6UDSt8hU-016|So let's look at the wave function, which the square is kind of shown here.
fiJ6UDSt8hU-019|The angle phi gets bigger as you go away from the positive x-axis.
fiJ6UDSt8hU-020|So this angle phi equals 0 is the x-axis.
fiJ6UDSt8hU-022|So anywhere along the x-axis or the z-axis, the wave function has to be 0.
fiJ6UDSt8hU-023|And that's what we have in this case.
fiJ6UDSt8hU-024|The wave function is 0 for values of phi that are equal to 0.
fiJ6UDSt8hU-025|If I let phi get bigger, what I'll find is, well, I'll start to sweep into these positive values of the wave function.
fiJ6UDSt8hU-026|The wave function should go positive.
fiJ6UDSt8hU-029|That's as big as it's going to get along the positive y-axis.
fiJ6UDSt8hU-030|So that should be a maximum in the wave function and the square of the wave function.
fiJ6UDSt8hU-031|And I can continue around.
fiJ6UDSt8hU-032|I can let phi go to pi.
fiJ6UDSt8hU-033|I'll come along to the negative x-axis, and of course, anywhere along the x-axis the function has to go to 0, and of course, the square of the wave function goes to 0.
oqN1aemQnu0-000|The three common phases of matter are solid, liquid, and gas.
oqN1aemQnu0-001|And we can plot where they occur on a pressure-temperature plot.
oqN1aemQnu0-002|And when we do, we call that a phase diagram.
oqN1aemQnu0-003|On a phase diagram, at low pressures, you'd expect the gas phase to exist.
oqN1aemQnu0-008|This is a plot for carbon dioxide.
oqN1aemQnu0-009|And here's the liquid-gas equilibrium line.
oqN1aemQnu0-012|Every point on this line, the solid and the liquid are in equilibrium.
oqN1aemQnu0-013|Now, for carbon dioxide, it turns out that common equilibrium that we observe is the solid gas, or sublimation equilibrium.
oqN1aemQnu0-016|And if I bring it here to the desktop, that's solid carbon dioxide.
oqN1aemQnu0-017|And that vibrating that you hear is the gas subliming off the carbon dioxide.
oqN1aemQnu0-018|That gas phase provides a cushion of gas.
oqN1aemQnu0-020|That's the sublimation of carbon dioxide.
oqN1aemQnu0-021|Now, let's talk about some other regions on the phase diagram.
oqN1aemQnu0-022|Interestingly, there's a point where all three of the solid, liquid, and gas phases intersect.
oqN1aemQnu0-023|This is called the triple point.
oqN1aemQnu0-027|A particle can move from the solid phase to the liquid phase, liquid to solid, or solid to gas independently.
oqN1aemQnu0-028|All three are equally likely at the triple point.
oqN1aemQnu0-029|We also have the critical point.
oqN1aemQnu0-035|Now, water is one of the only substances where the slope of the solid-liquid slopes backwards.
oqN1aemQnu0-036|Now, that's interesting.
oqN1aemQnu0-037|That says if you compress solid water, you go up in pressure-- so here's a solid water point.
oqN1aemQnu0-038|I go up in pressure.
oqN1aemQnu0-039|It should turn into the liquid.
oqN1aemQnu0-040|And that's counterintuitive.
oqN1aemQnu0-041|Most things, when you increase the pressure, you should continue to favor the solid, the more tightly compact phase of matter.
oqN1aemQnu0-042|Turns out for water, the solid is not the most tightly compact phase.
oqN1aemQnu0-043|The solid is actually slightly less dense than the liquid.
oqN1aemQnu0-044|And we know that.
oqN1aemQnu0-045|We've seen solid ice float on top of liquid ice.
oqN1aemQnu0-046|In fact, it's the most common solid-liquid phase transition we know about.
oqN1aemQnu0-047|So when we think about it, you'd think, oh, maybe all solids float on their liquids.
oqN1aemQnu0-048|It's actually the reverse.
oqN1aemQnu0-049|Just because that's a common substance, water, and we see the solid floating on the liquid, we get the impression that that's common.
oqN1aemQnu0-050|It's actually the least common interaction.
oqN1aemQnu0-051|Most solids sink in their own liquid.
oqN1aemQnu0-055|So very near 0 Celsius, but a fraction of an atmosphere.
oqN1aemQnu0-056|There are 760 torr in one atmosphere.
OlehBTSyPDA-000|Let's do a calculation that's a titration calculation.
OlehBTSyPDA-002|The question is, what is the pH after those two are mixed?
OlehBTSyPDA-003|So I can think about this as a titration.
OlehBTSyPDA-005|The question is, how far along on the titration curve does this addition of strong base take me?
OlehBTSyPDA-006|Does it take me to the center of the buffer region?
OlehBTSyPDA-007|Is it enough to take me to the equivalence point?
OlehBTSyPDA-009|We have HAc plus water, forming acetic acid and H3O plus.
OlehBTSyPDA-010|That's one of the equilibria we have to deal with for this problem.
OlehBTSyPDA-011|But we want to find out, well, what are the initial conditions?
OlehBTSyPDA-012|How much HAc, how much Ac minus, and how much H3O plus are there at a point?
OlehBTSyPDA-013|And then we'll allow the equilibria to expand.
OlehBTSyPDA-014|And this is how you do equilibrium calculations.
OlehBTSyPDA-015|You find a point that you can nail down.
OlehBTSyPDA-016|Say, OK, these will be my starting conditions.
OlehBTSyPDA-017|And then you apply the equilibrium to see how it shifts from that one point.
OlehBTSyPDA-018|And that's the great thing about equilibria, they approach the same equilibria no matter where you start from.
OlehBTSyPDA-019|So as long as you can nail down one point that you know and then apply equilibrium from there, you're going to get to the correct equilibrium condition.
OlehBTSyPDA-023|Well, molarity times volume is the number of moles.
OlehBTSyPDA-024|So 0.01 moles of acetic acid.
OlehBTSyPDA-025|Now this is assuming no dissociation.
OlehBTSyPDA-026|That's fine.
OlehBTSyPDA-027|I can choose any point along my equilibrium to be my starting point.
OlehBTSyPDA-033|And I choose that because the weak acid reacts completely with the strong base.
OlehBTSyPDA-034|That's a very strong reaction to favor the products.
OlehBTSyPDA-038|So that's not really an equilibrium, that one just goes all the way towards products.
OlehBTSyPDA-039|So if we start there and say, what happens after that occurs?
OlehBTSyPDA-040|Are there an equilibrium in effect, that will calculate the pH of the solution.
OlehBTSyPDA-041|So let's do that.
OlehBTSyPDA-042|We'll say, these two will react completely.
OlehBTSyPDA-043|So every mole of base will react with a mole of acid.
OlehBTSyPDA-044|And I'll have-- this is essentially half of this.
OlehBTSyPDA-045|So half of my base-- excuse me.
OlehBTSyPDA-046|Half of my acid will react with this base, and I'll form 0.05 moles of the conjugate base.
OlehBTSyPDA-047|So the acid reacts with the base to form the conjugate base.
OlehBTSyPDA-049|Here, the acid reacting with the base to form the conjugate base.
OlehBTSyPDA-050|In this case, I've added half the number of moles of HA that I originally had.
OlehBTSyPDA-052|Half of these react, so I have equal concentrations essentially of this and this at this point in the titration.
OlehBTSyPDA-053|So equal concentrations of these two.
OlehBTSyPDA-054|Half of my acid has been converted into its conjugate base.
OlehBTSyPDA-055|My OH minus concentration, it's gone to zero.
OlehBTSyPDA-056|I've used up one for one.
OlehBTSyPDA-057|One of those react with one of those, producing one of these.
OlehBTSyPDA-058|So let's treat these now as the initial concentrations in an equilibrium calculation and see where that gets us.
OlehBTSyPDA-063|But I know they're both in about 100 millimeters.
OlehBTSyPDA-064|So I can convert these easily into concentrations, moles per liter.
OlehBTSyPDA-070|Now this initial point is quite a ways into our calculation, but it's just the first point where I'm going to allow the equilibria to occur.
OlehBTSyPDA-071|In all the previous steps, I've said they go to completion.
OlehBTSyPDA-072|So now I'm at a point where everything has gone 100%, and I'm going to let the equilibria start to take over.
OlehBTSyPDA-073|And I do that at this point.
OlehBTSyPDA-074|And I can do that for equilibria at any point.
OlehBTSyPDA-075|Choose a point, nail it down, say, this is where I'm going to start, and then let your equilibria table take over.
OlehBTSyPDA-076|So the change to get to equilibrium is a little of this.
OlehBTSyPDA-077|We'll dissociate.
OlehBTSyPDA-078|And actually at this point, I'm not even sure.
OlehBTSyPDA-079|I'm not sure if the equilibrium will go this way or this way.
OlehBTSyPDA-080|I'm just going to assume a little of this goes away.
OlehBTSyPDA-081|And if it does, it'll produce a little of this.
OlehBTSyPDA-082|Since there's equal amounts, who knows?
OlehBTSyPDA-083|Maybe it'll shift back this way.
OlehBTSyPDA-084|I suspect it'll go this way though, because I think HAc is a strong enough acid to dissociate a little bit more.
OlehBTSyPDA-085|That will produce a little H3O plus.
OlehBTSyPDA-086|And I'm saying initially there was none of this.
OlehBTSyPDA-093|So that's 10 to the minus 4.75.
OlehBTSyPDA-094|4.75 is the Ka for acetic acid.
OlehBTSyPDA-102|That is the H3O plus concentration.
OlehBTSyPDA-103|And notice that I left the k like this, because I knew after I get the H3O plus concentration, I'm going to take minus log of it anyway.
OlehBTSyPDA-104|So I'm going to go right back to the exponent.
OlehBTSyPDA-105|So if I calculate the pH, that's minus log of 10 to the minus 4.75, which is 4.75.
OlehBTSyPDA-106|So what I found is that the pH is numerically equal to the pKa.
OlehBTSyPDA-107|That's just coincidental.
OlehBTSyPDA-108|That occurs at the midpoint of titrations, halfway between the start and the equivalence point.
OlehBTSyPDA-109|So where am I on my titration curve?
OlehBTSyPDA-111|My titration curve, I go for my weak acid to its conjugate base.
OlehBTSyPDA-112|And in this calculation, I've added enough to convert half of my initial acid into its conjugate base.
OlehBTSyPDA-113|I'm right smack in the middle here of this buffer region.
OlehBTSyPDA-114|I'm half way towards the equivalence point.
OlehBTSyPDA-116|That's just a feature of titration curves.
OlehBTSyPDA-117|It's interesting numerical equivalence.
OlehBTSyPDA-119|And that's numerically equal to the pKa.
BBqjKt0Qfd4-000|Now we understand how multiple electrons can exist around an atom.
BBqjKt0Qfd4-001|They go into the various orbitals, and they fill the various orbitals following some fairly specific rules.
BBqjKt0Qfd4-010|Here's the first few elements on the periodic table.
BBqjKt0Qfd4-011|Hydrogen has one electron.
BBqjKt0Qfd4-012|It will enter the 1s, the lowest available orbital.
BBqjKt0Qfd4-014|It's called a paramagnetic species.
BBqjKt0Qfd4-015|Hydrogen is paramagnetic in its atomic state.
BBqjKt0Qfd4-016|Helium has paired electrons, because the next electron will go in the 1s orbital paired up.
BBqjKt0Qfd4-017|The 2s orbital is higher in energy, so it's cheaper to spend that little energy to pair than to put the energy in the higher 2s orbital.
BBqjKt0Qfd4-025|So it's cheaper to put 2s paired than to put an electron in the higher energy 2p.
BBqjKt0Qfd4-026|So we're going to have 1s 2, 2s 2 for beryllium.
BBqjKt0Qfd4-027|And that will not be magnetic.
BBqjKt0Qfd4-029|Now, we can continue.
BBqjKt0Qfd4-030|Here's carbon and nitrogen, oxygen, fluorine.
BBqjKt0Qfd4-031|And you'll see that carbon has maximum multiplicity.
BBqjKt0Qfd4-032|The 2p orbitals are the same energy.
BBqjKt0Qfd4-033|So the spins go in parallel.
BBqjKt0Qfd4-034|Nitrogen, same thing-- three unpaired electrons.
BBqjKt0Qfd4-035|So 1s 2, 2s 2, 2p 3-- 3 times as magnetic as carbon, which is more magnetic than boron.
BBqjKt0Qfd4-036|In fact, that's how we can tell Hund's rule is being followed.
BBqjKt0Qfd4-037|Nitrogen is more magnetic than carbon, which is more magnetic than boron.
BBqjKt0Qfd4-038|So the spins are going in parallel.
BBqjKt0Qfd4-039|We can continue with fluorine, neon.
BBqjKt0Qfd4-040|And we see that the electrons start to pair in the 2s and 2p orbitals.
BBqjKt0Qfd4-041|So neon now has 2p 6.
BBqjKt0Qfd4-042|All the electrons are paired in the p shell.
BBqjKt0Qfd4-043|Neon is not magnetic.
BBqjKt0Qfd4-044|But sodium, the next element on the periodic table, will have one more electron.
BBqjKt0Qfd4-045|And now we can start to use a shorthand in our electronic configurations.
BBqjKt0Qfd4-046|We can say sodium has all the electrons that neon has plus a 3s 1.
BBqjKt0Qfd4-047|And of course, an unpaired electron gives you a magnetic state.
BBqjKt0Qfd4-048|So what we're doing is we're filling up the periodic table.
HBsY9ZOsWvY-000|Multiple equilibria occur when there's steps in the chemical reaction or reactions can happen sequentially.
HBsY9ZOsWvY-002|That has equilibrium constant K3 and delta H, delta H3.
HBsY9ZOsWvY-003|So how are K1, K2, and K3 related?
HBsY9ZOsWvY-005|Delta H3 is delta H1 plus delta H2.
HBsY9ZOsWvY-011|That's the equilibrium constant K2.
HBsY9ZOsWvY-012|And I have B over A. That's the equilibrium constant K1.
HBsY9ZOsWvY-013|So K3 is actually K1 times K2.
HBsY9ZOsWvY-014|So sequential reactions or sums of reactions, I multiply.
a2TpE4LATkc-000|Let's look at a different kind of titration, a solubility titration.
a2TpE4LATkc-001|I'm going to take 0.1 molar NaX solutions, where X is either chlorine, bromine, or iodine.
a2TpE4LATkc-002|So sodium chloride, sodium bromide, and sodium iodide are all very soluble.
a2TpE4LATkc-003|So I'll get a 100% dissociation.
a2TpE4LATkc-004|And I'll have chloride ions, or bromide ions, or iodide ions at 0.1 molar in solution.
a2TpE4LATkc-005|I'm going to add silver nitrate.
a2TpE4LATkc-006|Now, silver chloride, silver iodide, and silver bromide are low soluble salts.
a2TpE4LATkc-007|They have a very small Ksp value.
a2TpE4LATkc-008|So silver chloride, silver bromide, and silver iodide will precipitate out of solution.
a2TpE4LATkc-009|The question I have is, from this curve, can you tell which is the least soluble salt?
a2TpE4LATkc-019|We're looking at the titration of sodium salt, sodium chloride, bromide, or iodide, with a silver solution.
a2TpE4LATkc-020|And the silver salts, silver chloride, silver bromide, and silver iodide, are insoluble.
a2TpE4LATkc-021|So they precipitate out of solution.
a2TpE4LATkc-023|Well, the higher the pX, the lower the X concentration.
a2TpE4LATkc-024|Remember, it's minus log of the concentration that gives you the pX, so minus log of X concentration.
a2TpE4LATkc-025|So if the pX is high, then you have 10 to the minus high number versus 10 to the minus low number.
a2TpE4LATkc-026|10 to the minus high number is a smaller number.
a2TpE4LATkc-027|So this represents the smallest concentration of iodine.
DzmtbC7bYUI-000|Let's look at some various kinetic rate laws and various plot to determine the characteristic plot for various rate laws.
DzmtbC7bYUI-001|So for instance, zero order kinetics means the rate is independent of any concentration, so the rate is just a constant.
DzmtbC7bYUI-008|Let's look at first order kinetics.
DzmtbC7bYUI-009|First order kinetics, the rate is k times A to the first power.
DzmtbC7bYUI-010|And if I write that versus time, that gives me a exponential.
DzmtbC7bYUI-011|It's the concentration is e to the minus kt versus time.
DzmtbC7bYUI-018|And I can write that versus time that concentration versus time goes as this expression here.
DzmtbC7bYUI-019|And I have a half life that's dependent on 1 over the initial concentration.
DzmtbC7bYUI-024|So when you find the linear plot, you determine the kinetics.
DzmtbC7bYUI-025|So that's a summary of our characteristic kinetic plot and the parameters for zero order, first order, and second order kinetics.
1N6F0E1ReHk-000|Let's look at the molecular orbitals in benzene.
1N6F0E1ReHk-008|We're talking about the molecular orbitals in benzene.
1N6F0E1ReHk-009|Now, benzene is one of these conjugated, alternating double-bond systems, where each carbon is SP2 hybridized.
1N6F0E1ReHk-010|So it has an available p orbital to form extended molecular orbitals.
1N6F0E1ReHk-011|The question is which combination is the highest energy.
1N6F0E1ReHk-012|Again, we correlate the number of nodes with energy.
1N6F0E1ReHk-013|As the number of nodes increase, the energy goes up.
1N6F0E1ReHk-014|So A has a single node.
1N6F0E1ReHk-015|B has one two three nodes-- and C, two nodes.
1N6F0E1ReHk-016|So the highest energy orbital is the one with the most nodes-- in this case, B.
7A5UuW3EJPc-000|Let's look at a gas phase equilibrium, the equilibrium between NO2, a brown gas, and N2O4, the dimer, which is a clear gas.
7A5UuW3EJPc-001|They're in equilibrium here in this flask, and you can see the brown gas.
7A5UuW3EJPc-003|So since the equilibrium constant is larger than 1, the products are slightly favored, in this case, over the reactants.
7A5UuW3EJPc-004|So the products slightly favored.
7A5UuW3EJPc-005|Even though those products are slightly favored, you can see there are reactants-- that's the brown gas that we can see-- are still present at equilibrium.
7A5UuW3EJPc-006|There's an interchange between these two.
7A5UuW3EJPc-008|The difference in free energy between N2O4 and NO2 at equilibrium is 0.
7A5UuW3EJPc-009|There's no free energy penalty to switch-- interchange-- between products and reactants.
7A5UuW3EJPc-010|And so they do so-- back and forth-- with no energy penalty.
7A5UuW3EJPc-011|That's the nature of equilibrium.
7A5UuW3EJPc-012|You can go between products and reactants because the free energy doesn't change as you do that.
7A5UuW3EJPc-013|Delta G is 0 for equilibrium systems.
7A5UuW3EJPc-014|This reaction happens to be exothermic, and we could have predicted that because what's happening here is a bond is being formed.
7A5UuW3EJPc-015|You're going from a monomer to a dimer, a nitrogen-nitrogen bond turns out to be what is exactly formed here.
7A5UuW3EJPc-016|But even without knowing where the bond is, you can tell there's a bond formation, and bond formations are always exothermic.
7A5UuW3EJPc-017|It always takes energy to break bonds, and energy is always released when I make bonds.
7A5UuW3EJPc-018|So this is a bond making, a bond formation.
7A5UuW3EJPc-023|Because we're about 25 degrees C here in the studio.
7A5UuW3EJPc-025|They'll remain at equilibrium because at equilibrium, no free energy difference between the products and reactants.
7A5UuW3EJPc-026|They interchange freely.
7A5UuW3EJPc-027|It's a dynamic equilibrium, but it's one where k is constant.
7A5UuW3EJPc-028|And remember, regardless of the starting conditions, I'll achieve this k.
7A5UuW3EJPc-030|That's the nature of equilibrium, a balance between free energy and a constant value we call the equilibrium constant.
dEFsl3YYwuM-000|In this lesson, we're going to talk about bonding.
dEFsl3YYwuM-001|One of the kinds of bonds, ionic bonds, are formed when a positively-charged ion is attracted to a negatively-charged ion.
dEFsl3YYwuM-002|That is, the electron is physically transferred from one atom to the other.
dEFsl3YYwuM-003|An example is sodium chloride, where the electron leaves sodium and lands on the chlorine.
dEFsl3YYwuM-004|Now energetically, it takes more energy to remove the electron from sodium than you get back when the electron lands on the chlorine.
dEFsl3YYwuM-005|So it's the resulting plus-minus interaction that's very favorable, that makes the overall bond formation possible.
dEFsl3YYwuM-008|We're going to talk about all the kinds of bonds, though, today in the spectrum of covalent to polar covalent to ionic bonds.
hFWhPUDIhNY-000|When gasses deviate from ideal behavior, we can still calculate P, V and T using equations, and there's a variety of them.
hFWhPUDIhNY-001|The one I've listed here is called a Van der Waals equation.
hFWhPUDIhNY-002|A Van der Waals equation looks a lot like the ideal gas expression.
hFWhPUDIhNY-008|Let's look at some of these A and B parameters.
hFWhPUDIhNY-009|Here's helium.
hFWhPUDIhNY-015|It has Van der Waals dispersion interactions, it has dipole-dipole interactions, and hydrogen bonding interactions.
hFWhPUDIhNY-016|That leads to the largest value for A.
hFWhPUDIhNY-017|Notice the value for b is about the same for all these.
hFWhPUDIhNY-018|The volume of the particle on the grand scale is about the same order of magnitude for all these particles.
hFWhPUDIhNY-020|So when the ideal gas behavior no longer applies, I can apply the Van der Waals equation, and still calculate P, V, and T.
L348PufMnqw-000|Let's do a calculation involving a sparingly soluble salt, silver chloride, and a very soluble salt, sodium chloride.
L348PufMnqw-001|The Ksp, the solubility product, for silver chloride is 1.6 times 10 to the minus 10.
L348PufMnqw-002|For sodium chloride, we don't report a Ksp because it's very soluble.
L348PufMnqw-003|Sodium and chloride ions dissociate completely in solution, and the Ksp would be very much bigger than one.
L348PufMnqw-004|In fact, we really only report Ksp, solubility products, for salts that are sparingly soluble that have Ks less than 1.
L348PufMnqw-005|So all sodium salts, really, and all chlorides are very soluble.
L348PufMnqw-006|So they dissociate completely, and you can assume safely 100% dissociation in water.
L348PufMnqw-014|So x times x equals Ksp.
L348PufMnqw-019|Doesn't matter if I put more solid silver chloride in the flask.
L348PufMnqw-020|The equilibrium has been reached, and the product of the silver ions and the chlorine ions always has to be 1.6 times 10 to the minus 10.
L348PufMnqw-024|The sodium ions don't do anything.
L348PufMnqw-025|They're not involved in this equilibrium.
L348PufMnqw-026|And when sodium and chlorine ions are in solution, they stay as sodium chloride, the separate ions.
L348PufMnqw-027|But when silver and chloride ions are together in solution, they tend to precipitate to form silver chloride.
L348PufMnqw-028|So let's see that happening.
L348PufMnqw-034|That's because this equilibrium constant is very small so this reaction doesn't go very far towards the products.
L348PufMnqw-035|So these concentrations, x, are small, and I think they'll be very small with respect to 0.2.
L348PufMnqw-036|So this 0.2 plus x is essentially 0.2.
L348PufMnqw-037|It's 0.2 plus a tiny amount.
L348PufMnqw-041|That's 10 to the 4th times less soluble than without this chloride ions in solution.
L348PufMnqw-042|So this common ion reduces the solubility of silver chloride by a factor of 10 to the 4th.
L348PufMnqw-043|That's the common ion effect, and this is how we do common ion effect calculations.
49GgQNfExf8-000|Conversion factors are used commonly in chemical calculations.
49GgQNfExf8-001|You convert one set of units to another set of units and almost, anything can be used as a conversion factor.
49GgQNfExf8-002|So any expression of equality can be used as a conversion factor.
49GgQNfExf8-007|Now I can take this, that has value of 1 and multiply it.
49GgQNfExf8-008|It's the multiplicative identity 1.
49GgQNfExf8-009|I can multiply it by anything and not change the value, but the units will change.
49GgQNfExf8-010|So I can use conversion factors to change my units or dimensions from one unit or one dimension to another.
49GgQNfExf8-012|And I can use this as a conversion factor.
49GgQNfExf8-013|So let's try that.
EmCdAb0yiBM-000|Let's do a calculation involving a hydrogen atom where we assume the hydrogen atom behaves like a particle in a box.
EmCdAb0yiBM-001|And that's not a bad assumption.
EmCdAb0yiBM-002|The hydrogen atom has an electron that's bound around a nucleus, that's an electron that has boundaries on it.
EmCdAb0yiBM-003|And remember, when you take a wave like property, and that's the electron, and put boundaries on it, you naturally get quantized energy levels.
EmCdAb0yiBM-004|So let's do a quantum mechanical calculation on a hydrogen atom.
EmCdAb0yiBM-005|Already at this point, in Chem 1, we can do a quantum mechanical calculation.
EmCdAb0yiBM-008|So the particle energy levels depend only on the length of the box and the mass of the particle.
EmCdAb0yiBM-009|We know both those things, so let's do the calculation.
EmCdAb0yiBM-010|150 picometer box, an electron mass that we know, n we know, h we know.
EmCdAb0yiBM-011|We can simply say, well, if it's a transition, I have to subtract n3 from n equal 2.
EmCdAb0yiBM-012|I can do that.
EmCdAb0yiBM-014|Always use meters, kilograms, and seconds-- joules for energy.
EmCdAb0yiBM-017|Lambda the wavelength is what we want, and we can calculate that.
EmCdAb0yiBM-018|So if this energy is hc over lambda, then lambda is hc over the energy.
EmCdAb0yiBM-021|So this is a very hand-wavy, approximate calculation.
EmCdAb0yiBM-022|But you get kind of in the same ballpark, within a factor of one order of magnitude.
EmCdAb0yiBM-023|A factor of about tenish of the wavelength, just doing a simple particle-in-the-box calculation.
EmCdAb0yiBM-024|That's the beauty of quantum mechanics.
EmCdAb0yiBM-025|It has a lot of power to express the very tiny properties of matter.
vQ_ujXXGRZs-000|Let's look at a system where we take copper metal and immerse it in sulfuric acid solution.
vQ_ujXXGRZs-008|We're talking about immersing some copper metal in sulfuric acid solution.
vQ_ujXXGRZs-009|So we want to know what is the more stable state-- copper ions, hydrogen gas, or will no reaction occur?
vQ_ujXXGRZs-012|So the favored state is copper metal over hydrogen gas.
vQ_ujXXGRZs-013|And in fact, this half cell would effectively cause the hydrogen half cell to run in reverse.
vQ_ujXXGRZs-015|So essentially, we're already at the favored state.
vQ_ujXXGRZs-016|We're at the product state already-- copper metal and hydrogen ions.
vQ_ujXXGRZs-017|So this reaction will not proceed any further.
vQ_ujXXGRZs-018|The free energy difference here is negative K is greater than 1.
vQ_ujXXGRZs-019|The products are favored.
vQ_ujXXGRZs-020|And the products are essentially where we started here.
5t3xzqSPgMo-000|Let's look at a chemical reaction and determine what temperature range it will be favorable for.
5t3xzqSPgMo-001|So where is delta G negative?
5t3xzqSPgMo-002|The chemical reaction, carbon dioxide and calcium oxide forming calcium carbonate, the formation of limestone essentially.
5t3xzqSPgMo-004|The question I have-- what temperature range is this reaction spontaneous as written?
5t3xzqSPgMo-013|We're looking at carbon dioxide and calcium oxide forming calcium carbonate.
5t3xzqSPgMo-015|So kilojoules and jewels, a factor of 1,000 between the two.
5t3xzqSPgMo-016|Now, delta S is negative.
5t3xzqSPgMo-017|So if I'm talking about the free energy, I have a minus T delta S contribution.
5t3xzqSPgMo-018|The minus T delta S contribution with a minus delta S will have a positive overall contribution.
5t3xzqSPgMo-024|For temperatures less than 1,000, then the T delta S term is less than magnitude the delta H term.
5t3xzqSPgMo-025|So this positive contribution is smaller than this negative contribution, so overall delta G is negative.
5t3xzqSPgMo-026|So this reaction is spontaneous for temperatures less than 1,000 degrees.
5t3xzqSPgMo-027|The correct answer here is A.
AE8u8V-Pl-c-000|We've talked about photons as the smallest particle of light.
AE8u8V-Pl-c-001|And we said when you get down to the photon level, if you split that photon further, it loses the properties of that light.
AE8u8V-Pl-c-002|And, indeed, splitting a blue photon into other photons is possible, but it no longer has the properties of the blue light it originally had.
AE8u8V-Pl-c-003|It's the smallest particle of blue light that you could have, but splitting it is still possible.
AE8u8V-Pl-c-004|So let's talk about that.
AE8u8V-Pl-c-005|A photon at 400 nanometers is split into two.
AE8u8V-Pl-c-006|One of the photons that comes out is at 1,200 nanometers.
AE8u8V-Pl-c-007|What is the wavelength of the other photon?
AE8u8V-Pl-c-008|So let's consider that.
AE8u8V-Pl-c-009|Is it A, 200 nanometers?
AE8u8V-Pl-c-010|B, 600?
AE8u8V-Pl-c-011|Or C, 800 nanometers?
AE8u8V-Pl-c-021|When we split a photon into two other photons, the total energy that we start with can't be lost.
AE8u8V-Pl-c-022|So energy will be conserved.
AE8u8V-Pl-c-023|So the two smaller photon energies must add to the original photon energy.
AE8u8V-Pl-c-026|So indeed, two lower energy photons add to give the high energy photon.
G9YGwK0mP_g-000|Let's talk about molecular structure and geometry.
G9YGwK0mP_g-001|A good starting point to talk about it would be the Louis dot structure.
G9YGwK0mP_g-004|Molecules, of course, have all of three dimensions to work with.
G9YGwK0mP_g-005|So rather than being stuck square planar, a molecule can use a third dimension.
G9YGwK0mP_g-007|So you want to use all three dimensions to do that.
G9YGwK0mP_g-008|One thing the Lewis dot structure can do, though, is tell you what you have to separate in space.
G9YGwK0mP_g-009|Because the Lewis dot structure does give you the overall connectivity-- who's connected to what-- in the molecule.
G9YGwK0mP_g-010|For instance, this carbon is connected to 1, 2, 3 hydrogens and another carbon.
G9YGwK0mP_g-011|This oxygen has to accommodate 1, 2 lone pairs, this carbon atom, and that hydrogen.
G9YGwK0mP_g-012|So we'll define the steric number as the number of things that each atom has to accommodate-- lone pairs plus bonded atoms.
G9YGwK0mP_g-013|We'll count up lone pairs-- 1, 2-- and bonded atoms-- 1, 2-- to come up with a steric number of 4 for this oxygen.
G9YGwK0mP_g-014|When things have a steric number of 4, we'll see, they arrange themselves not in a square planar configuration, but in a tetrahedral configuration.
G9YGwK0mP_g-016|So using all three dimensions to separate those four things as far away from each other in space as possible.
qsNst-6SB0c-000|What we're looking for is a thermodynamic parameter that will predict the direction of physical processes and chemical reactions.
qsNst-6SB0c-001|We know enthalpy isn't sufficient.
qsNst-6SB0c-002|Reactions can go with releasing energy and absorbing energy.
qsNst-6SB0c-003|So we need another parameter.
qsNst-6SB0c-004|Well, let's look at this physical process.
qsNst-6SB0c-005|This is a gas.
qsNst-6SB0c-006|And we're going to let it expand into a vacuum.
qsNst-6SB0c-009|When this process goes, expanding against a vacuum, no work is done.
qsNst-6SB0c-010|It's an isothermal process.
qsNst-6SB0c-011|No heat is absorbed or released.
qsNst-6SB0c-012|No energy changes.
qsNst-6SB0c-013|So this process has three thermodynamic parameters.
qsNst-6SB0c-014|The energy change, the work and the heat all zero.
qsNst-6SB0c-015|There is no indicator among our thermodynamic parameters that this process will even proceed.
qsNst-6SB0c-016|Nothing changes thermodynamically, as far as we can see.
qsNst-6SB0c-017|So we need another thermodynamic parameter.
qsNst-6SB0c-018|Since the reverse never occurs, there has to be something, some driving force that makes it go in that direction.
qsNst-6SB0c-019|Yet, it's not energy, heat, or work.
qsNst-6SB0c-020|Well, it turns out a statistical argument is what we're looking for.
qsNst-6SB0c-021|And let's consider this situation in a very simple case.
qsNst-6SB0c-022|Instead of a mole of particles, let's just take two.
qsNst-6SB0c-025|So there's two equally likely energy states for the one on either side, the equal distribution.
qsNst-6SB0c-028|And we can track them.
qsNst-6SB0c-029|Because that actually tracks mathematically as binomial coefficients.
qsNst-6SB0c-030|Or we can use a Pascal's Triangle to track the number of ways you can arrange the system with equal distribution.
qsNst-6SB0c-031|The way you do a Pascal's Triangle is you take one and one.
qsNst-6SB0c-032|And then for two particles, you add the two numbers to get the lower number.
qsNst-6SB0c-033|So essentially, there's a zero here.
qsNst-6SB0c-034|So I would add zero and one to get one, one and one to get two, and one and zero to get one.
qsNst-6SB0c-035|This tells us, for two particles, what we already know.
qsNst-6SB0c-036|If you have both on one side, there's one way to do that.
qsNst-6SB0c-037|If you have one on each side, there's two ways to do that.
qsNst-6SB0c-038|And essentially, both on the other side, there's one way to do that.
qsNst-6SB0c-039|So twice as likely to see the one on each side solution.
qsNst-6SB0c-040|And you can expand the Pascal's Triangle easily, just keep doing that addition.
qsNst-6SB0c-047|So here are six particles.
qsNst-6SB0c-048|And I've made them distinguishable.
qsNst-6SB0c-049|So we can tell, three on each side, there's 20 ways to arrange this.
qsNst-6SB0c-050|And you could go through there.
qsNst-6SB0c-051|There's only 20.
qsNst-6SB0c-052|It would take a few minutes.
qsNst-6SB0c-053|But you could say, well, it could be the green over here, or the yellow over here with the brown and the blue.
qsNst-6SB0c-054|And you could keep going and find all 20 of those arrangements.
qsNst-6SB0c-055|The point is though, all those are equally likely.
qsNst-6SB0c-056|And there's 20 of them versus the one way to have them all on one side.
qsNst-6SB0c-058|Now, as you go to more particles, that becomes even more pronounced.
qsNst-6SB0c-060|So 100 trillion times more likely to see them equally distributed as all on one side, that's dramatic, for just 50 particles.
qsNst-6SB0c-063|In fact, I would call it bigger than astronomically large.
qsNst-6SB0c-064|Because this is bigger than the total number of particles in the universe.
qsNst-6SB0c-065|So it's just astronomically, bigger than astronomically, more likely to find the particles equally distributed between both sides.
qsNst-6SB0c-066|We call that, in fact, statistical inevitability.
qsNst-6SB0c-067|If you look at this system every second for a million lifetimes, the likely case is the one you'll see.
qsNst-6SB0c-068|It's trillions and trillions and trillions of times more likely that you'll see them equally distributed.
qsNst-6SB0c-069|So that's the likely case you'll see.
qsNst-6SB0c-070|No one has ever observed them all over on one side.
qsNst-6SB0c-072|So this is actually a measure of the direction that the universe likes to go.
qsNst-6SB0c-073|That is, the natural progression of things is towards the most likely arrangement.
qsNst-6SB0c-074|The most likely arrangement is the one with the most possible ways, the most possible microstates, that are all equal.
qsNst-6SB0c-076|That's a lot of ways, a very large number of ways to arrange that particular arrangement.
qsNst-6SB0c-077|It's the most likely arrangement.
qsNst-6SB0c-081|And this is our new thermodynamic parameter.
qsNst-6SB0c-082|Because when the entropy increases, that's the favored direction of the system.
qsNst-6SB0c-083|Systems move towards this distribution of states.
qsNst-6SB0c-084|The more ways I can arrange energy is the important facet of the universe.
qsNst-6SB0c-085|I go towards distributing energy among the most possible number of ways.
qsNst-6SB0c-086|So large numbers of microstates are favored by the universe.
qsNst-6SB0c-090|So this is a thermodynamic parameter that is not conserved.
qsNst-6SB0c-094|Spontaneous is the natural direction of the universe.
qsNst-6SB0c-095|And it occurs when entropy of the universe increases.
qsNst-6SB0c-096|Those are also called irreversible processes.
qsNst-6SB0c-097|That is, we never observe the reverse process happening by itself.
qsNst-6SB0c-098|Now, that doesn't mean we can't make the reverse process happen.
qsNst-6SB0c-099|You know we could create a system that would have a piston, and push all these gases back to one side, and evacuate this side.
qsNst-6SB0c-100|But in doing so, we'd have to bring energy in from the surroundings.
qsNst-6SB0c-104|That is, the systems and surroundings just balance each other out.
qsNst-6SB0c-105|Or there there's no change in entropy in the system and the surroundings.
qsNst-6SB0c-106|That would be saying that each side of a chemical reaction is equally likely.
qsNst-6SB0c-107|That is, there's no entropy penalty for being on one side or the other side.
qsNst-6SB0c-108|I can be a reactant or I can be a product.
qsNst-6SB0c-109|Going between the two, there's no entropy change of the universe.
qsNst-6SB0c-110|So it's equally likely that I'm at either side.
qsNst-6SB0c-111|That's the definition of equilibrium.
qsNst-6SB0c-112|If it's equally likely for me to be here or here, no energy penalty, then I'll switch between the two freely.
qsNst-6SB0c-113|I'll be at equilibrium.
qsNst-6SB0c-114|Now, entropy decreasing in the universe, those processes are not possible.
qsNst-6SB0c-115|So now, we have a thermodynamic parameter that will tell us the direction of things.
qsNst-6SB0c-117|It's entropy that determines the favored direction in the universe.
wFY9ZvUmM5o-000|So we've learned that light has a particle nature.
wFY9ZvUmM5o-001|There's a wave property associated with light and electromagnetic radiation, and also a particle nature-- packets of energy being carried along the wave.
wFY9ZvUmM5o-002|A brilliant experiment that demonstrates that property is the photoelectric effect.
wFY9ZvUmM5o-003|And here's how it works.
wFY9ZvUmM5o-004|You take a piece of metal.
wFY9ZvUmM5o-005|Now metal is an array of metal atoms, and each of those atoms holds on rather loosely to its outer electrons.
wFY9ZvUmM5o-006|That's why the metal conducts electricity.
wFY9ZvUmM5o-007|Those electrons are rather free to move about the surface.
wFY9ZvUmM5o-008|Now, if you shine light on that surface, what happens?
wFY9ZvUmM5o-012|And even if you make the light very bright, very intense, nothing happens.
wFY9ZvUmM5o-015|It's as if photons are striking electrons and kicking them off the metal.
wFY9ZvUmM5o-016|You bring in a more intense green light, and you get more electrons.
wFY9ZvUmM5o-017|They don't go away faster.
wFY9ZvUmM5o-018|You just get more electrons with a brighter light, all with the same kinetic energy.
wFY9ZvUmM5o-020|And again, the same correlation with brightness-- if you make the light brighter, you get more electrons per second released.
wFY9ZvUmM5o-025|So it doesn't matter if there's more of them.
wFY9ZvUmM5o-028|More energy still in blue photons ejects electrons with even more kinetic energy.
wFY9ZvUmM5o-029|So it's as if the photons of light are coming in and jostling electrons.
wFY9ZvUmM5o-030|I'm holding onto this tennis ball.
wFY9ZvUmM5o-031|Photons are coming in and jostling them.
wFY9ZvUmM5o-033|Can I have my tennis ball back?
wFY9ZvUmM5o-035|Now brightness doesn't matter.
wFY9ZvUmM5o-036|We said, well, brightness is just more photons per second.
wFY9ZvUmM5o-037|That's just peh, peh, peh, peh, peh, peh, peh, peh, peh, peh, peh, peh, peh, peh-- but not enough energy-- one photon per electron to eject any single electron.
wFY9ZvUmM5o-038|But big photons, high energy, blue light, say, comes in and slams that metal and really sends the electron flying.
wFY9ZvUmM5o-039|[SOUND OF BALL HITTING OBJECTS] CREW: Ow!
wFY9ZvUmM5o-041|[LAUGHS] Sorry, guys.
wFY9ZvUmM5o-042|[LAUGHS] High energy is what we have.
wFY9ZvUmM5o-043|So we can actually plot it.
wFY9ZvUmM5o-051|So all the excess energy of the photon goes into kinetic energy.
wFY9ZvUmM5o-052|So you can write the both energies in terms of photons.
wFY9ZvUmM5o-053|And you can realize there is a minimum photon energy required to eject the electron from the metal.
wFY9ZvUmM5o-054|If you look at different metals, different metals have different threshold frequencies.
wFY9ZvUmM5o-055|For instance, you could have a metal that's described by a blue photon is the minimum photon that ejects an electron.
wFY9ZvUmM5o-058|And it's actually Albert Einstein, a genius, who looked at this problem.
wFY9ZvUmM5o-060|But increasing the intensity-- very bright light-- didn't do anything.
wFY9ZvUmM5o-061|And when you could eject electron, increasing the intensity didn't increase the energy of the electrons.
wFY9ZvUmM5o-062|You just got more electrons coming off with the same energy.
wFY9ZvUmM5o-063|Well, it takes a genius, often, to look at a very troubling problem and see it in a whole new light.
wFY9ZvUmM5o-064|And that's what Einstein did.
wFY9ZvUmM5o-065|He said, well, that looks like the light's behaving like particles.
wFY9ZvUmM5o-066|It looks like little bits of light are coming in.
wFY9ZvUmM5o-067|So a bright light is just lots of bits.
wFY9ZvUmM5o-068|But they all have the same energy.
wFY9ZvUmM5o-069|So those lots of bits eject lots of electrons, each electron with the same energy.
wFY9ZvUmM5o-070|So the photoelectric effect and Albert Einstein have helped us understand the particle nature of light.
_56-KofIBng-000|Let's do a calculation with electromagnetic radiation, the properties of wavelength and frequency.
_56-KofIBng-001|What frequency and designation of radiation, with wavelength 8.83 picometers, is emitted from Technetium 99 during its nuclear decay?
_56-KofIBng-002|So, we understand a wavelength, 8.83 picometers.
_56-KofIBng-003|We can change that into a frequency knowing that the waves travel at the speed of light, c.
_56-KofIBng-010|So very, very, very high frequency.
_56-KofIBng-011|Very short wavelength, as we'd expect.
_56-KofIBng-012|What region of the electromagnetic spectrum does that correspond to?
_56-KofIBng-014|So we have gamma waves emitted from technetium 99 during its nuclear decay.
j7ROvrT6W1M-000|The polymerization of ethylene to polyethylene is taking individual monomers of ethylene and making a polymer.
j7ROvrT6W1M-001|Polymer means many elements.
j7ROvrT6W1M-002|That term poly means many.
j7ROvrT6W1M-015|So that's a wash, equal energy for breaking and making.
j7ROvrT6W1M-016|So if I break, let's say, just three carbon-carbon double bonds, how many single bonds do have to make to hook the chain together?
j7ROvrT6W1M-018|You're breaking N carbon-carbon double bonds and making 2 N, twice as many carbon-carbon single bonds.
j7ROvrT6W1M-019|So let's see how that energy balance works.
j7ROvrT6W1M-022|I get back 700.
j7ROvrT6W1M-024|The difference between those is a net release of energy.
1lHCPo6GH3I-000|Let's look at forming some molecular orbitals from atomic orbitals.
1lHCPo6GH3I-004|Negative signs and positive signs in the wave function gives zero somewhere in the middle.
1lHCPo6GH3I-005|So that mathematical combination will give me a node right between the two atoms.
1lHCPo6GH3I-006|That's not very good for bonding, that will be high energy.
1lHCPo6GH3I-009|So the highest energy will be the most nodes, the node right between the nuclei.
mKzVNK6FeKQ-000|Let's look at a common process-- a gas expanding adiabatically-- quickly-- against a constant pressure.
mKzVNK6FeKQ-001|If the gas expands adiabatically against constant pressure, what happens to the temperature of the gas?
mKzVNK6FeKQ-007|We're talking about the adiabatic expansion of an ideal gas against a constant pressure.
mKzVNK6FeKQ-008|And this system is like discharging an aerosol can.
mKzVNK6FeKQ-009|The gas expands out the nozzle of the can and pushes back the atmosphere.
mKzVNK6FeKQ-013|The system does work, uses its internal energy.
mKzVNK6FeKQ-014|If I use my internal energy, my internal energy goes down.
mKzVNK6FeKQ-015|If I'm a gas, a drop in internal energy always is accompanied by a drop in the temperature.
mKzVNK6FeKQ-016|So here, the temperature decreases.
mKzVNK6FeKQ-017|And you can try this at home.
mKzVNK6FeKQ-018|Discharge an aerosol can and feel the top of the can.
mKzVNK6FeKQ-019|It will be colder.
qhg-pZ-f-PM-000|Biology and chemistry are intimately related, because we understand biology now in terms of the molecules of biology.
qhg-pZ-f-PM-001|And molecules are, of course, the purview of chemistry.
qhg-pZ-f-PM-002|So let's look at some biological molecules.
qhg-pZ-f-PM-003|Here's ATP-- adenosine triphosphate, one of the most important and ubiquitous molecules in your cells.
qhg-pZ-f-PM-006|So let's look at ATP more carefully.
qhg-pZ-f-PM-007|ATP is adenosine triphosphate.
qhg-pZ-f-PM-014|Now, in biological solution, where you're around pH 7, these protons will be removed.
qhg-pZ-f-PM-016|So when you're on the basic side of the pKa, above the pKa, in pH, the basic form predominates.
qhg-pZ-f-PM-017|So at pH 7, ATP is a highly charged molecule.
qhg-pZ-f-PM-018|In fact, that's one reason that it's a good energy storage and transfer molecule.
qhg-pZ-f-PM-019|These highly charged nature means breaking it back apart separates those charges and that's downhill in energy.
qhg-pZ-f-PM-020|So ATP is a good energy storage and transfer molecule.
qhg-pZ-f-PM-021|Now, ATP has a "high energy"-- I put that in quotes-- phosphate bond, because you read that sometimes in textbooks.
qhg-pZ-f-PM-022|But high energy phosphate bond is actually something of a misnomer.
qhg-pZ-f-PM-023|It implies that if you break that bond, energy is released.
qhg-pZ-f-PM-024|And of course, as chemists, we know nothing could be further from the truth.
qhg-pZ-f-PM-025|It always requires energy to break bonds.
qhg-pZ-f-PM-026|You have to put energy in to break a bond.
qhg-pZ-f-PM-027|You're pulling the bond apart.
qhg-pZ-f-PM-029|But it's not the breaking of this bond that releases energy, it's the forming of other, more stable, and in fact, higher energy, bonds.
qhg-pZ-f-PM-030|So let's look at that.
qhg-pZ-f-PM-039|So protons will be produced in the reaction too, and in general those protons just jump right onto the phosphate that's formed.
qhg-pZ-f-PM-042|And that is a stronger bond than this one that breaks.
qhg-pZ-f-PM-047|Overall, you have a release in energy when you hydrolyze ATP.
7ylCr2MXVBU-000|Let's look at the photoelectric effect again.
7ylCr2MXVBU-001|Which combination of a photon striking a metal ejects an electron with the highest kinetic energy?
7ylCr2MXVBU-002|So I have two metals represented and several different photon energies.
7ylCr2MXVBU-003|So is it a yellow photons striking metal one, a green photon striking metal one, or a blue photon striking metal number two?
7ylCr2MXVBU-011|So which combination of photon and metal gives the electron with the highest kinetic energy?
7ylCr2MXVBU-012|Let's look at all three.
7ylCr2MXVBU-013|Yellow light striking metal one-- well, I've outlined about where the frequencies are.
7ylCr2MXVBU-014|But, of course, a single frequency can't encompass the whole band of green and the whole band of blues.
7ylCr2MXVBU-015|So the largest yellow is somewhere below the green.
7ylCr2MXVBU-016|That's all we know.
7ylCr2MXVBU-017|So it's somewhere in here.
7ylCr2MXVBU-019|What about green striking metal one?
7ylCr2MXVBU-021|Striking metal one will give a higher kinetic energy.
7ylCr2MXVBU-022|What about blue photons striking metal two?
7ylCr2MXVBU-023|Well, here's the blue region.
7ylCr2MXVBU-026|So a blue photon, even though it's the highest energy, is striking a higher threshold metal, resulting in lower kinetic energy photo electrons.
7ylCr2MXVBU-027|So green light on metal one will give the highest energy electrons ejected from this metal system.
pfpNo10xcTw-000|In oxidation-reduction reactions, electrons flow from reducing agents to oxidizing agents.
pfpNo10xcTw-001|And those electrons tend to flow even if the reactants and products are in separate beakers.
pfpNo10xcTw-002|We call the separate beakers half cells.
pfpNo10xcTw-003|And we can catalog the various half cells with respect to each other in terms of the potential to transfer electrons.
pfpNo10xcTw-007|How do I physically set that up?
pfpNo10xcTw-009|The one atmosphere and the 1 molar are to ensure that I'm at the standard state.
pfpNo10xcTw-010|Now, I'll put a platinum electrode in there, and the platinum electrode acts as an inert conduit for the electrons to flow.
pfpNo10xcTw-011|Now, I can pair this electrode with any of my other half cell electrodes.
pfpNo10xcTw-012|So let's look at that.
pfpNo10xcTw-015|That 0.34 volt we attribute entirely to the copper ion, copper metal half cell.
pfpNo10xcTw-016|We say that the hydrogen ion, hydrogen gas half cell has a zero potential.
pfpNo10xcTw-019|So the opposite potential.
pfpNo10xcTw-020|Electrons flow in the opposite direction in this case.
pfpNo10xcTw-021|So zinc ion, zinc metal electrode has a lower potential than the hydrogen ion, hydrogen gas electrode.
pfpNo10xcTw-022|Now, I can do this for a wide variety of electrodes, but you can see already this allows me to take different pairings.
pfpNo10xcTw-023|Now I know copper metal, copper ion relative to zinc metal, zinc ion.
pfpNo10xcTw-025|Well, we can set up the standard notation.
pfpNo10xcTw-026|When we do this, we always take the electrode on the right minus the electrode on the left.
pfpNo10xcTw-027|So in that case, the copper electrode minus the zinc electrode, and we subtract those two voltages.
pfpNo10xcTw-028|And you get a standard voltage for the copper-zinc system of 1.1 volts.
0Ot2WbfIeMQ-000|In oxidation reduction reactions, electrons flow from a reducing agent to an oxidizing agent.
0Ot2WbfIeMQ-001|And they flow because the free energy of the reactants are higher than the free energy of the products, so it's a down hill in free energy.
0Ot2WbfIeMQ-002|You can think of it as a gravitational potential.
0Ot2WbfIeMQ-004|So there's an electrical potential between the reducing agent and the oxidizing agent.
0Ot2WbfIeMQ-005|And that electrical potential exists whether those are in physical contact or not.
0Ot2WbfIeMQ-008|Now we know that the electrons want to flow from the zinc metal to the copper ions.
0Ot2WbfIeMQ-009|That's the natural direction, the downhill direction of this system.
0Ot2WbfIeMQ-010|But they're in physically separate beakers.
0Ot2WbfIeMQ-011|The zinc metal is not in contact with the copper ions.
0Ot2WbfIeMQ-017|And a wire connects the two.
0Ot2WbfIeMQ-024|This set up I have here is called a standard galvanic cell.
0Ot2WbfIeMQ-025|An anode and a cathode, and electron flow between them.
0Ot2WbfIeMQ-026|One way that I remember this is oxidation and anode both start with a vowel, and reduction and cathode both start with a consonant.
0Ot2WbfIeMQ-027|So that helps me keep straight, the oxidation occurs at the anode in a galvanic cell, and the reduction occurs at the cathode.
0Ot2WbfIeMQ-028|Electrons flow from the anode to the cathode.
0Ot2WbfIeMQ-030|So the flow is from an anode, negatively charged, to a cathode, positively charged.
0Ot2WbfIeMQ-035|And that describes this half cell here.
0Ot2WbfIeMQ-036|The zinc metal in contact with zinc ions.
0Ot2WbfIeMQ-037|Now there's a double bar here.
0Ot2WbfIeMQ-038|That's another connection.
0Ot2WbfIeMQ-039|And what this connection is is a salt bridge.
0Ot2WbfIeMQ-040|Now we need the salt bridge.
0Ot2WbfIeMQ-041|And why is that?
0Ot2WbfIeMQ-042|If we want this electrons to continue to flow, that continued flow would cause charge to build up on one side.
0Ot2WbfIeMQ-043|What the salt bridge does is allow that charge difference to be equalized.
0Ot2WbfIeMQ-045|So that's what our salt bridge does.
0Ot2WbfIeMQ-047|So I actually have this cell that I've described here, this galvanic cell and it's two half cells right here on the bench top.
0Ot2WbfIeMQ-051|And the potential in this case is 1.1 volts.
0Ot2WbfIeMQ-052|That's the potential for electrons to flow in a galvanic cell to separated half cells in a redox reaction.
f-SJBvBHpuM-000|Let's look at sets of four quantum numbers.
f-SJBvBHpuM-001|When I give you four quantum numbers, I define an electron in an atom.
f-SJBvBHpuM-002|I give you an n, an l, an m sub l, and an m sub s.
f-SJBvBHpuM-003|That tells you where in the atom you'll find that electron.
f-SJBvBHpuM-004|So when I have a set of four quantum numbers, some are allowed and some aren't.
f-SJBvBHpuM-006|And for values of l, only certain values of m sub l are allowed.
f-SJBvBHpuM-007|So certain quantum states are not allowed.
f-SJBvBHpuM-009|So I have 4, 2, -1, 1/2.
f-SJBvBHpuM-010|That's a set of four going in the sequence n, l, m sub l, m sub s.
f-SJBvBHpuM-011|So n equal 4, l equal 2.
f-SJBvBHpuM-012|That's fine.
f-SJBvBHpuM-013|L can be any value in integers between 0 and n -1.
f-SJBvBHpuM-014|So when l is smaller than n, we're in good shape.
f-SJBvBHpuM-015|So so far, OK.
f-SJBvBHpuM-016|-1, that's a possible value of m sub l, that's fine.
f-SJBvBHpuM-017|Because when l is 2, m sub l goes from -2 to 2 in integer steps.
f-SJBvBHpuM-018|-2, -1, 0, 1, and 2 are all possible values.
f-SJBvBHpuM-019|And -1 is certainly available.
f-SJBvBHpuM-020|1/2 is one of the possible values of m sub s, so that's fine.
f-SJBvBHpuM-024|We have n equal 5, l equal 0.
f-SJBvBHpuM-025|Those are both fine.
f-SJBvBHpuM-026|But for m sub l we have -1.
f-SJBvBHpuM-027|Now when l is 0, all m sub l can be is 0.
f-SJBvBHpuM-028|So this quantum number here is out of line.
f-SJBvBHpuM-029|M sub l can only be zero if l is 0.
f-SJBvBHpuM-030|So that's not an allowed set of quantum numbers.
f-SJBvBHpuM-031|This final one here, we have n equal 4.
f-SJBvBHpuM-032|We have l equal 4.
f-SJBvBHpuM-033|Now the maximum value of l is n -1.
f-SJBvBHpuM-038|So you have to very carefully apply your rules to see which quantum numbers are allowed.
f-SJBvBHpuM-039|That determines which orbitals exist around an atom.
R2TrrmdGM2A-000|Chemical bonds are characterized in a variety of ways.
R2TrrmdGM2A-001|Ionic bonds, for instance, are where the electron is transferred from one element to the other, and it's a plus-minus Coulombic interaction that holds the bond together.
R2TrrmdGM2A-002|That occurs when an electronegativity difference is very large between the two elements.
R2TrrmdGM2A-004|Lithium fluoride is an example of that.
R2TrrmdGM2A-005|Now you can have bonds where the electron is shared more equally between the two.
R2TrrmdGM2A-006|You can imagine a perfect sharing.
R2TrrmdGM2A-007|For instance, covalent bonding is where sharing occurs, and chlorine Cl2 is an example where you have a homonuclear bond.
R2TrrmdGM2A-008|That is, the electron has a preference for neither chlorine.
R2TrrmdGM2A-009|chlorines are identical, the electron has equal preference, there's an equal probability of finding the electron on either chlorine atom.
R2TrrmdGM2A-010|So that bond is perfectly covalent, an equal sharing.
R2TrrmdGM2A-011|But it's actually more likely that you'll have bonds between different elements.
R2TrrmdGM2A-013|So those are called polar covalent bonds.
R2TrrmdGM2A-014|An example, hydrogen chloride.
R2TrrmdGM2A-015|Here, the chlorine is the more electronegative element.
R2TrrmdGM2A-016|So it draws the electron toward itself and there's a higher probability of finding the electron around the chlorine.
R2TrrmdGM2A-017|That gives the chlorine a slight negative charge and the hydrogen a slight positive charge.
R2TrrmdGM2A-018|So the hydrogen end of the molecule, slightly positive; the chlorine end of the molecule, slightly negative, because there's a higher probability of finding the electron around the chlorine.
R2TrrmdGM2A-020|A dipole moment is a vector quantity, it has size-- the size of the charge separation-- and the distance between the two.
R2TrrmdGM2A-021|So in this case, we draw a dipole moment vector from the positive to the negative end of the band.
R2TrrmdGM2A-022|Now, virtually every bond is going to have a dipole moment because it's very rare you find perfectly equal sharing of electrons.
R2TrrmdGM2A-023|So if all bonds have dipole moments, it's likely that all molecules will have dipole moments.
R2TrrmdGM2A-024|But there will be cases where the symmetry of the molecule allows dipole moment vectors to cancel.
R2TrrmdGM2A-026|There are other symmetry examples where that can happen.
R2TrrmdGM2A-028|And usually, the molecules will have a permanent dipole moment.
R2TrrmdGM2A-030|And I can demonstrate the dipole moment in water, because the dipole moment creates a charged end of the molecule that will be attracted to a charged surface.
R2TrrmdGM2A-031|So let's look at that.
R2TrrmdGM2A-032|What I have here is a setup for a stream of water, I'm going to bring a stream of water from this burette.
R2TrrmdGM2A-034|So I'll start a water flow from this burette.
R2TrrmdGM2A-038|That's a very dramatic demonstration of dipole moment in molecules.
1JrO7QQ1o1M-000|If you take a particle that behaves like a wave, and you trap it, you put boundaries on it.
1JrO7QQ1o1M-001|You trap it in a box, it turns out it can only have certain energies.
1JrO7QQ1o1M-002|It's not like when you take a marble, and you put it in a box, and you shake the marble around.
1JrO7QQ1o1M-003|It can have any old kinetic energy it wants.
1JrO7QQ1o1M-004|It can be going fast, it can be going slow.
1JrO7QQ1o1M-005|It can be going in a continuous amount of velocities in between.
1JrO7QQ1o1M-006|That's not true for a particle that behaves like a wave.
1JrO7QQ1o1M-007|Particles that behave like a wave can have only certain energies.
1JrO7QQ1o1M-008|The energies are quantized.
1JrO7QQ1o1M-009|Remember wave plus boundaries gives you quantization.
1JrO7QQ1o1M-010|So I have some quantized levels written here, and quantization is the fundamental property that allows absorption and emission of light.
1JrO7QQ1o1M-011|Because if you're going to make the system go from one level to the next, you can't absorb just any old wavelength.
1JrO7QQ1o1M-012|You have to absorb the wavelength that actually fits, that gives you the exact amount of energy to go from low state to high-energy state.
1JrO7QQ1o1M-013|All other energy levels, and all other energies that the system experiences will be ignored.
1JrO7QQ1o1M-017|So that is transmitted.
1JrO7QQ1o1M-018|Here's a wavelength that exactly matches this high-energy transition, that one will be absorbed.
1JrO7QQ1o1M-024|Some wavelengths are too big, some are too little, some are just right.
1JrO7QQ1o1M-025|So you have another, just right.
1JrO7QQ1o1M-026|That will be absorbed and perhaps you'll find a third just right energy level that will be absorbed.
1JrO7QQ1o1M-027|So you get an absorption spectrum that says, oh, this blue, this yellow, and this red are removed from the continuous light that's hitting this object.
1JrO7QQ1o1M-028|Or you could have the light emit by the system.
1JrO7QQ1o1M-029|The system could emit the energy of the highway of length, and then you'd have a single emission.
1JrO7QQ1o1M-030|It can't emit all wavelengths of light, only certain wavelengths of light are emitted.
OSts9bfX6cA-001|Do you think that will look like A, B, or C?
OSts9bfX6cA-008|We're talking about the conductometric titration of barium hydroxide with H2SO4.
OSts9bfX6cA-009|So we want to know how does the light intensity change with added H2SO4.
OSts9bfX6cA-010|Will it stay bright?
OSts9bfX6cA-011|Will it get dim initially and return?
OSts9bfX6cA-012|Or will it get dim and stay dim?
OSts9bfX6cA-013|Well, we have this on the bench top.
OSts9bfX6cA-014|So let's actually do the experiment.
OSts9bfX6cA-015|Here I have barium hydroxide in this lower flask.
OSts9bfX6cA-016|And I have H2SO4 in this burette.
OSts9bfX6cA-026|So the solution has very low ionic strength and very low conductivity.
OSts9bfX6cA-029|Now, the light returns because H2SO4 forms ions in solution.
OSts9bfX6cA-030|So initially, I use ions up.
OSts9bfX6cA-031|And after all the barium is used up, then the extra sulfate that goes into solution stays in the solution and allows the solution to conduct.
OSts9bfX6cA-033|So H2SO4 lends itself to making ions in solution and gives us a bright light.
OSts9bfX6cA-034|And after the solid is formed, these ions conduct.
OSts9bfX6cA-035|The correct answer here is B.
oYVD8MSW8NE-000|Glucose is oxidized in your body to produce energy.
oYVD8MSW8NE-001|Now you can oxidize glucose on the benchtop.
oYVD8MSW8NE-002|You could burn it in oxygen, the combustion of glucose.
oYVD8MSW8NE-003|You'd burn glucose in oxygen, you'd form carbon dioxide and water, energy would be released.
oYVD8MSW8NE-004|In your body, the same thing happens.
oYVD8MSW8NE-005|You oxidize glucose, (INHALING) breathe in oxygen to react with the glucose, (EXHALING) you breathe out carbon dioxide and water.
oYVD8MSW8NE-006|So the products and reactants are the same.
oYVD8MSW8NE-008|So thermodynamically, the energetics are the same basically.
oYVD8MSW8NE-009|Free energy, entropy, enthalpy, they're state functions.
oYVD8MSW8NE-010|They only depend on the initial and final state.
oYVD8MSW8NE-011|I have glucose and oxygen, and I form carbon dioxide and water, about the same state, the same energy is released.
oYVD8MSW8NE-013|In your body, the oxidation occurs more like a battery.
oYVD8MSW8NE-014|The oxidation occurs in one compartment and the reduction in another compartment, like an electrochemical cell.
oYVD8MSW8NE-015|The compartment where the oxidation occurs is inside the mitochondria of your cell.
oYVD8MSW8NE-019|Now in solution, the hydrophobic ends associate and the hydrophilic ends are out towards the water.
oYVD8MSW8NE-021|Inside the mitochondria, the oxidation of glucose occurs in the absence of oxygen. It's a formal oxidation.
oYVD8MSW8NE-022|I remove a pair of electrons.
oYVD8MSW8NE-027|These are large molecules-- I've drawn them as ovals here-- that span the membrane.
oYVD8MSW8NE-028|They go from one side of the membrane to the other.
oYVD8MSW8NE-029|So small molecules take our products of our oxidation, our electrons and our protons, to the membrane spanning protein.
oYVD8MSW8NE-033|So as an electrochemical cell, you might call this the cathode of the electrochemical cell.
oYVD8MSW8NE-034|This membrane spanning protein acts as a cathode where the reduction occurs.
oYVD8MSW8NE-036|So what you have is a pH gradient across the mitochondria.
sFbI9fAOsKE-000|Let's look at a chemical reaction that's at equilibrium and then we disturb it.
sFbI9fAOsKE-001|So which is true at time 1 for the chemical reaction A and B go to C in the aqueous phase?
sFbI9fAOsKE-002|And you have equilibrium and then a perturbation of one of the concentrations.
sFbI9fAOsKE-003|Which is true?
sFbI9fAOsKE-010|We're looking at a reaction that was at equilibrium.
sFbI9fAOsKE-011|Macroscopically, all the concentrations had stopped changing.
sFbI9fAOsKE-012|And then a perturbation occurred, and it looks like that perturbation was to increase the concentration of B.
sFbI9fAOsKE-015|So in this case, Q would be less than K.
sFbI9fAOsKE-016|What do we expect would happen here?
sFbI9fAOsKE-017|Well, we've increased a reactant concentration.
sFbI9fAOsKE-018|We expect the equilibrium to shift towards products.
sFbI9fAOsKE-019|So that numerator in the reaction quotient would get larger until Q is equal to K again and equilibrium is re-established.
sFbI9fAOsKE-020|But just after I add that reactant, Q is less than K, and the correct answer here is C.
MOyr35CZxF8-000|When a proton is transferred in a chemical reaction from one compound to another, that's called an acid base reaction.
MOyr35CZxF8-001|The compound that gives up the proton is the acid, the compound that accepts the proton is the base.
MOyr35CZxF8-002|Electrons can also be transferred in chemical reactions from one compound to another.
MOyr35CZxF8-003|I've written possible electron transfer reactions here.
MOyr35CZxF8-004|When electrons are transferred, we call them reduction oxidation reactions or redox reactions.
MOyr35CZxF8-005|Here's copper ions and zinc metal or zinc ions and copper metal.
MOyr35CZxF8-006|The question is, where do the electrons prefer to reside?
MOyr35CZxF8-007|That is, will zinc transfer a pair of its electrons to copper and make copper metal?
MOyr35CZxF8-008|If it did, that would be a reduction.
MOyr35CZxF8-010|Zinc would be oxidized in that process.
MOyr35CZxF8-011|Its oxidation number would go from zero, increasing to plus two.
MOyr35CZxF8-015|So in this process, copper ions act as an oxidizer.
MOyr35CZxF8-018|Once you donate a proton as an acid, you become a base.
MOyr35CZxF8-019|Same thing here.
MOyr35CZxF8-020|Once you oxidize by accepting electrons, you have the potential to be a reducer and donate electrons.
MOyr35CZxF8-021|We can look at the zinc as the reducer here.
MOyr35CZxF8-022|It's adding electrons, donating electrons to the copper to form copper metal.
MOyr35CZxF8-023|In the process, it becomes oxidized.
MOyr35CZxF8-025|It would accept electrons from the copper, itself would be reduced, while the copper is being oxidized.
MOyr35CZxF8-026|Now this reaction, as I've written, is spontaneous in this direction.
MOyr35CZxF8-027|That is, the standard state free energy difference is less than zero.
MOyr35CZxF8-029|That standard state free energy difference less than zero indicates a k, an equilibrium constant, greater than one.
MOyr35CZxF8-030|And in this case, it's a large equilibrium constant.
MOyr35CZxF8-031|It strongly favors the products.
MOyr35CZxF8-032|And we can see that happen.
MOyr35CZxF8-039|And where does the copper metal end up?
MOyr35CZxF8-042|On this side, there's no reaction.
MOyr35CZxF8-043|This represents the reverse reaction-- zinc ions and copper metal.
MOyr35CZxF8-045|Now we can look at half reactions.
MOyr35CZxF8-046|We can split up the chemical reaction into two half reactions involving electrons.
MOyr35CZxF8-049|That happens, in this case, simultaneously with a reduction of copper ions.
MOyr35CZxF8-050|The copper oxidation number goes from plus two to zero.
MOyr35CZxF8-051|That's a reduction in oxidation number.
MOyr35CZxF8-052|We break these up into what are called half reactions to help us study redox reactions.
MOyr35CZxF8-054|This will help us study redox chemistry as a whole.
aDm54rB3l2I-000|Let's look at these properties of ionization energy and electron affinity in more detail.
aDm54rB3l2I-003|That is, if you add an electron to an element, it generally accepts that element, goes to a more stable state releasing energy.
aDm54rB3l2I-004|So electrons are generally accepted.
aDm54rB3l2I-005|Free electron and an atom, the most stable state is that electron attached to the atom.
aDm54rB3l2I-011|You can see neon, at the edge of the periodic table, is in the noble gases, a relatively unreactive element.
aDm54rB3l2I-012|And it's so unreactive even-- won't even pick up free electrons.
aDm54rB3l2I-013|Sodium and, excuse me, nitrogen is a half filled o orbital.
aDm54rB3l2I-014|And the half filled p orbital is relatively stable.
aDm54rB3l2I-015|It has three parallel spins.
aDm54rB3l2I-016|When it accepts another electron, that electron has to go in anti-parallel and spin pair with one of the electrons there.
aDm54rB3l2I-017|And that's just enough perturbation to make that electron affinity not so favorable.
aDm54rB3l2I-018|Now, the ionization energies are all positive.
aDm54rB3l2I-019|It always requires energy to pull electrons away from atoms.
aDm54rB3l2I-021|Carbon, nitrogen, oxygen, fluorine.
OTQOaH4nAto-000|Let's look at the molecular orbitals and the bonding in butadiene.
OTQOaH4nAto-002|Each carbon is sp2 hybridized.
OTQOaH4nAto-005|So you know that each carbon has to have four total bonds.
OTQOaH4nAto-006|So this carbon here at this vertex has one, two, three, so there must be a hydrogen here, four.
OTQOaH4nAto-007|So shorthand notation for writing butadiene looks like this.
OTQOaH4nAto-008|But with these sp2 hybridized carbons, we'll have p orbitals left over on each carbon.
OTQOaH4nAto-012|So here's the carbon.
OTQOaH4nAto-013|Now I have each carbon and notice 120 degree bond angles.
OTQOaH4nAto-014|That's the sp2 hybridized.
OTQOaH4nAto-015|sp2 leaves behind a p orbital and here they are above and below the plane of the molecule.
OTQOaH4nAto-016|So a p orbital left over on each carbon.
OTQOaH4nAto-017|I can form from these one, two, three, four atomic orbitals four molecular orbitals.
OTQOaH4nAto-018|And now this is going to be interesting because the orbital that I form, rather than just spanning two atoms, can span four atoms.
OTQOaH4nAto-019|So I'll form a molecular orbital that looks like this.
OTQOaH4nAto-020|A long, extended, delocalized orbital.
OTQOaH4nAto-021|And delocalized I mean it's over several nuclei.
OTQOaH4nAto-022|The electron can delocalize.
OTQOaH4nAto-023|There's a probability of finding it all the way along the length of the molecule.
OTQOaH4nAto-024|So a second possibility is to have molecular orbitals that look like this.
OTQOaH4nAto-025|Here's another possibility where you have a node now in the middle.
OTQOaH4nAto-026|That's a higher energy.
OTQOaH4nAto-027|Remember, more nodes, higher energy.
OTQOaH4nAto-028|And I can continue to a two node molecular orbital and a three node molecular orbital.
OTQOaH4nAto-029|As you go up in number of nodes, of course you go up in energy.
OTQOaH4nAto-030|And we're going to scale these as pi and pi star.
OTQOaH4nAto-032|So we'll just take the upper half as antibonding and the lower half as bonding.
OTQOaH4nAto-033|Now this is interesting.
OTQOaH4nAto-034|I said it looks like the electron can delocalize over the length of the molecule.
OTQOaH4nAto-035|It's as if the electron is trapped in a little box that's the length of the molecule.
OTQOaH4nAto-036|So the electron has probability all along that molecule, just like the particle in a box that we talked about before.
OTQOaH4nAto-041|So let's just remind ourselves of the particle in a box wave functions.
OTQOaH4nAto-042|Here they are.
OTQOaH4nAto-046|So the energy levels for a particle in a box, we should remember a couple of things.
OTQOaH4nAto-047|One is as the box gets bigger, the energy levels go as n squared over l squared.
OTQOaH4nAto-048|Will l is the length, so longer boxes, the energy levels start to come closer together.
OTQOaH4nAto-049|They get closer together.
OTQOaH4nAto-050|And when you get to a very big box, the energy levels are stacked right on top of each other and virtually become continuous.
OTQOaH4nAto-051|And that's where you'd make the transition from quantum mechanical to classical.
OTQOaH4nAto-052|In a very large box or big particles you'd have continuous energy levels.
OTQOaH4nAto-053|You wouldn't have quantized velocities for a marble rolling around in a box.
OTQOaH4nAto-054|It can go at any velocity.
OTQOaH4nAto-055|A quantum mechanical particle has quantized energies.
OTQOaH4nAto-056|So small boxes, I have spacing between my energy levels.
OTQOaH4nAto-060|So a big particle in a box, a macroscopic particle or a marble rolling around in a box, you could bring to a stop at its lowest energy level.
OTQOaH4nAto-061|Quantum mechanical particles don't stop.
rCVJLXWD1ng-000|Let's talk about ionization energies now that we understand a little bit about effective nuclear charge and shielding.
rCVJLXWD1ng-001|Which atom or iron has the lowest ionization energy-- a hydrogen in the excited 2p state?
rCVJLXWD1ng-002|A helium plus in the excited 3p state?
rCVJLXWD1ng-003|Or lithium plus 2 in the excited 4p state?
rCVJLXWD1ng-010|We're talking about three species, and we're trying to determine the ionization energy.
rCVJLXWD1ng-013|So let's just do that.
rCVJLXWD1ng-014|For hydrogen in the 2p state, z is 1, charge in the nucleus, n is 2, so we'll have a quarter of a Rydberg to ionize.
rCVJLXWD1ng-017|We have 9/16 of a Rydberg, a little more than half a Rydberg to ionize.
luB5a39kGkA-000|Particles can have wave-like properties.
luB5a39kGkA-001|If the momentum is small enough, a moving particle displays the properties of a wave.
luB5a39kGkA-002|We can actually demonstrate it.
luB5a39kGkA-005|So electrons that behave like waves plus boundaries gives you quantization.
luB5a39kGkA-006|Now we need to look at that in an interesting situation.
luB5a39kGkA-007|We've looked at a particle in a one-dimensional box.
luB5a39kGkA-008|What about an atom?
luB5a39kGkA-009|An atom is an electron bound by the nucleus.
luB5a39kGkA-010|An electric charge holds the electron about the nucleus.
luB5a39kGkA-011|So that's a particle that behaves like a wave, plus boundaries on where it has to be in space.
luB5a39kGkA-012|That should lead to quantized energy levels for the electron.
luB5a39kGkA-013|So let's look at that carefully.
luB5a39kGkA-014|First, we'll have to describe three-dimensional space in terms of easy parameters for quantum mechanical calculations.
luB5a39kGkA-015|We'll do that using something called spherical polar coordinates.
luB5a39kGkA-018|So we'll get regions of space, just like in the particle in a box, where the particle is more likely to be located.
luB5a39kGkA-019|So let's look at these spherical polar coordinates.
luB5a39kGkA-020|How do we describe where an electron is about an atom?
luB5a39kGkA-021|Well, it's somewhere about the nucleus in three-dimensional space.
luB5a39kGkA-022|We'll draw a vector from the center of the coordinate system, which is where the nucleus will be, out to the electron.
luB5a39kGkA-023|And it'll have length r.
luB5a39kGkA-024|That will be our first coordinate, the length r.
luB5a39kGkA-025|Then we'll take the angle from the positive z-axis out to that vector r.
luB5a39kGkA-026|That will be the angle theta.
luB5a39kGkA-027|That's our second coordinate.
luB5a39kGkA-028|And then we'll take a third coordinate, the angle of the projection into the xy plane of the vector, and that'll be our third coordinate, phi.
luB5a39kGkA-029|So r, theta, phi.
luB5a39kGkA-030|Three coordinates for three dimensions of space.
luB5a39kGkA-033|For three dimensions, we'll get three quantum numbers, one for each dimension.
luB5a39kGkA-034|So those quantum numbers will be n, l, and m sub l.
luB5a39kGkA-035|Let's look at those quantum numbers and how they relate to the properties of the wave function.
luB5a39kGkA-036|So we'll look at the quantum number values and the orbital property.
luB5a39kGkA-037|Now I'm going to start to use the term orbital.
luB5a39kGkA-038|What I mean is wave function, or square of the wave function.
luB5a39kGkA-039|The orbital is the region in space where the electron can exist.
luB5a39kGkA-040|So that's what the wave function describes.
luB5a39kGkA-041|I'll use those two terms interchangeably.
luB5a39kGkA-042|So quantum number.
luB5a39kGkA-043|First, the principle quantum number, n, just like in a particle in a box, has values 1, 2, 3, and describes the overall energy of the system.
luB5a39kGkA-044|And just like in the particle in a box, when you have the total number of nodes, n minus 1, you know how high your energy is.
luB5a39kGkA-045|More nodes, more energy.
luB5a39kGkA-046|Now-- or a higher energy system.
luB5a39kGkA-047|Now remember node.
luB5a39kGkA-048|That's an area where the wave function goes to zero.
luB5a39kGkA-049|The square of the wave function is zero.
luB5a39kGkA-052|So the larger n, the more values of l you have.
luB5a39kGkA-053|So l starts at zero.
luB5a39kGkA-054|Now it's our first zero value quantum number.
luB5a39kGkA-055|And it counts up in integers up to n minus 1.
luB5a39kGkA-056|So if n is 3, l could be 0, 1, or 2.
luB5a39kGkA-057|That's n minus 1.
luB5a39kGkA-058|Now for l, we'll also use letter designations.
luB5a39kGkA-059|Now I know sometimes in chemistry you're thinking, you're throwing this nomenclature at me just to make it confusing.
luB5a39kGkA-060|And unfortunately for the s, p, and d designations, I don't have a good answer for you.
luB5a39kGkA-061|They're historical designations of these values of l, and you just have to memorize them.
luB5a39kGkA-062|It is kind of almost a confusion factor that we're going to throw in on top of this.
luB5a39kGkA-063|Sometimes the quantum numbers will be a number, and sometimes it's going to be a letter.
luB5a39kGkA-064|So memorize these values for l.
luB5a39kGkA-065|L tells you the overall shape of the orbital, the shape of the wave function.
luB5a39kGkA-066|Is it dumbbell shaped?
luB5a39kGkA-067|Is it spherical shape?
luB5a39kGkA-068|Is it elongated?
luB5a39kGkA-069|Is it narrow?
luB5a39kGkA-070|Those kind of things come from l.
luB5a39kGkA-072|And it will have values ranging from minus l to l.
luB5a39kGkA-073|So if l is 2, you'll go minus 2, minus 1, 0, 1, 2.
luB5a39kGkA-074|From minus l to l in integer steps.
luB5a39kGkA-075|So there could be many values of m sub l for a given value of l.
luB5a39kGkA-076|And it's going to tell you something about the orientation.
luB5a39kGkA-077|If I have a dumbbell shaped, maybe along the x-axis, maybe I'm along the z-axis.
luB5a39kGkA-078|M sub l will help you determine that.
luB5a39kGkA-079|So we're not going to look at the actual values of the wave function, the mathematical formulas.
luB5a39kGkA-080|What we will look at is the values of the quantum numbers.
luB5a39kGkA-081|Each set of quantum numbers describes an individual wave function.
luB5a39kGkA-082|Just like in a particle in a box when you took n equal 1, that was a single hump function.
luB5a39kGkA-084|We're going to learn how those orbitals relate to the various wave function quantum numbers.
luB5a39kGkA-085|So that if I give you a set of quantum numbers, you'll be able to tell me the orientation and shape of that orbital.
luB5a39kGkA-086|So we'll use pictures to describe the orbitals rather than the wave functions themselves.
luB5a39kGkA-087|The wave functions themselves are a complex mathematical formula.
luB5a39kGkA-088|What we want to think about is how are the electrons actually distributed about the atom.
luB5a39kGkA-089|Wave functions and quantum mechanics tell us that with a high, high degree of precision.
luB5a39kGkA-090|This is a beautiful method to describe atoms.
luB5a39kGkA-095|So quantum mechanics, incredibly precise, incredibly accurate, and a beautiful description of the atom.
jjOzgw5nAZk-000|There's another property that we can look at in terms of its periodic nature on the periodic table and see if there's a trend.
jjOzgw5nAZk-001|And that's electronegativity.
jjOzgw5nAZk-002|Electronegativity is the general tendency to draw electrons towards yourself.
jjOzgw5nAZk-003|If you're strongly electronegative, then electrons move towards you in a bond.
jjOzgw5nAZk-004|Now, electronegativity isn't something that we can do an experiment to measure.
jjOzgw5nAZk-005|That is, we don't have, as in ionization energy, an obvious experiment-- have the atom, pull off an electron, see how much energy that takes.
jjOzgw5nAZk-006|For electronegativity, what we do is we look at a bunch of bonds.
jjOzgw5nAZk-007|You take an element and you see how it bonds to other elements.
jjOzgw5nAZk-008|And you see, are those bonds polar?
jjOzgw5nAZk-009|Are the electrons in that bond spending more time around one atom than the other atom?
jjOzgw5nAZk-010|If one of the atoms tends to draw the electrons towards itself, then that atom is more electronegative.
jjOzgw5nAZk-011|So you can do this by cataloging hundreds of bonds between different elements and seeing which elements tend to draw the electrons toward themselves.
jjOzgw5nAZk-012|And it does go in a trend.
jjOzgw5nAZk-013|You find that electronegativity increases as you go from the bottom corner to the top corner of the periodic table.
jjOzgw5nAZk-016|And you can judge the strength of the polarity.
jjOzgw5nAZk-017|Polarity increases.
jjOzgw5nAZk-018|In fact, it can increase so much that the electronegative element actually ionizes the other element.
jjOzgw5nAZk-019|It takes the electron to itself all together.
jjOzgw5nAZk-020|Then you're held together by the extreme of polarity-- one negatively charged atom being attracted to a positively charged atom.
jjOzgw5nAZk-021|The electronegative element has completely captured an electron from the other element.
jjOzgw5nAZk-022|That's called an ironic bond.
jjOzgw5nAZk-023|And what holds it together is a Coulombic interaction-- the positive attracted to the negative.
jjOzgw5nAZk-024|You can catalog a continuum of bonds by looking at differences in electronegativity.
jjOzgw5nAZk-029|That's the strongest kind of bond you can have.
jjOzgw5nAZk-030|You can have intermediate differences in electronegativity.
jjOzgw5nAZk-031|Here is partially ionic hydrogen chloride where the difference is modest.
jjOzgw5nAZk-035|That's perfect sharing of electrons.
jjOzgw5nAZk-036|But for small differences in electronegativity, you'll have covalent bonds.
jjOzgw5nAZk-037|So covalent, partially covalent, ionic, very ironic-- separated in polarity.
jjOzgw5nAZk-038|The more polar, the stronger the bond.
1FpJl_Rg4AY-000|Let's look at some various forms of equilibrium.
1FpJl_Rg4AY-001|Now equilibrium is when products and reactants have stopped changing in their macroscopic concentrations.
1FpJl_Rg4AY-002|So A and B and C and D macroscopically aren't changing.
1FpJl_Rg4AY-003|But of course, A and B is still making C and D.
1FpJl_Rg4AY-004|But every time it makes some C and D, some of that reacts back to form A and B.
1FpJl_Rg4AY-005|So macroscopically, there's no change in concentrations.
1FpJl_Rg4AY-006|But dynamically, the forward and reverse reactions are going on.
1FpJl_Rg4AY-007|It's just their rates are equivalent.
1FpJl_Rg4AY-011|Water, liquid water, in equilibrium with water gas.
1FpJl_Rg4AY-012|Now this doesn't have to be at the boiling point of water.
1FpJl_Rg4AY-013|There's always an equilibrium between liquid water and gaseous water.
1FpJl_Rg4AY-014|In this case, the equilibrium vapor pressure of water is just 0.03 of an atmosphere.
1FpJl_Rg4AY-015|So there's 0.03 of an atmosphere of water at 25 degrees C, the equilibrium vapor pressure, over the gas.
1FpJl_Rg4AY-018|N2O4 and NO2 brown in equilibrium.
1FpJl_Rg4AY-020|Macroscopically, these all look very static.
1FpJl_Rg4AY-021|And that's what equilibrium is on the macroscopic scale-- static.
1FpJl_Rg4AY-022|But on the microscopic scale, NO2 is still converting to N2O4, and N2O4 are breaking down into NO2.
1FpJl_Rg4AY-025|This color intensity won't change, because as the forward and reverse reactions occur, they balance each other out.
1FpJl_Rg4AY-026|You could also have heterogeneous equilibrium-- a solid in equilibrium with its aqueous ions.
1FpJl_Rg4AY-028|So as a little dissolves, a little of the ions precipitate.
1FpJl_Rg4AY-029|And that dynamic equilibrium exists.
1FpJl_Rg4AY-031|So various forms of equilibrium, all of them dynamic, but all of them, from a macroscopic sense, appear to be static.
1FpJl_Rg4AY-032|That's the nature of equilibrium.
PvCyqGnM-j0-000|Let's look at ionization energies when we're not all atoms.
PvCyqGnM-j0-001|That is, I have chlorine minus, argon, and potassium plus here, a mixture of atoms and ions.
PvCyqGnM-j0-002|Can we deduce which of these will have the lowest ionization energy?
PvCyqGnM-j0-012|Now interestingly, these three all have 18 electrons.
PvCyqGnM-j0-013|Cl minus, argon, and potassium plus have the same ground state electronic configuration.
PvCyqGnM-j0-014|They're called isoelectronic, same number of electrons.
PvCyqGnM-j0-016|Fewer nuclear charges will allow it to expand and be easier to ionize.
PvCyqGnM-j0-017|So among these three, it's the one with the smallest nuclear charge that's the easiest to ionize.
PvCyqGnM-j0-018|And that's chlorine.
PvCyqGnM-j0-019|Chlorine minus easier to ionize than argon or potassium plus.
-tJR8S6uJEI-000|Let's look at the reaction of oxygen atoms to form oxygen molecules, and think about what happens to the total paramagnetism of the sample.
-tJR8S6uJEI-001|Paramagnetic oxygen atoms making paramagnetic oxygen molecules.
-tJR8S6uJEI-009|We're looking at the reaction of action atoms to form oxygen molecules and the total paramagnetism of the sample.
-tJR8S6uJEI-011|And they're diparamagnetic.
-tJR8S6uJEI-014|And here the molecular orbitals from these P atomic orbitals.
-tJR8S6uJEI-015|And you fill them with 1, 2, 3, 4, 5, 6, 7, 8 P electrons.
-tJR8S6uJEI-016|So I'll take these eight P electrons and make eight molecular orbital electrons.
-tJR8S6uJEI-017|1, 2, 3, 4, 5, 6, 7, 8.
-tJR8S6uJEI-018|And I find oxygen, the molecule, is also paramagnetic.
-tJR8S6uJEI-019|So I take atoms, each of which is paramagnetic, and form a paramagnetic molecule.
-tJR8S6uJEI-020|So the total number of unpaired electrons in the sample decreases.
-tJR8S6uJEI-021|I'll have twice as many unpaired electrons when I have oxygen atoms as I have oxygen molecules.
-tJR8S6uJEI-022|So total paramagnetism in this case decreases.
jMj7Vse19_I-000|Combustion of glucose, the oxidation in your body, releases around 3,000 kilojoules per mole.
jMj7Vse19_I-008|We're talking about capturing the energy from the combustion of glucose as ATP.
jMj7Vse19_I-010|How much can be captured as ATP?
jMj7Vse19_I-011|Well, let's look back.
jMj7Vse19_I-018|So that's around 40%.
jMj7Vse19_I-019|Your body operates at about 40% efficiency in transferring the energy from the catabolism of glucose and storing it as ATP.
1yoOSIcsCyI-000|When we're making molecular orbitals, we'd like to take atomic orbitals that have the right geometry for the problem.
1yoOSIcsCyI-001|You'll have a tetrahedral arrangement in my molecule, I'd like to have atomic orbitals that point to the vertices of a tetrahedron.
1yoOSIcsCyI-002|I can do that by a process called hybridisation-- taking a combination of the s and p orbitals on my atom and making new orbitals, hybrid orbitals.
1yoOSIcsCyI-011|And that electronic configuration will be sp2 now.
1yoOSIcsCyI-012|sp2 with 3 electrons, and 1 electron left in the p.
1yoOSIcsCyI-015|So the electronic configuration-- 2 sp3 with 4 electrons.
OCK36oYeAFA-000|Let's do a calculation involving standard enthalpies of formation and enthalpies of reactions.
OCK36oYeAFA-001|So I want to calculate the enthalpy of formation for N2O5 gas starting with some information.
OCK36oYeAFA-004|Because I know enthalpies of formation are the formation of the compound from the elements in their standard states.
OCK36oYeAFA-005|And this compound contains nitrogen and oxygen.
OCK36oYeAFA-006|Nitrogen in its standard state is N2 gas.
OCK36oYeAFA-007|Oxygen in its standard state is O2 gas.
OCK36oYeAFA-008|So I can write N2 gas plus O2 gas goes to two NO, just balancing that chemical equation.
OCK36oYeAFA-009|Of course, the enthalpy information is for a single mole.
OCK36oYeAFA-010|I've formed two moles here, so this enthalpy is the enthalpy for 1/2 this chemical reaction.
OCK36oYeAFA-011|I've got 2 here, so I'll just double this enthalpy.
OCK36oYeAFA-013|181 kilojoules.
OCK36oYeAFA-014|Now I want to find the enthalpy of formation for N2O5.
OCK36oYeAFA-015|I can write that because it's the enthalpy of formation.
OCK36oYeAFA-018|So I'd like to sum these to form this one.
OCK36oYeAFA-019|And I noticed right off, well, this reaction here, that would give me two moles of N2O5 on the product side.
OCK36oYeAFA-020|I only want one mole, so I'm going to divide this one through by 2.
OCK36oYeAFA-021|If I divide this reaction through by 2, I have to divide the enthalpy by 2.
OCK36oYeAFA-022|So let's do that.
OCK36oYeAFA-023|And then, when we add down, you'll see that these three reactions add to give the reaction we're looking for.
OCK36oYeAFA-024|So it's a Hess's law situation.
OCK36oYeAFA-025|I can add these three.
OCK36oYeAFA-033|It's slightly endothermic.
2hohw6e19_s-000|We can calculate the pH in the buffer region using the Henderson-Hasselbalch expression.
2hohw6e19_s-002|So here, I've expanded the buffer region.
2hohw6e19_s-003|And we know that, at half equivalence, halfway to the equivalence point, the concentrations of the acid and base form are equal.
2hohw6e19_s-004|So this ratio is 1, the log term is 0, and the pH is numerically equal to the pKa at half equivalence.
2hohw6e19_s-007|This term would be negative.
2hohw6e19_s-008|And of course, if the base form predominates, you'd expect this term to be positive.
2hohw6e19_s-010|So the Henderson-Hasselbalch expression allows us to calculate the pH throughout this region.
2hohw6e19_s-013|The buffer region would be from 3.75 to 5.75, one unit around the pKa.
2hohw6e19_s-014|So let's look at that in more detail.
2hohw6e19_s-023|So a buffer resists change in pH from adding acid, from adding base, and from dilution.
-zXxTQt2slA-000|Let's look at hybridization in benzene.
-zXxTQt2slA-001|Benzene, C6H6, I've drawn it like this with this circle representing the alternating double bonds.
-zXxTQt2slA-007|We're talking about the hybridization of carbons in benzene.
-zXxTQt2slA-008|Now, we drew benzene schematically, but that's enough to determine the steric number of each carbon, and that's what you need to determine hybridization.
-zXxTQt2slA-009|Carbon in benzene has to accommodate three things-- two other carbons and a hydrogen.
-zXxTQt2slA-010|So to accommodate three things with steric number 3, I have hybridize three atomic orbitals-- an s and two p's.
-zXxTQt2slA-011|That leads to sp2 hybridization, 3 equivalent orbitals that have a 120 degree bond angle.
-zXxTQt2slA-013|All the carbons are equivalent, so these carbons here will also be sp2 hybridized.
-zXxTQt2slA-016|So the correct answer here-- sp2 hybridisation for the carbons in benzene.
y97cegGFECA-000|When salts dissolve in water there's a wide variation in the solubility.
y97cegGFECA-001|Some salts like sodium chloride dissolve quite well and they form high concentrations of sodium ions and chloride ions in solution.
y97cegGFECA-002|Others like barium sulfate are not very soluble.
y97cegGFECA-004|So a small K means I favor the reactant side, the solid as opposed to the ions.
y97cegGFECA-007|It kind of means that barium and sulfate ions don't like to exist together in solution.
y97cegGFECA-008|The more stable state is the barium sulfate solid.
y97cegGFECA-009|So if barium comes from one source and sulfate from another, those two ions seek each other out.
y97cegGFECA-017|So the K for this reaction, the reverse, is 1 over 10 to the minus 10, or 10 to the plus 10.
y97cegGFECA-018|A large K means I strongly favor the barium sulfate solid.
y97cegGFECA-020|So both these reactions strongly favor the solid and water.
y97cegGFECA-022|So the ion concentration drops if you mix these two solutions.
y97cegGFECA-024|The barium and hydroxide ions would conduct electricity across this gap and allow this light to come on.
y97cegGFECA-025|When there's high ionic concentration, that light will be bright.
y95SN6mDh0Y-000|Now the intensity of light, or electromagnetic radiation, may be more of an intuitive property.
y95SN6mDh0Y-001|Let's look at it this way.
y95SN6mDh0Y-003|So I0 going through a pair of identical filters.
y95SN6mDh0Y-004|Would that reduce the intensity down to a quarter, an eighth, or 0 of the original intensity?
y95SN6mDh0Y-011|So we've been talking about intensity, the brightness of light, in terms of attenuating it, or reducing it, with a filter.
y95SN6mDh0Y-013|Well if the first filter reduces it by one-half, so it's half as bright here.
y95SN6mDh0Y-014|A second filter would make this half as bright.
y95SN6mDh0Y-015|So you'd go from half the original intensity to a quarter of the original intensity after two filters.
aoAdE8L2ASw-000|Let's look at our liquid-to-gas transition for water and ask, what is the standard state free energy difference?
aoAdE8L2ASw-002|What is the standard state free energy difference for this physical change at 100 degrees C?
aoAdE8L2ASw-010|We're talking about the liquid-to-gas phase transition.
aoAdE8L2ASw-011|And at 100 degrees C, what's the standard state free energy difference?
aoAdE8L2ASw-014|So we can talk about those states of matter at different temperatures.
aoAdE8L2ASw-015|And in this case, we're talking about it at 100 degrees C.
aoAdE8L2ASw-017|And I think you could say, well, that's what boiling is.
aoAdE8L2ASw-018|Boiling occurs when there's a vapor pressure of one atmosphere above the liquid.
aoAdE8L2ASw-019|So that is the equilibrium point.
aoAdE8L2ASw-021|So what we have here is, where is the free energy difference between the one atmosphere of gas and liquid exactly 0?
aoAdE8L2ASw-025|If you went to a different temperature and you were talking about the standard state, you would still have one atmosphere of gas.
aoAdE8L2ASw-026|So that's why you say, well, if I go to a lower temperature and I still have one atmosphere of gas, well, then of course, I favor the liquid.
aoAdE8L2ASw-027|That's a higher amount of gas, one atmosphere, than the vapor pressure at these lower temperatures.
aoAdE8L2ASw-028|And as I go to higher temperatures, then I have the one atmosphere of gas is the favored state of the system.
aoAdE8L2ASw-029|So delta G standard is a function of temperature.
aoAdE8L2ASw-030|In this case, I'm right at the equilibrium where the gas and the liquid are equally favored.
aoAdE8L2ASw-031|In this case, the correct answer is B.
Hx9dFbQ2zhk-000|Let's look at both the quantum mechanical description and the Lewis dot structure description of a molecule.
Hx9dFbQ2zhk-001|We'll pick oxygen, O2.
Hx9dFbQ2zhk-002|Oxygen, each atom has 6 valence electrons in principle quantum level 2.
Hx9dFbQ2zhk-004|Each oxygen obeys the octet, it has 2, 4, 6, 8 electrons around it.
Hx9dFbQ2zhk-005|So that's our good Lewis electron dot structure.
Hx9dFbQ2zhk-006|Let's look at the quantum mechanical description.
Hx9dFbQ2zhk-016|So those are the s orbitals, let's take the p orbitals.
Hx9dFbQ2zhk-017|There are 6 p orbitals, 3 on each atom.
Hx9dFbQ2zhk-024|Again, sigma bonding, sigma antibonding, energetically distributed.
Hx9dFbQ2zhk-025|Now, let's look at the px and the py.
Hx9dFbQ2zhk-026|The px and the py lie in front and behind and above and below the internuclear axis.
Hx9dFbQ2zhk-027|And you can see that on a model.
Hx9dFbQ2zhk-030|It doesn't really matter which one I look at, they're both identical in energy.
Hx9dFbQ2zhk-035|And the minus combination gives me antibonding, pi orbitals, two of those.
Hx9dFbQ2zhk-036|One from the minus combination of px and one from the minus combination of py.
Hx9dFbQ2zhk-038|Each oxygen had an s and three p's.
Hx9dFbQ2zhk-039|So 4 orbitals on each oxygen, 8 total, make 8 molecular orbitals.
Hx9dFbQ2zhk-040|I need to fill those with the 12 valence electrons from oxygen.
Hx9dFbQ2zhk-043|So the bond order, I have a sigma bonding and antibonding orbital to start.
Hx9dFbQ2zhk-044|Two electrons in each, those cancel each other out, zero bond order from that.
Hx9dFbQ2zhk-046|So 6 minus 2 is 4 bonding electrons divided by 2-- 2 formal bonds.
Hx9dFbQ2zhk-047|So I have a formal bond order of 2 in oxygen, and that is consistent with the double bond that my Lewis dot structure predicted.
Hx9dFbQ2zhk-048|But here's something that the Lewis dot structure didn't predict-- oxygen has two unpaired electrons.
Hx9dFbQ2zhk-049|Oxygen is paramagnetic.
Hx9dFbQ2zhk-051|And if you do the experiment, you find that oxygen is actually paramagnetic and you'll see that in the demonstration lab.
Hx9dFbQ2zhk-052|So the molecular orbital picture more powerful in determining a molecular property than the Lewis electron dot structure.
VabAZE71dH0-000|Let's look at the chemical reaction hydrogen plus oxygen go into water under three different sets of circumstances.
VabAZE71dH0-001|One, will ignite the hydrogen and oxygen. Two, will add platinum.
VabAZE71dH0-002|And three, will just let the reaction go.
VabAZE71dH0-011|We're looking at the chemical reaction hydrogen and oxygen forming water under three sets of circumstances.
VabAZE71dH0-012|One, we ignite the reaction.
VabAZE71dH0-013|Two, we add platinum.
VabAZE71dH0-014|And three, we allow the reaction to proceed.
VabAZE71dH0-015|Now, we've seen each of these in the demo lab.
VabAZE71dH0-016|And when we ignite the chemical reaction it goes very rapidly.
VabAZE71dH0-017|When you add platinum, the reaction goes at room temperature.
VabAZE71dH0-018|Platinum is a catalyst for this chemical reaction and it lowers the activation energy.
VabAZE71dH0-021|Reaction C and reaction A have the same activation energy.
VabAZE71dH0-022|This one will go faster because we add energy, but their activation energies are the same.
X0nCoOX0LAg-000|Let's look at a real gas that we're all familiar with, water vapor.
X0nCoOX0LAg-006|Above that temperature, you'll always have a single phase.
X0nCoOX0LAg-007|That is, you can compress the gas down.
X0nCoOX0LAg-008|That phase might get very dense, but I won't see distinct liquid and gas phase.
X0nCoOX0LAg-009|Below the critical temperature, I'll see distinct liquid and gas phases.
X0nCoOX0LAg-013|That's 100 degrees C.
X0nCoOX0LAg-014|Where does this flat line occur at 100 degrees C?
X0nCoOX0LAg-015|Well, I think you can guess.
X0nCoOX0LAg-016|It's going to be one atmosphere of pressure.
X0nCoOX0LAg-020|So it's the partial pressure of the gas phase water above the liquid for that temperature.
X0nCoOX0LAg-021|The normal boiling point is defined as where the vapor pressure equals atmospheric pressure.
X0nCoOX0LAg-026|That's near room temperature.
X0nCoOX0LAg-027|So the vapor pressure of water at room temperature is [? 0.31 ?] of an atmosphere.
X0nCoOX0LAg-028|So that's the partial pressure of water above liquid water at room temperature.
X0nCoOX0LAg-029|So the total pressure, if the sample is open to the atmosphere, you'll have a atmosphere of the atmospheric gases, nitrogen and oxygen.
X0nCoOX0LAg-030|And [? 0.31 ?] of an atmosphere of water vapor pressure.
X0nCoOX0LAg-031|So the vapor pressure can be defined as the pressure at which the gas spontaneously condenses for a given temperature.
X0nCoOX0LAg-032|Or it can be defined as the partial pressure of the gas above a liquid sample.
100n-E7o9Ug-000|When we make molecular orbitals from atomic orbitals, we take the atomic orbital wave functions and use linear combinations.
100n-E7o9Ug-001|We add and subtract them with constant coefficients to make molecular orbitals.
100n-E7o9Ug-005|So a helium molecule has four total electrons, and they need to occupy the molecular orbitals.
100n-E7o9Ug-009|And I find that both the bonding and antibonding orbital for this molecule are completely full.
100n-E7o9Ug-010|Now the bonding orbital is favorable for bonding.
100n-E7o9Ug-011|The antibonding orbital is unfavorable for bonding.
100n-E7o9Ug-012|A lower energy interaction and a higher energy interaction.
100n-E7o9Ug-013|So those two cancel each other out.
100n-E7o9Ug-017|Now I'm looking at a 2s electron.
100n-E7o9Ug-018|A 2s electron I can take the 2s orbitals and use this same procedure.
100n-E7o9Ug-019|Add and subtract them to take a linear combination, form a sigma 2s and a sigma 2s star antibonding orbital.
100n-E7o9Ug-020|Those molecular orbitals are filled with the electrons from lithium.
100n-E7o9Ug-021|Lithium has one electron in its 2s orbital, so two lithiums, two electrons, they fill up the bonding orbital for the lithium molecule.
100n-E7o9Ug-022|Li2 we would predict to be a stable molecule.
100n-E7o9Ug-023|It has a bond order of one.
100n-E7o9Ug-024|Now the bond order will define as the sum of the bonding electrons minus the antibonding electrons.
100n-E7o9Ug-027|Be2 has four electrons.
100n-E7o9Ug-028|I'll put two electrons in the bonding of sigma and the two electrons in the antibonding sigma star orbital.
100n-E7o9Ug-029|I'll calculate the bond order, two electrons here for bonding minus two antibonding electrons.
100n-E7o9Ug-030|The bond order is zero.
100n-E7o9Ug-031|So Be2, the beryllium 2 molecule, doesn't form.
100n-E7o9Ug-032|So my molecular orbital theory has predictive power and it can predict the bond order and whether or not bonds form in molecules.
c-yoRAiBQ1k-001|So to which energy level scheme-- A, B, or C-- does this emission spectrum correspond?
c-yoRAiBQ1k-007|Let's look at the relationship between emission spectra and energy-level spacing in the matter.
c-yoRAiBQ1k-008|So if you have an energy-level spacing that looks like A, what would the emission spectrum look like?
c-yoRAiBQ1k-009|Well, we have to look at every possible transition in the system.
c-yoRAiBQ1k-010|And here you can see three high energy level transitions-- those would give you high frequency lines-- and three low.
c-yoRAiBQ1k-011|You could have this tiny transition, this tiny transition, and this tiny transition.
c-yoRAiBQ1k-012|These two of equal energy, so they're degenerate.
c-yoRAiBQ1k-013|They would fall right on top of each other and give you only one line even though there's two transitions.
c-yoRAiBQ1k-014|But the two transitions have the same energy, so we can't resolve them in terms of energy.
c-yoRAiBQ1k-015|So you get just two lines, one for this transition and one corresponding to both of these transitions.
c-yoRAiBQ1k-016|So that doesn't look like the right answer.
c-yoRAiBQ1k-018|They would fall right on top of each other.
c-yoRAiBQ1k-019|So you'd have a very high energy level, a medium and a medium at the same energy, and a low in your high band.
c-yoRAiBQ1k-020|That's a total of three from these four transitions since two are exactly the same.
c-yoRAiBQ1k-021|And then two tiny low energy transitions, but again, they are of equal energy, so that would give you one line.
c-yoRAiBQ1k-022|That looks like the spectrum we've seen, and if you look at C, that, of course, isn't anything like what we see.
c-yoRAiBQ1k-024|So this pair gives you one line, this pair gives you one line, two intermediate lines, and then a small energy transition.
c-yoRAiBQ1k-028|And remember, you can't look at tiny little matter in your microscope.
c-yoRAiBQ1k-030|So here the correct answer is B.
Ie1qr8XGi6g-000|When you form compounds from elements, there'll be an enthalpy associated with that.
Ie1qr8XGi6g-001|So let's start with the elements in their standard states and form compounds.
Ie1qr8XGi6g-002|What are the standard states of the elements?
Ie1qr8XGi6g-003|Well, that's how you would find an element at 1 atmosphere of pressure and 25 degrees C.
Ie1qr8XGi6g-004|So some elements from the periodic table will be a solid.
Ie1qr8XGi6g-005|Some will be liquids.
Ie1qr8XGi6g-006|Some will be gases.
Ie1qr8XGi6g-007|Some will be diatomic.
Ie1qr8XGi6g-012|Those are lower on enthalpy hill.
Ie1qr8XGi6g-013|Overall, those compounds are a little more likely to be found because they're lower in energy than the elements.
Ie1qr8XGi6g-014|So we have standard enthalpies of formation are the formation of compounds from their elements in their standard states.
9sDdIaBhtgk-000|Let's look at weak acid, weak base, strong acid, and strong base solutions.
9sDdIaBhtgk-004|So the pH is about one.
9sDdIaBhtgk-006|The HAc weak acid dissociates to a lesser extent.
9sDdIaBhtgk-007|So the pH is higher, indicating a lower concentration of H3O plus.
9sDdIaBhtgk-008|Water, concentrations of H3O pluses at around 10 to the minus 7.
9sDdIaBhtgk-009|NH3.
9sDdIaBhtgk-010|Now we're getting to a base.
9sDdIaBhtgk-011|So the H3O plus concentration drops even further, so I get a higher pH, 10 to the 10.
9sDdIaBhtgk-013|The OH minus and the H3O plus concentrations, the product is always 10 to the minus 14 in water.
9sDdIaBhtgk-014|So I can look at these, and we can put a little indicator in each.
9sDdIaBhtgk-016|There's my strong acid solution.
9sDdIaBhtgk-017|Here's my weak acid solution.
9sDdIaBhtgk-018|Let's add a little more so that we get a little darker color.
9sDdIaBhtgk-019|Here's my strong acid solution.
9sDdIaBhtgk-020|Here's my weak acid solution.
9sDdIaBhtgk-021|I already added a little indicator to the water, so we'll skip over that one.
9sDdIaBhtgk-022|And my weak base solution.
9sDdIaBhtgk-023|So the base turns this indicator blue.
9sDdIaBhtgk-024|That's a nice little mnemonic to remember, because the bases are blue, both starting with the same letter.
9sDdIaBhtgk-026|So that indicator in the strong acid, weak acid, water, weak base, and strong base solution.
9sDdIaBhtgk-027|Now, the number of ions in each of these solutions is indicative of the acid and base strength, as well.
9sDdIaBhtgk-028|HCl, remember, completely disassociates.
9sDdIaBhtgk-029|So 0.1 molar HCl makes 0.1 molar H plus ions and 0.1 molar Cl minus ions.
9sDdIaBhtgk-030|So I have 0.2 molar total ion concentration.
9sDdIaBhtgk-031|And that ion concentration will make the solution conductive.
9sDdIaBhtgk-032|So if I test the conductivity, just kind of qualitatively with this light bulb, I'll see that HCl, there's ions in there.
9sDdIaBhtgk-033|Ions conduct, and I'll have a strongly conductive solution.
9sDdIaBhtgk-035|And in general, water, pure, is not a conductive solution.
9sDdIaBhtgk-036|So let's look at the weak acid.
9sDdIaBhtgk-037|Now, the weak acid dissociates much less strongly than the strong acid.
9sDdIaBhtgk-039|We should see the same correlation between the weak base and the strong base.
9sDdIaBhtgk-045|A correlation between pH, color, and ionic strength for acid and base solutions.
hTimaCA_A88-000|Let's look at the phase diagram of carbon dioxide and see if we can predict where the sublimation of carbon dioxide occurs at 1 atmosphere.
hTimaCA_A88-004|We're trying to find the sublimation line, a sublimation at constant pressure, transition for carbon dioxide.
hTimaCA_A88-005|Well, if it's a constant pressure transition, it has to be a horizontal line.
hTimaCA_A88-006|And if it's a sublimation, it must cross the solid/gas transition line.
hTimaCA_A88-007|Sublimation is the conversion from solid to gas.
hTimaCA_A88-009|The correct answer here is A.
--imKPteSwQ-000|When I burn the hydrocarbon butane, heat is released.
--imKPteSwQ-009|We're talking about burning butane in oxygen, and in each case oxygen is in excess.
--imKPteSwQ-010|So butane limits the amount of energy released.
--imKPteSwQ-014|But the speed of the reaction and the overall energy release are independent.
--imKPteSwQ-015|I can release a lot of energy over a long time, or release a small amount of energy quickly.
-dpZipe4Hm0-000|Let's look at some of the molecular orbitals involved in forming multiple bonds.
-dpZipe4Hm0-001|So when we have a carbon that's sp2 hybridized-- and I've drawn one here-- I have three equivalent sp2 orbitals.
-dpZipe4Hm0-002|And I have a model that looks like that here.
-dpZipe4Hm0-003|The three bonds here at 120 degrees from each other are sp2 hybrid orbitals.
-dpZipe4Hm0-004|Now, sp2 hybrids leave behind one p.
-dpZipe4Hm0-005|Sp2 hybrids are formed from an s and two p's.
-dpZipe4Hm0-006|So there's an unchanged p orbital that I represented here by this plus minus lobe.
-dpZipe4Hm0-012|I'll form a sigma bond between the two carbons.
-dpZipe4Hm0-014|Those two atomic hybrid orbitals will form a sigma bonding and antibonding molecular orbital.
-dpZipe4Hm0-015|So two atomic orbitals, both of them sp2, form two molecular orbitals, a sigma bonding and antibonding orbital.
-dpZipe4Hm0-016|Now, the p orbital left over can overlap to form a pi bond, a second bond.
-dpZipe4Hm0-017|And remember, pi bonds are bonds that are above and below the internuclear axis, usually formed by p orbitals.
-dpZipe4Hm0-019|There.
-dpZipe4Hm0-020|There is the sp sigma bond.
-dpZipe4Hm0-021|And now the p orbitals can overlap and form a bond above and below the axis, a second bond.
-dpZipe4Hm0-022|So I have a sigma and a pi bond.
-dpZipe4Hm0-023|The first bond formed between atoms is always sigma.
-dpZipe4Hm0-024|Other bonds are pi.
-dpZipe4Hm0-030|So multiple bonds, a sigma and a pi bond, formed from a combination of hybrid orbitals and leftover p orbitals on carbon.
oTCKWNiWdus-000|Just like the entropy and the enthalpy, the free energy is a state function.
oTCKWNiWdus-002|Now, here's a table of standard free energies of formations.
oTCKWNiWdus-004|That's because it wouldn't be fair to take the reactants at, say, two or three atmospheres of pressure and compare that to the products at half an atmosphere of pressure.
oTCKWNiWdus-005|We compare everything across the board in a standard state of conditions.
oTCKWNiWdus-006|So in a sense, it makes it a little artificial.
oTCKWNiWdus-008|And that's not necessarily how you set it up in the laboratory, but it'll still help you determine whether overall the products or the reactants are favored.
oTCKWNiWdus-009|If delta g is overall negative, then the products are favored.
oTCKWNiWdus-010|So here we have the tables now of enthalpy, entropy, and free energy.
oTCKWNiWdus-011|So we can use these to calculate free energies, standard state enthalpies, and standard entropy differences.
ma-MKQ2TAGk-000|Our bodies use the metabolism of glucose as an energy source.
ma-MKQ2TAGk-002|Now, that oxidation that's down hill in free energy, and that's coupled to an uphill in free energy reaction.
ma-MKQ2TAGk-003|The phosphorylation, the addition of a phosphate group, to adenosine diphosphate to form adenosine triphosphate.
ma-MKQ2TAGk-007|So ATP hydrolysis from ATP to ADP, that releases energy.
ma-MKQ2TAGk-008|And that can be coupled with building something in your cells.
ma-MKQ2TAGk-009|So for instance, taking free amino acids and making them into proteins.
ma-MKQ2TAGk-010|So free amino acids are small individual molecules.
ma-MKQ2TAGk-011|They're linked together to form proteins.
ma-MKQ2TAGk-012|And proteins are molecules that are catalysts in our body and they're structural in your body.
ma-MKQ2TAGk-013|They're a very important molecule in your body.
ma-MKQ2TAGk-015|For protein synthesis, it turns out that about four moles of ATP are hydrolyzed for every peptide bond that's formed.
ma-MKQ2TAGk-016|Now, the peptide bond is the bond between the amino acids.
ma-MKQ2TAGk-017|So as you build up a protein, you add amino acids.
ma-MKQ2TAGk-018|For each amino acid you add, you need to form a peptide bond.
ma-MKQ2TAGk-019|And for each peptide bond you form, you hydrolyze about four moles of ATP.
ma-MKQ2TAGk-020|So that's an overall accounting of how energy is used in your cells.
cLXnZlwrU04-000|Let's look at a gas phase equilibrium.
cLXnZlwrU04-001|Here's the reaction of nitrogen and hydrogen to form ammonia.
cLXnZlwrU04-002|Now, this reaction is important industrially, because nitrogen, although it's abundant-- there's 70% of it in the air we breathe-- it's unreactive.
cLXnZlwrU04-003|There's a triple bond in N2.
cLXnZlwrU04-004|It's an unreactive molecule.
cLXnZlwrU04-005|But nitrogen is required for all life.
cLXnZlwrU04-006|It's in our proteins and DNA.
cLXnZlwrU04-007|So getting nitrogen from this unreactive form to a reactive form is important in both nature and industry.
cLXnZlwrU04-008|In industry, to make fertilizers, and in nature, to fix the nitrogen so plants and animals can grow.
cLXnZlwrU04-009|This nitrogen fixation reaction is exothermic.
cLXnZlwrU04-010|It releases 92 kilojoules per mole of nitrogen that reacts.
cLXnZlwrU04-014|Raising the temperature increases the motion of the molecules and accelerates the kinetics of the reaction.
cLXnZlwrU04-015|But you're fighting thermodynamics, because as you raise the temperature, an exothermic chemical reaction will tend to favor the reactants.
cLXnZlwrU04-016|So K decreases.
cLXnZlwrU04-017|So you start to favor reactants as the temperature increases.
cLXnZlwrU04-018|So you want to fight that, and one way to do it is to raise the pressure.
cLXnZlwrU04-019|Le Chatelier's principle tells you if you raise the pressure, you're going to favor the ammonia side.
cLXnZlwrU04-020|And why is that?
cLXnZlwrU04-021|Well, raising the pressure, you have 4 moles of gas on this side and 2 moles of gas on this side.
cLXnZlwrU04-022|So the product side is favored.
cLXnZlwrU04-023|It would shift towards products to relieve that high pressure situation, shift towards the fewer molecules.
cLXnZlwrU04-026|That will efficiently balance the kinetics, the speed of the reaction, and the thermodynamics that are tending to make it favor reactants.
6DGPuhFoiJI-000|Let's take a week acid, acetic acid, originally at pH 3, dilute it by a factor of 10, and ask what is the new pH?
6DGPuhFoiJI-001|So acetic acid, pH 3, dilute with water a factor of 10.
6DGPuhFoiJI-012|Well, the pH is 3, so the H3O plus concentration is 10 to the minus 3.
6DGPuhFoiJI-016|This reaction favors the reactants.
6DGPuhFoiJI-017|So to get a product concentration of 10 to the minus 3, we have to have a much higher reactant concentration.
6DGPuhFoiJI-018|We think that's to be the case.
6DGPuhFoiJI-019|Now we dilute it.
6DGPuhFoiJI-020|When you dilute by a factor of 10, initially, this H3O plus concentration drops from 10 to the minus 3 to 10 to the minus 4.
6DGPuhFoiJI-021|If you look at this reaction, what you're doing is you're adding water.
6DGPuhFoiJI-022|If you're adding water, that shifts the reaction towards the product side.
6DGPuhFoiJI-023|If you have a greater volume, you make more ions to fill that volume.
6DGPuhFoiJI-024|So a shift towards the product side.
6DGPuhFoiJI-029|So the answer here is less than 4.
6DGPuhFoiJI-030|Now that's interesting.
6DGPuhFoiJI-031|If you did the strong acid, and this was HCl, that totally dissociates.
6DGPuhFoiJI-032|So you have H3O plus, 10 to the minus 3 originally, you dilute by a factor of 10.
6DGPuhFoiJI-033|It goes to 10 to the minus 4.
6DGPuhFoiJI-034|And there's no more H's to be had.
6DGPuhFoiJI-035|HCl completely dissociates in one shot.
6DGPuhFoiJI-036|So the pH can't change after the initial dilution.
6DGPuhFoiJI-038|In this case the pH less than 4.
jUhvGL5Jwl4-000|Let's look at a reaction with iron oxide that produces molten iron.
jUhvGL5Jwl4-001|So, which of the following-- carbon, iron, or aluminum-- should I react with iron oxide-- and I'll give you a hint, it will be an exchange reaction.
jUhvGL5Jwl4-010|Well, let's look at the possible oxides of carbon, iron, and aluminum with respect to the elements in their standard states.
jUhvGL5Jwl4-019|So, let's look at that.
jUhvGL5Jwl4-020|Here is aluminum oxide on our table relative to the elements in their standard state.
jUhvGL5Jwl4-023|So if I convert between one and another, I can tell what the entropy change would be.
jUhvGL5Jwl4-026|So the correct answer here-- aluminum to form aluminum oxide.
--pzKBW13FE-000|PolyProtic acids are compounds with more than one acidic proton.
--pzKBW13FE-001|Some examples are carbonic acid with two acidic protons, phosphoric with three, and sulfuric acid with two acidic protons.
--pzKBW13FE-002|When you titrate a PolyProtic acid, in general, the protons come off stepwise.
--pzKBW13FE-003|Another way to say that is the pKas are often separated by many units.
--pzKBW13FE-004|So you'll have an independent equilibrium for each of the acidic protons.
--pzKBW13FE-008|At pH 6.1 there would be an equal mixture of both of those.
--pzKBW13FE-013|And notice I use the terminology, the proton comes off.
--pzKBW13FE-015|So I can sketch out a titration curve.
--pzKBW13FE-016|Here's a titration curve with two buffer regions and two equivalence points.
--pzKBW13FE-019|PHs below that, the acid form here dominates, doubly protonated.
--pzKBW13FE-020|Above that you have the singly protonated species, one proton removed.
--pzKBW13FE-021|And that species, they'll be 100% that species at the first equivalence point.
--pzKBW13FE-023|As you pass through pH 10.2, then the doubly deprotonated, the doubly basic form predominates.
--pzKBW13FE-024|And then you'll get to an equivalence point where it's 100% essentially of the doubly deprotonated form.
--pzKBW13FE-025|So by looking at the pH and the various pKas, you can predict which forms are present in solution.
--pzKBW13FE-026|That's the titration curve for a PolyProtic acid.
upr1m4DR00M-000|Let's look at the pressure-volume relationship for a real gas.
upr1m4DR00M-001|Now, if the temperature is high enough, real gases behave like ideal gases.
upr1m4DR00M-002|That is, they have a PV plot that looks like PV is a constant.
upr1m4DR00M-005|The critical temperature defines the relationship between ideal and real gas behavior.
upr1m4DR00M-006|Above the critical temperature gases behave like ideal gases, and below the critical temperature they behave like real gases.
upr1m4DR00M-007|So what is real gas behavior?
upr1m4DR00M-013|Well, how do I decrease volume and not increase the pressure?
upr1m4DR00M-014|I can't change the temperature.
upr1m4DR00M-015|We're on an isotherm.
upr1m4DR00M-016|So the only other parameter left is the number of particles, but this is a fixed cylinder.
upr1m4DR00M-017|How do I decrease the number of particles in that cylinder?
upr1m4DR00M-018|Well, I can't decrease the number of particles, but I can change their phase.
upr1m4DR00M-019|I can take some particles from the gas phase to the liquid phase, where they occupy very, very small volume by comparison.
upr1m4DR00M-020|So right here, what's happening is particles are moving from the gas phase to the liquid phase as I decrease the volume.
upr1m4DR00M-021|Since the number of particles decrease, the pressure can stay the same.
upr1m4DR00M-023|So ideal gases, as you compress them, the compressibility decreases.
upr1m4DR00M-024|Real gases, as you compress them below their critical temperature, will undergo a phase transition.
upr1m4DR00M-025|Phase transition is the critical separating point between real and ideal gases.
U6uWvuiO1rA-000|Let's look at lactic acid and try to decide which atomic orbitals make up the molecular orbitals in a bond.
U6uWvuiO1rA-001|We'll look at the carbon-oxygen bond shown here.
U6uWvuiO1rA-011|We're talking about the molecular orbitals that are formed between the carbon and the oxygen in lactic acid.
U6uWvuiO1rA-012|So the carbon has to accommodate this oxygen, this oxygen, and this carbon.
U6uWvuiO1rA-013|It'll be sp2 hybridized, accommodate three things with three equivalent orbitals.
U6uWvuiO1rA-014|The oxygen has to accommodate the carbon, the hydrogen, and two lone pairs, four things.
U6uWvuiO1rA-015|So it'll have four equivalent orbitals or sp3's.
U6uWvuiO1rA-016|So the molecular orbital is going to be an overlap of the sp3 on the oxygen with the sp2 on the carbon.
U6uWvuiO1rA-017|Those will form a sigma bonding, an antibonding orbital, to make up this sigma bond.
U6uWvuiO1rA-018|So the correct answer here sp2 on carbon to sp3 on oxygen.
KtxKNXYhpy4-000|We can use the fact that enthalpy is a state function, and it only depends on the initial and final state to track enthalpies and add enthalpies.
KtxKNXYhpy4-002|That process is known as an application of Hess's law.
KtxKNXYhpy4-003|So let's just do that.
KtxKNXYhpy4-004|Here's a process, methane burning and oxygen to give carbon dioxide and liquid water.
KtxKNXYhpy4-005|The enthalpy for that is 890 kilojoules.
KtxKNXYhpy4-014|We can look at that a little differently.
KtxKNXYhpy4-015|We can look at the relative enthalpy scale.
KtxKNXYhpy4-019|So the enthalpy of vaporization is added here and takes this final state to a higher state by the enthalpy of vaporization of water.
KtxKNXYhpy4-022|So I can add this enthalpy and this enthalpy to get to this enthalpy-- two known steps to get to an unknown step.
KtxKNXYhpy4-023|That's an application of Hess's law.
9j22V8j_x0M-004|This rate will react to a temperature change more dramatically than the forward rate.
9j22V8j_x0M-007|Now, let's look at the effect of a catalyst.
9j22V8j_x0M-008|Here's the both reactions under the effect of a catalyst.
9j22V8j_x0M-009|A catalyst lowers activation energies.
9j22V8j_x0M-013|So you can run a catalyzed chemical reaction and the same amount of energy will be released or absorbed when that reaction proceeds.
9j22V8j_x0M-014|So this is a summary of the thermodynamic and kinetic variables in a chemical reaction.
-AOyf8eA1ic-000|We understand the relationship between free energy and temperature.
-AOyf8eA1ic-001|The standard state free energy is delta H minus T delta S.
-AOyf8eA1ic-007|Now, I have a linear function of natural log k and 1 over T.
-AOyf8eA1ic-010|So this slope I expect to be positive.
-AOyf8eA1ic-013|Heat is a reactant, so if I raise the temperature, I start to favor products.
-AOyf8eA1ic-014|For exothermic chemical reactions, heat is a product.
-AOyf8eA1ic-015|So as I add heat, I start to favor reactants.
-AOyf8eA1ic-016|k gets smaller.
-AOyf8eA1ic-017|Now, this helps me in another way.
-AOyf8eA1ic-018|This k is something that's relatively easy to measure.
-AOyf8eA1ic-019|I can mix reactants together, let them go to equilibrium-- essentially, wait some time.
-AOyf8eA1ic-020|When the macroscopic properties stop changing, I can measure the concentrations or pressures.
-AOyf8eA1ic-021|When the macroscopic pressures and concentration stop changing, I'm at equilibrium.
-AOyf8eA1ic-022|So I can measure equilibrium constants.
-AOyf8eA1ic-023|I can also measure temperatures.
-AOyf8eA1ic-026|Measure k versus T, plot it out, and you can measure entropies, which are very difficult to measure in the laboratory.
-AOyf8eA1ic-027|So the relationship between k and temperature, linear in lnK versus 1 over T, is displayed here.
fCGDtkmkMgU-000|Let's look at a plating reaction, where an ion from solution plates out onto a metal.
fCGDtkmkMgU-008|If I look at my standard reduction potentials, I can find that silver, as a reduction, has a higher potential than copper.
fCGDtkmkMgU-009|So silver, when it's reduced, can form oxidized copper.
fCGDtkmkMgU-012|Silver metal will plate out on to copper.
fCGDtkmkMgU-014|Now, you could also make gold plate out if you had the correct combination of ions.
fCGDtkmkMgU-015|So if you said silver or gold in this case, you are correct.
B-HgZ3G95oA-000|Let's look at the equilibrium situation between liquid water and gaseous water at 25 degrees C.
B-HgZ3G95oA-001|What can we determine about the free energy difference and K?
B-HgZ3G95oA-002|Now, here I'm talking about the free energy without the standard sign.
B-HgZ3G95oA-003|So the free energy difference under the situation that exists right now.
B-HgZ3G95oA-012|So delta G without the standard state is 0.
B-HgZ3G95oA-013|It's the free energy difference between these actual concentrations at equilibrium.
B-HgZ3G95oA-014|So at equilibrium you may remember that the partial pressure of gaseous water is around 0.03 of an atmosphere.
B-HgZ3G95oA-015|So the free energy difference between gaseous water at 0.03 of an atmosphere and liquid water at 25 degrees C is 0.
B-HgZ3G95oA-016|K is the equilibrium constant for that value, which is 0.03 of an atmosphere.
B-HgZ3G95oA-017|So K is less than 1, 0.03, and delta G is 0.
B-HgZ3G95oA-018|The correct answer here is B.
KT_N4_kKHnQ-000|Gases are an ideal system to look at energy changes, because gases change their energy in direct proportion to temperature changes.
KT_N4_kKHnQ-001|Remember, the energy change of a gas is 3/2 RT for one mole, 3/2 nRT for a given number of moles.
KT_N4_kKHnQ-002|But if the temperature changes, the energy changes.
KT_N4_kKHnQ-003|So an energy change always corresponds to a temperature change, so the gas is kind of transparent that way.
KT_N4_kKHnQ-004|It can't hide an energy change because the temperature always changes.
KT_N4_kKHnQ-005|So when work is done on a gas, that's an energy change and could be reflected in a temperature change.
KT_N4_kKHnQ-006|If heat is absorbed by gas, that's an energy change, could be reflected in the temperature change.
KT_N4_kKHnQ-009|So I have to keep track of both the heat and the work to manage energy changes.
KT_N4_kKHnQ-010|Let's look at the work.
KT_N4_kKHnQ-011|If I do work for a gas, I push a cylinder up in a gas, I change the height of this cylinder.
KT_N4_kKHnQ-012|Work is just force times distance, or in this case, force times the height.
KT_N4_kKHnQ-016|If I expand, the system is also doing the work, and that should take some of the system's energy to do that work.
KT_N4_kKHnQ-020|If a gas is compressed, then I do work on the gas.
KT_N4_kKHnQ-023|I'm the surroundings, the system is the gas, I compress it, the energy of the system tends to go up, work is positive in those circumstances.
KT_N4_kKHnQ-024|So that's how work, the change in volume, and external pressure are involved in calculating energy changes for gases.
V-FBHIdXk20-000|Let's look at the hydrogen-like energy systems.
V-FBHIdXk20-001|Now when I say hydrogen-like, I mean the one-electron systems.
V-FBHIdXk20-002|Hydrogen, a proton and an electron, that's a complete hydrogen atom.
V-FBHIdXk20-004|Or you could go lithium plus 2.
V-FBHIdXk20-005|As you have one-electron systems, they all behave just like hydrogen atoms.
V-FBHIdXk20-006|So let's look at those.
V-FBHIdXk20-007|Which ion does the energy-level diagram X correspond to?
V-FBHIdXk20-008|So here's hydrogen and near it, some unknown ion, X.
V-FBHIdXk20-019|Here we have hydrogen and some other ion.
V-FBHIdXk20-020|Now, what lines up here?
V-FBHIdXk20-021|We know that the energies go as minus Z squared over n squared.
V-FBHIdXk20-022|And we see in this system that level 2 in the unknown system lines up with level 1 in the hydrogen system.
V-FBHIdXk20-023|And that energy 1 for this system is 1 squared over 1 squared times i infinity.
V-FBHIdXk20-024|That's the hydrogen atom.
V-FBHIdXk20-025|For this one, it has to be something over something to still give me minus R infinity.
V-FBHIdXk20-026|The value of Z is undetermined here because we don't know which ion it is.
V-FBHIdXk20-027|But the value of n is 2.
V-FBHIdXk20-031|So if it has two protons-- I don't care what else it has in it-- it is a helium nucleus.
V-FBHIdXk20-032|So a helium plus is the ion that would have this configuration on its electronic energy scale.
V-FBHIdXk20-033|Z will be 2 for helium.
z76W5579a5Y-000|The decomposition of hydrogen peroxide, H2O2, is exothermic.
z76W5579a5Y-001|Now, if you look carefully at this reaction, what can you determine about the structure of H2O2 simply knowing that that reaction is exothermic?
z76W5579a5Y-012|So energetically, that doesn't tell us anything about whether this reaction is endothermic or exothermic.
z76W5579a5Y-013|What about the oxygen bond?
z76W5579a5Y-014|On this side, I'm forming oxygen gas, so O2 double-bonded.
z76W5579a5Y-015|And I form half a mole of those.
z76W5579a5Y-016|On this side, I have to break a mole of oxygen-oxygen bonds.
z76W5579a5Y-017|The question is, are they double bonds or single bonds?
z76W5579a5Y-019|And we know it's not endothermic, so we can already rule out C and B.
z76W5579a5Y-020|Now, what about if it's a single bond?
z76W5579a5Y-022|That is, the double bonds have to be more than twice as energetic to be stable as the single bonds.
z76W5579a5Y-025|So that will give us a net exothermic reaction, and the correct answer here is A, the single-bonded structure.
1iUgRsNvJkA-000|The isomerisation of propanol to isopropanol is endothermic.
1iUgRsNvJkA-001|Now what does that tell you about burning propanol and isopropanol?
1iUgRsNvJkA-002|Which burns cooler?
1iUgRsNvJkA-010|Propanol and isopropanol both burn in oxygen to form carbon dioxide in water.
1iUgRsNvJkA-011|And they are the same chemical formula, but different structure.
1iUgRsNvJkA-012|They're structural isomers.
1iUgRsNvJkA-013|Does that structurally isomerisation cause a different enthalpy of burning?
1iUgRsNvJkA-018|But they go to the same products.
1iUgRsNvJkA-019|So isopropanol burning to give the same products this same state is a greater drop in enthalpy.
1iUgRsNvJkA-020|So there's a greater release in energy on the burning of isopropanol.
1iUgRsNvJkA-021|So propanol burns cooler.
1iUgRsNvJkA-022|The correct answer here is A.
qpmTRXIJhFo-000|When we talk about ionization energy, we're talking about how easy is it to remove an electron from an atom?
qpmTRXIJhFo-001|Well, what holds the electron about the atom?
qpmTRXIJhFo-002|It's the positive charge on the nucleus.
qpmTRXIJhFo-003|So the more positive charge that an electron can see, the more difficult it is to ionize.
qpmTRXIJhFo-004|And by contrast, the more of that charge that's shielded, the easier it is to remove that electron.
qpmTRXIJhFo-005|And easier to remove means higher energy.
qpmTRXIJhFo-006|Those energy levels go up closer to the 0 state.
qpmTRXIJhFo-007|Remember, the 0 state for the nucleus and the electron is the ionized state.
qpmTRXIJhFo-008|So you get closer to the ionized state, the more charge you can shield from outer electrons.
qpmTRXIJhFo-009|Let's look at that.
qpmTRXIJhFo-010|So if we have electrons, hydrogen and helium plus, those are both one electron systems, and they're very easy to understand.
qpmTRXIJhFo-011|Their ionization energy will just be z squared-- the charge on the nucleus-- over n-- whichever orbital the electron happens to be in.
qpmTRXIJhFo-013|And that's identically four times that.
qpmTRXIJhFo-014|They're one electron systems.
qpmTRXIJhFo-015|The charge on the nucleus is plus 2 for helium.
qpmTRXIJhFo-016|Plus 2 squared is 4, so you have four times the ionization energy for helium, as opposed to hydrogen.
qpmTRXIJhFo-017|Now, what if you start to put some electrons in the helium?
qpmTRXIJhFo-018|We can look at helium in the 1s2 state.
qpmTRXIJhFo-020|If it's sufficient to ionize, the electrons will be ejected.
qpmTRXIJhFo-021|And if you put even more energy in, that electron goes-- that energy goes into kinetic energy of the electron.
qpmTRXIJhFo-022|So let's go ahead and look at helium in the 1s2 state.
qpmTRXIJhFo-023|Now the electric ionization energy is given by the effective charge squared over n squared.
qpmTRXIJhFo-026|Well, the answer is no.
qpmTRXIJhFo-027|Two s electrons don't shield each other fully.
qpmTRXIJhFo-029|So the s electrons-- there's a little bit of shielding going on, but it's not a full nuclear charge that shielded.
qpmTRXIJhFo-030|How about if one of those electrons is promoted, is excited, out to a 2 state, the 2p, for instance?
qpmTRXIJhFo-031|Now, two p electrons are very well shielded by s electrons, because the electrons have access to the nucleus.
qpmTRXIJhFo-032|There's no nodes in an s orbital at the nucleus.
qpmTRXIJhFo-033|But the p electrons do have a node at the nucleus, an angular node.
qpmTRXIJhFo-036|So shielding is very effective of low l-- l equals 0-- s electrons on p and higher electrons.
pJ9_mvvB98Y-000|I'd like to show you two simple chemical reactions, the dissolving of salts in water.
pJ9_mvvB98Y-014|But the energy was not a good predictor.
pJ9_mvvB98Y-015|One absorbed energy as it proceeded forward, the other released energy as it proceeded forward.
pJ9_mvvB98Y-016|So there has to be another thermodynamic parameter that will predict what's the forward direction for chemical reactions?
pJ9_mvvB98Y-017|What's the preferred way that chemical reactions will go?
pJ9_mvvB98Y-018|We need to find a thermodynamic parameter, and that's the subject of this lesson.
Mh8MzONj6sM-000|Let's look at the reaction between two weak acids, HA1 and HA2.
Mh8MzONj6sM-001|And p Ka1 is smaller than p Ka2.
Mh8MzONj6sM-003|What is the expression for the equilibrium constant?
Mh8MzONj6sM-004|Is it KA1 plus KA2, KA1 times KA2, or KA1/KA2?
Mh8MzONj6sM-011|We're talking about the reaction of HA1, a weak acid, and A 2-, the conjugate base of another weak acid.
Mh8MzONj6sM-012|So how do those two react?
Mh8MzONj6sM-013|Well, here's HA1 reacting with water.
Mh8MzONj6sM-017|But what we want is HA1 reacting with A 2-.
Mh8MzONj6sM-018|So let's reverse this lower reaction.
Mh8MzONj6sM-023|Now, you can tell which is the stronger acid to which side this equilibrium will lie.
Mh8MzONj6sM-024|And you'd probably say that intuitively.
Mh8MzONj6sM-025|You'd say if HA1 is the stronger acid, then we'll favor the products here.
Mh8MzONj6sM-026|If HA2 is the stronger acid, the reaction would be pushed back this way, and we'd favor the reactants.
Mh8MzONj6sM-027|And this expression tells you the same thing, because if HA1 is the stronger acid, then KA1 is larger than KA2.
Mh8MzONj6sM-028|And that gives you a K bigger than 1, or a reaction that favors the products.
Mh8MzONj6sM-029|If HA2 is the stronger acid, then KA2 is larger.
Mh8MzONj6sM-030|And this has a K less than 1, or one that favors the reactants.
FCnePMnxfF0-000|Paramagnetic species are species that have at least one unpaired electron.
FCnePMnxfF0-001|For which species does the paramagnetism increase upon ionization?
FCnePMnxfF0-002|Carbon, nitrogen, or neon?
FCnePMnxfF0-009|We're looking for a species that increases its paramagnetism upon ionization.
FCnePMnxfF0-011|Neon, all the electrons are paired.
FCnePMnxfF0-012|Nitrogen, three unpaired electrons.
FCnePMnxfF0-013|Carbon, two unpaired electrons.
EgZZXsWKM7o-000|Molecular geometry and individual bond dipole moments often lead to molecules having dipole moments.
EgZZXsWKM7o-008|We're determining dipole moments for molecules.
EgZZXsWKM7o-010|So I've done those steps for these molecules.
EgZZXsWKM7o-011|And I've presented just the final molecular geometry.
EgZZXsWKM7o-012|So for CHCl2, a tetrahedral confirmation, and this tetrahedral confirmation is going to lead to a dipole moment.
EgZZXsWKM7o-013|The Cl atoms are more electronegative than the carbon.
EgZZXsWKM7o-014|There'll be a dipole moment pointing away from the Cl atoms.
EgZZXsWKM7o-015|And the H atoms, there will also be a small dipole moment.
EgZZXsWKM7o-016|Those won't cancel out.
EgZZXsWKM7o-017|The whole molecule will have a dipole moment.
EgZZXsWKM7o-021|And for BrF3, now this molecule is not trigonal.
EgZZXsWKM7o-022|If it were perfectly trigonal and symmetric, the dipoles might cancel out.
EgZZXsWKM7o-026|And indeed, it's common-- more common for molecules to have dipole moments than to not have them, because remember, every individual bond has a dipole moment.
EgZZXsWKM7o-027|And for them to cancel out, you need a very exact symmetry in the molecule.
EgZZXsWKM7o-028|These molecules, not symmetric, so they have dipole moments.
Q9oGYNqs5TU-000|Let's look at some bond angles in two different molecules.
Q9oGYNqs5TU-002|Well, looking at it here, it looks like those angles are both 90 degrees.
Q9oGYNqs5TU-003|And of course, that's just because we've drawn them straight out on paper.
Q9oGYNqs5TU-004|In order to determine the actual bond angle, you have to go to a good Lewis dot structure, get steric numbers, and go to the molecular geometry.
Q9oGYNqs5TU-005|We can actually do that.
Q9oGYNqs5TU-006|So about the carbon in question, I've drawn the appropriate Lewis dot structure.
Q9oGYNqs5TU-007|I haven't drawn the full Lewis dot structure, but I've drawn the full one, and then taken just the important part.
Q9oGYNqs5TU-008|That allows me to determine the steric number around this carbon and this carbon.
Q9oGYNqs5TU-009|So the steric number here around this carbon, 1, 2, 3, 4.
Q9oGYNqs5TU-010|That carbon has to accommodate four things.
Q9oGYNqs5TU-011|This carbon has to accommodate 1, 2, 3 things.
Q9oGYNqs5TU-012|Now, again, it's not the number of bonds.
Q9oGYNqs5TU-013|It's the number of things bonded to that carbon, an oxygen, a hydrogen, and a carbon.
Q9oGYNqs5TU-014|This has two hydrogens, an oxygen, and a carbon.
Q9oGYNqs5TU-015|Steric number four here, so the bond angle is going to be tetrahedral.
Q9oGYNqs5TU-017|So the bond angle increases as I do that oxidation.
6V6JgmfTpyY-000|Let's think about hydrocarbon combustion in terms of a ChemQuiz.
6V6JgmfTpyY-001|I'm going to take a hydrocarbon, burn it in oxygen, and take the product, inject it into my mass spectrometer, and generate a mass spectrum.
6V6JgmfTpyY-002|The question I have is, which of these hydrocarbons would give this mass spectrum after reaction with oxygen?
6V6JgmfTpyY-009|Let's look at each of those hydrocarbons, their reaction with oxygen, and the mass spectrum of the products.
6V6JgmfTpyY-014|So the mass spectrum of the products would show a one to one peak height at 18 and 44 from the carbon dioxide in the water.
SswF_wnX1Pg-000|In electrochemistry, we tabulate the various half-cell reactions relative to the standard hydrogen electrode.
SswF_wnX1Pg-001|And here I've done that.
SswF_wnX1Pg-003|I've arranged them from high potential to low potential, from large positive potential to large negative potential.
SswF_wnX1Pg-004|In doing that, that tells me a few things.
SswF_wnX1Pg-006|That means copper ions are relatively strong oxidizing agents.
SswF_wnX1Pg-007|They're willing to accept electrons.
SswF_wnX1Pg-009|When they withdraw electrons, they're oxidizing that metal.
SswF_wnX1Pg-010|The copper ions themselves are being reduced.
SswF_wnX1Pg-011|Conversely, this zinc system would work the opposite way.
SswF_wnX1Pg-012|When zinc metal is in contact with ions, the tendency is for zinc metal to give up electrons to those ions.
SswF_wnX1Pg-013|So the zinc metal is a reducing agent.
SswF_wnX1Pg-014|It tends to donate electrons, to give up its electrons, and thereby reduce something while the zinc metal itself is oxidized.
SswF_wnX1Pg-016|So I can write down now the reaction.
SswF_wnX1Pg-018|I could look at it the other way and say the zinc metal is going to donate electrons to that copper ion.
SswF_wnX1Pg-024|So the potential for this reaction to go should remind you of another thermodynamic parameter-- the free energy.
SswF_wnX1Pg-025|The potential for this reaction to go, the fact that it favors the products here, sounds like the free energy.
SswF_wnX1Pg-026|And indeed, the potential is related to the standard state free energy difference.
SswF_wnX1Pg-027|The standard state free energy difference is given by minus nF delta E, the potential difference.
SswF_wnX1Pg-028|In this case, n is the number of electrons transferred in the process, in that case, two electrons.
SswF_wnX1Pg-029|F is Faraday's constant, that's the charge on a mole of electrons.
SswF_wnX1Pg-030|So what we see is the free energy difference related to the voltage in a way we understand.
SswF_wnX1Pg-031|A positive voltage, the potential for this reaction to go as positive, relates to a negative free energy difference.
fryL1jN2OHY-000|Certain bacteria can live around oceanic thermal vents.
fryL1jN2OHY-002|High pressure means water has a higher boiling point, and you can have liquid water above 100 degrees C.
fryL1jN2OHY-003|Now, temperatures above 100 degrees C normally kill bacteria.
fryL1jN2OHY-004|In fact, you can use hot water and steam to disinfect surfaces, to remove bacteria.
fryL1jN2OHY-005|These bacteria, however, thrive there.
fryL1jN2OHY-021|Well, what about at higher temperatures?
3tl5gAwHj6w-000|Molecular shape and geometry is motivated by trying to keep everything apart in space.
3tl5gAwHj6w-001|So if you have lone pairs or if you have a bonded atom, you'd like to arrange them as far apart as possible.
3tl5gAwHj6w-002|It's called valence shell electron pair repulsion.
3tl5gAwHj6w-003|That is, you have electron pairs and either bonds or lone pairs and they repel each other, and you'd like to distribute them in space as far apart as possible.
3tl5gAwHj6w-004|So how do you do that?
3tl5gAwHj6w-005|Well, you know the steric number is the number of lone pairs and the number of bonded atoms-- that tells you how many things you have to distribute in space.
3tl5gAwHj6w-006|We can look at that for several examples.
3tl5gAwHj6w-007|For instance, if we have six things distributed in space, here are six balloons distributed around the central point.
3tl5gAwHj6w-008|And the balloons touching each other demonstrates the steric interaction, how everything is trying to spread itself out as far as possible in space.
3tl5gAwHj6w-009|And what you get is this kind of arrangement.
3tl5gAwHj6w-010|In this arrangement, all the positions are equivalent, even though I've got red and yellow balloons.
3tl5gAwHj6w-011|The yellow balloons or the green balloons could be on top, each position is identical.
3tl5gAwHj6w-012|But what you have here is an octahedral shape.
3tl5gAwHj6w-013|An octahedral shape is six things arranged around a central point-- six vertices.
3tl5gAwHj6w-018|What if I have five things?
3tl5gAwHj6w-019|Well, with a balloon, it's easy to go from six to five.
3tl5gAwHj6w-024|So here's a 90-degree bond angle between green and yellow, but the green greens are 120 degrees.
3tl5gAwHj6w-028|We'll lose one more of our balloons.
3tl5gAwHj6w-029|Four things.
3tl5gAwHj6w-030|Now here's a tetrahedral shape.
3tl5gAwHj6w-031|Notice that these four things don't naturally arrange themselves as a square plane.
3tl5gAwHj6w-032|I could probably force them into a square planar configuration, but I have to hold them there.
3tl5gAwHj6w-039|So you can see this VSEPR arrangement is better energetically, the electron pairs have bigger angles, more space between them.
AV5nRtwehD0-000|Let's see if you can predict some bond orders.
AV5nRtwehD0-001|Here we'll take nitrite going to nitrate-- NO2- to NO3-.
AV5nRtwehD0-002|In that conversion, it's an oxidation.
AV5nRtwehD0-008|We're looking at NO2- and NO3- and how the bond order changes when I do that oxidation reaction.
AV5nRtwehD0-009|Now, the way to do this is to draw a Lewis dot structure and all the resonance structures for NO2- and NO3- and determine the bond order.
AV5nRtwehD0-010|Now I know you can do that, so I'm going to show you a shortcut that also works.
AV5nRtwehD0-011|What we're going to recognize is that NO2- is isoelectronic with something we already know-- ozone.
AV5nRtwehD0-012|Ozone has 3 oxygen atoms, each has 6 valence electrons, so that's 18 electrons.
AV5nRtwehD0-013|Ozone is an 18-electron system.
AV5nRtwehD0-014|What about NO2-?
AV5nRtwehD0-015|Well, let's look at the periodic table.
AV5nRtwehD0-016|NO2-, here's oxygen with its 6 valence electrons.
AV5nRtwehD0-017|Nitrogen with 5.
AV5nRtwehD0-019|So in total, we have 6 electrons from each oxygen, that's 12, 5 from the nitrogen, that 17, 1 from the charge, that's 18.
AV5nRtwehD0-020|So an 18-electron system.
AV5nRtwehD0-022|So if I have the same number of atoms, the same number of electrons, they'll bond together in the same way.
AV5nRtwehD0-025|We already know the carbonate bond order is 1 and 1/3.
AV5nRtwehD0-026|So going from 1 and 1/2 to 1 and 1/3 in that oxidation reaction decreases the bond order.
AV5nRtwehD0-027|The correct answer here-- decreases.
BSnGqJsLCZA-000|Acid strength is dependent on a variety of factors.
BSnGqJsLCZA-001|Certainly one of them is the strength of the bond holding the proton on the molecule.
BSnGqJsLCZA-004|And that equilibrium constant is determined not just by enthalpic or bond strength considerations.
BSnGqJsLCZA-005|It's also entropic considerations.
BSnGqJsLCZA-006|Remember, K for this reaction depends on the standard state free energy difference between the products and reactants.
BSnGqJsLCZA-007|In fact, we define weak acids as having a K less than 1 or a positive free energy difference.
BSnGqJsLCZA-008|And strong acids as a K greater than 1 for this reaction or a spontaneous or negative standard state free energy difference.
BSnGqJsLCZA-009|So it's an enthalpic and an entropic term.
BSnGqJsLCZA-013|HF, it takes 543 kilojoules to break that bond.
BSnGqJsLCZA-014|Where HBr, only 354 kilojoules to break that bond.
BSnGqJsLCZA-018|In this case, depended upon the strengths of the bond holding the proton on the molecule.
t-WkmDPN8Ws-000|Let's look at another physical equilibrium, the solid gas sublimation equilibrium for iodine.
t-WkmDPN8Ws-001|I've written the reaction here, iodine solid going to iodine gas.
t-WkmDPN8Ws-002|The question I have is what happens when I add more iodine solid to this equilibrium?
t-WkmDPN8Ws-003|What happens to the intensity of the I2 gas?
t-WkmDPN8Ws-011|We're talking about the sublimation of iodine going from the solid to the gas.
t-WkmDPN8Ws-013|So an equilibrium that's dynamic, but from a macroscopic sense it appears static.
t-WkmDPN8Ws-014|Let's look at the equilibrium expression if we write products over reactants.
t-WkmDPN8Ws-015|For this we would have the partial pressure of the I2 gas, I2 solid, that's a pure solid.
t-WkmDPN8Ws-016|And pure liquids and solids don't appear in our equilibrium expressions.
t-WkmDPN8Ws-019|If it's independent of the solid, then as I add solid I shouldn't expect this equilibrium to change.
t-WkmDPN8Ws-020|Doesn't appear in the equilibrium expression, the solid, so the solid doesn't affect the equilibrium expression.
t-WkmDPN8Ws-021|Let's actually see if that happens.
t-WkmDPN8Ws-022|What I have here is the solid and vapor in equilibrium.
t-WkmDPN8Ws-031|The equilibrium constant is a function only of the vapor pressure.
t-WkmDPN8Ws-035|But at the same time, I've added that more solid, that gives more surface area for the vapor that's there to condense back on the solid.
t-WkmDPN8Ws-036|So I don't change the position of equilibrium.
-DM5LRsQNbA-000|Let's look at the equilibrium between NO2 gas and N2O4 if we change the temperature.
-DM5LRsQNbA-001|So given this equilibrium, what happens to the color intensity when I raise the temperature?
-DM5LRsQNbA-002|Will it increase, decrease, or stay the same?
-DM5LRsQNbA-009|We're talking about the equilibrium between NO2 and N2O4.
-DM5LRsQNbA-010|NO2, a brown gas, N2O4, a clear gas.
-DM5LRsQNbA-011|Now we know that this is an exothermic chemical reaction, so heat is a product.
-DM5LRsQNbA-012|If we add heat, then we'd expect the equilibrium to shift back towards reactants to relieve that stress by Le Chatelier's principle.
-DM5LRsQNbA-014|So let's see that happen.
-DM5LRsQNbA-015|I have two flasks here, both with the N2O4 and the NO2 gas.
-DM5LRsQNbA-016|The NO2 gas is the brown gas, the reactant.
-DM5LRsQNbA-017|I'm going to heat one and cool the other to see the effect of temperature.
-DM5LRsQNbA-018|We expect that the hotter, the higher temperature will be darker.
-DM5LRsQNbA-019|Let's see that happen.
-DM5LRsQNbA-026|So that's a beautiful effect.
-DM5LRsQNbA-027|And in fact, it's common for equilibrium constants to behave like this.
-DM5LRsQNbA-028|If you have an equilibrium constant for an exothermic reaction, the equilibrium constant will decrease with temperature.
-DM5LRsQNbA-032|Now this is actually the equilibrium constant changing.
-DM5LRsQNbA-033|So this is not a shift where q changes, it's actually k that changes.
-DM5LRsQNbA-034|So I go to a k that smaller, a smaller k means bigger denominator, more reactants.
-DM5LRsQNbA-035|The color increases, k gets smaller.
rYgOBHeQyjA-000|Let's look at a titration where we dilute the solution and then titrate again.
rYgOBHeQyjA-001|So I'm going to titrate the strong acid, nitric acid, HNO3, with the strong base, potassium hydroxide.
rYgOBHeQyjA-002|I'm then going to dilute it and do the titration again.
rYgOBHeQyjA-003|So shown here in yellow is the initial titration.
rYgOBHeQyjA-004|I'm going to dilute by a factor of 10, just add water, titrate again.
rYgOBHeQyjA-012|We're looking at two titrations, one where we've diluted by a factor of 10.
rYgOBHeQyjA-013|If you dilute a strong acid solution by a factor of 10, you change the H3O+ concentration by a factor of 10, and you increase the pH by one unit.
rYgOBHeQyjA-014|So the initial pH should be one unit higher than the undiluted.
rYgOBHeQyjA-015|But in dilution, you're just adding water.
rYgOBHeQyjA-016|You're not adding any more moles of acid.
rYgOBHeQyjA-017|So the same number of moles of acid are there.
rYgOBHeQyjA-018|So it requires the same number of moles of base to reach equivalence point.
nyp9rT6pRXo-001|That's a fancy way of saying water absorbs energy as it boils.
nyp9rT6pRXo-002|The question I have for you is, which is more dangerous at 100 degrees C?
nyp9rT6pRXo-003|Which is more dangerous to come in contact with?
nyp9rT6pRXo-014|That means as you go from steam back to water, that 44 kilojoules will be released.
nyp9rT6pRXo-015|So when you come in contact with steam, it can condense back to water and release 44 kilojoules and burn your skin.
nyp9rT6pRXo-016|And then you have water at 100 degrees C still on your skin.
nyp9rT6pRXo-017|So steam has 44 extra kilojoules to burn you with over water at 100 degrees C.
nyp9rT6pRXo-018|So steam is the more dangerous of these combinations.
nyp9rT6pRXo-019|Don't come in contact with steam over boiling water.
nyp9rT6pRXo-020|It's more hazardous than the boiling water itself.
YsyMQYr4h_4-000|When you talk about electromagnetic waves, the waves are properties of an electric and magnetic field.
YsyMQYr4h_4-001|And they can be very long, like radio waves, or they can be very short like gamma waves.
YsyMQYr4h_4-002|But there's no sense that we can perceive them.
YsyMQYr4h_4-003|We can't see them or smell them or touch them or understand the wave behavior.
YsyMQYr4h_4-004|When we look at ocean waves, it's obvious that there is a wave property.
YsyMQYr4h_4-005|And we can see their speed and their wavelengths and their frequencies.
YsyMQYr4h_4-006|But with electromagnetic radiation, you can't.
YsyMQYr4h_4-007|So it would be nice to be able to demonstrate the wave property in an experiment.
YsyMQYr4h_4-008|Well, with ocean waves, water waves, it's well-known, when those waves hit obstacles, how they behave.
YsyMQYr4h_4-009|So it would be neat to arrange an experiment where an electromagnetic wave hits some obstacles and display those same kind of wave characteristics.
YsyMQYr4h_4-010|So that's what we're going to do.
YsyMQYr4h_4-011|So consider this.
YsyMQYr4h_4-013|So I think you understand this.
YsyMQYr4h_4-014|This is like poking a hole in a pie plate and shining a flashlight through it onto a wall.
YsyMQYr4h_4-015|You know what you'd see.
YsyMQYr4h_4-016|You'd see an image of the hole that you poked, and it would be diffuse around the edges, and it would be bright in the center.
YsyMQYr4h_4-017|That's because the light is diffracting and spreading out around the edges of the hole you poked.
YsyMQYr4h_4-024|So what you have, though, is the same thing-- a bright spot getting dim at the edges that I've plotted out here.
YsyMQYr4h_4-025|So this intensity distribution is just like a probability distribution if it were particles.
YsyMQYr4h_4-026|That is, let's say I shot bullets or BBs or something through a hole.
YsyMQYr4h_4-027|Some of them would tick off the sides here, and they'd spread out, and they'd end out at the fringes.
YsyMQYr4h_4-028|But a lot of them would go right through and hit directly across from the hole.
YsyMQYr4h_4-029|So you'd have a high probability of a particle hitting directly across from the hole and slightly lower probability as you move away from the whole.
YsyMQYr4h_4-032|So two slits together is where it gets interesting.
YsyMQYr4h_4-033|When waves interact, that's how we can tell they're waves.
YsyMQYr4h_4-034|And you probably know this by looking at waves in the bathtub or waves on the beach.
YsyMQYr4h_4-035|If there's something in the middle, and waves hit them, there's a pattern that's predictable as those waves moves away from the obstacle.
YsyMQYr4h_4-036|So this is like an obstacle to light waves.
YsyMQYr4h_4-039|And it turns out, for some wavelengths and for some orientation of the slits, that's what you get.
YsyMQYr4h_4-040|But if you arrange the slits appropriately, you get a very distinct pattern.
YsyMQYr4h_4-042|So you see bright spot, dark spot, bright spot, dark spot, bright spot, dark spot alternating away from the center of the slits.
YsyMQYr4h_4-043|The brightest spot is actually right between the slits.
YsyMQYr4h_4-044|So that doesn't look like a particle-like property at all.
YsyMQYr4h_4-046|So this is a property only of waves.
YsyMQYr4h_4-047|And how can we understand it?
YsyMQYr4h_4-048|Well, it's actually pretty clear.
YsyMQYr4h_4-049|If you think about it, a wave has to travel from the slits to the screen.
YsyMQYr4h_4-050|And if a light has to travel from the slits to the screen-- so here's a wave coming from each slit hitting the center.
YsyMQYr4h_4-051|Notice that the distance that this wave has to travel and the distance that that wave has to travel will be the same.
YsyMQYr4h_4-054|Now what about a dim spot up here.
YsyMQYr4h_4-056|So the one that has the farther path is going to be out of phase, somewhat, with the one that has the shorter path.
YsyMQYr4h_4-057|So think about that.
YsyMQYr4h_4-058|One wave travels up, down, up, down, up.
YsyMQYr4h_4-059|The other one has a longer path up, down, up, down, up, down.
YsyMQYr4h_4-060|So one arrives in an up phase.
YsyMQYr4h_4-061|The other arrives in a down phase.
YsyMQYr4h_4-062|And when you add those two intensities together, what you get is destructive interference.
YsyMQYr4h_4-063|So the waves add to give zero intensity.
YsyMQYr4h_4-064|It's actually really cool.
YsyMQYr4h_4-065|And it's something you can see in the experiments that we'll show you later on.
YsyMQYr4h_4-066|We'll shine a laser through a slit, and we'll actually show you this pattern of light and dark spots.
YsyMQYr4h_4-067|This shows us that electromagnetic radiation is actually a wave.
7mSVntGkeSk-000|When we talk about ideal gases, we're talking about a construct that doesn't actually exist.
7mSVntGkeSk-001|We're talking about particles that don't interact with each other.
7mSVntGkeSk-002|And the particles themselves don't have any value.
7mSVntGkeSk-003|So you theoretically could compress an ideal gas down to zero value.
7mSVntGkeSk-004|Or you could cool it down to zero volume.
7mSVntGkeSk-005|In fact, that's how we define the absolute zero in temperature.
7mSVntGkeSk-006|The zero case for the volume of an ideal gas.
7mSVntGkeSk-007|But those are limiting conditions and we don't often go there.
7mSVntGkeSk-014|That is, the compressibility decreases.
7mSVntGkeSk-015|It gets harder to compress the gas.
7mSVntGkeSk-017|That is pressure increases become very large for very small decreases in the volume.
7mSVntGkeSk-018|It's very difficult to compress an ideal gas down near zero volume.
7mSVntGkeSk-019|What does the temperature is different?
7mSVntGkeSk-020|Well I can look at different temperatures.
7mSVntGkeSk-021|Here is higher temperatures for an ideal gas.
7mSVntGkeSk-022|But ideal gas temperature variation, they still behave the same way.
7mSVntGkeSk-023|You'll still reach that point where decreasing the volume corresponds to very high increases in the pressure.
7mSVntGkeSk-024|These are the features of an ideal gas.
sxWKXP-mHsA-000|There's another kind of stereo isomerism.
sxWKXP-mHsA-006|Here I have two models where if I placed a mirror plane right here, these molecules represent mirror images.
sxWKXP-mHsA-007|That is, there's a direct reflection here to here, here to here, here to here, here to here.
sxWKXP-mHsA-008|So these molecules, mirror images, are they the same?
sxWKXP-mHsA-009|If I take this molecule, let's bring it over and try to superimpose it on this one, we find out they're not.
sxWKXP-mHsA-010|Here yellow and yellow match up, but green and purple don't.
sxWKXP-mHsA-011|These two molecules, though at first glance appear identical, are actually stereoisomers of each other.
sxWKXP-mHsA-012|They are stereoisomers called and enantiomers.
sxWKXP-mHsA-013|When I have this kind of non-superimposable mirror image, the isomers are called enantiomers and the molecules are called chiral.
sxWKXP-mHsA-014|Now chirality is the same property your hands have.
sxWKXP-mHsA-015|Your hands are mirror images of each other, but they're not superimposable.
sxWKXP-mHsA-016|I can't make my thumb line up perfectly and get all the fingers to line up.
sxWKXP-mHsA-017|So my hands have this property.
sxWKXP-mHsA-020|They're very difficult to distinguish and it often takes a sophisticated spectroscopy experiment to distinguish them.
sxWKXP-mHsA-022|These are different molecules.
sxWKXP-mHsA-023|This is important because in nature many of the molecules in your body are chiral.
sxWKXP-mHsA-026|Here's a carbon with ammonia group, a carboxyl group, a hydrogen, and a methyl group attached, and here is its mirror image.
sxWKXP-mHsA-027|They are different molecules in a very subtle way.
sxWKXP-mHsA-028|They have just a handedness.
sxWKXP-mHsA-029|And there's a handedness that predominates in nature.
sxWKXP-mHsA-033|So this handedness is interesting.
sxWKXP-mHsA-034|It appears in your body and you might say, well if these molecules are so similar, how does your body know that it needs only L?
sxWKXP-mHsA-035|In fact, it's completely true if you were to feed a person all of D amino acids it would be protein.
sxWKXP-mHsA-036|You could eat it, but you would actually die because your body can't process the other handedness.
sxWKXP-mHsA-037|And here's a little simple example why.
sxWKXP-mHsA-038|If your hand is right and you go to shake a hand, you can detect whether you're shaking a right hand.
sxWKXP-mHsA-039|A right hand fits into your right hand.
sxWKXP-mHsA-040|Your hand is chiral and it can detect I need that same chirality to shake hands.
sxWKXP-mHsA-041|If a left hand comes in, they don't fit together right.
sxWKXP-mHsA-042|So there's a handedness that predominates, and it can be detected in nature.
sxWKXP-mHsA-044|So this achiral system doesn't have that property of I need to have one isomer to have that particular fit.
sxWKXP-mHsA-045|Chirality, a very subtle form of isomorphism, is important in nature right down to the very molecules that make up our bodies.
2wza6DXebFQ-001|Now, the pressure comes from the particles impacting the wall of the flask.
2wza6DXebFQ-002|And when they impact the wall of the flask, they impart a momentum.
2wza6DXebFQ-003|A change in momentum is a force.
2wza6DXebFQ-004|And we want the force per unit area.
2wza6DXebFQ-005|So there's a momentum and then there's how frequently the particles hit the wall.
2wza6DXebFQ-010|Their speeds, v. If there's more speed, everything else being equal, the particles are going to hit the wall more often.
2wza6DXebFQ-011|But it will be inversely proportional to the volume.
2wza6DXebFQ-012|Because everything else being equal, a larger volume, you'll have fewer collisions with the wall.
2wza6DXebFQ-013|So this frequency of collisions term depends on those three factors.
2wza6DXebFQ-014|The product of these two is going to give us a measure of the pressure.
2wza6DXebFQ-019|Now, that factor of three comes from the fact that this frequency here we kind of over account because not all the particles are heading towards the wall.
2wza6DXebFQ-020|Some are moving parallel to the wall, some are moving away from the wall.
2wza6DXebFQ-021|So we over count it a little bit, so that compensates for this.
2wza6DXebFQ-022|So we have a pressure in terms of the number of particles, their masses, their velocity squared, and the volume.
2wza6DXebFQ-025|So this expression for the pressure, the average mean square speed, the number of particles, their masses, and the volume.
rvPHWWNYC0I-009|We're talking about the structural isomers of pentane.
rvPHWWNYC0I-010|Now, there's no neat way to go from the chemical formula, C5H12, to the number of isomers.
rvPHWWNYC0I-011|You pretty much have to scratch them out on paper and rearrange the atoms and then count up the unique combinations.
rvPHWWNYC0I-012|What I've done here is I started with a straight carbon chain, the 5 carbons, I haven't drawn the hydrogens in.
rvPHWWNYC0I-013|And you'll find that you don't really need to draw the hydrogens in these straight chain hydrocarbons to come up with the isomers.
rvPHWWNYC0I-017|Now, if this branch is here or here, that's actually the same molecule.
rvPHWWNYC0I-018|So what you have-- if I move this carbon over here but I flip the whole molecule over, it's the same molecule.
rvPHWWNYC0I-019|So those are not two unique isomers.
Z08pNaGQPv8-000|When more than one electron is around an atom, those electrons have a set of four quantum numbers.
Z08pNaGQPv8-001|And they'll occupy the available orbitals around an atom, but they'll do it in very specific ways, they'll follow very specific rules.
Z08pNaGQPv8-002|One of those rules is of the four quantum numbers, n, l, m sub l, and m sub s, each electron must have a unique set.
Z08pNaGQPv8-003|No two electrons can have exactly the same quantum numbers.
Z08pNaGQPv8-004|That's called the Pauli exclusion principle.
Z08pNaGQPv8-005|Now, there's other rules that they follow.
Z08pNaGQPv8-006|For instance, if there's degenerate energy levels-- and we know when we have the p orbitals, there's three that are of the same energy.
Z08pNaGQPv8-007|How do the electrons choose to occupy those?
Z08pNaGQPv8-008|Well, there's several possibilities.
Z08pNaGQPv8-009|They could go into the same orbital, spin-up and spin-down.
Z08pNaGQPv8-010|That would give them different quantum numbers.
Z08pNaGQPv8-011|m sub s would be different, although n, l, and m sub l would be the same.
Z08pNaGQPv8-012|They could have different m sub l values and different m sub s values.
Z08pNaGQPv8-018|And indeed, if you have electrons anti-parallel, that's a little bit higher energy.
Z08pNaGQPv8-019|They like to be parallel.
Z08pNaGQPv8-020|So it turns out that the electrons like to spread out in space.
Z08pNaGQPv8-021|That makes sense, because they have negative charges.
Z08pNaGQPv8-022|You wouldn't want to cram them both in the same orbital if you didn't have to.
Z08pNaGQPv8-026|They're not disallowed by any quantum mechanical rules, but they're higher in energy.
Z08pNaGQPv8-027|So this is called Hund's rule.
Z08pNaGQPv8-028|And it turns out electrons enter degenerate orbitals the same way people get on a bus.
Z08pNaGQPv8-029|You know, if you get on a bus, there's a lot of seats and they each hold two people.
Z08pNaGQPv8-030|But if there's someone sitting in one seat, do you go in and sit right next to them in the same seat?
Z08pNaGQPv8-031|[LAUGHS] No.
Z08pNaGQPv8-032|You go sit in a seat far away from them.
Z08pNaGQPv8-033|Do you go and sit in a seat near them and face them?
Z08pNaGQPv8-035|You go and sit in the same direction facing away on the bus.
Z08pNaGQPv8-036|Electrons enter orbitals the same way people enter seats on a bus.
Z08pNaGQPv8-037|Let's look at that.
Z08pNaGQPv8-038|I've chosen to use as my model for electrons the universal eating implement, the spork.
Z08pNaGQPv8-039|I do that because sporks have a preferred orientation.
Z08pNaGQPv8-040|I can tell this spork is pointing up, and this spork is pointing down.
Z08pNaGQPv8-041|So let's bring in some sporks as electrons to these two energy levels.
Z08pNaGQPv8-043|If there's no other spins there, then it doesn't matter.
Z08pNaGQPv8-045|But the way the first one goes in determines how the second one goes in.
Z08pNaGQPv8-048|The next spin that comes in now is forced to sit next to somebody, because these are the only degenerate energy levels.
Z08pNaGQPv8-049|There may be higher energy levels up here, but these are lower energy.
Z08pNaGQPv8-051|So the next electron will go in anti-parallel.
Z08pNaGQPv8-054|Electrons are fermions.
Z08pNaGQPv8-055|They're spin 1/2, and fermions follow this Pauli exclusion principle, where no two can have exactly the same quantum numbers.
Z08pNaGQPv8-056|They can't occupy the same space at the same time.
Z08pNaGQPv8-057|You can think of that m sub s quantum number as a time quantum number.
Z08pNaGQPv8-058|It says, I'll occupy the same space, the same orbital, but I'll do it at different times.
Z08pNaGQPv8-059|M sub s is different, so we can both sit there, we just won't see each other in time.
Z08pNaGQPv8-060|It's interesting, there's another class of particles called bosons, that can have the same quantum numbers.
Z08pNaGQPv8-061|They can all collapse into exactly the same quantum state.
Z08pNaGQPv8-062|And when that happens-- and this has been proven, a Nobel Prize was awarded for this-- where a Bose-Einstein condensate occurs.
Z08pNaGQPv8-063|All the quantum particles collapse into exactly the same quantum state.
Z08pNaGQPv8-064|They exist in the same space at the same time.
Z08pNaGQPv8-065|Matter collapsed upon itself.
Z08pNaGQPv8-066|And you can Google this on the web and see some beautiful animations of a Bose-Einstein condensate.
Z08pNaGQPv8-067|Particles existing in the same space at the same time.
Z08pNaGQPv8-068|It's a quantum feature of particles.
Z08pNaGQPv8-069|They're wavelike, they can exist in the same space in the same time, if they're bosons.
Z08pNaGQPv8-070|They have an integer spin.
Z08pNaGQPv8-073|So this set of energy levels is full, I can occupy no more.
jv9_JbddC7Y-000|Equilibria are dynamic.
jv9_JbddC7Y-002|That dynamic nature allows the equilibrium to respond to a stress, to shift towards reactants or shift towards products.
jv9_JbddC7Y-003|In the case of this dissolution, let's say we had an extra source of sulfate ions.
jv9_JbddC7Y-004|So we're trying to dissolve barium sulfate, but there is some sulfate ions already in solution.
jv9_JbddC7Y-005|A common ion already exists in solution.
jv9_JbddC7Y-006|Le Chatelier's principle would predict that would make the salt less soluble.
jv9_JbddC7Y-007|Let's see if we can calculate that.
jv9_JbddC7Y-008|Let's say we dissolve barium sulfate in 0.1 molar of a sulfate solution.
jv9_JbddC7Y-012|The barium concentration times the sulfate concentration has to equal Ksp.
jv9_JbddC7Y-016|So you can essentially neglect it.
jv9_JbddC7Y-017|So you have 0.1x equals 10 to the minus 10, or x equals 10 to the minus 9.
jv9_JbddC7Y-018|That's a very small concentration.
jv9_JbddC7Y-019|If you had barium sulfate in pure water, then the concentration, you can figure that would be x squared equals Ksp.
jv9_JbddC7Y-020|The concentration would be 10 to the minus 5th.
jv9_JbddC7Y-021|So adding just a little sulfate reduces the barium sulfate concentration by a factor of 10 to the 4.
SpwSZfpYOLY-000|Let's talk more about the strength of ionic interactions.
SpwSZfpYOLY-001|There's two components to an ionic interaction.
SpwSZfpYOLY-002|One is the charge-- in this case, plus 1 and minus 1.
SpwSZfpYOLY-003|The other is the distance between them.
SpwSZfpYOLY-004|As the distance gets smaller, that ionic interaction gets stronger and that forms a stronger bond.
SpwSZfpYOLY-006|When chlorine approaches a lithium or a sodium or potassium.
SpwSZfpYOLY-007|The chlorine wants to pull an electron.
SpwSZfpYOLY-008|The sodium and lithium and potassium want to give up an electron.
SpwSZfpYOLY-009|The question is at what distance does it make sense for them to do so?
SpwSZfpYOLY-010|When is it energetically efficient for the electron to move from the potassium or the sodium or the lithium over to the chlorine?
SpwSZfpYOLY-011|Let me ask you this question.
SpwSZfpYOLY-012|The R, the distance between nuclear centers, where an electron jumps from one element to the other varies.
SpwSZfpYOLY-020|We're looking at the formation of chlorides and we're trying to decide when the electron jumps from either sodium, potassium, or lithium onto the chlorine.
SpwSZfpYOLY-021|We know when it does, there'll be an ionic bond formation.
SpwSZfpYOLY-022|So we're balancing two things.
SpwSZfpYOLY-023|We get a bunch of energy out when we form the ionic bond.
SpwSZfpYOLY-024|But it takes a bunch of energy to transfer that electron from the potassium, sodium, or lithium over to the chlorine.
SpwSZfpYOLY-025|The question is, how do I balance that distance?
SpwSZfpYOLY-026|The closer the distance, the stronger the bond.
SpwSZfpYOLY-027|So I get more and more energy released the closer I get.
SpwSZfpYOLY-028|If I'm far apart, I don't get as much energy released.
SpwSZfpYOLY-029|So if I'm far apart, I need it to be easy to peel the electron off the sodium, potassium, or lithium.
SpwSZfpYOLY-030|Whereas if I get close together, I can peel the electron off something that's less willing to give it up.
SpwSZfpYOLY-031|So it's the ionization energy of the sodium or lithium or potassium that I have to look at.
SpwSZfpYOLY-036|I can do so at a greater distance.
SpwSZfpYOLY-037|Potassium's the right answer in this case.
SpwSZfpYOLY-038|Now there's a lot going on in this chem quiz.
SpwSZfpYOLY-039|We had to talk about ionization.
SpwSZfpYOLY-040|We had to talk about coulombic interaction.
SpwSZfpYOLY-041|We had to talk about ionization energy.
SpwSZfpYOLY-042|If you didn't get everything here, don't worry about that.
SpwSZfpYOLY-043|We'll have chem quizzes like this where we bring everything together, and it might be really tough.
SpwSZfpYOLY-044|The point is, stop and think about it.
SpwSZfpYOLY-045|Talk about it with your friends.
SpwSZfpYOLY-046|If you get it, great.
SpwSZfpYOLY-047|If you don't, listen to the explanation again, and see if you can.
BceZjaeLCsw-000|Here's a chemical reaction-- the reaction of iron with oxygen to form iron oxide.
BceZjaeLCsw-001|The question is, what's the enthalpy change for that chemical reaction?
BceZjaeLCsw-007|So we can find the enthalpy for this reaction by looking up on a table the enthalpy of formation of iron oxide.
VYDCI3l22Q0-000|Let's do a calculation involving oxidation numbers determined from Lewis dot structures.
VYDCI3l22Q0-004|Well, in order to do that, we need a Lewis electron dot structure for each molecule.
VYDCI3l22Q0-005|I can draw those here.
VYDCI3l22Q0-006|And I've simplified them a little, because we're interested only in this carbon.
VYDCI3l22Q0-007|So I haven't drawn the full Lewis electron dot structure, but the relevant parts around that carbon.
VYDCI3l22Q0-008|Now, to do an oxidation number, I have to assign the electrons based on relative electronegativities.
VYDCI3l22Q0-009|I'm going to assign electrons to this carbon based on electronegativities on the things it's bonded to.
VYDCI3l22Q0-010|This carbon is more electronegative than the hydrogen. So it will get all of these electrons.
VYDCI3l22Q0-011|It's less electronegative than this oxygen.
VYDCI3l22Q0-012|So it won't get any of those.
VYDCI3l22Q0-013|It's equally electronegative with this carbon.
VYDCI3l22Q0-014|So to count them up, the carbon gets two electrons from that bond.
VYDCI3l22Q0-015|It gets two electrons from this bond, being more electronegative than the hydrogen.
VYDCI3l22Q0-017|So two, four, five electrons around carbon in that molecule.
VYDCI3l22Q0-018|Carbon, as an atom, has four electrons around it.
VYDCI3l22Q0-019|This carbon has one more than that.
VYDCI3l22Q0-020|So it has an oxidation number of minus one.
VYDCI3l22Q0-021|How about this molecule?
VYDCI3l22Q0-022|Again, we'll do the same thing.
VYDCI3l22Q0-023|It'll get one electron here, because it's equally electronegative with the carbon next to it.
VYDCI3l22Q0-024|And it will get two electrons from that hydrogen, because that bond, both electrons will be assigned to the more electronegative carbon.
VYDCI3l22Q0-025|And none in this double bond, because those electrons will be assigned to the oxygen.
VYDCI3l22Q0-026|So one, two, three is the number of electrons here.
VYDCI3l22Q0-027|Normally, it has four.
VYDCI3l22Q0-030|So the oxidation number changes two steps in this oxidation reaction.
hXLiubyd58Y-001|Now, the reaction goes because the reaction of the weak acid and the strong base forms water.
hXLiubyd58Y-002|And the formation of the water is very favored.
hXLiubyd58Y-003|So this reaction lies very strongly towards the products.
hXLiubyd58Y-005|Let's see what that reaction looks like.
hXLiubyd58Y-006|We have a typical titration curve plotted out here.
hXLiubyd58Y-007|And at several points in the titration curve, the pH is relatively easy to calculate.
hXLiubyd58Y-011|But we can also talk about here point A, the beginning of the titration, the initial point.
hXLiubyd58Y-012|That's when I have simply a solution of the weak acid.
hXLiubyd58Y-013|Haven't added any base yet.
hXLiubyd58Y-015|If you think about what's happening at point A, it really is just a solution of a weak acid.
hXLiubyd58Y-017|So we've converted half of the HA to A minus.
hXLiubyd58Y-018|And these concentrations are about equal.
hXLiubyd58Y-019|In fact, you could find a point directly in the center where they are equal, the HA concentration is equal to the A minus concentration.
hXLiubyd58Y-020|This region is called the buffer region.
hXLiubyd58Y-021|When HA concentrations and A minus concentrations are about equal, the solution resists changes in pH.
hXLiubyd58Y-022|And you can see the line is relatively flat in this region.
hXLiubyd58Y-023|That resistance to change in pH is called buffering.
hXLiubyd58Y-024|And the property of weak acids to act like buffers is very important.
hXLiubyd58Y-028|And I get to what's called the end point or the equivalence point or the stoichiometric point.
hXLiubyd58Y-029|All different names for the same thing, where you've added a mole of base for every mole of acid that you originally had.
hXLiubyd58Y-030|So these are the major points along the titration curve.
hXLiubyd58Y-031|Out here at point D, it's simply really a solution of just a strong base.
hXLiubyd58Y-032|You've used up all your weak acid, and now the pH rapidly changes because you continue to add strong base.
hXLiubyd58Y-033|And then the strong base dominates the pH.
hXLiubyd58Y-034|So these are the major points along a titration curve of a weak acid by a strong base.
BwPxen6eBV8-006|We're trying to determine which of these structural isomers is chiral.
BwPxen6eBV8-007|And remember, chirality involves non-superimposable mirror images, and that's a very subtle thing.
BwPxen6eBV8-008|But the basis of chirality is an asymmetric center, so a carbon with four different things attached, and that's the key, that's what we have to look for in our molecules.
BwPxen6eBV8-009|It's much too difficult to draw the molecules and try to imagine them superimposing in their mirror images.
BwPxen6eBV8-010|The thing that we can really latch onto is a carbon with four things attached.
BwPxen6eBV8-011|So in molecule A, if you look through all the carbons, each one has at least two substituents the same.
BwPxen6eBV8-015|Four different things attached to one carbon makes that asymmetric center and lends chiropody to the entire molecule.
BwPxen6eBV8-016|So the correct answer here is B.
Rde6Bvnbp9A-000|Let's look at the equilibrium constant for the autodissociation of water, Kw.
Rde6Bvnbp9A-001|Here I've plotted Kw for water versus temperature.
Rde6Bvnbp9A-002|So what is the pH of pure hot water?
Rde6Bvnbp9A-010|We're talking about the pH of hot water based on Kw for water, the autodissociation constant.
Rde6Bvnbp9A-016|So these two concentrations will be equal in pure water, and both of them 10 to the minus 6.5, approximately.
Rde6Bvnbp9A-017|So the pH is minus log of H3O+.
Rde6Bvnbp9A-018|So minus log of 10 to the minus 6.5 is 6.5.
Rde6Bvnbp9A-019|So the pH in hot water is lower.
Rde6Bvnbp9A-022|Hot water, lower pH than cool water.
6CRbRJsgI-A-000|Let's look at a couple fluorine species and try to determine which of the paramagnetic species has the weakest bond.
6CRbRJsgI-A-002|The way to do this is to look at them like their orbital structure, find the bond orders to determine which has the weakest bond, and then check to see if they're paramagnetic.
6CRbRJsgI-A-006|Four orbitals from each fluorine makes fluorine the molecule have eight molecular orbitals.
6CRbRJsgI-A-007|Eight atomic orbitals should form eight molecular orbitals.
6CRbRJsgI-A-008|So those molecular orbitals look like this.
6CRbRJsgI-A-012|So 1,2,3,4,5,6,7,8 orbitals.
6CRbRJsgI-A-013|I have to fill those orbitals with the seven electrons from fluorine plus any charge.
6CRbRJsgI-A-014|So let's look at the number of electrons, the bond order, and whether it's paramagnetic.
6CRbRJsgI-A-020|What's the bond order?
6CRbRJsgI-A-021|Well, the sigma 2s and the sigma star 2s, those cancel each other out, that gives zero bond order.
6CRbRJsgI-A-023|Is it per magnetic?
6CRbRJsgI-A-024|Yes there is an unpaired electron.
6CRbRJsgI-A-030|The difference is 1 divided by 2, a bond order of 1/2.
6CRbRJsgI-A-031|So a weaker bond here in F2 minus.
6CRbRJsgI-A-032|Is it paramagnetic?
6CRbRJsgI-A-033|Yeah, there is an unpaired electron.
6CRbRJsgI-A-034|What about F2 minus 2?
6CRbRJsgI-A-038|So a bond order of zero.
6CRbRJsgI-A-039|And I'm not paramagnetic anymore, every electron is paired.
6CRbRJsgI-A-040|So this has the weakest bond order, F2 minus 2, would fall apart as a molecule.
6CRbRJsgI-A-041|Picking up two electrons goes to bond order zero, very unfavorable for F2.
6CRbRJsgI-A-042|The paramagnetic species that has the weakest bond is F2 minus.
6CRbRJsgI-A-043|So among these three, two paramagnetic species, and the weaker bond F2 minus.
6iT8qY0GRaY-000|Another property of gases that we can talk about is their compressibility.
6iT8qY0GRaY-001|How easy is it to squeeze them into a smaller volume.
6iT8qY0GRaY-002|We define that as the ratio of the change in volume over the change in pressure.
6iT8qY0GRaY-003|Now, we use this symbol delta to indicate a change, so a change in volume over a change in pressure.
6iT8qY0GRaY-004|So let's look at a couple instances.
6iT8qY0GRaY-007|A gas is easy to compress at high volume.
6iT8qY0GRaY-012|So a large increase in pressure for a small change in volume, this is not very compressible gas.
6iT8qY0GRaY-013|This compressibility factor is very small for low volumes.
6iT8qY0GRaY-014|So compressibility tells us how much can I change the volume for a given change in pressure.
e27ZIIsvuys-000|Hi.
e27ZIIsvuys-001|Let's talk about electromagnetic radiation.
e27ZIIsvuys-002|Electromagnetic radiation is a wave-like property.
e27ZIIsvuys-003|It has alternating magnetic and electric fields.
e27ZIIsvuys-004|Now we're not going to talk about the electric and magnetic fields too much, but the wave nature is important to us.
e27ZIIsvuys-005|What are the properties of waves?
e27ZIIsvuys-006|Well, we're all familiar with waves-- we've been to the beach, we've seen waves come in from the ocean-- what are the properties?
e27ZIIsvuys-007|Well, there's a speed that the wave moves.
e27ZIIsvuys-008|If you see a surfer and he catches a wave, the speed that the surfer moves riding the crest is the speed of the wave.
e27ZIIsvuys-011|When we talk about electromagnetic radiation, it moves at the speed of light, as opposed to the speed of a surfer.
e27ZIIsvuys-012|So the speed is the speed of light.
e27ZIIsvuys-016|And their speed will be equal and their distance will stay about the same.
e27ZIIsvuys-017|Their distance between them, the wavelength.
e27ZIIsvuys-021|And I'll often use the term light and electromagnetic radiation interchangeably.
e27ZIIsvuys-024|So different wavelengths have different properties and for some electromagnetic radiation, our eyes are sensitive to them.
e27ZIIsvuys-025|So we have different colors for different wavelengths.
e27ZIIsvuys-026|Another property of waves are the frequency.
e27ZIIsvuys-027|The frequency of the wave is how often a wave crests past you.
e27ZIIsvuys-028|So if three wave crests passed you per second the frequency would be 1 over 3, or 1/3 of a second.
e27ZIIsvuys-029|Now we say reciprocal seconds, 1 over 3 seconds, we give that a special unit called hertz.
WYPgOKP-a78-004|He entering the system is positive and causes a positive enthalpy change.
WYPgOKP-a78-005|The enthalpy change is the state function associated with constant pressure energy changes.
WYPgOKP-a78-006|Heat is absorbed.
WYPgOKP-a78-007|That means endothermic processes.
WYPgOKP-a78-008|Heat comes into the system.
WYPgOKP-a78-009|Now, one classic endothermic process is the boiling of water, water going from liquid water to steam.
WYPgOKP-a78-010|That's an endothermic process.
WYPgOKP-a78-011|Heat is absorbed.
WYPgOKP-a78-012|And here's where you have to be careful, because your intuition may tell you, oh, when water is boiling on the stove, that's hot.
WYPgOKP-a78-013|It feels like heat is coming out, and that's true.
WYPgOKP-a78-015|I have to absorb that heat, overcome intermolecular interactions, and become a gas.
WYPgOKP-a78-020|q is less than 0, and the enthalpy change less than 0.
WYPgOKP-a78-021|Now, applying a sign to heat doesn't mean that heat comes as positive or negative.
WYPgOKP-a78-022|Heat is just a quantity of energy.
WYPgOKP-a78-023|The sign says which direction is that heat flowing.
WYPgOKP-a78-024|Is it flowing out of the system?
WYPgOKP-a78-025|That's negative.
WYPgOKP-a78-026|Or into the system?
WYPgOKP-a78-027|That's positive.
WYPgOKP-a78-028|It's just like speed.
WYPgOKP-a78-029|You could have speed in one direction or speed in another direction, and you could associate this with the positive direction and this with a negative direction.
WYPgOKP-a78-030|So positive or negative speed, same kind of thing.
WYPgOKP-a78-032|Energy coming out of the system into the surroundings is often more dramatic.
WYPgOKP-a78-035|When we talk about a chemical reaction and the concurrent energy change, we set a standard state.
WYPgOKP-a78-036|So I'll measure the difference in enthalpy between products and reactants by measuring the enthalpy change at a very specific set of conditions.
WYPgOKP-a78-038|Or if they're gases, their pressure is 1 atmosphere, and the temperature of the whole system is 25 degrees C.
WYPgOKP-a78-039|That's our standard state reference point for measuring enthalpies, and it'll allow us to change enthalpies and compare different chemical reactions.
WYPgOKP-a78-040|That is, it's not fair to compare one chemical reaction to another chemical reaction, if the concentrations and pressures are different.
WYPgOKP-a78-041|So our standard state helps us compare chemical reactions and compare enthalpies across chemical reactions.
WYPgOKP-a78-042|Standard states, endothermic processes, exothermic processes, that's thermal chemistry.
W7aJk7yrbUE-000|Let's look at three different processes and see if we can figure out which has the greatest entropy change.
W7aJk7yrbUE-001|So A, I'm going to take solid ice water at minus 10 to liquid water at plus 10 Celsius.
W7aJk7yrbUE-002|B, liquid water at 0 degrees Celsius up to liquid water at 100 degrees Celsius.
W7aJk7yrbUE-011|We're talking about the heat required to do some physical and temperature changes on water.
W7aJk7yrbUE-016|And that requires the heat of vaporization of a mole of water at 100 degrees C, which is around 40 kilojoules.
W7aJk7yrbUE-017|So by far, the most amount of energy is required step C.
W7aJk7yrbUE-018|Now, we can see that reflected in our heating curve.
DnySy87He-A-000|Many compounds have several acidic protons, so-called polyprotic acids.
DnySy87He-A-002|The titration curve of polyprotic acids look stepwise.
DnySy87He-A-003|The protons are titrated individually.
DnySy87He-A-004|Usually in a polyprotic system, the pKa of the first proton is significantly different than the pKa of the second proton.
DnySy87He-A-005|Now I'm using pKas to describe protons, which is common parlance.
DnySy87He-A-006|We know, of course, that the pKas refer to K's, which are equilibrium constants.
DnySy87He-A-007|But I talk about the pKa of the first proton as a property of the first proton.
DnySy87He-A-008|It's the pH at which that proton is removed from the molecule in a titration.
DnySy87He-A-009|Let's look at that.
DnySy87He-A-010|So here's the titration curve of a polyprotic acid.
DnySy87He-A-013|So I always get a point in the titration curve for free where the pH is equal to the pKa at halfway to the equivalence point.
DnySy87He-A-014|In this case, it's halfway to the first equivalence point, the first proton coming off.
DnySy87He-A-015|So I can look at this and I say, well, below the first pKa, the acid form predominates.
DnySy87He-A-016|Above the first pKa, the base form predominates.
DnySy87He-A-017|So at the equivalence point, half equivalence point, I have an equal mixture of the acid and base form.
DnySy87He-A-018|So let's go along the titration curve, mostly acid form to start.
DnySy87He-A-019|As I get to half equivalence, equal concentration of the acid and base.
DnySy87He-A-020|As I go beyond that, the base form starts to predominate.
DnySy87He-A-021|And when I get to the first equivalence point, I'm all the base form.
DnySy87He-A-022|But now I can start another titration of this proton.
DnySy87He-A-024|So let's look at the species that are available.
DnySy87He-A-025|So right here at the equivalence point, I had all HCO 3-.
DnySy87He-A-026|And then HCO 3- starts to act like an acid and its proton is titrated.
DnySy87He-A-027|So as I go along the titration curve, the predominant form is the acid form.
DnySy87He-A-028|I'm below the second pKa.
DnySy87He-A-029|As I get to the second pKa, equal amounts of the acid and base form.
DnySy87He-A-030|And as I pass through the second pKa, the base form predominates, CO3 -2.
DnySy87He-A-033|Let's look at a polyprotic acid, carbonic acid, buffering your blood.
DnySy87He-A-034|Now your blood is buffered with carbonic acid, but only the first pKa-- that is, the buffer around the first pKa-- is used.
DnySy87He-A-035|And that first pKa is around 6.
DnySy87He-A-036|So how is it done?
DnySy87He-A-037|Well, carbon dioxide is produced in yourselves by metabolism and then exhaled through your lungs.
DnySy87He-A-042|Now the base form, HCO 3- is maintained at about 24 millimolar by your kidneys.
DnySy87He-A-043|Your kidneys filter your blood, and they have the remarkable property of being able to remove selectively specific ions.
DnySy87He-A-044|So they can maintain the H3O- concentration at 24 millimolar.
DnySy87He-A-045|So what does that mean for the pH of your blood?
DnySy87He-A-046|Well, you can use and apply your Henderson-Hasselbalch expression.
DnySy87He-A-051|So if you solve, you get a pH of around 7.4.
DnySy87He-A-052|So pH is maintained in your blood by a buffer system, and it's maintained around pH 7.4.
DnySy87He-A-053|And that turns out to be vitally important.
DnySy87He-A-054|If your pH changes by just half a unit in either direction, it's essentially fatal.
DnySy87He-A-055|And it's fatal because you changed the protonation states.
DnySy87He-A-056|Remember, the pH determines which forms are present in solution, the acid or the base form.
DnySy87He-A-057|If you arbitrarily change that, that changes the potential difference and the selectivity of the membranes in your system, in your body.
DnySy87He-A-058|It changes the structure of the proteins in your body.
DnySy87He-A-059|So very dramatic changes can occur by just a small change in pH.
DnySy87He-A-060|So it's vital to buffer your blood and resist changes in pH.
DnySy87He-A-062|If you reduce the partial pressure of carbon dioxide in your lungs, your blood will become slightly alkaline.
DnySy87He-A-063|You're reducing the acid form.
DnySy87He-A-064|So if your blood becomes slightly alkaline by hyperventilating, that can be dangerous.
DnySy87He-A-065|But you know how to cure that.
DnySy87He-A-066|If you find someone that's hyperventilating, what do you do?
DnySy87He-A-067|Well, you have them breathe into a bag.
zsk1S2r2r_o-005|Because the intercept is determined by delta-H.
zsk1S2r2r_o-006|So an endothermic reaction will have a positive intercept.
zsk1S2r2r_o-007|The slope is determined by delta-S.
zsk1S2r2r_o-008|So if delta-S is positive, the slope will be negative.
zsk1S2r2r_o-012|So for an endothermic reaction-- that's a positive delta-H-- you get a negative slope.
zsk1S2r2r_o-015|Let's summarize the relationship between the free energy in temperature and the equilibrium constant in temperature.
zsk1S2r2r_o-016|We can plot them both in a linear fashion if we plot the free energy versus temperature directly and natural log of the equilibrium constant versus 1 over temperature.
zsk1S2r2r_o-017|There's a couple of situations to consider depending on the relative sign of the enthalpy and the entropy.
zsk1S2r2r_o-021|And delta-S negative gives me a negative intercept.
zsk1S2r2r_o-022|We can continue.
zsk1S2r2r_o-026|Negative intercept for delta-G because delta-H determines the intercept.
zsk1S2r2r_o-027|Negative intercept for lnK versus 1 over T because delta-S determines the intercept.
zsk1S2r2r_o-035|So that's a summary of the various lnK versus 1 over T and delta-G versus T situations.
MySri1HHJ_0-000|When a chemical reaction occurs, there's an enthalpy change, or an energy change.
MySri1HHJ_0-002|Either way, the enthalpy of a reaction is a state function.
MySri1HHJ_0-003|It only depends on where I start and where I finish, and not the pathway to go between the products and the reactants.
MySri1HHJ_0-008|It's downhill.
MySri1HHJ_0-009|Energy is released.
MySri1HHJ_0-013|So if I'm going from reactant to their atoms, that's adding energy for all the bonds in the reactants.
MySri1HHJ_0-014|Now, I can just deform all the bonds in the products.
MySri1HHJ_0-015|And that will always release energy.
MySri1HHJ_0-016|So there'll always be an exothermic step, and endothermic step to break, and then exothermic step to make the bonds.
MySri1HHJ_0-017|Now, it just depends.
MySri1HHJ_0-018|Do I get more energy back when I make the product bonds than when I broke the reactant bonds?
MySri1HHJ_0-019|That will determine whether the overall reaction is exothermic-- in this case, I put in some energy to break the react in bonds.
MySri1HHJ_0-020|But the product bonds were more stable overall.
MySri1HHJ_0-021|So I got an overall release in energy.
MySri1HHJ_0-022|But you could imagine a case where you break the reactant bonds and form the product bonds, but you don't get as much energy back.
MySri1HHJ_0-023|And that would be an overall endothermic reaction.
MySri1HHJ_0-028|Hydrogen, as in hydrogen gas, in its standard state-- diatomic molecules.
MySri1HHJ_0-032|Simply accounting, to keep track of the entropy of a chemical reaction.
ancOLDlrc88-001|So how does K vary with temperature for water gas going to water liquid?
ancOLDlrc88-009|We're talking about the equilibrium between water liquid and water gas at various temperatures.
ancOLDlrc88-010|K, the equilibrium constant for chemical reactions and physical processes, varies with temperature.
ancOLDlrc88-011|So how does it vary for this reaction?
ancOLDlrc88-014|And this is varying with temperature.
ancOLDlrc88-015|So this line in the phase diagram corresponds to the equilibrium values.
ancOLDlrc88-016|These are the values of the equilibrium constant with temperature.
ancOLDlrc88-017|So the equilibrium constant looks just like that line as it varies with temperature.
ancOLDlrc88-018|Now, another way you could have done this is said, well, I know as temperature increases, this favors the products.
ancOLDlrc88-019|So K should get larger.
ancOLDlrc88-020|It's more likely to have gaseous water at equilibrium at high temperatures than the liquid water.
ancOLDlrc88-021|So you could also say well, K increases with T because I know something about this physical process.
ancOLDlrc88-022|Either way, you arrive at the same answer, C.
eoFw8xEy-m4-000|We're talking about the orbitals about a hydrogen atom, and we're using the three quantum numbers-- n, l, and m sub l-- to describe wave functions.
eoFw8xEy-m4-001|Each wave function describes an orbital.
eoFw8xEy-m4-002|We're going to use those terms interchangeably-- wave function, orbital.
eoFw8xEy-m4-003|So we're up to n equal 2, and one possible value for n equal 2 is l equal 1.
eoFw8xEy-m4-007|So the value of l equal 1, we give the letter designation p.
eoFw8xEy-m4-008|So these are going to be p orbitals.
eoFw8xEy-m4-009|How about the values of m sub l?
eoFw8xEy-m4-014|An angular node is a node that has a planar shape rather than a spherical shape for the radial nodes.
eoFw8xEy-m4-017|The square of the wave function is large around the x-axis for the x.
eoFw8xEy-m4-019|That helps you find the angular node.
eoFw8xEy-m4-020|Now, there's one other value of m sub l, equal 0.
eoFw8xEy-m4-021|And that will give you the pz orbital, where the square of the wave function has its maximum along the z-axis.
eoFw8xEy-m4-024|And the m sub l equals 0 already neatly lies along the z-axis.
eoFw8xEy-m4-027|And we'll call them the 2px, 2py, and 2pz orbitals.
eoFw8xEy-m4-029|These are the p orbitals.
eoFw8xEy-m4-030|Now, we can continue.
eoFw8xEy-m4-032|It'll have two nodes now, two radial nodes, because the total number of nodes is n minus 1.
eoFw8xEy-m4-033|We'll have some 3p orbitals-- that is, n equal 3, l equals 1.
eT6IA89Vt4Y-000|We're comparing two acids, HF, a weak acid and HBr, a strong acid.
eT6IA89Vt4Y-001|We know the relative bond enthalpies go like this.
eT6IA89Vt4Y-004|Now, that's really the indicator of what we're looking for here because if K is less than 1, then the standard state free energy difference must be positive somewhere.
eT6IA89Vt4Y-005|It must have some positive values.
eT6IA89Vt4Y-007|If you dissolve HBr in water, that's an exothermic reaction.
U0N3Vqy1QtQ-001|As energy is dispersed over more equal energy microstates, the entropy increases.
U0N3Vqy1QtQ-002|Now there's another definition, the thermodynamic definition, that says as heat is transferred that entropy increases.
U0N3Vqy1QtQ-003|So can I rationalize those two and bring them together as the same definition?
U0N3Vqy1QtQ-007|Let's look at those both in terms of a system, say like a particle in a box.
U0N3Vqy1QtQ-008|A very simple system where there are several microstates available to the system.
U0N3Vqy1QtQ-010|Now more microstates are available to the system.
U0N3Vqy1QtQ-011|That is, I can distribute the energy over many more microstates.
h-7G6XgIPGY-000|Let's look at a specific equilibrium-- water liquid and water gas.
h-7G6XgIPGY-001|Now, here I have water liquid and water gas at about 25 degrees C, and they're in equilibrium.
h-7G6XgIPGY-002|There's a little bit of vapor above this liquid.
h-7G6XgIPGY-003|We could write a reaction quotient for this.
h-7G6XgIPGY-004|And when we do, that reaction quotient is the pressure of H2O.
h-7G6XgIPGY-005|That's because the liquid water is pure liquid water, and pure liquids and solids do not appear in equilibrium expressions.
h-7G6XgIPGY-006|So products over reactants, products is the partial pressure of the gas, the reactants are liquid water.
h-7G6XgIPGY-007|But it's a pure liquid, so that doesn't appear in the equilibrium expression.
h-7G6XgIPGY-008|So this reaction quotient is just equivalent to the partial pressure of the gaseous water.
h-7G6XgIPGY-009|Now that's the vapor pressure of water.
h-7G6XgIPGY-010|When this is in equilibrium, that will become the vapor pressure.
h-7G6XgIPGY-011|So the vapor pressure is the equilibrium constant.
h-7G6XgIPGY-013|And we could find those equilibrium vapor pressures.
h-7G6XgIPGY-015|Notice here I've plotted several different temperatures for this gas.
h-7G6XgIPGY-016|So as the temperature increases, the vapor pressure increases.
h-7G6XgIPGY-017|And you can see that.
h-7G6XgIPGY-028|So in this case, we'd have the equilibrium constant K prime is 1 over K for this forward reaction that we've been talking about.
h-7G6XgIPGY-029|So equilibrium expressions-- when we write Qs and Ks, we omit pure liquids and solids.
h-7G6XgIPGY-030|In these physical equilibria, vapor pressure and K are equivalent, and they are functions of temperature that we already stand.
h-7G6XgIPGY-031|If we reverse a reaction, we invert the equilibrium constant.
h-7G6XgIPGY-032|Those are the properties of Q and K.
XwD8___Mm1I-000|When an electron is promoted from one energy level to another energy level in an atom, an electronic transition occurs, and there's many orbitals in an atom that an electron can occupy.
XwD8___Mm1I-001|Some are low energy some are high energy, but they have specific energies between them.
XwD8___Mm1I-002|They are quantized energy levels and an electron will absorb a photon of a specific energy to go from one energy level to another.
XwD8___Mm1I-003|Let's look at those photon energies.
XwD8___Mm1I-004|So the energy levels in the atom go as Z squared over n squared R infinity.
XwD8___Mm1I-005|This quantity tells us the energy level based on n, the principle quantum level.
XwD8___Mm1I-008|So if I start here and I emit a photon, the energy difference can be calculated by this expression.
XwD8___Mm1I-009|This will allow us to take photons that we see being emitted from an atom and correlate those to transitions within the electronic structure.
JAMNm0dG7jk-000|Let's look at the pH gradient across the mitochondrial membrane.
JAMNm0dG7jk-010|We're talking about the pH gradient across the mitochondrial membrane during ATP synthesis.
JAMNm0dG7jk-012|And that's accomplished by the oxidation of glucose.
JAMNm0dG7jk-013|This high concentration of H plus on the outside means the pH is lower on the outside than the inside.
JAMNm0dG7jk-014|It's more basic on the inside and more acidic on the outside of the mitochondria.
ss4JSV-YRec-000|Let's try to predict which species is present in the titration of carboning.
ss4JSV-YRec-001|So carbonic acid all three is titrated.
ss4JSV-YRec-002|And when we get to pH 7, which species predominates?
ss4JSV-YRec-014|We go from this form to this form, at around pH equal to pKa 1.
ss4JSV-YRec-015|At pH 6, there's an equal mixture of these two.
ss4JSV-YRec-018|We're at pH 7, which is between those two-- one greater than the first pKa and 3 less than the second pKa.
ss4JSV-YRec-019|So we have titrated along the titration curve.
ss4JSV-YRec-020|We've passed this point where there's an equal mixture where the pH is equal to the pKa.
s7PIqFpYXJA-001|Standard conditions and standard states.
s7PIqFpYXJA-002|So 25 degrees C, that will give us a standard condition.
s7PIqFpYXJA-003|And standard states, one atmosphere of pressure, pure liquids and solids, 1 molar concentrations.
s7PIqFpYXJA-004|So I have pure liquid water.
s7PIqFpYXJA-010|We're talking about water liquid, water gas, in equilibrium.
s7PIqFpYXJA-011|But we've made specific conditions.
s7PIqFpYXJA-012|We've said standard conditions.
s7PIqFpYXJA-013|25 degrees C and standard states.
s7PIqFpYXJA-014|So 1 atmosphere of pressure of gas and pure liquids.
s7PIqFpYXJA-015|So that means I have 1 atmosphere of gas and pure liquid water.
s7PIqFpYXJA-016|What's the situation with this equilibrium?
s7PIqFpYXJA-018|And at equilibrium, that's the vapor pressure of water for that temperature.
s7PIqFpYXJA-022|So if Q is bigger than K, I'm going to go back towards reactants.
s7PIqFpYXJA-023|And I think you'd predict that what's happened.
s7PIqFpYXJA-024|If you had a whole atmosphere of pressure of water at 25 degrees C, that's too high a pressure.
s7PIqFpYXJA-026|So the answer here, Q is larger than K.
Vzy2wd5oIKs-000|Let's look at some energy changes under very specific situations.
Vzy2wd5oIKs-001|The first is I'll fix the volume.
Vzy2wd5oIKs-002|So if I fix the volume of the system, what can happen?
Vzy2wd5oIKs-003|Well, energy is still heat plus work.
Vzy2wd5oIKs-004|But in a fixed situation, the system can't expand and do any work, or contract and have work done on it.
Vzy2wd5oIKs-005|So work is zero in this case.
Vzy2wd5oIKs-006|That means all the energy exchanges have to come from heat flow.
Vzy2wd5oIKs-007|That means, in this specific case, heat kind of behaves like a state function.
Vzy2wd5oIKs-008|In general, heat depends on the path.
Vzy2wd5oIKs-009|But we're saying this specific path, a fixed volume pathway, heat behaves like a state function.
Vzy2wd5oIKs-010|It's equal to the energy change.
Vzy2wd5oIKs-011|So that energy change will be reflected in a temperature change.
Vzy2wd5oIKs-015|Then the heat flow is not a good measure of the energy change.
Vzy2wd5oIKs-016|And that's actually convenient to just measure one parameter to get the energy change.
Vzy2wd5oIKs-017|So what we do is we define a new state function called the enthalpy.
Vzy2wd5oIKs-018|The enthalpy is the energy change plus the pressure volume constant.
Vzy2wd5oIKs-022|Again, for an ideal gas, the enthalpy changes will be directly proportional to temperature changes.
Vzy2wd5oIKs-025|And that's proportional to an energy change for an ideal gas as well-- constant volume, constant pressure, energy changes.
kTJEs_u1fkQ-000|Let's look at the relationship between the macroscopic properties for a sample of gas, the pressure, the volume, the temperature, the number of moles of gas.
kTJEs_u1fkQ-012|This is the ideal gas law, the macroscopic properties of a gas, in one neat equation.
RffOkFeAkEE-001|What's the entropy change?
RffOkFeAkEE-002|Well, we know there's various ways to calculate entropy changes.
RffOkFeAkEE-011|So I choose the middle equation here, the reversible heat over the temperature change.
RffOkFeAkEE-012|So let's just do that.
RffOkFeAkEE-016|So freezing requires a release of heat, so freezing is endothermic.
RffOkFeAkEE-018|This is a positive heat.
RffOkFeAkEE-019|And I find that that's 22 joules per Kelvin mole of water.
RffOkFeAkEE-020|But we're not talking about a mole of water.
RffOkFeAkEE-021|We're talking about just 10 grams of water.
uwL_j9yXkN8-000|Let's look at protein synthesis and how it's coupled to the oxidation of glucose.
uwL_j9yXkN8-001|So how many grams of glucose are required to make a mole of serum albumin?
uwL_j9yXkN8-002|Now, serum albumin is around 100 amino acids.
uwL_j9yXkN8-003|So I need about 100 peptide bonds per mole of serum albumin.
uwL_j9yXkN8-012|We're talking about forming peptide bonds as we oxidize glucose.
uwL_j9yXkN8-014|That's transferred via the conversion of ADP to ATP to the formation of peptide bonds.
uwL_j9yXkN8-015|And you get about four moles of peptide bonds for every mole of glucose that you metabolize.
FmTLXnrPBPQ-000|In oxidation reduction chemistry, oxidations and reductions always occur in pairs.
FmTLXnrPBPQ-001|If you are going to be oxidized, that is you're going to give up some electrons, those electrons have to have a place to land.
FmTLXnrPBPQ-002|They have to reduce some other compound.
FmTLXnrPBPQ-003|So, let's look at a redox reaction and break it up into half cells, and determine the oxidation and reduction pairs.
FmTLXnrPBPQ-004|And use our tables of standard reduction potentials to help us balance that reaction.
FmTLXnrPBPQ-007|So, let's look at those two half cells.
FmTLXnrPBPQ-008|We can go back to our table.
FmTLXnrPBPQ-009|Here's our table of standard reduction potentials.
FmTLXnrPBPQ-014|Here's the permanganate, and here's the bromine.
FmTLXnrPBPQ-018|So we can balance the number of electrons, whenever oxidation and reduction occurs.
FmTLXnrPBPQ-019|Of course, the number of electrons are conserved.
FmTLXnrPBPQ-020|If I reduce you by two electrons, I must accept those two electrons.
FmTLXnrPBPQ-027|Now, we can also calculate the relative cell potential for that.
FmTLXnrPBPQ-028|We can go back to our table.
FmTLXnrPBPQ-032|So, that overall cell potential, 0.43.
FmTLXnrPBPQ-033|Now, notice again I didn't multiply my cell potentials, my standard reduction potentials, by two and five.
FmTLXnrPBPQ-034|That's because cell potential is an intensive property, it's independent of the extent of the system.
FmTLXnrPBPQ-036|That's downhill in free energy.
WvG46BwP-XY-001|So if we take these three atomic orbitals, we can make three new atomic orbitals.
WvG46BwP-XY-004|Now, I've drawn them off-center here.
WvG46BwP-XY-005|So obviously, they're all about the same nucleus.
WvG46BwP-XY-006|So let's collapse those together on the same nucleus.
WvG46BwP-XY-007|And that will obscure the little negative lobes.
WvG46BwP-XY-008|But the three positive lobes are appropriate for steric number three and a bond angle of 120 degrees.
WvG46BwP-XY-009|Let's say we want to accommodate four, steric number four, or tetrahedral symmetry.
WvG46BwP-XY-010|I think you can guess the pattern here.
WvG46BwP-XY-011|It's actually possible to take the s and all three p orbitals.
WvG46BwP-XY-012|Those four orbitals make four linear combinations and make four new atomic orbitals.
WvG46BwP-XY-015|109 degrees, I'll bring them all to the same center here four sp3 orbitals.
WvG46BwP-XY-016|Again, the negative lobes obscured.
WvG46BwP-XY-017|But they'll accommodate a steric number of four and a bond angle of 109 degrees.
zsB5ZigPQHY-000|Let's look at a situation where we're bracketing an oxidation-reduction reaction.
zsB5ZigPQHY-001|So which of the following will be reduced by copper but not by bromide?
zsB5ZigPQHY-009|We're looking for a compound that can oxidize copper metal, but not bromide.
zsB5ZigPQHY-010|So if we look at our table of standard production potentials, we see nitrate is above the copper half cell.
zsB5ZigPQHY-011|That means nitrate can force the oxidation of copper, but not the oxidation of bromide.
Sh42Wpvs42k-000|Light has both the properties of a wave and the properties of a particle.
Sh42Wpvs42k-001|It's a wave and a particle at the same time.
Sh42Wpvs42k-002|There's a duality to the wave particle relationship.
Sh42Wpvs42k-004|Particle properties that are moving, the most important property is the momentum, it's mass times its velocity.
Sh42Wpvs42k-005|So how do we reconcile the two?
Sh42Wpvs42k-006|Well light, photons of light, carry no mass.
Sh42Wpvs42k-007|So how can they carry momentum?
Sh42Wpvs42k-012|So using those two relationships we can derive that the momentum is Planck's constant divided by the wavelength.
Sh42Wpvs42k-016|Wave, particle, acting together, sometimes the properties of the particle exert themselves.
Sh42Wpvs42k-017|Sometimes the properties of the wave exert themselves.
Sh42Wpvs42k-018|They both exist together all the time, particle and wave, nature of light.
DP4HnlNRR_4-000|Nitrogen molecules, N2, have a triple bond.
DP4HnlNRR_4-001|That's predicted both by molecular orbital theory and the Lewis dot structure.
DP4HnlNRR_4-002|What I'd like to think about is the ionization N2 molecules-- that is, withdrawing an electron-- and adding an electron, electron capture by N2.
DP4HnlNRR_4-003|That will form a positive and a negative ion.
DP4HnlNRR_4-004|The question I have is, what's the relative bond strength?
DP4HnlNRR_4-010|We're talking about ionization and electron capture in N2 molecules.
DP4HnlNRR_4-012|And I filled them with the p electrons from nitrogen. Each nitrogen has 3 p electrons for a total of 6 in the molecule.
DP4HnlNRR_4-016|Adding an electron to the next available orbital is an antibonding orbital.
DP4HnlNRR_4-017|So electron capture into this antibonding orbital is going to reduce the bond order and the relative bond strength.
DP4HnlNRR_4-019|So the bond order reduces to 2 and 1/2.
DP4HnlNRR_4-022|And if you calculate the bond order here, again, 2 and 1/2.
DP4HnlNRR_4-023|But interestingly, both electron capture and ionization reduce the bond order of nitrogen by about the same amount.
DP4HnlNRR_4-024|So you'd predict that the bond strengths of N2+ and N2- are about the same.
s_xlDaR53v0-000|Let's look at a broad range of temperatures on the absolute scale.
s_xlDaR53v0-001|Here I've listed several.
s_xlDaR53v0-002|The interior of the Sun for instance-- 10 to the 8th Kelvin, a very high temperature.
s_xlDaR53v0-003|As we go down in temperature, we can go past the surface of the Sun, tens of thousands of degrees.
s_xlDaR53v0-004|The melting point of tungsten-- about 1,000 degrees Kelvin.
s_xlDaR53v0-011|That's a very ideal gas.
s_xlDaR53v0-013|So that's a gas that behaves ideally over a broad range of temperatures.
s_xlDaR53v0-016|If you slow down particles, you reduce their temperature.
s_xlDaR53v0-017|We'll always associate the speed of molecules in a sample with their temperature.
s_xlDaR53v0-018|Higher temperature-- more speed.
s_xlDaR53v0-019|Some of the coldest temperatures ever achieved are in a Bose-Einstein condensation.
s_xlDaR53v0-020|That's where you do laser cooling and magnetic trapping, and get molecules to virtually stop.
inLhIqLOouY-000|When an acid reacts with water, it forms its conjugate base.
inLhIqLOouY-001|Let's look at that.
inLhIqLOouY-010|Again, products over reactants, water doesn't appear, pure water, in equilibrium constant expressions.
inLhIqLOouY-011|Now, I know the stronger the conjugate acid, the weaker the conjugate base intuitively.
inLhIqLOouY-012|The stronger this acid is, the more the equilibrium lies towards Ac-.
inLhIqLOouY-013|So the more Ac- is likely to be in solution, here, the Ac- is favored.
inLhIqLOouY-014|Here the Ac- is favored.
inLhIqLOouY-015|Here it's a product.
inLhIqLOouY-016|Here it's a reactant.
inLhIqLOouY-017|But the stronger the acid, the more that ion should be favored.
inLhIqLOouY-018|Now, let's look at the two.
inLhIqLOouY-019|If I look at Ka and Kb, they're not the reciprocals of each other.
inLhIqLOouY-020|That is Ka is not 1/Kb.
inLhIqLOouY-026|So now, I have analytically that the product Ka times Kb is a constant.
inLhIqLOouY-027|So if Ka gets larger, a stronger acid, Kb must get smaller.
inLhIqLOouY-028|So I have an analytical expression for strong acid means a weaker conjugate base.
ZpzjU4NfAhw-000|So let's look at the various orbitals of a hydrogen atom.
ZpzjU4NfAhw-001|Now, remember each orbital is a specific wave function, and that wave function will be described by three quantum numbers-- n, l, and m sub l.
ZpzjU4NfAhw-002|Once I give you three quantum numbers, you should be able to tell me the orbital.
ZpzjU4NfAhw-003|So let's look at them in sequence.
ZpzjU4NfAhw-004|For n equal 1-- that's the ground state, the lowest possible energy level-- we'll have one value of l.
ZpzjU4NfAhw-005|Now, remember l goes from 0 to n minus 1.
ZpzjU4NfAhw-006|So if n is 1, the only possible value of l is 0.
ZpzjU4NfAhw-007|So for n equal 1, we can have l equal 0.
ZpzjU4NfAhw-008|And we're also going to use the designation s for the quantum number l.
ZpzjU4NfAhw-009|So if I say l is s or if I say l is 0, I'm saying the same thing.
ZpzjU4NfAhw-010|You have to memorize that little connection.
ZpzjU4NfAhw-011|Now, for l equals 0, the only possible values of m sub l are 0.
ZpzjU4NfAhw-012|Remember, m sub l goes in integer steps from minus l to l.
ZpzjU4NfAhw-014|So now we have three quantum numbers that describes an orbital, and we're going to call this orbital the 1s orbital.
ZpzjU4NfAhw-015|We're going to use the designations of the quantum numbers to describe the orbitals.
ZpzjU4NfAhw-016|What does that look like?
ZpzjU4NfAhw-017|Well, it turns out it's spherically symmetric.
ZpzjU4NfAhw-018|The 1s orbital-- in fact, all s orbitals are spherical symmetric.
ZpzjU4NfAhw-019|Remember, the l quantum number-- in this case, 0 equal to s-- that gives you something to understand the shape of the orbital.
ZpzjU4NfAhw-020|In this case, the shape is spherical.
ZpzjU4NfAhw-021|In that case, the s might be a little bit helpful.
ZpzjU4NfAhw-022|s spherical l equals 0 is spherical.
ZpzjU4NfAhw-023|So the 1s orbital is spherical, and I've written this plus sign to indicate there's no nodes.
ZpzjU4NfAhw-024|The wave function does not change sign.
ZpzjU4NfAhw-027|And of course, the number of nodes is n minus 1, so 1 minus 1 is 0.
ZpzjU4NfAhw-028|There's no nodes in the 1s orbital.
ZpzjU4NfAhw-029|Well, can we go to n equal 2?
ZpzjU4NfAhw-030|We're done with n equal 1.
ZpzjU4NfAhw-031|There's no other possibilities of the quantum numbers.
ZpzjU4NfAhw-032|We go to n equal 2.
ZpzjU4NfAhw-033|For n equal 2, one possible value of l now is 0.
ZpzjU4NfAhw-034|So l can be 0 or 1.
ZpzjU4NfAhw-035|But let's look at the 0 value first.
ZpzjU4NfAhw-036|So that's s.
ZpzjU4NfAhw-037|m sub l is still 0.
ZpzjU4NfAhw-038|So what we have here is another spherically symmetric orbital, but in this case, it's a 2s orbital, but n equals 1.
ZpzjU4NfAhw-039|So the total number of nodes and minus 1 is 1.
ZpzjU4NfAhw-042|It had a positive sign in this region, a negative sign in that region.
ZpzjU4NfAhw-043|The square of the wave function is always positive, but the wave function changed sign, giving you this radial node.
7Nv8QLeF3yY-000|Let's look at that NO2 to NO3- reaction, and this time think about the oxidation number change.
7Nv8QLeF3yY-001|So how does the oxidation number of nitrogen change in the transition NO2- to NO3-?
7Nv8QLeF3yY-010|If we assign the electrons in the bonds to the more electronegative element, this nitrogen has only two electrons around it in a lone pair.
7Nv8QLeF3yY-014|So in NO2-, it has three fewer than it normally would for a plus 3 oxidation number.
7Nv8QLeF3yY-015|In this case, NO3-, nitrogen has five fewer electrons than it would as a neutral atom.
7Nv8QLeF3yY-016|So an oxidation number of plus 5.
-vJXlNkMqYE-003|Now, this reaction profile doesn't tell us anything about how fast that energy will be released.
-vJXlNkMqYE-007|But intuitively, we know that time is a parameter that we can look at.
-vJXlNkMqYE-008|Some reactions go quickly and some go slowly.
-vJXlNkMqYE-009|Here's a reaction of zinc and acid.
-vJXlNkMqYE-010|Now, the zinc and acid, same amount of zinc in both of these flasks.
-vJXlNkMqYE-012|But since there's the same amount of zinc, zinc limits the reaction.
-vJXlNkMqYE-013|When the zinc is consumed, the reaction is over.
-vJXlNkMqYE-014|So the same amount of energy is released in both cases.
-vJXlNkMqYE-017|So this time parameter is the realm of chemical kinetics.
-vJXlNkMqYE-018|We know that when we mix H2 and O2 together, they can sit there almost indefinitely.
-vJXlNkMqYE-019|And the reason is the collisions between the hydrogen and the oxygen often don't have sufficient energy to go from products to reactants.
-vJXlNkMqYE-021|That barrier says you need energy to go from reactants to products.
-vJXlNkMqYE-023|Everything else being equal-- concentrations, temperatures, pressures.
-vJXlNkMqYE-025|So you can see along a reaction profile you have two separate domains-- the domain of thermodynamics and the domain of kinetics.
myr7i1nYj-4-000|Let's do a chem quiz involving isotopes.
myr7i1nYj-4-003|I'm going to let them react to form oxygen molecules.
myr7i1nYj-4-011|Our problem was oxygen atoms in a flask.
myr7i1nYj-4-012|Two different isotopes, oxygen 16, and oxygen 18.
myr7i1nYj-4-013|They're going to react to form oxygen molecules.
myr7i1nYj-4-014|So I'll get oxygen molecules with three different masses.
myr7i1nYj-4-015|If an oxygen 16 reacts with an oxygen 16, that gives me an oxygen molecule of mass 32.
myr7i1nYj-4-016|If an auction 16 reacts with an auction 18, an oxygen molecule of mass 34.
myr7i1nYj-4-017|And if an 18 reacts with an 18, an oxygen molecule of mass 36.
myr7i1nYj-4-018|So what's the distribution of masses?
myr7i1nYj-4-025|It turns out it's actually twice as likely that you get the 16-18 combination.
myr7i1nYj-4-026|It's the same thing in this physical example, where I have blue and red chocolate-covered candies.
myr7i1nYj-4-027|If I mix them together-- equal numbers of each-- and then pick them out at random, what kind of combinations would I get?
myr7i1nYj-4-028|Well, let's pick out one with each hand.
myr7i1nYj-4-029|If I reach in and I pick out a blue, and I reach in with the other hand, there's two possible combinations.
myr7i1nYj-4-030|I could pick out a red, or I could reach in and pick out a blue.
myr7i1nYj-4-031|But what if this hand first picked out a red?
myr7i1nYj-4-032|Well, the same thing could occur.
myr7i1nYj-4-033|I could pick out a blue, or I could pick out a red.
myr7i1nYj-4-034|So the combination of red-blue, occurred both times, regardless of what was in this hand.
myr7i1nYj-4-035|It's twice as likely that the red-blue combination occurs, just like in an equimolar mixture of oxygen atoms.
myr7i1nYj-4-038|And it's always likely that chocolate-covered candies are incredibly delicious.
ZFZ2J8gwIkE-000|Equilibrium can occur in a variety of situations.
ZFZ2J8gwIkE-001|A heterogeneous equilibrium would be one between a gas and a liquid, two different phases, or a solid and a liquid.
ZFZ2J8gwIkE-002|If the phases are different, we call it a heterogeneous equilibrium.
ZFZ2J8gwIkE-006|If you wait long enough these two values will come to equilibrium, the quotient will stop changing.
ZFZ2J8gwIkE-007|It will become a constant.
ZFZ2J8gwIkE-011|When it's sealed, there's a high pressure of carbon dioxide gas.
ZFZ2J8gwIkE-012|A high pressure because by Le Chatelier's principle that drives the reaction towards the dissolved carbon dioxide.
ZFZ2J8gwIkE-013|So you can get more carbon dioxide gas dissolved at high pressure.
ZFZ2J8gwIkE-014|When you open it, you reduce the pressure, lowering the pressure here causes the equilibrium to shift back this way.
ZFZ2J8gwIkE-015|So aqueous carbon dioxide comes out of solution, and it spontaneously comes out of solution all through the solution.
ZFZ2J8gwIkE-016|That's what the bubbles are.
ZFZ2J8gwIkE-017|They're coming out of solution and then they bubble to the surface.
0Ps7ux68RMU-000|Let's look at a phase diagram for water and try to predict where the 373 kelvin isotherm is.
0Ps7ux68RMU-006|We're trying to position the 373 kelvin isotherm on the phase diagram for water.
0Ps7ux68RMU-007|Now, the 373 isotherm is a 100 degrees Celsius, so that's the normal boiling point.
0Ps7ux68RMU-008|The normal boiling point is the transition between the liquid and the gas water.
0Ps7ux68RMU-009|So if there are two phases present, I must be below the critical temperature.
0Ps7ux68RMU-010|Remember the critical temperature, the definition is, above the critical temperature, I have a single phase regardless of the pressure.
0Ps7ux68RMU-011|If I have below the critical point, this critical temperature, then I can have phase transitions.
0Ps7ux68RMU-012|So the phase transition at 100 degrees C implies that I'm below the critical temperature.
0Ps7ux68RMU-014|If you decrease the pressure, you go from liquid to gas, or if you increase the pressure, you go from the gas to the liquid.
_7s29Q76st0-000|Let's look at three molecules and see if we can determine if they have a permanent dipole moment.
_7s29Q76st0-001|Three molecules, carbon monoxide, carbon dioxide, and acetone.
_7s29Q76st0-002|Which has no permanent dipole moment?
_7s29Q76st0-003|It's interesting to note that no permanent dipole moment means a molecule will be transparent to microwaves.
_7s29Q76st0-004|It turns out the permanent dipole moment is what allows molecules to absorb microwaves.
_7s29Q76st0-005|It's what allows microwave spectroscopy to work, and even your microwave oven.
_7s29Q76st0-006|So while you're thinking about which of these has no permanent dipole moment, you can think about which of these would be transparent to microwaves at the same time.
_7s29Q76st0-013|We're looking at three molecules and trying to determine their electric dipole moments.
_7s29Q76st0-016|Carbon monoxide.
_7s29Q76st0-017|There's a diatomic molecule, two atoms, unequal electronegativities between oxygen and carbon.
_7s29Q76st0-018|So there'll be an electric dipole moment in carbon monoxide.
_7s29Q76st0-019|Carbon dioxide.
_7s29Q76st0-020|Now here we have two bonds to consider.
_7s29Q76st0-021|Remember, we need to consider each bond in the molecule.
_7s29Q76st0-022|Each bond will be polar.
_7s29Q76st0-023|Each will have a tiny dipole moment that will add to give the total dipole moment for the molecule.
_7s29Q76st0-024|In the case of carbon dioxide, the dipole moments are equal and opposite, so they'll identically cancel out.
_7s29Q76st0-025|How about acetone?
_7s29Q76st0-027|Now those are not equal and opposite.
_7s29Q76st0-028|In fact, they'll point in different directions and have different magnitudes, and they will not cancel out.
_7s29Q76st0-030|So the answer here, carbon dioxide has no permanent dipole moment.
_7s29Q76st0-031|Now those of you that are looking at this very closely might say, wait a minute, isn't oxygen more electronegative than carbon?
_7s29Q76st0-032|Shouldn't this electric dipole moment point the other way?
_7s29Q76st0-033|Shouldn't there be a higher probability of finding electrons around the more electronegative element, oxygen, in this diatomic molecule?
_7s29Q76st0-034|Well, your initial impression would be correct.
_7s29Q76st0-035|Your feeling from electronegativities is that this dipole moment should point in the other direction.
_7s29Q76st0-036|And in fact, it does point from the oxygen to the carbon.
zdupTfmI3ME-000|Let's look at the catalyzed reaction of hydrogen and oxygen to form water.
zdupTfmI3ME-013|Now, we weren't given the reagents hydrogen and oxygen in that ratio.
zdupTfmI3ME-014|In fact, you hardly ever are.
zdupTfmI3ME-015|You don't provide things in exact chemical ratio and when you don't one thing will run out before the other thing.
zdupTfmI3ME-016|One thing will be in excess.
zdupTfmI3ME-017|And in this case, I think you can see we have much more oxygen than we need to react with one mole of hydrogen.
zdupTfmI3ME-018|To react with one mole of hydrogen we'd only need half a mole of oxygen and we have two.
zdupTfmI3ME-022|Hydrogen and water will produce in one to one ratios.
zdupTfmI3ME-023|They have the same stoichiometric coefficient.
zdupTfmI3ME-024|So one mole of hydrogen, if there's enough oxygen, which there is, will produce one mole of water.
zdupTfmI3ME-025|But there's oxygen left over.
zdupTfmI3ME-026|We call hydrogen here the limiting reagent because when it ran out it limited how much product I could form.
WH4YuPwH9l8-003|So which of these transitions in thymol blue has the greatest absorbance at pH 10?
WH4YuPwH9l8-011|We're looking at the indicator thymol blue, which is a polyprotic acid, and it has several color variations.
WH4YuPwH9l8-015|So we're talking about thymol blue in the region where we have a pH of 10.
WH4YuPwH9l8-016|So at pH 10, the blue form predominates.
WH4YuPwH9l8-017|Now, if a solution appears blue, where does it absorb?
WH4YuPwH9l8-018|Remember, solutions that appear blue absorb other regions.
WH4YuPwH9l8-019|And something that absorbs strongly in the red will pass blue colors and appear blue.
5_H82ywrQIY-000|Let's look at dissolving barium sulfate, a sparingly soluble solid, in water.
5_H82ywrQIY-001|If we try to dissolve it and a little, tiny speck remains, say about a milligram remains in the bottom, what's the best way to dissolve that additional milligram?
5_H82ywrQIY-009|We're looking at barium sulfate in water, and dissolving a tiny, milligram speck, and the most efficient way to do that.
5_H82ywrQIY-010|Should we add some acid?
5_H82ywrQIY-011|H2SO4?
5_H82ywrQIY-012|Should we add barium chloride, BaCl2, or water?
5_H82ywrQIY-013|Well, if you look at the reaction for barium sulfate solid dissolving in water, barium sulfate forms barium ions and sulfate ions.
5_H82ywrQIY-018|What you want is more ions to get this to dissolve.
5_H82ywrQIY-019|Sulfuric acid, H2SO4, the same thing, you're going to add more sulfate ions to solution.
5_H82ywrQIY-021|So in this case, the best strategy is simply add more water, increase the concentration.
5_H82ywrQIY-022|That gives you diluting these concentrations, making them smaller.
5_H82ywrQIY-023|By Le Chatelier's principle, I'll shift to make these concentrations higher again.
5_H82ywrQIY-024|So in this case, add more water to dissolve a one milligram speck of barium sulfate.
aNEDU6EL8jc-000|Let's look at the course of chemical reactions.
aNEDU6EL8jc-006|And I can calculate this value Q at any time.
aNEDU6EL8jc-013|So I can calculate this number Q at any time.
aNEDU6EL8jc-016|You can see they flatline here.
aNEDU6EL8jc-017|There is no change with time for the concentrations.
aNEDU6EL8jc-018|So if you wait long enough, these values of concentration stop changing and that means Q stops changing.
aNEDU6EL8jc-019|It becomes a constant after some time.
aNEDU6EL8jc-021|And your reaction quotient stops changing.
aNEDU6EL8jc-022|It becomes a constant.
aNEDU6EL8jc-024|That is, A and B are still changing into C and D. It's just at the same time, C and D are changing back into A and B.
aNEDU6EL8jc-025|So macroscopically, things don't change, but the reverse and forward rates have equalized.
aNEDU6EL8jc-026|So it's a dynamic equilibrium.
aNEDU6EL8jc-029|So it's indeed a constant, and it's a characteristic of the chemical reaction.
aNEDU6EL8jc-030|So if you measure lots of these K's you can predict where the reaction's going to go.
aNEDU6EL8jc-032|The numerator is the products.
aNEDU6EL8jc-033|So if these products are too small-- because Q is too small-- I should go make more of them.
aNEDU6EL8jc-034|So I should go towards products.
aNEDU6EL8jc-035|So comparing Q that you measure at any time with your known constant gives you predictive power.
aNEDU6EL8jc-036|Which way is that reaction going to go?
aNEDU6EL8jc-037|Same thing, if you measure Q and its bigger than K, well, that says, this numerator's too large.
aNEDU6EL8jc-038|Too many products, I should go back to reactants, because I know Q's always go back to the K.
aNEDU6EL8jc-039|If I wait long enough, reactions will proceed.
aNEDU6EL8jc-040|So Q, the instantaneous concentrations, stops changing at this value of K.
aNEDU6EL8jc-041|And it will progress towards K in a relatively straightforward monotonic fashion.
aNEDU6EL8jc-042|So here we have Q equal K at equilibrium.
aNEDU6EL8jc-043|Now, the approach from Q to K can be monotonic, as I've written it, but it actually could also oscillate.
aNEDU6EL8jc-044|Whatever it is, it'll get to K eventually.
aNEDU6EL8jc-045|So Q equals K at equilibrium.
aNEDU6EL8jc-046|And what I have is a comparison between Q and K will tell me which direction I have to go to get equilibrium.
aNEDU6EL8jc-047|I can measure this value of K. What I need to do is measure it at different temperatures because K will change with temperature.
aNEDU6EL8jc-048|So if I do this reaction at 25 degrees C, I might get a different value of K than I would at 50 degrees C.
zez2R4EHKtc-000|We can measure the free energy difference between products and reactants at any stage in a chemical reaction.
zez2R4EHKtc-002|But the product and reactant concentrations can change.
zez2R4EHKtc-004|So as reactions proceed, the free energies of the products and reactants change.
zez2R4EHKtc-005|At the beginning, the reactant free energy will be high, and the product free energy is low.
zez2R4EHKtc-007|So the difference between the free energy of the products and the reactants is zero.
zez2R4EHKtc-008|And that's the equilibrium state where I can switch in between product and reactant with no free energy penalty.
zez2R4EHKtc-009|The playing field has been leveled.
zez2R4EHKtc-012|Remember it's products minus reactants.
zez2R4EHKtc-013|So if you have a higher reactant free energy than the products, then delta G will be negative.
zez2R4EHKtc-014|That's the same case for Q less than K. A Q less than K says Q is too small.
zez2R4EHKtc-015|The denominator, the reactants, is too big.
zez2R4EHKtc-016|I should go toward products.
zez2R4EHKtc-017|Delta G less than zero says the same thing.
zez2R4EHKtc-021|The playing field has been leveled, the free energy of the products, the free energy of the reactants are the same.
zez2R4EHKtc-022|I can interchange between the two, the macroscopic properties won't change, Q and K are the same.
zez2R4EHKtc-023|I can also think about the situation where delta G is positive.
zez2R4EHKtc-024|That would mean the product free energies are higher.
zez2R4EHKtc-029|As the reaction proceeds, the reactant free energies change, the product free energies change until they come to equilibrium.
zez2R4EHKtc-030|Notice that delta G standard is a constant.
zez2R4EHKtc-031|That's the delta G difference for everything at one molar or one atmosphere of pressure.
zez2R4EHKtc-032|That's a constant for a given temperature.
zez2R4EHKtc-033|So delta G varies based on that standard state and the concentrations.
beq0Y7dvEKI-000|Let's talk about the intermolecular forces between atoms and molecules.
beq0Y7dvEKI-001|They're the forces that lead to condensation of the gas phase into the liquid phase.
beq0Y7dvEKI-002|Now, there's various kinds of intermolecular forces.
beq0Y7dvEKI-003|But all of them depend on a plus-minus interaction-- somewhere in the sample, a positive charge attracted to a negative charge.
beq0Y7dvEKI-004|We classify them in terms of their energy.
beq0Y7dvEKI-005|The lowest energy, we call Van der Waals dispersion interactions.
beq0Y7dvEKI-006|Now, those are interactions between molecules that on the surface appear very symmetric.
beq0Y7dvEKI-007|So there's no plus or minus interaction that you'd predict.
beq0Y7dvEKI-008|But what happens is those electronic clouds around those molecules can be distorted.
beq0Y7dvEKI-009|And when they're distorted, just transiently, that leads to a little plus-minus interaction.
beq0Y7dvEKI-010|And that transient dipole moment can actually induce dipole moments in nearby molecules.
beq0Y7dvEKI-011|So you have a cooperative effect.
beq0Y7dvEKI-012|We call these transient or induced forces, dispersion Van der Waals forces.
beq0Y7dvEKI-013|They look like this.
beq0Y7dvEKI-016|I can go to slightly higher energy.
beq0Y7dvEKI-017|Slightly higher energy is when I have a permanent dipole moment in the particle.
beq0Y7dvEKI-018|Now, if I have a permanent dipole moment, there is a clear positive and a clear negative end of the particle, which can be attracted to each other.
beq0Y7dvEKI-019|Those are higher energy, because they're permanent.
beq0Y7dvEKI-020|And they require a higher temperature-- more kinetic energy to overcome.
beq0Y7dvEKI-021|So particles with only Van der Waals interactions, they will boil at low temperatures.
beq0Y7dvEKI-030|Now, it's not an actual chemical bond.
beq0Y7dvEKI-031|In fact, it's about 100 times weaker.
beq0Y7dvEKI-032|But it's actually hundreds of times stronger than these interactions.
beq0Y7dvEKI-033|So the strongest type of intermolecular interaction-- hydrogen bonding-- occurs in special cases.
beq0Y7dvEKI-034|Water is the classic case.
beq0Y7dvEKI-035|And we'll talk about water and hydrogen bonding more later in this series.
PyIiTK0eXrA-000|Let's look at an example of oxidation reduction reactions for the autoionization of water.
PyIiTK0eXrA-001|Now, the autoionization of water is important in our acid base chemistry.
PyIiTK0eXrA-002|In fact, we saw water the reverse reaction auto ionizing to form OH minus and H3O plus.
PyIiTK0eXrA-003|This forward reaction, the way it's written, is very spontaneous.
PyIiTK0eXrA-004|The K, the equilibrium constant, 10 to the 14th, so very much favors water.
PyIiTK0eXrA-005|And we should expect a positive cell potential, if we write this in terms of cell potentials.
PyIiTK0eXrA-006|Now, there's nothing being oxidized or reduced in this reaction.
PyIiTK0eXrA-007|But we can cast it in terms of oxidation reduction of hydrogen gas.
PyIiTK0eXrA-015|The hydrogen gas cancels out.
PyIiTK0eXrA-017|So I'd have a cell at pH 14 connected to a cell at pH 0.
PyIiTK0eXrA-018|The difference in cell potential, 0.83 volts.
PyIiTK0eXrA-019|Now, it's interesting the potential of this cell is a function of the pH.
PyIiTK0eXrA-020|So that kind of tells you how a pH meter could work.
PyIiTK0eXrA-022|So you could correlate the voltage to a pH.
PyIiTK0eXrA-023|And that's indeed how pH meters work.
PyIiTK0eXrA-024|So our understanding of electrochemistry helps us understand something very important about acid base chemistry.
aVRWeXk4oK8-000|Let's look at the combustion of acetylene, C2H2, and a couple of different flavors.
aVRWeXk4oK8-009|We're burning acetylene in oxygen in a few different ways.
aVRWeXk4oK8-017|We have the same type of conversion here.
aVRWeXk4oK8-018|We have liquid water and gaseous water.
aVRWeXk4oK8-024|But could I have come up with just observing what was happening?
aVRWeXk4oK8-025|And indeed, I can, because enthalpies of vaporization involve breaking of intermolecular interactions.
aVRWeXk4oK8-026|Which of these two molecules-- water or acetylene-- would have the stronger intermolecular interactions?
aVRWeXk4oK8-027|Well, water is a molecule that has hydrogen bonding in the liquid state.
aVRWeXk4oK8-028|And hydrogen bonding is the strongest form of intermolecular interaction.
aVRWeXk4oK8-029|Acetylene doesn't have hydrogen bonding.
aVRWeXk4oK8-030|So the enthalpy of vaporization for water will be much greater than the enthalpy of vaporization for acetylene.
aVRWeXk4oK8-031|So now I can compare easily.
aVRWeXk4oK8-033|So burning from gas to gas is the lower enthalpy change.
aVRWeXk4oK8-034|The correct answer here-- delta-H1 is smaller than delta-H2.
lRYn99xxj3I-000|Let's look at the real behavior.
lRYn99xxj3I-001|For the real behavior, we'll use the van der Waals expression with a correction to the pressure and volume turns.
lRYn99xxj3I-013|For carbon dioxide, a small deviation from ideality at this low pressure.
lRYn99xxj3I-019|So high pressures, low volumes, you expect greater deviations from ideality.
lRYn99xxj3I-020|And that's what we see here-- a pressure of 67 versus 163 atmospheres for the carbon dioxide case.
mcQ1eivjgEs-000|Let's use conversion factors to estimate just how big a mole is.
mcQ1eivjgEs-001|So let's say you won a mole of dollars in a lottery.
mcQ1eivjgEs-002|And you could spend at a billion dollars per second-- an incredible spending rate.
mcQ1eivjgEs-003|How long in years would it take you to spend your mole of dollars?
mcQ1eivjgEs-009|So what I have is dollars per year leftover.
mcQ1eivjgEs-010|So I have 3.15 times 10 to 16th dollars per year.
mcQ1eivjgEs-011|Now I still have a mole of dollars to spend.
mcQ1eivjgEs-014|So at that rate, here the dollars cancel and leave me the number of years, 1.91 times 10 to the 7th years.
mcQ1eivjgEs-015|So almost 20 million years to spend your mole of dollars, even at a billion dollars per second.
mcQ1eivjgEs-016|That's just how large a mole is.
k09wNUM2xkA-000|Entropy is a state function.
k09wNUM2xkA-001|That means it only depends on the final and initial state of the system.
k09wNUM2xkA-003|We'd like to be able to do that same thing with entropies.
k09wNUM2xkA-004|So we need standard molar entropies of substances.
k09wNUM2xkA-005|Now, these aren't going to be entropies of formation.
k09wNUM2xkA-007|That's where there's only one possible arrangement and no thermal energy.
k09wNUM2xkA-008|So the entropy of those systems is zero.
k09wNUM2xkA-009|Then we'll take and add all the entropies for warming them and changing the phase till we get to the standard state.
k09wNUM2xkA-025|Notice as well, as you compare hydrogen and oxygen, oxygen has a larger standard molar entropy than hydrogen.
k09wNUM2xkA-026|And that's because the heat capacity of oxygen is greater.
k09wNUM2xkA-027|Remember, the heat capacity is involved in all these warming steps.
k09wNUM2xkA-028|And oxygen has a higher heat capacity because oxygen-- some of the energy goes into tumbling the oxygen molecule.
k09wNUM2xkA-029|And it requires more energy to tumble an oxygen, because of the heavier atoms, than a hydrogen.
k09wNUM2xkA-030|So oxygen has a higher heat capacity.
k09wNUM2xkA-031|Each joule of heat you add to oxygen goes into rotating that molecule.
k09wNUM2xkA-033|Here are some compounds.
k09wNUM2xkA-036|And we can calculate the entropy change for a chemical reaction using standard molar entropies.
t6cGw1scLvc-000|Another way to do a process would be to seal it in a thermal container so heat can't flow.
t6cGw1scLvc-001|When heat can't flow, we call those processes adiabatic.
t6cGw1scLvc-002|For adiabatic compression and expansion, we can also talk about energy changes.
t6cGw1scLvc-006|Now it can't absorb a joule of heat to replace the joule of work it did to re-energize itself.
t6cGw1scLvc-007|So I do a joule of work, I use a joule of energy.
t6cGw1scLvc-008|Do a joule of work, use a joule of energy.
t6cGw1scLvc-009|My energy drops.
t6cGw1scLvc-010|And for an ideal gas, if the energy drops, the temperature drops.
t6cGw1scLvc-011|So ideal gas is expanding adiabatically, work is negative, and the change in energy, negative.
t6cGw1scLvc-012|It can also be zero because I could expand against a vacuum and not have to do work.
t6cGw1scLvc-013|What about compression?
t6cGw1scLvc-014|If I compress the gas, now work is being done on the gas.
t6cGw1scLvc-015|And the gas is saying I'm getting this energy from work, but I can't release it as heat, q has to be zero, so my energy just goes up.
t6cGw1scLvc-017|Practically, it usually means you just do the process quickly because you can quickly do work, but heat flow always takes time.
t6cGw1scLvc-018|So a rapid process is often an adiabatic process.
t6cGw1scLvc-019|And we can expand either adiabatically, or we can be compressed adiabatically.
t6cGw1scLvc-020|For ideal gases these are the summary.
3T5TXTUaA_E-000|Let's look at a couple acids reacting with each other.
3T5TXTUaA_E-001|So, here's acid HA1, just a generic acid, and HA2, generic acid.
3T5TXTUaA_E-002|They have pKa's, where the pKa1 is less than pKa2.
3T5TXTUaA_E-010|We're looking at two weak acids reacting together.
3T5TXTUaA_E-015|So, HA1 is the stronger acid.
3T5TXTUaA_E-016|So, if it's a HA1 versus HA2, which will have the higher concentration in solution?
3T5TXTUaA_E-017|Well, HA1 is the stronger acid.
3T5TXTUaA_E-018|It will dissociate more than HA2.
3T5TXTUaA_E-023|I encourage you to do that.
3T5TXTUaA_E-024|Write out the acid equilibrium expressions.
3T5TXTUaA_E-026|HA1, stronger acid, forces this towards products.
3T5TXTUaA_E-027|That's a K larger than 1.
WMAnyBjtj1Y-000|In the laboratory, we need a relationship between the microscopic properties-- numbers of particles-- and the macroscopic properties-- the mass that we can measure.
WMAnyBjtj1Y-001|And we find that using relative masses.
WMAnyBjtj1Y-003|So 12 grams of carbon-12 has Avogadro's number, 6.02 times 10 to the 23rd particles of carbon-12.
WMAnyBjtj1Y-004|Hydrogen is 1/12 as massive.
WMAnyBjtj1Y-005|So one gram of hydrogen has 6.02 times 10 the 23rd particles-- a 12 to 1 ratio.
WMAnyBjtj1Y-009|So if I wanted to react oxygen and hydrogen atoms in a 1 to 1 ratio, I should keep the mass ratio 16 to 1.
WMAnyBjtj1Y-010|What if I wanted to react them in a 2 to 1 ratio-- two hydrogens for every oxygen to form, say, water?
WMAnyBjtj1Y-012|I can do this with the molecules themselves.
WMAnyBjtj1Y-013|In fact, I can add up the atomic relative masses to get the molecular relative masses.
WMAnyBjtj1Y-017|Here's a couple of molecules.
WMAnyBjtj1Y-018|H2O and CO2, water and carbon dioxide.
WMAnyBjtj1Y-019|Water, relative mass 18.
WMAnyBjtj1Y-022|So how do I know one is bent and one is linear, both three-atom molecules?
WMAnyBjtj1Y-023|Well, that's the nature of the quantum mechanical interaction of the electrons that form the bonds in these molecules.
WMAnyBjtj1Y-024|And we'll study that in detail in this course.
WMAnyBjtj1Y-028|That's, say, about this much water in the liquid phase.
WMAnyBjtj1Y-029|If it were in the gas phase, the volume would be about 1,000 times as large.
WMAnyBjtj1Y-030|But here is 18 grams of water.
WMAnyBjtj1Y-031|That's not very much-- one mole.
WMAnyBjtj1Y-032|So let's look at 55-mole samples of some elements and molecules.
WMAnyBjtj1Y-033|Here's water.
WMAnyBjtj1Y-034|Now, water, 55 moles, is 55 times 18 grams of water.
WMAnyBjtj1Y-035|But the cool thing is, here's 55 times 18 grams of water, but I know how many particles are there.
WMAnyBjtj1Y-036|There's 55 times Avogadro's number, 55 times 6.02 times 10 to the 23rd particles, in here.
WMAnyBjtj1Y-037|Now, I also have carbon.
WMAnyBjtj1Y-038|Here's 55 moles of carbon.
WMAnyBjtj1Y-039|And this has the same number of particles as water particles.
WMAnyBjtj1Y-040|They're both 55 moles.
WMAnyBjtj1Y-041|And their masses, actually, they're quite similar.
WMAnyBjtj1Y-042|And we can look at the table and see that.
WMAnyBjtj1Y-043|Water has relative mass 18, carbon, relative mass 12.
WMAnyBjtj1Y-044|So those are very similar masses.
WMAnyBjtj1Y-045|Now, I also have aluminum here.
WMAnyBjtj1Y-046|Now, aluminum has mass about-- well, looks like twice carbon.
WMAnyBjtj1Y-047|And, yeah, I feel that.
WMAnyBjtj1Y-048|This is twice as massive, 55 moles of aluminum, about twice as massive as 55 moles of carbon.
WMAnyBjtj1Y-049|I also have lead.
WMAnyBjtj1Y-050|I can barely lift the 55 moles of lead.
WMAnyBjtj1Y-051|Here's 55 moles of lead.
WMAnyBjtj1Y-052|Very massive, because each lead atom is something like eight times the mass of the aluminum atoms.
WMAnyBjtj1Y-053|So very much more massive, 55 moles of lead, than 55 moles of aluminum.
WMAnyBjtj1Y-054|Both the same number of particles, but each lead particle is something like eight times heavier.
WMAnyBjtj1Y-055|So that's a lot of mass.
WMAnyBjtj1Y-056|Now, you may be curious.
WMAnyBjtj1Y-057|Why are these numbers for my relative masses not exactly integers?
WMAnyBjtj1Y-058|Well, that's because in nature every carbon, for instance, does not have a mass of 12.
WMAnyBjtj1Y-059|There are some carbon atoms out there that have a mass of 13.
WMAnyBjtj1Y-060|In fact, about 1% of all carbon atoms in nature have a mass of 13.
WMAnyBjtj1Y-064|We call them isotopes.
WMAnyBjtj1Y-065|So this is an average over all the isotopes of the various elements that have different masses.
Nb76my9vHLY-000|Let's try to draw some Lewis electron dot structures.
Nb76my9vHLY-001|Formaldehyde is a common compound, CH2O.
Nb76my9vHLY-002|The question I have is, which of these structures is the correct Lewis electron dot structure for formaldehyde?
Nb76my9vHLY-009|We're writing the Lewis electron dot structure for formaldehyde.
Nb76my9vHLY-010|Whenever you write Lewis electron dot structures, you first have to get the correct number of electrons.
Nb76my9vHLY-011|So you add up the valence electrons for all the participating atoms.
Nb76my9vHLY-012|So we'll take carbon, which has four valence electrons.
Nb76my9vHLY-013|We'll take an oxygen with six valence electrons.
Nb76my9vHLY-014|And we need to hydrogens.
Nb76my9vHLY-015|Each has one valence electron.
Nb76my9vHLY-016|So that's a total of 12 valence electrons.
Nb76my9vHLY-017|So our structures each must have 12 valence electrons to hold them together.
Nb76my9vHLY-020|There aren't 14 valence electrons among those atoms, so that's not a proper Lewis electron dot structure.
Nb76my9vHLY-022|If you look at B, there is a subtle error here.
Nb76my9vHLY-023|This has 12 electrons two, four, six, eight, 10, 12.
Nb76my9vHLY-024|But carbon has too many.
Nb76my9vHLY-025|Carbon wants an octet, but it's sharing two, four, six, eight, 10 electrons.
Nb76my9vHLY-026|So there's no octet on carbon here.
m9XOfqalSgs-000|Let's look at the correlation between intermolecular forces and Van der Waals a parameters.
m9XOfqalSgs-004|And in this case, it's obvious that water has the largest intermolecular forces.
Dc69O-XLKtc-000|Let's construct an electrolytic cell with gold and zinc.
Dc69O-XLKtc-005|So if gold is reduced, electrons flow to the gold electrode and at the gold electrode, gold medal is plated out from ions.
Dc69O-XLKtc-006|So gold ions accept electrons-- they're reduced to gold metal.
Dc69O-XLKtc-007|At the same time, you'd have zinc metal being oxidized to zinc ions.
Dc69O-XLKtc-011|The mass of the gold electrode would increase, the mass of the zinc electrode would decrease.
Dc69O-XLKtc-012|We're going to make this go in the electrolytic direction though by providing a external voltage.
Dc69O-XLKtc-013|So the external voltage will force electrons in this direction.
Dc69O-XLKtc-014|So electrons will flow from the gold electrode to the zinc electrode.
Dc69O-XLKtc-015|To get electrons to flow in this direction, that means that gold metal has to be oxidized to gold ions.
Dc69O-XLKtc-019|So that's theoretically what's going to happen.
Dc69O-XLKtc-020|Let's do some mathematics-- a little algebra and arithmetic to figure out the actual voltages.
Dc69O-XLKtc-021|So here's gold and zinc half cells.
Dc69O-XLKtc-022|Here are their voltages.
Dc69O-XLKtc-023|Of course, in the electrolytic reaction, zinc will be oxidized while gold is reduced.
Dc69O-XLKtc-024|So I'll reverse the zinc reaction and change the sign of the potential.
Dc69O-XLKtc-025|I can then add these if I keep track of the number of electrons-- so two electrons here and two electrons here.
Dc69O-XLKtc-026|That gives me a two electron transfer process with a total voltage of the sum of 1.69 and 0.76 volts.
Dc69O-XLKtc-027|So the overall voltage of the standard galvanic direction would be 2.54.
Dc69O-XLKtc-028|I need to overcome that to go in the electrolytic direction.
Dc69O-XLKtc-029|So the minimum extra voltage I need is something greater than 2.45.
Dc69O-XLKtc-030|How much greater, than what kind of overvoltage I need, depends on the configuration of the system.
Dc69O-XLKtc-032|Minimum voltage of 2.54 volts is the minimum overvoltage we need.
Dc69O-XLKtc-034|So that's one coulomb per second for 3,600 seconds is 36,000 coulombs.
Dc69O-XLKtc-035|So 36,000 coulombs of charge are going to be transferred.
Dc69O-XLKtc-036|Well, how many electrons is that?
Dc69O-XLKtc-037|Well we can convert 36,000 coulombs of charge into moles of electrons using Faraday's constant.
Dc69O-XLKtc-038|Faraday's constant is the charge in coulombs on a mole of electrons.
Dc69O-XLKtc-040|Well, how much zinc metal does that 0.37 moles of electrons produce?
Dc69O-XLKtc-041|Let's look back at our chemical reaction-- it's gold ions and zinc producing zinc ions and gold metal.
Dc69O-XLKtc-042|It's a two electron process that I'm going to make go in reverse to make zinc metal.
Dc69O-XLKtc-044|So overall, my 0.37 moles of electrons produces 12 grams of zinc.
Dc69O-XLKtc-045|So this electrolytic process, over 10 hours, produces 12 grams of zinc metal.
NI0c8hjRWc0-000|The equilibrium of NO2 gas and N2O4 is dynamic.
NI0c8hjRWc0-002|N2O4 dimers are breaking down to form NO2.
NI0c8hjRWc0-004|The dynamic nature of that equilibrium allows the equilibrium to adjust if there's a change in the system.
NI0c8hjRWc0-005|And it does so in a logical way.
NI0c8hjRWc0-007|So let's look at that.
NI0c8hjRWc0-008|Here's the system in equilibrium, as I have it here.
NI0c8hjRWc0-010|But now, it's not at equilibrium anymore.
NI0c8hjRWc0-011|In fact, Q will be greater than K.
NI0c8hjRWc0-012|And you can prove that to yourself by putting in one half the partial pressures for the equilibrium quotient expression.
NI0c8hjRWc0-015|Now, what you'll have is a different partial pressure of NO2 and a different partial pressure of N2O4 then you started here.
NI0c8hjRWc0-017|What you'll notice here is you could have thought of this another way.
NI0c8hjRWc0-018|The system expanded and went to a larger volume.
NI0c8hjRWc0-019|So the system wants to expand to fill that volume.
NI0c8hjRWc0-020|Well, how does the system expand to fill that volume?
NI0c8hjRWc0-021|It goes to the side where there are more particles.
NI0c8hjRWc0-022|If some of these particles dissociate, it can fill the volume more effectively.
NI0c8hjRWc0-023|So I shift towards the sides with more particles when I go to larger volume.
NI0c8hjRWc0-024|So this is an example of Le Chatelier's principle, where a system moves in response to stress and re-establishes equilibrium.
1ZTXQhi9aW4-005|But I can also form all of the reactants from elements in their standard states.
1ZTXQhi9aW4-010|Now what if I want to find this enthalpy of reactants going to products?
1ZTXQhi9aW4-011|Well, the thing about enthalpy is it's a state function.
1ZTXQhi9aW4-012|So it doesn't matter the path that I go from reactants to products, it only matters that I start at reactants and end at products.
1ZTXQhi9aW4-013|The initial and final state are the only thing that determine the enthalpy change.
1ZTXQhi9aW4-015|This path.
1ZTXQhi9aW4-016|Now this path from reactants to products has the same enthalpy as this path.
1ZTXQhi9aW4-018|The negative sign here, because I've gone from reactants to elements in their standard state.
1ZTXQhi9aW4-019|I've taken the reverse reaction of heats of formation.
1ZTXQhi9aW4-021|A very powerful tool.
3L5bi-9jO74-000|Let's look at two gases mixed in a balloon.
3L5bi-9jO74-001|I'll take bromine and hydrogen gas, their relative masses shown here.
3L5bi-9jO74-002|And I'd like to know what are their average molar kinetic energies?
3L5bi-9jO74-003|Does bromine gas have the higher kinetic energy?
3L5bi-9jO74-004|Are they about equal?
3L5bi-9jO74-012|We're talking about hydrogen and bromine gas mixed together in a balloon, one mole of each.
3L5bi-9jO74-013|Which has the higher kinetic energy?
-LFgLSytl1U-000|We can apply our notions of heat flow to applications in the home.
-LFgLSytl1U-001|Let's say I'm going to bake a cake in my oven, and I know it takes half hour to bring my cake to 325 degrees in my oven.
-LFgLSytl1U-002|Now, that's the minimum time it takes.
-LFgLSytl1U-003|I can't heat it any faster than that.
-LFgLSytl1U-004|What if two cakes are placed in that same oven?
-LFgLSytl1U-005|After that same half an hour, what will the relative temperatures be?
-LFgLSytl1U-011|Let's look at baking cakes in an oven.
-LFgLSytl1U-012|Now, when you bake the first cake in an oven, the heat flow is similar to liquid flow.
-LFgLSytl1U-013|So what you have is a reservoir of heat in the oven that you deliver to the cakes.
-LFgLSytl1U-014|It's like delivering water through a spigot and the flow is limited by the spigot.
-LFgLSytl1U-021|And you may have heard this.
-LFgLSytl1U-023|If you put two cakes or five cakes in that same oven, the heat flow will be distributed between all the cakes, you'll effectively overload the oven.
-LFgLSytl1U-025|So thermodynamics has its place in the home when you're baking.
sihnvASQFSw-000|Let's look at radioactive decay in terms of kinetics.
sihnvASQFSw-002|Now, millicurie is just a measure of the amount of tritium, so it's like a concentration.
sihnvASQFSw-003|So if you plot the natural log of tritium concentration versus time, it decays away like this.
sihnvASQFSw-004|Natural log concentration versus time is linear.
sihnvASQFSw-012|We're talking about the decay of tritium over time.
sihnvASQFSw-013|And we're actually talking about the biological half-life.
sihnvASQFSw-014|If you ingest some tritium, the tritium acts just like hydrogen, and that's expelled from your body relatively quickly.
sihnvASQFSw-015|So if you plot the tritium concentration in your body over time, the natural log versus time, it's linear.
sihnvASQFSw-019|So in five half-lives, I'm down to less than a millicurie in my body.
sihnvASQFSw-020|So what if I had a 40 millicurie exposure?
sihnvASQFSw-022|So how long is a half-life?
WitrD0bFL24-000|We know electrons can absorb energy and change energy levels, go from one orbital to another, in an atom.
WitrD0bFL24-009|We're talking about electronic transitions, going from n equal 1 to n equal 2.
WitrD0bFL24-010|Which of these possible species can do that?
WitrD0bFL24-011|Let's look at their electronic configurations.
WitrD0bFL24-012|Here's fluorine, 1s2 2s2 2p5.
WitrD0bFL24-013|Fluorine minus has an extra electron, one more electron than fluorine, so it has electronic configuration 1s2 2s2 2p6.
WitrD0bFL24-016|So 1s2 2s2 2p6.
WitrD0bFL24-017|Notice fluorine minus and neon have the same electronic configuration, plus it has an electron in the three s orbital.
Zfx54dYEDBs-007|We're looking at the oxidation of NO2 minus to NO3 minus and looking at how the formal charge on nitrogen changes in that process.
Zfx54dYEDBs-012|And it'll share one of these and two of those.
Zfx54dYEDBs-013|So one, two, three, four, five electrons in a sharing mode for nitrogen in that molecule.
Zfx54dYEDBs-014|This molecule, nitrogen shares all the electrons around it.
Zfx54dYEDBs-015|It shares one from this bond, one from this bond, and two from this bond.
Zfx54dYEDBs-016|One, two, three, four.
Zfx54dYEDBs-017|Four electrons.
Zfx54dYEDBs-018|Nitrogen as an atom has five electrons.
4blDQFbly-Y-000|Let's look at the Gibbs free energy for a system.
4blDQFbly-Y-001|The Gibbs function is defined as the enthalpy minus the temperature times the entropy.
4blDQFbly-Y-004|You can talk about that for any system in general or you can talk about specifically the standard states of the system.
4blDQFbly-Y-014|But the free energy is always a predictor of the direction of the chemical reaction.
4blDQFbly-Y-019|So I have predictive power now, whether a reaction is likely to go based on the free energy of the system.
TPQjDIPwJ-A-000|Let's do a calculation involving the photoelectric effect, ejected photo electrons and photons of a certain energy hitting a metal.
TPQjDIPwJ-A-002|So this work function for chromium metal, 4.37 electron volts, is how strongly chromium holds on to its electrons.
TPQjDIPwJ-A-003|So we have to say, well, what is the photoelectric effect?
TPQjDIPwJ-A-004|The photoelectric effect, we have to balance the energies.
TPQjDIPwJ-A-006|So the metal holds on to the electron, but if we bring in a high enough energy photon we can eject an electron with a certain kinetic energy.
TPQjDIPwJ-A-007|So that's what we have to find.
TPQjDIPwJ-A-010|Now, we do that to keep all the things we multiply together consistent unit wise.
TPQjDIPwJ-A-011|We're dimensionally consistent.
TPQjDIPwJ-A-012|We always use kilograms for mass, so don't just put your masses into your equations with grams or pounds or some random mass unit.
TPQjDIPwJ-A-013|If you come across a mass, convert it to the kilograms.
TPQjDIPwJ-A-014|If you come across a distance, a length, convert that to meters.
TPQjDIPwJ-A-015|A time, seconds.
TPQjDIPwJ-A-016|An energy, joules.
TPQjDIPwJ-A-022|That's going to be an energy, a kinetic energy, so this will be joules.
TPQjDIPwJ-A-023|In fact, that's how I remember the SI units of joules.
TPQjDIPwJ-A-030|Tiny, tiny, tiny number of joules.
TPQjDIPwJ-A-031|Of course, it's an electron.
TPQjDIPwJ-A-035|Now electron volts in a unit of energy, it's the kinetic energy that an electron gains as you accelerate it across the potential of 1 volt.
TPQjDIPwJ-A-037|So we can do that product, so the work function in terms of joules, 7.04 times 10 to the minus 19th joules.
TPQjDIPwJ-A-046|And that's what we want.
TPQjDIPwJ-A-047|It's always good to check, do the units that we have left makes sense for the quantity that we're solving for?
TPQjDIPwJ-A-048|We're solving for wave length, a length.
TPQjDIPwJ-A-049|Do I have meters?
TPQjDIPwJ-A-050|In this case, I do.
TPQjDIPwJ-A-053|109 nanometers is, as we recall, in the UV, the ultraviolet region.
TPQjDIPwJ-A-054|We know visible went from 700 down to 400 nanometers.
wA_IvsdRjpM-000|Let's look at the root mean squared velocity for bromine, at 300 Kelvin in a sample of gas, in our Nuts and Boltz section.
wA_IvsdRjpM-001|And in honor of Maxwell Boltzmann distribution, we'll spell bolts with a z this time.
wA_IvsdRjpM-006|And I can solve for that.
wA_IvsdRjpM-007|I get 216 meters per second.
wA_IvsdRjpM-008|Now, 216 meters per second, that's about 800 feet per second.
wA_IvsdRjpM-011|That's amazing speeds for gas particles at modest temperatures.
ktySzUONBUU-000|So how do we determine the exact mass of an atom?
ktySzUONBUU-001|Well, one way is to use a mass spectrometer.
ktySzUONBUU-003|A mass spectrometer, you take a sample and you ionize it.
ktySzUONBUU-004|And by ionization, we mean stripping off electrons.
ktySzUONBUU-005|In this case, we strip off all of the electrons.
ktySzUONBUU-006|The sample is then accelerated towards an electric field and directed through a magnetic field.
ktySzUONBUU-007|Now when charged particles hit a magnetic field, they are deflected by the magnetic field.
ktySzUONBUU-008|And they're deflected according to their mass.
ktySzUONBUU-009|Heavier particles are not as deflected as lighter particles.
ktySzUONBUU-010|So you have this stream of particles and it hits that magnetic field, and it's fanned out based on their mass.
ktySzUONBUU-015|Let's look at a larger mass coming through the same system.
ktySzUONBUU-016|We're going to ionize it, it's not deflected as much.
ktySzUONBUU-017|So you have a distribution of masses that you can resolve using a mass spectrometer.
ktySzUONBUU-019|And the vertical axis will tell us about how many particles there were, and that's very useful.
ktySzUONBUU-020|Remember when we had our carbon, we said that one in hundred of these particles was a carbon-13.
ktySzUONBUU-021|The other 99 were carbon-12.
ktySzUONBUU-022|That's a difference of one mass unit.
ktySzUONBUU-024|So, let's look at a few masses.
ktySzUONBUU-025|We have, for instance, what would show up at mass 1?
ktySzUONBUU-029|So you can tell by the mass spectrometer you can get the mass.
ktySzUONBUU-030|But remember, mass does not determine the identity of the element-- certainly a clue, but it does not determine the identity.
ktySzUONBUU-031|We need to know the number of protons in the nucleus to get the identity of an element.
ktySzUONBUU-032|So mass three, that could be tritium or it could be a molecule of deuterium and hydrogen hooked together to form a hydrogen molecule.
ktySzUONBUU-036|And you can go on down the line.
ktySzUONBUU-037|Here is oxygen 16.
ktySzUONBUU-038|At mass 18, you could have a water molecule with oxygen 16 and two protons, two hydrogen atoms.
ktySzUONBUU-044|So a mass spectrometer, exquisitely sensitive makes.
ktySzUONBUU-046|The e equals mc squared mass loss can be determined by modern mass spectrometer.
ktySzUONBUU-047|That's how accurate they are.
ktySzUONBUU-048|Incredibly tiny mass losses can be determined by this method.
ktySzUONBUU-049|It's a beautiful instrument and it's very, very, very powerful for measuring atomic masses.
ksZoOde-7zY-000|We noticed that the trend in ionization energies is to kind of increase across a row in the periodic table.
ksZoOde-7zY-001|Let's look at that in more detail.
ksZoOde-7zY-003|That's a 1s electron.
ksZoOde-7zY-004|For helium, about 2,400 kilojoules per mole for that ionization.
ksZoOde-7zY-006|So increasing that positive charge on the nucleus was much more dramatic than adding another electron.
ksZoOde-7zY-007|It's much harder to pull off an electron from helium than hydrogen.
ksZoOde-7zY-011|And lithium is remarkably easy to ionize.
ksZoOde-7zY-018|It's a general trend.
ksZoOde-7zY-019|And we can kind of rationalize that.
ksZoOde-7zY-020|Because what we're doing is we're adding more nuclear charge, but we're staying in the same principle quantum level.
ksZoOde-7zY-021|So electrons in about the same region of space but more positive charge holding them.
ksZoOde-7zY-022|Of course, those will be slightly more difficult to ionize.
ksZoOde-7zY-023|As you continue across, there's a slight dip here at oxygen.
ksZoOde-7zY-024|And I think we can understand that too in terms of the fact that nitrogen now is that half-filled state.
ksZoOde-7zY-025|Oxygen has that paired electron.
ksZoOde-7zY-026|So what you're doing is saying I have three p electrons in the nitrogen state.
ksZoOde-7zY-027|When I go to oxygen, I add a paired electron.
ksZoOde-7zY-028|That's a little bit tough.
ksZoOde-7zY-029|Pairing electrons, there's a little bit of repulsion energy.
ksZoOde-7zY-031|So oxygen, slightly easier to ionize than its counterparts.
ksZoOde-7zY-033|And of course, then sodium, look how similar sodium is to lithium.
ksZoOde-7zY-043|And we understood that in terms of the fine structure of the electronic configurations.
ksZoOde-7zY-044|A stable half-filled shell or the pairing energy affecting ionization energies.
ksZoOde-7zY-045|So this trend is one that we can look at in detail and with the quantum mechanics of a very good understanding of the trends.
e7Ofj4bDdqI-000|Let's look at several molecules and see if we can predict their relative boiling or condensation points.
e7Ofj4bDdqI-001|Here's three molecules-- H2O, H2S, or NH3.
e7Ofj4bDdqI-011|Now, all these particles have the three classic intermolecular interactions.
e7Ofj4bDdqI-012|Dispersion forces, they have dipole-dipole interactions, and hydrogen bonding.
e7Ofj4bDdqI-013|Whenever hydrogen bonding is involved though, that's the dominant force.
e7Ofj4bDdqI-014|It's orders of magnitude greater than the other two forces.
e7Ofj4bDdqI-015|So what we have to determine is, which of these has the strongest hydrogen bonding interaction?
e7Ofj4bDdqI-016|The strongest hydrogen bonding interaction will come from the greatest dipole.
e7Ofj4bDdqI-017|So is the OH bond, the SH bond, or the NH bond the greatest dipole moment?
e7Ofj4bDdqI-018|Well, we determine that by the difference in electronegativity.
e7Ofj4bDdqI-019|You know O is more electronegative than S is more electron give than N, nitrogen.
e7Ofj4bDdqI-023|Sulphur slightly smaller, and the NH bond slightly smaller yet.
e7Ofj4bDdqI-024|So the greatest dipole-dipole gives the greatest hydrogen bonding interaction, and in that case, that's water.
e7Ofj4bDdqI-025|So among these three, the highest boiling point is the one with the strongest intermolecular interaction.
e7Ofj4bDdqI-026|Remember, the intermolecular interaction has to be overcome by kinetic energy to boil.
e7Ofj4bDdqI-027|Higher kinetic energy, higher temperature.
e7Ofj4bDdqI-028|So more intermolecular interactions that are very strong means higher temperature to boil.
e7Ofj4bDdqI-029|So in this case, water is the largest intermolecular interaction, largest hydrogen bonding interaction, and that's the correct answer.
e7Ofj4bDdqI-032|So we have the hydrogen bonding interactions reflected in the boiling points and the size of the van der Waals A parameters.
QoHZH6OkzQs-000|Let's look at molecular oxygen.
QoHZH6OkzQs-001|Molecular oxygen is paramagnetic, but it can undergo a transition where it becomes diamagnetic.
QoHZH6OkzQs-002|The question I have for you is, which process involving a photon makes paramagnetic oxygen into diamagnetic oxygen?
QoHZH6OkzQs-011|Now these would be effectively paired and this would be a non-magnetic species, diamagnetic.
QoHZH6OkzQs-012|An ionization event won't allow you to go to a diamagnetic species.
QoHZH6OkzQs-013|If you remove an electron from a paramagnetic oxygen, you still have an unpaired electron and it's still paramagnetic.
QoHZH6OkzQs-015|Well I think you remember from filling electrons into orbitals, we put them in spin parallel because anti-parallel is the higher energy state.
QoHZH6OkzQs-016|Making those magnets line up anti-parallel is a higher energy interaction.
QoHZH6OkzQs-018|Now it's an absorption in the red region of the visible spectrum, a low energy red photon.
QoHZH6OkzQs-019|And oxygen, liquid oxygen will absorb red photons and it will appear blue.
QoHZH6OkzQs-020|Remember, things that absorb in the red will transmit to blues.
QoHZH6OkzQs-021|So liquid oxygen appears blue.
QoHZH6OkzQs-022|We can see that in the demonstration lab.
JXHl3rYqlHU-000|For an ideal gas, the product of the pressure and volume is constant for a fixed temperature.
JXHl3rYqlHU-001|What happens when we change the temperature?
JXHl3rYqlHU-010|Now, on this scale, as I go to zero in volume, that will represent an absolute zero, because I can't go negative in volume.
JXHl3rYqlHU-011|There's no such thing as negative volume.
JXHl3rYqlHU-012|So as I get to zero in volume, I reach a logical zero in temperature.
JXHl3rYqlHU-013|And we call that absolute zero because we can't measure a temperature on this scale below that temperature.
JXHl3rYqlHU-014|The Celsius temperature scale has a zero set at the freezing point of water.
JXHl3rYqlHU-015|And I can measure this zero on this temperature scale on the Celsius scale.
JXHl3rYqlHU-016|It turns out to be minus 273 Celsius.
JXHl3rYqlHU-017|So minus 273 Celsius is an absolute zero in temperature.
JXHl3rYqlHU-020|Our standard temperature and pressure will be 1 atmosphere of pressure and 0 Celsius, or 273 Kelvin.
JXHl3rYqlHU-021|The Kelvin unit, the Kelvin degree, and the Celsius degree are the same size.
JXHl3rYqlHU-022|The 0 is offset by 273 degrees.
JXHl3rYqlHU-023|Now, let's look at our temperature-volume relationship in a demonstration.
JXHl3rYqlHU-024|I can take a volume of gas and expose it to very low temperature.
JXHl3rYqlHU-025|Here's liquid nitrogen. This is at about 70 on the absolute temperature scale, 70 Kelvin.
JXHl3rYqlHU-026|As I expose a volume of gas to that very low temperature, the volume decreases.
JXHl3rYqlHU-027|And we understand that this volume is decreasing linearly.
JXHl3rYqlHU-028|As the temperature decreases, the volume decreases in a corresponding linear fashion.
JXHl3rYqlHU-029|I'm going to go down to very near, within 70 degrees of that absolute zero temperature.
JXHl3rYqlHU-030|That's where this liquid nitrogen is.
JXHl3rYqlHU-031|The volume of the balloon has gone very, very, very, very low.
JXHl3rYqlHU-032|Now, let's remove the balloon from liquid nitrogen and let it expand.
JXHl3rYqlHU-033|As the temperature increases, the volume of the gas increases.
JXHl3rYqlHU-034|The volume of the gas increases linearly as the temperature increases.
JXHl3rYqlHU-035|And you can see this volume comes back to near its original volume.
JXHl3rYqlHU-036|Linear relationship between temperature and volume.
JXHl3rYqlHU-042|Let's say I take a 1 mole sample of gas, and I look at just half of the mole, half of the gas, half a mole.
JXHl3rYqlHU-043|That half a mole of gas, those particles, they behave as if they have the whole volume to themselves.
JXHl3rYqlHU-044|And that's how gas particles always behave.
JXHl3rYqlHU-045|Gas particles don't interact with each other in the ideal gas sense.
JXHl3rYqlHU-046|So they behave like they have the whole volume.
JXHl3rYqlHU-047|Half a mole of a sample will exert half the pressure.
JXHl3rYqlHU-048|It'll occupy the whole volume, but exert half the pressure.
JXHl3rYqlHU-049|1/3 third of the particles will exert 1/3 of the pressure.
JXHl3rYqlHU-050|So we can look at fractional pressures and partial pressures in terms of partial number of particles.
fysCY3d9zAU-000|Equilibrium is the condition where products and reactants freely interchange.
fysCY3d9zAU-001|The free energy difference between the products and the reactants is 0, so there's no energy penalty to switch back and forth.
fysCY3d9zAU-002|And they do so freely.
fysCY3d9zAU-003|Now at equilibrium, the products or the reactants may be favored.
fysCY3d9zAU-005|So those two things are correlated.
fysCY3d9zAU-007|So if I looked at this in the standard state, I'd have 1 molar d and c, 1 molar a and b.
fysCY3d9zAU-008|And the free energy difference between those would be the standard free energy difference.
fysCY3d9zAU-009|Now that's not the equilibrium situation.
fysCY3d9zAU-010|Obviously, if I have 1 molar of everything, that's an arbitrary standard condition that I've chosen.
fysCY3d9zAU-013|K greater than 1 indicates products will be favored.
fysCY3d9zAU-014|K less than 1 indicates reactants will be favored.
fysCY3d9zAU-019|Products over reactants, if the product is larger than reactants, then that's greater than 1 in the ratio.
fysCY3d9zAU-020|And that's a K greater than 1.
fysCY3d9zAU-021|Products are favored.
fysCY3d9zAU-022|So if both of these favor products, spontaneous is the label we give them.
fysCY3d9zAU-023|Now notice that that label, spontaneous, means products are favored at equilibrium.
fysCY3d9zAU-024|It doesn't have anything to do with the rate of the reaction or how fast we get there.
fysCY3d9zAU-025|It just means that once we get to equilibrium, the products are favored.
fysCY3d9zAU-027|Those are called not spontaneous reactions or non spontaneous.
fysCY3d9zAU-028|Now just because a reaction is non spontaneous does not mean it doesn't form any products.
fysCY3d9zAU-029|It doesn't mean the reaction doesn't go.
fysCY3d9zAU-030|It just means when I get to equilibrium, it's the reactants that are favored.
fysCY3d9zAU-031|There will be some products, but the reactants are favored.
fysCY3d9zAU-034|And of course K would be equal to 1, because you would put in 1 for every pressure or every concentration, and you'd get a K equal to 1.
fysCY3d9zAU-036|They stay at 1.
fysCY3d9zAU-037|Standard free energy 0, K equal 1.
fysCY3d9zAU-040|K is less than 1, delta G greater than 0.
fysCY3d9zAU-041|K is greater than 1, delta G less than 0.
fysCY3d9zAU-042|So that behaves like a natural log function.
fysCY3d9zAU-043|Here, if you take the natural log of K, and you go to K is greater than one, that's positive.
fysCY3d9zAU-044|Put a negative sign in front of that, that would give you negative delta G's.
fysCY3d9zAU-045|So K's greater than 1 give you negative delta G's for this natural log function.
fysCY3d9zAU-046|And when you go to a fraction less than 1, the natural log function becomes negative.
fysCY3d9zAU-047|Multiply that times a negative, and you get a positive sign.
fysCY3d9zAU-048|So that accounts for this relationship.
fysCY3d9zAU-049|A K less than 1, a fraction, gives you a negative natural log and a positive delta G.
fysCY3d9zAU-053|So these two relationships allow us to switch between free energies and equilibrium constants analytically.
fysCY3d9zAU-055|So you can see the power of the thermodynamics as they relate to each other.
fysCY3d9zAU-056|Tables of standard state free energies give you equilibrium constants and tell you how a reaction is favored at equilibrium, products or reactants.
foJavVRkARU-000|Strong acids and strong base is completely dissociated in water.
foJavVRkARU-001|So if I take a mole of strong acid, HCl, hydrochloric acid, and I put it in water, I get a mole of H30 plus and Cl minus.
foJavVRkARU-002|That complete dissociation is what characterizes it as a strong acid.
foJavVRkARU-003|Some acids and bases don't completely dissociate, and we call those weak acids or weak bases.
foJavVRkARU-007|The acetate ion is actually H3CCOO minus.
foJavVRkARU-008|Just Ac minus here for convenience.
foJavVRkARU-014|So there's another difference between the strong acid and the weak acid.
foJavVRkARU-015|At equilibrium, the strong acid favors the products-- complete dissociation.
foJavVRkARU-016|The weak acid favors the reactants.
foJavVRkARU-017|You can look at it in terms of free energy.
foJavVRkARU-018|Since this K is less than one, the standard state free energy difference is positive.
foJavVRkARU-019|So this is a non-spontaneous reaction.
foJavVRkARU-020|Again, non-spontaneous does not mean it doesn't go at all.
foJavVRkARU-021|It just means at equilibrium the reactants are favored.
foJavVRkARU-022|For a strong acid, by contrast, the standard state free energy difference is negative.
foJavVRkARU-023|That's a spontaneous reaction.
foJavVRkARU-026|So this is a pKa of 4.7.
foJavVRkARU-027|And the smaller the pKa, the larger the Ka.
foJavVRkARU-028|So a small pKa means a relatively stronger acid.
foJavVRkARU-029|The smaller the pKa, the stronger the acid.
foJavVRkARU-030|Let's look at the conjugate base.
foJavVRkARU-034|The reverse of the acid reaction would be AC minus reacting with H3O plus.
foJavVRkARU-035|Here's AC minus reacting with H2O.
foJavVRkARU-036|So when I write the equilibrium constant for this, I'm writing a reaction with H2O and Ac minus.
foJavVRkARU-039|This is a weak base.
foJavVRkARU-040|This favors the reactants.
foJavVRkARU-041|You should also notice that Kas and Kbs, even though they're not reverse-- that is, Ka isn't one over Kb-- they are correlated.
foJavVRkARU-042|The larger the acid dissociation constant, the smaller the base dissociation constant.
foJavVRkARU-043|And that should be obvious because the stronger the acid, the more this equilibrium lies towards Ac minus in solution.
foJavVRkARU-044|So Ac minus should be favored in solution.
foJavVRkARU-045|That means this equilibrium constant must be small.
01ZxF6TcY-A-000|Let's talk about the photoelectric effect in terms of ChemQuiz.
01ZxF6TcY-A-001|If we shine a beam of light on a certain metal, it has no effect.
01ZxF6TcY-A-002|The question I have for you is, what change in that beam should I make in my best hope to eject electrons?
01ZxF6TcY-A-011|If you have a photon that won't eject an electron, what do you need?
01ZxF6TcY-A-012|You need photons with more energy.
01ZxF6TcY-A-013|You have to increase the energy of the photon.
01ZxF6TcY-A-014|Increasing the number of photons won't do.
01ZxF6TcY-A-015|You'll just strike a lot of electrons with a little bit of energy.
01ZxF6TcY-A-023|So here you could have answered B or C, and you would have got an electron ejected from that metal.
2hikObHZrXg-000|Let's try to sketch the titration curve for a polyprotic weak acid.
2hikObHZrXg-001|So I've chosen here histidine.
2hikObHZrXg-005|There are three of them, with pKa1, 1.81, pKa2, 6, and pKa3, 9.85.
2hikObHZrXg-006|The associated sites are as follows.
2hikObHZrXg-010|PKa2 is associated with this site here.
2hikObHZrXg-013|The final pKa of 9.85 is associated with this amino group here.
2hikObHZrXg-018|So here I've drawn the molecule again.
2hikObHZrXg-019|And I want to start by just talking about the first proton to be removed from solution.
2hikObHZrXg-021|So these two molecules, you noticed they're identical except for this proton has been removed.
2hikObHZrXg-022|And these two forms will be equal in concentration at pHs equal to the pKa of 1.81.
2hikObHZrXg-025|That's something I just defined.
2hikObHZrXg-026|I said, let's start at a low pH and go to a high pH.
2hikObHZrXg-027|So the solution is prepared at low pH and the initial pH will be less than 1.8.
2hikObHZrXg-028|And what I'm going to do is start adding base, raising the pH.
2hikObHZrXg-029|And as I get to pHs around the first pKa, I'll have an equal mixture of these two forms.
2hikObHZrXg-034|Now I can continue to add base.
2hikObHZrXg-035|And when I do, I'll go over to the equivalence point of the first proton.
2hikObHZrXg-036|So I'll completely remove the first proton.
2hikObHZrXg-037|Remember at half equivalent, I have equal concentrations of these.
2hikObHZrXg-038|As I keep adding base, I get more and more of the base form until all of the acid form is used up.
2hikObHZrXg-039|So at this point, the first equivalence point, all of the acid form, is converted into the base form.
2hikObHZrXg-040|But notice, it's only for this first titration, this first proton.
2hikObHZrXg-041|These other two protons are still attached.
2hikObHZrXg-046|So this second plateau occurs at pH 6.
2hikObHZrXg-047|So let's look at that.
2hikObHZrXg-050|The first proton stays gone, it's completely gone.
2hikObHZrXg-051|And now the second proton is being titrated.
2hikObHZrXg-055|I'll get to the second equivalence point.
2hikObHZrXg-056|At that second equivalence point, I've been converted to all this form.
2hikObHZrXg-057|All these protons are gone.
2hikObHZrXg-058|And that's the second endpoint, or the second equivalence point.
2hikObHZrXg-059|It's funny to say endpoint in a multiple titration-- equivalence point is certainly the better term.
2hikObHZrXg-060|So the second equivalence point, it's going to occur at less than pH 9, but more than pH 6.
2hikObHZrXg-061|So somewhere between 9 and 6, the two pKas, I'll have the second equivalence point.
2hikObHZrXg-062|At that second equivalence point, the form in solution is this.
2hikObHZrXg-063|That is, the second proton has been completely removed.
2hikObHZrXg-064|So let's continue.
2hikObHZrXg-068|Notice this proton is now being titrated, or this proton is being removed.
2hikObHZrXg-069|And when the pH is exactly equal to the pKa, there'll be an equal mixture of these two species in solution.
2hikObHZrXg-070|Now I can continue to the third endpoint.
2hikObHZrXg-071|And the third endpoint is where I've removed all of these protons, and the form in solution is exclusively completely deprotonated now.
2hikObHZrXg-072|And that pH will be above 9.8.
2hikObHZrXg-075|We've sketched out the titration curve.
2hikObHZrXg-076|And what we'd like to say is, at pH 8 which species are present in solution.
2hikObHZrXg-077|Well, pH 8 is between 6 and 10.
2hikObHZrXg-078|It's well above this initial pH.
2hikObHZrXg-079|So pKa1, that proton is certainly fully removed by the time we're at pH 8.
2hikObHZrXg-080|So this is a negative ion right now.
2hikObHZrXg-081|The other two species, we have to worry about.
2hikObHZrXg-082|Now this proton here, this won't be removed until we're above pH 9.9.
2hikObHZrXg-083|So that one is certainly going to be attached, so we don't have to worry about this pKa.
2hikObHZrXg-084|We just have to worry about this pKa6.
2hikObHZrXg-085|We're about two pH units above that.
2hikObHZrXg-086|So at two pH units above this pKa, what species is present in solution?
2hikObHZrXg-090|This proton either attached or removed.
2hikObHZrXg-091|The pKa is 6 for that, but we're at pH 8, so the difference between pH and pKa is 2.
2hikObHZrXg-092|So the difference is 2, so 2 has to be log of this concentration ratio.
2hikObHZrXg-095|So these two forms, about a 100 to 1.
2hikObHZrXg-096|So clearly, mostly the base form of this equilibrium is present at pH 8.
2hikObHZrXg-097|Now there's also some sodium ions from the sodium hydroxide that we've been adding.
2hikObHZrXg-098|But there are no hydroxide ions, per se.
2hikObHZrXg-099|Very, very low concentration of hydroxide ions because all the hydroxide ions have been used up in converting the acid form to the base form.
2hikObHZrXg-101|Those are the ions present in highest concentration at pH 8.
hOwQQryixJM-000|Let's look at the pi bonding orbitals in butadiene.
hOwQQryixJM-001|Can you determine which is the lowest energy pi bonding orbital?
hOwQQryixJM-002|A, B, or C?
hOwQQryixJM-013|We're looking for the lowest energy pi molecular orbital in butadiene, and we have three choices.
hOwQQryixJM-014|This first choice actually isn't a pi orbital.
hOwQQryixJM-015|It's a low energy orbital, if this were to form.
hOwQQryixJM-016|Because it's along the internuclear axis though, we would call it a sigma orbital.
hOwQQryixJM-017|This orbital has a node along the internuclear axis.
hOwQQryixJM-018|And that's common for pi orbitals.
hOwQQryixJM-019|They have internuclear axis nodes, and they have electron density above and below the axis.
hOwQQryixJM-020|So this is a pi orbital and this is a pi orbital.
hOwQQryixJM-021|And among these two, this has the lower number of nodes.
hOwQQryixJM-022|So it's the lower energy.
hOwQQryixJM-023|So in this case, B is the lowest energy pi molecular orbital among those choices.
nE480ADhZsY-000|The free energy change, or delta G, is what determines whether a process is favored.
nE480ADhZsY-001|Negative delta Gs indicate the favored direction for a chemical reaction.
nE480ADhZsY-002|A reaction is spontaneous if delta G is negative.
nE480ADhZsY-003|Delta G is composed of delta H and delta S and the temperature.
nE480ADhZsY-004|So we need to look at three factors to determine where our negative delta G, or favored reactions, occur.
nE480ADhZsY-010|Because a negative delta S makes the minus T delta S term positive.
nE480ADhZsY-011|Remember, temperatures are always positive.
nE480ADhZsY-012|So minus T delta S will be positive for negative delta S.
nE480ADhZsY-013|So if we look at the overall space and say, well, what about the regions where delta S is negative and delta H is positive?
nE480ADhZsY-017|No conditions of temperature will make that spontaneous.
nE480ADhZsY-018|Let's look at a couple other conditions.
nE480ADhZsY-020|Both of these favor delta G being negative.
nE480ADhZsY-021|Of course, exothermic, delta H is negative.
nE480ADhZsY-022|And an increase in entropy, that's a positive delta H times a positive temperature.
nE480ADhZsY-023|With this minus sign, this term will always be negative, and this term will always be negative.
nE480ADhZsY-024|So when delta S is positive and delta H is negative, that's favorable for all temperatures.
nE480ADhZsY-031|But you can overcome it as long as you have a high enough temperature and a positive delta S.
u2xEJdSJZOs-000|Now we can look at that trend in atomic radius in a little more detail.
u2xEJdSJZOs-003|And then I've gone from sodium to argon across the next row of the periodic table.
u2xEJdSJZOs-005|So lithium at 152 picometers, fluorine at 71.
u2xEJdSJZOs-006|We have a decrease as I go across.
u2xEJdSJZOs-007|And we understand that now.
u2xEJdSJZOs-008|We understand that we're going into the same principle quantum level, but we're adding more nuclear charge.
u2xEJdSJZOs-010|So those 2p orbitals that start out rather large at boron are contracting and getting smaller and smaller by the time I get over to fluorine.
u2xEJdSJZOs-011|That's because a bigger charge on the nucleus.
u2xEJdSJZOs-013|So we understand why ionization energy and atomic radii mirror each other in the trend because of the quantum mechanical structure of the atoms.
u2xEJdSJZOs-014|We can understand trends in the periodic table because of our understanding of quantum mechanics.
z24IYbHtphM-000|Matter absorbs or emits light based on its fine electronic structure.
z24IYbHtphM-001|That is, you have electrons behaving like waves, they're bounded in the matter.
z24IYbHtphM-003|Now let's take this.
z24IYbHtphM-004|Let's say a certain piece of matter has this energy level scheme that's shown here, and it has an emission spectrum.
z24IYbHtphM-005|Which of these three emission lines arises from the transition of high energy to low energy 3 to 1 emission in this system?
z24IYbHtphM-018|We're trying to get from the energy level spacings in some matter to the actual emission spectrum that we observe.
z24IYbHtphM-019|So if you look at the possible energy transitions, a transition between level 3 and 2 is possible.
z24IYbHtphM-020|That would be the lowest energy of the possible.
z24IYbHtphM-021|That's the smallest spacing here.
z24IYbHtphM-022|So that's the smallest energy.
z24IYbHtphM-025|The other transition from 3 to 1 is the highest possible for this system.
Yzz2eluxRYc-000|The natural direction for a process is determined by the overall increase in entropy.
Yzz2eluxRYc-001|If there's an overall increase in entropy between the system and the surroundings, then that's the favored direction for that process.
Yzz2eluxRYc-003|So what's the difference between a process that has a natural direction and a process that has equilibrium?
Yzz2eluxRYc-004|Both directions are equally favored.
Yzz2eluxRYc-005|Well, it's a balance between the entropy change.
Yzz2eluxRYc-008|I know if I go from the liquid to the gas, that's an increase in entropy.
Yzz2eluxRYc-009|I go from the relatively constrained liquid to the many microstates of the gas.
Yzz2eluxRYc-010|And of course, that's correct.
Yzz2eluxRYc-011|But the error there is we're only thinking about the system.
Yzz2eluxRYc-012|We need to think about the system and the surroundings to calculate the total entropy.
Yzz2eluxRYc-016|You know as you evaporate the gas, as you go from the liquid to the gas, that you have to absorb energy.
Yzz2eluxRYc-017|So the surroundings gives up some heat.
Yzz2eluxRYc-018|As that surrounding gives up heat to the system, the entropy of the surroundings decrease.
Yzz2eluxRYc-019|So system entropy increasing, surroundings entropy decreasing, the overall balance is 0.
Yzz2eluxRYc-020|And there's no net tendency for the reaction to go one direction or the other.
Yzz2eluxRYc-025|So a careful balance between the system and surrounding entropies can give you equilibrium situations.
Yzz2eluxRYc-026|And if you're very careful, that equilibrium can exist between the three phases.
Yzz2eluxRYc-027|And that's what happens at the triple point of water.
FiAyWh9kb9w-000|Our understanding of gases and expansions and contractions help us understand something about the atmosphere, which is a big layer of gas.
FiAyWh9kb9w-002|As you go up in altitude, the pressure decreases.
FiAyWh9kb9w-003|There is less atmosphere above you, fewer molecules, lower pressure.
FiAyWh9kb9w-006|Now warm gases near the surface of the Earth tend to rise.
FiAyWh9kb9w-007|And as they rise through that decreasing pressure they expand against that pressure.
FiAyWh9kb9w-008|That expansion against that pressure is an adiabatic process.
FiAyWh9kb9w-009|It's adiabatic because there's no heat.
FiAyWh9kb9w-010|There's no source of heat to return that energy it took to expand.
FiAyWh9kb9w-011|So that adiabatic process, an adiabatic expansion is a cooling process.
FiAyWh9kb9w-012|I use some of my energy to expand.
FiAyWh9kb9w-013|I can't get it back as heat, so the gas cools off.
FeJ1LQe6Zh4-000|Let's look at a cell potential of 0.5 volts for the oxidation of bromide by permanganate.
FeJ1LQe6Zh4-001|The question I have is what pH would cause this voltage?
FeJ1LQe6Zh4-011|We're looking at the oxidation of bromide by permanganate.
FeJ1LQe6Zh4-012|And we find that cell is operating at 0.5 volts.
FeJ1LQe6Zh4-013|So how can that happen?
FeJ1LQe6Zh4-014|Well, let's look back at the standard reaction.
FeJ1LQe6Zh4-017|Well, you could raise the reactant concentration.
FeJ1LQe6Zh4-018|Raising the reactant concentrations would increase the potential for the reaction to go.
FeJ1LQe6Zh4-019|So let's look at that.
OkoFU7hocpc-000|If I take hydrogen molecules and oxygen molecules and put them in a balloon, I can pretty much wait forever and they don't form water molecules.
OkoFU7hocpc-001|That reaction is favorable, but there is a barrier in between that prevents the rapid room temperature conversion.
OkoFU7hocpc-002|It turns out though that hydrogen and oxygen can be made to form water, release the same amount of energy at room temperature.
OkoFU7hocpc-003|You can do with a process called catalysis.
OkoFU7hocpc-004|Catalysis lowers this energy barrier.
OkoFU7hocpc-005|When you lower that energy barrier, that allows particles-- maybe even at room temperature kinetic energies-- to get over the barrier and form products.
OkoFU7hocpc-008|So catalysis lowers the energy barrier between products and reactants.
OkoFU7hocpc-009|It allows a low road path between the products and reactants.
OkoFU7hocpc-010|And allows reactions to go at much lower temperatures and much more quickly.
wcMCDbnuNms-000|When an electron is bound around an atom, an electron has wave-like properties.
wcMCDbnuNms-001|It's bounded, and when you bounce something with a wave-like property, you naturally get quantization, different energy levels.
wcMCDbnuNms-002|Now, our energy levels are described by wave functions with three quantum numbers-- n, l, and m sub l.
wcMCDbnuNms-003|And there's a myriad of possibilities.
wcMCDbnuNms-004|So what we want to do is catalog them and see which energy levels an electron will exist in about an atom.
wcMCDbnuNms-005|So let's start with the n equal 1 state.
wcMCDbnuNms-006|n gives the total energy of the system.
wcMCDbnuNms-007|Now, you can also have m equal 2 and keep on going, up to the ionized state, n equal infinity.
wcMCDbnuNms-011|The lowest energy orbital about an atom is n equal 1, and the only possible value of l is s.
wcMCDbnuNms-013|And m sub l is also 0.
wcMCDbnuNms-014|And the only possibility is 0, so we just leave that off.
wcMCDbnuNms-015|If you go to n equal 2, you can also have an s.
wcMCDbnuNms-016|So n equals 1 and n equals 2 both give rise to a spherically symmetric s orbital.
wcMCDbnuNms-017|But when n equals 2, you can also have l equal 1.
wcMCDbnuNms-018|l equal 1 has three values of m sub l-- m sub l minus 1, 0, and 1.
wcMCDbnuNms-019|These are the three equivalent p orbitals.
wcMCDbnuNms-023|Now, we can continue and go to principal quantum level 3.
wcMCDbnuNms-024|And a 3 s orbital is also possible.
wcMCDbnuNms-025|It'll also be spherically symmetric.
wcMCDbnuNms-027|So the number of nodes increase, but the overall spherical symmetry remains the same.
wcMCDbnuNms-028|And of course, when n equals 2, you can still have l equal 1.
wcMCDbnuNms-029|So you'll have a set of 3s orbitals-- 3px, 3py, and 3pz.
wcMCDbnuNms-030|But when n equals 3, you can also have l equal 2.
wcMCDbnuNms-032|That gives rise to a set of five different shape and orientation, but equal in energy orbitals.
wcMCDbnuNms-033|And I have some models of them right here.
wcMCDbnuNms-034|So one, two, three, four, five d orbitals.
wcMCDbnuNms-036|And again, when you look in your textbook, these will be given geometric notations-- z squared, x squared minus y squared, et cetera, rather than the m sub l integer values.
wcMCDbnuNms-037|We do that for purely geometric reasons.
wcMCDbnuNms-038|The important thing is there's five equivalent orbitals in a set of d orbitals.
wcMCDbnuNms-039|So you can see an electron can exist in a myriad of different states around the atom.
wcMCDbnuNms-040|And when you have more than one electron they need to share these states.
wcMCDbnuNms-041|So that's what we'll look at next.
ER7lUmsy9m4-000|So let's do that phosphorus oxide reaction in a vessel where we have a fixed amount of oxygen and we add differing initial amounts of phosphorus.
ER7lUmsy9m4-010|You add a little bit more phosphorus, you'll produce a little bit more phosphorus oxide.
ER7lUmsy9m4-011|You've used a little bit more of your oxygen.
ER7lUmsy9m4-013|You cannot produce any more phosphorus oxide.
ER7lUmsy9m4-014|Oxygen limits the reaction.
ER7lUmsy9m4-015|It's called the limiting reagent.
ER7lUmsy9m4-016|So, product will build up until all the oxygen is consumed.
lItaSuF6Uqg-000|Let's do a calculation where we calculate ionization energy.
lItaSuF6Uqg-001|We're going to look at a system where we bathe rubidium atoms in ultraviolet light.
lItaSuF6Uqg-002|That will eject electrons and those electrons will travel off with a certain kinetic energy, and we'll measure that.
lItaSuF6Uqg-003|So knowing the kinetic energy of the electrons and the amount of energy from the photons of light, we can calculate the kinetic energy.
lItaSuF6Uqg-005|Let's do a calculation where we calculate ionization energy.
lItaSuF6Uqg-006|What we're going to do is we're going to take rubidium atoms and bathe them in ultraviolet light.
lItaSuF6Uqg-007|That ultraviolet light will ionize the rubidium atoms.
lItaSuF6Uqg-008|Electrons will leave with a certain kinetic energy.
lItaSuF6Uqg-009|We can measure that kinetic energy.
lItaSuF6Uqg-010|Now the ultraviolet radiation that comes in has to do two things.
lItaSuF6Uqg-011|Some of the energy goes into ionizing the rubidium atoms.
lItaSuF6Uqg-012|The rest of the energy goes into the kinetic energy.
lItaSuF6Uqg-013|So I know the sum of the kinetic energy and the ionization energy is the photon energy.
lItaSuF6Uqg-017|So the photon is split into two things, ionizing plus kinetic energy.
lItaSuF6Uqg-020|Now this is called a photo electron spectroscopy experiment.
lItaSuF6Uqg-021|Ultraviolet radiation measure the kinetic energy of electrons emitted.
lItaSuF6Uqg-022|So I know that kinetic energy and the ionization energy add to give the photon energy.
lItaSuF6Uqg-023|So I can calculate the ionization energy from the photon energy and the kinetic energy.
lItaSuF6Uqg-024|Kinetic energy of electron at 2,450 kilometers per second, we've done these kind of things before.
lItaSuF6Uqg-025|1/2 half times the mass of the electron times the velocity squared.
lItaSuF6Uqg-026|And I write the velocity in meters per second to always operate in kilometers, excuse me, meters, kilograms, and seconds.
lItaSuF6Uqg-029|So from that and the photon energy, I can calculate the ionization energy.
lItaSuF6Uqg-030|So the photon energy, I'll cast in terms of the wavelength because that's the parameter I have.
lItaSuF6Uqg-032|And now I can calculate.
P_e4n9XSjUw-000|Particles, electrons, small atoms have wave-like properties.
P_e4n9XSjUw-001|Let's look at that in terms of a chem quiz.
P_e4n9XSjUw-003|The sodium atom, impacted by the wavelength of light, behaving like a particle, behaving like a photon.
P_e4n9XSjUw-004|About how many will stop that?
P_e4n9XSjUw-005|About 1, about 100, about 10,000?
P_e4n9XSjUw-014|Matter has both wave and particle characteristics.
P_e4n9XSjUw-015|We've seen the momentum and the wavelength are related by the de Broglie relationship, and we calculated it for several different objects.
P_e4n9XSjUw-016|Now we're saying, well, a sodium atom is going to be traveling at about 300 meters per second.
P_e4n9XSjUw-017|It's going to encounter photons of 600 nanometer wavelength.
P_e4n9XSjUw-018|About how many photons have to strike those sodium atoms to get them to slow down, and about stop?
P_e4n9XSjUw-021|Those photons of yellow light coming in at 600 nanometers.
P_e4n9XSjUw-022|We can say, well, I want the momenta of these two systems to be equal.
P_e4n9XSjUw-023|I want to transfer enough momenta from these waves to stop the particle.
P_e4n9XSjUw-024|So the wave has a particle nature, it has a tiny little momentum.
P_e4n9XSjUw-025|I need to transfer it to this larger momentum.
P_e4n9XSjUw-026|And here comes the sodium atom, I want to bap, bap, bap, bap, bap, bap, bap, bap, bap.
P_e4n9XSjUw-027|Keep hitting it with photons until-- bap, bap, bap, bap, bap, bap-- that sodium about stops.
P_e4n9XSjUw-028|And I can do that by just calculating the momenta of each.
P_e4n9XSjUw-029|And as I do so, I see that the momenta are related by their wavelength by the de Broglie relationship.
P_e4n9XSjUw-033|The experimenters who did this were awarded a Nobel Prize for what's called laser cooling.
P_e4n9XSjUw-035|And then using this additional method, using lasers to trap the atoms, and bounce photons off them 'til they virtually come to a stop.
P_e4n9XSjUw-036|The lowest temperatures ever achieved by what's called laser cooling.
P_e4n9XSjUw-037|The correct answer for our laser cooling experiment is C, about 10,000 yellow photons to stop that sodium atom.
cy4vUW9x62s-003|But don't be fooled.
cy4vUW9x62s-014|I have those two here.
cy4vUW9x62s-015|Here's the tetrahedral arrangement and the trigonal bipyramidal arrangement.
cy4vUW9x62s-016|But remember, when we're talking about the actual molecular geometry, we ignore the lone pairs.
cy4vUW9x62s-017|So for these two molecules here, there are two lone pairs.
cy4vUW9x62s-018|So you have the tetrahedral geometry, but it's as if you're going to ignore two of the arms.
cy4vUW9x62s-019|If you ignore two of the arms, you get a molecule that just looks bent.
cy4vUW9x62s-020|And that's what we say here.
cy4vUW9x62s-022|That angle is going to be something like 104 degrees, 105.
cy4vUW9x62s-023|104 and 1/2, two lone pairs, and two bonding pairs in a tetrahedral configuration generally gives you about 105 degrees.
cy4vUW9x62s-024|For our ICl2, we have five things.
cy4vUW9x62s-025|Three of them along the equatorial positions are lone pairs, so we can ignore those.
cy4vUW9x62s-026|Taking off the three lone pairs in the equatorial position just leaves us the two axial chlorines.
cy4vUW9x62s-028|So of these three, ICl2- is the linear ion.
3WNdFxNR9tY-001|So I'll mix solutions 1 and 5, equal volumes, equal concentration.
3WNdFxNR9tY-002|So 0.1 molar HCl with an equal volume of 0.1 molar NaOH.
3WNdFxNR9tY-003|Can you predict will the solution be acidic, neutral, or basic?
3WNdFxNR9tY-015|We're going to mix a strong acid and a strong base and try to predict will it be acidic, neutral, or basic and the ionic strength.
3WNdFxNR9tY-016|So let's do the experiment.
3WNdFxNR9tY-017|Here I have my strong acid or my strong base, and I'm going to mix it with my strong acid.
3WNdFxNR9tY-018|Now I think we can tell already from the color that the pH is going to be about neutral.
3WNdFxNR9tY-019|It's very similar to my neutral water.
3WNdFxNR9tY-020|And that's true.
3WNdFxNR9tY-022|And a salt solution, sodium chloride solution, I think you understand is neutral.
3WNdFxNR9tY-023|Neither the sodium nor chloride ions contribute H3O pluses or OH minuses.
3WNdFxNR9tY-024|Neither of them is a good acid or a base.
3WNdFxNR9tY-025|So the pH is simply the pH of neutral water.
3WNdFxNR9tY-026|What about the ionic strength?
3WNdFxNR9tY-027|Well, to do that, we can insert our ionic tester, our light bulb, and see the strength of the light intensity.
3WNdFxNR9tY-028|I'll put this in and bright light.
3WNdFxNR9tY-029|So that's interesting.
3WNdFxNR9tY-030|We mix the strong acid and the strong base, and we got a bright light.
3WNdFxNR9tY-031|Why is that?
3WNdFxNR9tY-032|Because this salt that's produced, sodium chloride, completely ionizes to form sodium and chlorine ions.
3WNdFxNR9tY-033|They stay in solution at a neutral pH but give you a bright light.
3WNdFxNR9tY-034|So in this case, the answer neutral pH, relatively strong ionic strength-- lots of ions in solution, sodium and chloride.
3WNdFxNR9tY-035|The answer's B and A.
M6AdsBPkJSU-000|Let's look at making a buffer solution.
M6AdsBPkJSU-001|Which of the following when added to NH3, ammonia, the base, forms a basic buffer at pH greater than 7?
M6AdsBPkJSU-002|Is it the strong base, sodium hydroxide, sodium chloride, the salt, or the strong acid, hydrochloric acid, HCl.
M6AdsBPkJSU-009|We're talking about making a buffer from NH3, ammonia, the weak base.
M6AdsBPkJSU-010|Now to make a buffer, you need a base and it's conjugate acid or an acid in its conjugate base.
M6AdsBPkJSU-011|They need to be in about equal concentrations.
M6AdsBPkJSU-012|So if I'm talking about NH3, I need it's conjugate acid.
M6AdsBPkJSU-013|It's a base already.
M6AdsBPkJSU-014|I need it's conjugate acid.
M6AdsBPkJSU-015|So to form it's conjugate acid, I would add a strong acid and convert some of my NH3 to NH4 plus.
M6AdsBPkJSU-016|That would form the buffer.
M6AdsBPkJSU-018|So the pKa of the acid NH4 plus is 9.3.
M6AdsBPkJSU-019|So I'll form a buffer of NH4 plus and NH3, the ammonium ion and ammonia.
M6AdsBPkJSU-020|The pH will be around 9.3.
M6AdsBPkJSU-021|And in order to form it, I actually have to add a strong acid to convert some of my base to the acid form.
M6AdsBPkJSU-022|So in this case, the correct answer is I add some HCl to get equal concentrations of an acid and it's conjugate base.
DCcIg-TtH7U-000|Let's look at the ionization of helium plus.
DCcIg-TtH7U-001|Helium plus is a one-electron system like hydrogen.
DCcIg-TtH7U-002|The question I have is, can a 30 nanometer photon ionize He+ from its ground state?
DCcIg-TtH7U-003|And I give you this information about hydrogen that we've been talking about.
DCcIg-TtH7U-004|The 1 to 2 transition in hydrogen is in the ultraviolet at 120 nanometers.
DCcIg-TtH7U-005|The question is, given that, can we deduce whether helium plus, also a one-electron system, will be ionized?
DCcIg-TtH7U-010|So we're kind of doing a little spectroscopy here.
DCcIg-TtH7U-011|We have a transition in hydrogen. It's 120 nanometers.
DCcIg-TtH7U-012|And we do the calculation, and we say, well, that transition energy is about 3/4 of a Rydberg.
DCcIg-TtH7U-013|That's the final state minus the initial state-- R minus 1/4R is 3/4R.
DCcIg-TtH7U-014|So 120 nanometer photon corresponds to about 3/4 of a Rydberg.
DCcIg-TtH7U-015|We have a 30 nanometer photon, which is four times as energetic.
DCcIg-TtH7U-018|It's z squared over n squared.
DCcIg-TtH7U-019|So z is 2 for helium.
DCcIg-TtH7U-020|So it's minus 4 Rydbergs.
DCcIg-TtH7U-021|So a photon with an energy of about 3 Rydbergs is not sufficient to ionize He+, which requires an energy of about 4 Rydbergs.
DCcIg-TtH7U-022|So not enough energy in a 30 nanometer photon to ionize helium plus.
VL_TnSkqNYY-000|Let's look at the equilibrium between NO2, a brown gas, and N2O4, a clear gas.
VL_TnSkqNYY-001|Let's say I start, as I have it here, with a bulb of the equilibrium mixture.
VL_TnSkqNYY-002|I open a valve and let it expand to twice its volume.
VL_TnSkqNYY-003|Now, initially that I do that, obviously it'll get lighter because it'll expand and be less dense but then the new equilibrium will be established.
VL_TnSkqNYY-011|What I'm going to do here is write the expression for Q for that chemical reaction, that dimerization.
VL_TnSkqNYY-012|The products, N2O4, over the reactants squared, because the stoichiometric coefficient of the NO2 was two.
VL_TnSkqNYY-013|So if I double the volume for this, what that will do is reduce the pressures instantaneously by a factor of two.
VL_TnSkqNYY-018|So Q will be larger than K, it will be 2 times bigger than the equilibrium constant and I want to go back towards equilibrium.
VL_TnSkqNYY-019|So I need to make Q smaller.
VL_TnSkqNYY-020|It's two times too big, I need to make it smaller.
VL_TnSkqNYY-021|How do I make it smaller?
VL_TnSkqNYY-022|Well, I have to shift back to reactants.
VL_TnSkqNYY-023|The products are on top, making the products bigger would make Q larger.
VL_TnSkqNYY-024|Making products smaller and reactants bigger would make Q smaller.
VL_TnSkqNYY-025|And I want Q to get smaller to go back towards K so I'll shift back toward the reactants and the reactants or the brown gas.
VL_TnSkqNYY-026|So I'll get darker as I shift towards equilibrium, in this case.
VL_TnSkqNYY-027|The correct answer here is increase.
D4R7ho3YnSo-000|Let's calculate the pH at a specific point on the titration curve.
D4R7ho3YnSo-002|And we've actually done this before, but let's go through the calculation.
D4R7ho3YnSo-009|And solve for x.
D4R7ho3YnSo-010|Now, to solve for x the easiest thing to do is assume that x is small with respect to 0.1.
D4R7ho3YnSo-011|And I do that because I see that k is a relatively small number.
D4R7ho3YnSo-012|And if k is small, then this reaction doesn't go very far an x's should be small.
D4R7ho3YnSo-013|I should favor this side, the 0.1 side.
D4R7ho3YnSo-014|So if only a small fraction of those 0.1 initial moles react, x will be small.
D4R7ho3YnSo-016|x is the H3O plus concentration.
D4R7ho3YnSo-017|So I can calculate the pH directly from that.
D4R7ho3YnSo-018|The pH is about 2.88.
D4R7ho3YnSo-020|I haven't done any base.
D4R7ho3YnSo-021|The pH can be calculated just as if it were a weak acid solution.
D4R7ho3YnSo-022|Weak acid solution of acetic acid, the initial pH of 0.1 molar acetic acid has an initial pH of 2.88.
-fHWcbYcLeE-000|When you react phosphorous with oxygen, you get a brilliant white light and the production of a phosphorus oxide.
-fHWcbYcLeE-001|Let's look at that in terms of a ChemQuiz.
-fHWcbYcLeE-002|31 grams of phosphorus will react with excess oxygen to form 71 grams of a phosphorus oxide.
-fHWcbYcLeE-003|The question I have for you is, what's the empirical formula of that phosphorus oxide?
-fHWcbYcLeE-011|When the phosphorus and oxygen react to form the phosphorus oxide, 31 grams of phosphorus in excess oxygen produce 71 grams of the phosphorus oxide.
-fHWcbYcLeE-012|So that difference of 40 grams must have been the oxygen.
-fHWcbYcLeE-013|So it's 31 grams of phosphorus react with 40 grams of oxygen for a mass ratio of 3 to 4.
-fHWcbYcLeE-014|But when we write chemical formulas, we do not use mass ratios.
-fHWcbYcLeE-015|We use mole ratios.
-fHWcbYcLeE-019|We could denote that with the subscript 2.5, but we like to have integer subscripts in our molecular formulas.
-fHWcbYcLeE-020|So let's multiply through by 2 and use the mole ratios 2 to 5.
-fHWcbYcLeE-021|So the empirical formula of this phosphorus oxide is P2O5.
-fHWcbYcLeE-022|The correct answer is A, P2O5.
-fHWcbYcLeE-023|It turns out the actual molecule is P4O10.
CNAU31Q7cJg-001|We can calculate the energy difference and calculate the wavelength of the photon that's emitted.
CNAU31Q7cJg-002|So the energy levels go as minus z squared over n squared times the Rydberg constant.
CNAU31Q7cJg-004|We can put in all those numbers.
CNAU31Q7cJg-005|Here is z, we're talking about a hydrogen atom, so the charge on the nucleus z is 1.
CNAU31Q7cJg-006|The two states from the n equal 2 to n equal 4 state.
CNAU31Q7cJg-007|I'm doing a transition where it's emission.
CNAU31Q7cJg-008|I'm starting out in 4 and going to the n equal 2 state.
omeeRx3Zl84-000|Now let's write the electronic configuration of rubidium atoms.
omeeRx3Zl84-001|And give all the quantum numbers of the outermost electron that was ejected in that photo electron spectroscopy experiment.
omeeRx3Zl84-002|So when you're writing electronic configurations you need to know the total number of electrons.
omeeRx3Zl84-003|And in a neutral atom, the total number of electrons is equal to the positive charges, the number of neutrons, the atomic number of the atom.
omeeRx3Zl84-004|Rubidium atomic number 37, 37 protons in its nucleus.
omeeRx3Zl84-005|It has 37 electrons in its neutral state.
omeeRx3Zl84-006|Now, we're going to take all the possible orbitals.
omeeRx3Zl84-007|And we're going to write them out in kind of a diagrammatic way here.
omeeRx3Zl84-008|Remember, as you add electrons, the energy levels of the orbitals shift around a bit.
omeeRx3Zl84-009|So there's a little device, you can use to remember the actual order, which energy level is the highest.
omeeRx3Zl84-010|So you write them all out like this, the 1s, the 2s, all the threes, all the fours, all the fives, and you can keep going, sixes et cetera.
omeeRx3Zl84-011|We won't need that many orbitals for rubidium.
omeeRx3Zl84-012|And then you make diagonal lines.
omeeRx3Zl84-013|This is the order that the electrons fill in.
omeeRx3Zl84-014|It's the energy ordering of orbitals when there are many electrons present.
omeeRx3Zl84-015|So 1s is the first orbital to fill, then 2s.
omeeRx3Zl84-016|And you keep running diagonal lines.
omeeRx3Zl84-017|Then the 2p, then the 3s.
omeeRx3Zl84-019|That just goes by principle quantum level.
omeeRx3Zl84-020|Then the 3p, then the 4s, then the 3d, then the 4p, then the 5s.
omeeRx3Zl84-021|And that would continue.
omeeRx3Zl84-022|You would do 4d, 5p, 6s.
omeeRx3Zl84-024|So now we've to do is go through that and count up to 37 electrons.
omeeRx3Zl84-025|So we can do that 1s is the first orbital we'll fill.
omeeRx3Zl84-026|And the 1s will hold two electrons, so 1s, two.
omeeRx3Zl84-027|We can keep total electron count, so we remember when we get to 37.
omeeRx3Zl84-028|So far we have two electrons.
omeeRx3Zl84-029|Next we'll fill the 2s.
omeeRx3Zl84-030|It also holds two, bringing our count to four.
omeeRx3Zl84-031|Then we fill the 2p and the 3s.
omeeRx3Zl84-032|The 2p hold six electrons.
omeeRx3Zl84-033|Remember, three equivalent p orbitals, each holding two electrons is six.
omeeRx3Zl84-034|Brings our total count to 10.
omeeRx3Zl84-035|Then the 3s we'll fill with it two bringing our count to 12.
omeeRx3Zl84-036|The 3p and the 4s we'll fill.
omeeRx3Zl84-037|3p holds six.
omeeRx3Zl84-038|The 4s holds two.
omeeRx3Zl84-039|20 total electrons.
omeeRx3Zl84-040|Now, rather than fill the 4p, we're going to fill the 3d next.
omeeRx3Zl84-043|Then the 4p accommodates six electrons.
omeeRx3Zl84-044|Then the 5s can accommodate two, but we're already up to 36.
omeeRx3Zl84-045|All we need is 37.
omeeRx3Zl84-047|Quite a mouthful of electronic configuration.
omeeRx3Zl84-048|What are the quantum numbers for that outermost electron.
omeeRx3Zl84-049|Well, the quantum numbers we generally write as n, l, m sub l, and m sub s.
omeeRx3Zl84-050|So n for the outermost electron is five.
omeeRx3Zl84-051|It's in the fifth principle quantum level, so five.
omeeRx3Zl84-052|L is zero because its an s orbital.
omeeRx3Zl84-053|Remember, the designation for l equals zero is s.
omeeRx3Zl84-054|So we'll change that s to a zero for l.
omeeRx3Zl84-055|M sub l will be zero because that's the only possible value of m sub l for l equals zero.
omeeRx3Zl84-059|It can either be in the plus one half or the minus one half state.
omeeRx3Zl84-060|So there are our electronic configuration and our quantum numbers for the outermost electron for rubidium atoms.
AlwmlHuZRUU-000|When we make molecular orbitals from atomic orbitals we want to make molecular orbitals that have the right geometry for the molecular problem we're solving.
AlwmlHuZRUU-001|If we have a tetrahedral molecule, or a linear molecule.
AlwmlHuZRUU-002|So what we need to do is look at our atomic orbitals and see if they satisfy that.
AlwmlHuZRUU-005|Two orbitals 180 degrees from each other.
AlwmlHuZRUU-006|How do we do that?
AlwmlHuZRUU-007|Well, let's start by taking the s and the p orbital, the s and the pz for instance, from an atom.
AlwmlHuZRUU-008|If I take the s and the pz, that's two atomic orbitals.
AlwmlHuZRUU-012|Now again, I'm adding, so green areas are positive amplitude.
AlwmlHuZRUU-013|They'll get bigger.
AlwmlHuZRUU-014|Red areas are negative amplitude.
AlwmlHuZRUU-015|So when I add red and green, I should see a decrease in amplitude in those areas.
AlwmlHuZRUU-016|So adding those two together gives me an orbital that looks like this.
AlwmlHuZRUU-017|The green and the green added, giving me higher amplitude.
AlwmlHuZRUU-018|And the green and the red added, giving me lower amplitude.
AlwmlHuZRUU-019|And that's a new orbital.
AlwmlHuZRUU-020|It's a combination of an s and a p.
AlwmlHuZRUU-021|We'll call it an sp atomic orbital.
AlwmlHuZRUU-022|So that sp atomic orbital is going to have a pair that's only one.
AlwmlHuZRUU-023|I started with two atomic orbitals.
AlwmlHuZRUU-024|I need to create two new ones.
AlwmlHuZRUU-025|So I can take the opposite.
AlwmlHuZRUU-026|I can take the difference of those two.
AlwmlHuZRUU-027|And now I'll have high amplitude on this side.
AlwmlHuZRUU-028|This side will become green, this side will become red.
AlwmlHuZRUU-029|And I'll have high amplitude on this side, 180 degrees from this high amplitude.
AlwmlHuZRUU-030|This is a combination of s minus the pz.
AlwmlHuZRUU-036|I still have my remaining p orbitals that I didn't use.
AlwmlHuZRUU-037|I used the pz, so the px and the py are still available.
AlwmlHuZRUU-038|But I have a new set, two sps, and a px, and a py.
AlwmlHuZRUU-039|Those are the atomic orbitals that are appropriate for a 180 degree bond angle.
3CHic39yxjs-000|Let's look at a couple of filters and how they'll affect a blue object.
3CHic39yxjs-001|So three possible filters put in front of a blue object.
3CHic39yxjs-009|So we've got a blue object.
3CHic39yxjs-010|We're going to put filters between the blue object and us.
3CHic39yxjs-011|And we're going to see what we see.
3CHic39yxjs-012|So a blue object, wavelengths of blue light are being emitted from that object.
3CHic39yxjs-013|That's why it appears blue.
3CHic39yxjs-014|How do we make it appear black?
3CHic39yxjs-016|So we need to find a filter that absorbs strongly in the blue.
3CHic39yxjs-017|And that is filter number two.
SKDF7sz_wgI-000|Let's see if we can deduce an atomic radius from the information we have so.
SKDF7sz_wgI-001|Far we haven't really talked about the trend in atomic radius.
SKDF7sz_wgI-002|We've talked about ionization energy and electron affinity.
SKDF7sz_wgI-003|But you might be able to surmise how the atoms go based on our other atomic trends.
SKDF7sz_wgI-004|So let's take sulfur, chlorine, and potassium and ask, which has the lowest atomic radius?
SKDF7sz_wgI-008|We're trying to deduce atomic radius from other properties that have periodic trends that we understand.
SKDF7sz_wgI-009|So we're taking sulfur, chlorine, and potassium-- three atoms.
SKDF7sz_wgI-026|So for smallest atomic radii, your best bet, chlorine.
JIKXNx5TK3Q-000|Let's look at two titrations-- the same acid, but tenfold diluted.
JIKXNx5TK3Q-001|So I'm going to take the weak acid, acetic acid, and titrate it with a strong base, potassium hydroxide.
JIKXNx5TK3Q-002|That is plotted here in the dotted yellow line.
JIKXNx5TK3Q-003|Then I'm going to dilute by a factor of 10 and reperform the titration.
JIKXNx5TK3Q-011|We're looking at titrating HAc, acetic acid, two times.
JIKXNx5TK3Q-012|One titration, then dilute by a factor of 10, and titrate again.
JIKXNx5TK3Q-013|Now, if you dilute by a factor of 10, the weak acid, initially, the pH will change by one unit because the H3O+ concentration changes by a factor of 10.
JIKXNx5TK3Q-014|But remember, in a weak acid, there's plenty of undissociated HAc around that will dissociate after that dilution and lower the pH.
JIKXNx5TK3Q-017|So the total amount of acid is the same, so the total amount of base required to reach equivalence point is the same.
JIKXNx5TK3Q-018|In the buffer region, remember, buffers are also resistant to changes in dilution.
JIKXNx5TK3Q-019|You dilute a buffer by a factor of 10, but the ratio of acid to base is what's important in a buffer.
DGTtekW_2J8-000|Let's look at a point on the titration curve.
DGTtekW_2J8-002|At point C, I've added one mole of my strong base for every mole of weak acid that I originally had.
DGTtekW_2J8-004|So what you have here is a solution of weak base.
DGTtekW_2J8-005|So you can calculate the pH here at point C by using a calculation of a weak base.
DGTtekW_2J8-006|Let's do that.
DGTtekW_2J8-007|Here's the weak base, in this case the conjugate base of acetic acid, reacting with water, because these are the two species left at the endpoint of the titration.
DGTtekW_2J8-008|They form the conjugate acid and OH minus.
DGTtekW_2J8-009|So at this point in the titration, we've titrated a weak acid with a strong base.
DGTtekW_2J8-010|At the endpoint, where I've used up all the acid, the solution is actually slightly basic.
DGTtekW_2J8-011|And it's slightly basic because I've converted all of the weak acid into its conjugate base.
DGTtekW_2J8-014|And I say, well, I have approximately 0.1 moles per volume.
DGTtekW_2J8-015|And the volume doesn't change much during a titration, because we can say we still have about 0.1 molar Ac ions.
DGTtekW_2J8-016|A little of that reacts to go back and form HAc and some OH minus.
DGTtekW_2J8-022|If I know HAc, I know Ka times Kb is Kw.
DGTtekW_2J8-023|So I can calculate a Kb.
DGTtekW_2J8-024|Kb is small.
DGTtekW_2J8-025|Kb is small, so I'll assume that 0.1 minus x is essentially 0.1, that x is small compared to 0.1.
DGTtekW_2J8-026|When I do that, the math gets relatively easy.
DGTtekW_2J8-027|x squared is 0.1Kb.
DGTtekW_2J8-028|So x is 7.5 times 10 to the minus 6.
DGTtekW_2J8-029|Now, x is the OH minus concentration.
DGTtekW_2J8-030|If I want the pH, I need the H3O plus concentration.
DGTtekW_2J8-031|And of course, I can get that because H3O plus and OH minus in water are always a product to 10 to the minus 14.
DGTtekW_2J8-032|So H3O plus times OH minus is 10 to the minus 14th all the time.
DGTtekW_2J8-033|That's an equilibrium that's holding.
DGTtekW_2J8-034|At the same time, this equilibrium is holding.
DGTtekW_2J8-035|So we satisfy them both at the same time.
DGTtekW_2J8-037|And in this case, the pH, 8.88.
DGTtekW_2J8-039|And if you have just weak base, the pH will be slightly basic.
w_9DNwm1DFI-000|Let's look at the amino acid cysteine at a specific pH.
w_9DNwm1DFI-001|So here I've drawn cysteine, and I've shown you the pKa's of each group.
w_9DNwm1DFI-002|So the pKa of this proton, the pKa of one of these protons, and the pKa of this proton.
w_9DNwm1DFI-003|The question is at pH 5, what's the overall charge?
w_9DNwm1DFI-004|So what are the protonation states?
w_9DNwm1DFI-005|Is the overall charge plus 1, 0, or minus 1?
w_9DNwm1DFI-012|We're looking at the amino acid cysteine at pH 5.
w_9DNwm1DFI-014|So, for instance, here the pKa is 9, so the pH is less than the pKa.
w_9DNwm1DFI-015|If the pH is less than the pKa, you're on the acidic side, so the acidic form predominates.
w_9DNwm1DFI-016|In fact, you're four units away from this.
w_9DNwm1DFI-017|So you're four units to the acidic side, so this is all in its acidic form.
w_9DNwm1DFI-019|Now here, pKa 8.
w_9DNwm1DFI-020|Again, the pH is 5, well to the acidic side.
w_9DNwm1DFI-024|Here, pKa 2.
w_9DNwm1DFI-025|This is the carboxylic acid part of the molecule.
w_9DNwm1DFI-026|The pKa is 2.
w_9DNwm1DFI-027|Now, I am on the basic side.
w_9DNwm1DFI-028|The pH is 5.
w_9DNwm1DFI-029|The pKa is 2.
w_9DNwm1DFI-030|So I'm at a pH above the pKa.
w_9DNwm1DFI-031|On the basic side, so the basic form predominates.
w_9DNwm1DFI-032|The basic form is the form without the proton.
w_9DNwm1DFI-036|So in this case, the overall charge is 0.
w_9DNwm1DFI-037|Minus 1, plus 1, and 0 all sum to 0.
nbeE9b2JV5Q-000|Let's look at dissolving a solid.
nbeE9b2JV5Q-001|We'll have magnesium hydroxide, and we'll put it in water.
nbeE9b2JV5Q-003|Now, I'd like that to dissolve.
nbeE9b2JV5Q-004|What's the best strategy for dissolving that additional speck of magnesium hydroxide?
nbeE9b2JV5Q-005|Should I add 0.1 molar H2SO4, sulfuric acid?
nbeE9b2JV5Q-006|0.01 molar sodium hydroxide?
nbeE9b2JV5Q-014|We're dissolving magnesium hydroxide in water to form magnesium and hydroxide ions.
nbeE9b2JV5Q-015|Magnesium hydroxide, sparingly soluble, so the reaction favors the solid, and you can write the equilibrium expression.
nbeE9b2JV5Q-016|Magnesium ions and hydroxide ions squared, in this case, stoichiometric coefficient of 2.
nbeE9b2JV5Q-017|The pure solid doesn't appear in equilibrium expressions.
nbeE9b2JV5Q-018|So we have a Ksp, solubility product, for magnesium hydroxide.
nbeE9b2JV5Q-020|Well, you can think about this in a couple of ways.
nbeE9b2JV5Q-021|This A is a strong acid.
nbeE9b2JV5Q-022|Adding a strong acid will react with the OH minus quite strongly.
nbeE9b2JV5Q-023|They'll form water, H3O pluses and OH minuses form water, that will lower this concentration.
nbeE9b2JV5Q-024|If I lower that concentration, the reaction will shift, Le Chatelier's principle, shift towards the products to raise these concentrations.
nbeE9b2JV5Q-025|So the magnesium ion concentration will go up, and you'll dissolve more solid by adding some strong acid.
nbeE9b2JV5Q-026|Actually, the reverse occurs if you go with B. Adding NaOH adds sodium hydroxide, adds hydroxide ions.
nbeE9b2JV5Q-027|That will shift the reaction back towards product.
nbeE9b2JV5Q-028|Excuse me, back towards reactant, the solid.
nbeE9b2JV5Q-029|And the solid will increase.
nbeE9b2JV5Q-030|So this will actually make these guys come out of solution and precipitate to form the solid.
nbeE9b2JV5Q-031|So this is actually worst case scenario.
nbeE9b2JV5Q-032|If I add plain water, that will dilute the solution.
nbeE9b2JV5Q-034|The question is which goes faster?
nbeE9b2JV5Q-035|Adding some acid or adding the straight water?
nbeE9b2JV5Q-041|So this is just some math, but what it shows you, it shows you, well, I understand how equilibrias hold.
nbeE9b2JV5Q-042|They all hold at the same time.
nbeE9b2JV5Q-044|So Kw is H3O plus times OH minus, so I can use that equilibrium and this equilibrium simultaneously.
nbeE9b2JV5Q-045|When I do, I get an analytical expression for the magnesium ion concentration versus the acid concentration.
nbeE9b2JV5Q-046|And it's a strong function.
nbeE9b2JV5Q-047|It goes as the square of the acid concentration.
nbeE9b2JV5Q-048|So as H3O plus goes up, the magnesium goes up as well in a squared fashion.
nbeE9b2JV5Q-049|So a quadratic increase in my magnesium iron concentration by adding strong acid.
nbeE9b2JV5Q-050|In this case, the best answer is A.
jW15zq0_78Q-000|Let's do a calculation where we look at formal charges to determine the best Lewis electron dot structures.
jW15zq0_78Q-003|Now, in order to do that, we need Lewis electron dot structures.
jW15zq0_78Q-004|And to do that, we need to draw them all, all the various resonance structures and determine the best ones based on formal charges.
jW15zq0_78Q-005|So let's look at a few.
jW15zq0_78Q-006|Here's Lewis electron dot structure number one for our sulfate ion.
jW15zq0_78Q-007|You can see, I've distributed the electrons around the atoms and bonded this all together.
jW15zq0_78Q-008|All the rules for Lewis electron dot structures are followed for this molecule.
jW15zq0_78Q-009|Now, let me assign formal charges.
jW15zq0_78Q-011|So 1, 2, 3, 4, 5, 6, 7 electrons around the oxygen.
jW15zq0_78Q-012|Oxygen normally has 6 as an atom, has one extra, that's a minus 1 formal charge.
jW15zq0_78Q-013|All those oxygens are the same.
jW15zq0_78Q-014|So they all have minus 1 formal charge.
jW15zq0_78Q-017|So there, we've determined the formal charges for that Lewis electron dot structure.
jW15zq0_78Q-021|Is this one better than this one is our question.
jW15zq0_78Q-022|Well, let's, again, calculate formal charges.
jW15zq0_78Q-023|Here, oxygen-- 1, 2, 3, 4, 5, 6.
jW15zq0_78Q-024|Again, these two lone pairs are assigned to that oxygen, and this double bond has 4 electrons and I'm going to share them.
jW15zq0_78Q-025|So 2 electrons are assigned to the oxygen, plus these two plus these two-- 6 electrons are assigned to oxygen.
jW15zq0_78Q-026|It normally has 6 as an atom, so formal charge of 0 there.
jW15zq0_78Q-027|This oxygen-- identical, so it'll have a 0.
jW15zq0_78Q-028|And these oxygens look just like these oxygens, they have a formal charge of minus 1 again.
jW15zq0_78Q-029|The sulfur, let's calculate its formal charge.
jW15zq0_78Q-030|1, 2, 3, 4, 5, 6 electrons shared in the bonds around sulfur.
jW15zq0_78Q-031|6 electrons-- as an atom it has 6, so it has a formal charge of 0.
jW15zq0_78Q-035|We know these oxygens are similar to these oxygens, so they'll have formal charge 0 again.
jW15zq0_78Q-036|The sulfur shares 1 electron from each of these 8 bonds, that gives it 8 electrons around it.
jW15zq0_78Q-038|Now here, I have a lot of 0 formal charges, that's nice, but I have a minus 2 on a lesser electronegative element.
jW15zq0_78Q-039|When I compare sulfur and oxygen, oxygen is the more electronegative.
jW15zq0_78Q-040|I'd rather have the formal charges negative on the more electronegative elements.
jW15zq0_78Q-041|So this Lewis dot structure I don't like a lot because of the numbers and distribution of formal charges.
jW15zq0_78Q-042|This Lewis dot structure, I've got a negative charge on the less electronegative element.
jW15zq0_78Q-043|So those two I don't like as much as I like this one, which has low formal charges and the negative formal charges are on the more electronegative element.
jW15zq0_78Q-044|So if I calculate the bond order here, I think it's pretty obvious, the bond order is 1 in an S-O bond here.
jW15zq0_78Q-045|An S-O bond here, I can see two single bonds, two double bonds, I'll get a bond order of 1 and 1/2.
jW15zq0_78Q-046|Here, it's obvious that the bond order is going to be 2.
jW15zq0_78Q-047|All of them are identical and double bonds.
jW15zq0_78Q-048|So I'll have a bond order of 2 here.
jW15zq0_78Q-049|But I'm going to reject these two Lewis electron dot structures based on their formal charges.
eCc24p1bLig-000|Another way to form bonds and fill octets is to share the electrons rather than to transfer them from one element to another.
eCc24p1bLig-001|If I share electrons to fill octets that's called covalent bonding.
eCc24p1bLig-005|Same thing with chlorine.
eCc24p1bLig-006|Chlorine can come together and fill its octet of eight by sharing one valence electron, forming a covalent bond.
eCc24p1bLig-007|And of course, hydrogen chloride could form.
eCc24p1bLig-008|Hydrogen can share its valence electron with chlorine.
eCc24p1bLig-009|Chlorine gets its full octet by sharing.
eCc24p1bLig-010|Hydrogen gets its full outer shell of two by sharing.
eCc24p1bLig-011|And they form a stable bond.
eCc24p1bLig-012|So covalent bonding is sharing electrons to fill octets.
43YS58bxWuU-000|When an electronic transition occurs, a photon is either absorbed or emitted.
43YS58bxWuU-001|So if a photon is absorbed, an electron goes from a low energy state to a high energy state.
43YS58bxWuU-002|When a photon is emitted, that's an electron in an excited state dropping down to a lower energy state.
43YS58bxWuU-003|When we observe those things, we say, oh, there's blue light, or there's an ultraviolet photon being absorbed or emitted.
43YS58bxWuU-004|We can infer something about the electronic structure of that atom.
43YS58bxWuU-006|So let's look at that.
43YS58bxWuU-007|In the hydrogen atom, you have various energy levels that we're now familiar with.
43YS58bxWuU-008|And if you have transitions that end or start in principle quantum level 1, those transitions will all be in the ultraviolet.
43YS58bxWuU-009|Our eyes won't be sensitive to them.
43YS58bxWuU-010|You can design a detector that detects ultraviolet, but our eyes don't work to detect these.
43YS58bxWuU-011|They're too high energy of photons.
43YS58bxWuU-012|If transitions start or end-- that is they end in 2 if it was an emission.
43YS58bxWuU-013|They would start in 2 if there was an absorption of a photon.
43YS58bxWuU-014|We can have the first, which is the 3 to 2 transition, actually is in a visible region.
43YS58bxWuU-015|So it's about 657 nanometers.
43YS58bxWuU-016|That's a red photon.
43YS58bxWuU-017|And we can detect that with our eyes.
43YS58bxWuU-018|The 4 to 2 is in the green, about 487 nanometers.
43YS58bxWuU-021|The ultraviolet Lyman series of lines, or the Balmer series of lines-- those ending or starting in an equal 2.
43YS58bxWuU-022|Now, since there is these visible emission lines, we can detect those.
43YS58bxWuU-023|If we excite hydrogen, or any number of gases, there'll be emissions that are in the visible spectrum, and we can see them.
43YS58bxWuU-024|And if you do excitation in particular-- if you excite gases, and let the photons be emitted, you can actually see them.
43YS58bxWuU-025|And that's how neon signs work.
43YS58bxWuU-026|They excite gases.
43YS58bxWuU-027|And then those gases emit photons.
43YS58bxWuU-028|And those photons are visible to us.
43YS58bxWuU-029|So we can do that.
43YS58bxWuU-030|We have a mini neon tube.
43YS58bxWuU-031|And it's called neon generically, of course.
43YS58bxWuU-032|In our tubes, there's several different gases.
43YS58bxWuU-033|We have, in fact, all the gas of the atmosphere.
43YS58bxWuU-034|We have nitrogen, oxygen, neon, helium, hydrogen. All the gases that are found, to a certain extent, in the atmosphere.
43YS58bxWuU-035|When I excite them with an electric charge, the electrons will be promoted, and they will emit specific energies.
43YS58bxWuU-036|They'll emit them all at once, so the colors will be a mixture of these, they'll also be emitting that ultraviolet radiation.
43YS58bxWuU-037|We won't be able to see that, but it's emitted at the same time.
43YS58bxWuU-038|So let's look at the emission spectrum of atoms and molecules.
43YS58bxWuU-040|Let's look at that.
43YS58bxWuU-042|You can see the various colors, a lot of them bluish, where there is high energy photons.
43YS58bxWuU-043|And there's even photons, as we said, in the ultraviolet being emitted.
43YS58bxWuU-044|So we're doing kind of a spectroscopy experiment here.
43YS58bxWuU-046|If we know exactly the wavelength, we can calculate exactly the energy difference of the atomic orbitals in these atoms.
43YS58bxWuU-047|So you can see how powerful spectroscopy is to understand the properties and the structure of atoms.
SUkBTHfKqf4-000|Titration is the process of reacting acid and base together.
SUkBTHfKqf4-001|And I do that in a controlled way.
SUkBTHfKqf4-002|So I can take an acid solution and a base.
SUkBTHfKqf4-003|The acid solution will originally have low pH, a high acid concentration.
SUkBTHfKqf4-004|As you add base, that base will react with the acid and lower that acid concentration.
SUkBTHfKqf4-005|Lowering the acid concentration raises the pH.
SUkBTHfKqf4-006|The solution becomes more basic, which makes sense because you're adding a base.
SUkBTHfKqf4-007|Eventually, you'll use up all of the acid that's there.
SUkBTHfKqf4-008|And if you continue to add base, then that base will dominate the pH.
SUkBTHfKqf4-009|And you'll have a basic solution.
SUkBTHfKqf4-010|You can plot that in something that's called a titration curve.
SUkBTHfKqf4-013|This totally dissociates, forms 0.1 or 10 to the minus 1 moles of H3O+ moles per liter.
SUkBTHfKqf4-014|The pH is minus log of 10 to the minus 1 or 1.
SUkBTHfKqf4-015|So I start with pH 1.
SUkBTHfKqf4-016|I add some base.
SUkBTHfKqf4-017|And as I add moles of base, those react with my moles of acid to form water.
SUkBTHfKqf4-018|And that continues to happen until I use up all the original acid that is there.
SUkBTHfKqf4-019|When I've added 1 mole of base for every mole of acid in my original solution, I've used up all my acid, and I've used up all the base that I've added.
SUkBTHfKqf4-020|They've all completely reacted together.
SUkBTHfKqf4-022|As you continue beyond that, then the base that you're adding dominates the pH, and the pH rapidly goes to a basic level.
SUkBTHfKqf4-023|You can do the reverse.
SUkBTHfKqf4-024|You can start with a strong base solution or any base solution and add an acid solution.
SUkBTHfKqf4-025|Here the pH will start high because the H3O+ concentration is low.
SUkBTHfKqf4-026|And as you add acid, that pH will drop.
SUkBTHfKqf4-028|But a sodium chloride solution has pH 7.
SUkBTHfKqf4-029|The ions sodium and chlorine don't contribute to the pH.
SUkBTHfKqf4-030|We'll talk about that more later.
SUkBTHfKqf4-031|So you have a pH of 7, a neutral salt water solution.
SUkBTHfKqf4-032|As I continue to add acid, the acid will dominate the pH.
SUkBTHfKqf4-033|And you'll go to a low pH solution.
SUkBTHfKqf4-034|So this is a strong acid or strong base titration.
qvaEyckBI0g-000|So let's look at electrons in their various orbitals in forms of a ChemQuiz.
qvaEyckBI0g-001|When an electron spin flips, for which of the following is the energy not altered?
qvaEyckBI0g-009|So we're talking about nitrogen, oxygen, and fluorine, and we're trying to decide can we flip an electron spin and not change the overall energy of the atom?
qvaEyckBI0g-010|Nitrogen has the electronic configuration 1s2 2s2 2p3, three unpaired electrons in those 2p orbitals.
qvaEyckBI0g-011|Oxygen, 2p4, fluorine, 2p5.
qvaEyckBI0g-014|In oxygen and nitrogen, there's other electrons.
qvaEyckBI0g-015|So these electrons will try to be parallel.
qvaEyckBI0g-016|So the energy of this electron depends on the spin of this electron.
qvaEyckBI0g-017|But fluorine, that's not the case, everybody is paired up except for this unpaired electron, so fluorine can either have spin up or spin down.
7TZXu6s0KA0-000|Let's look at the Maxwell Boltzmann distribution for a pair of gasses.
7TZXu6s0KA0-001|I've plotted them here, and I've indicated where the root mean square velocity would fall for gas X at 400 Kelvin, and gas Y at 800 Kelvin.
7TZXu6s0KA0-002|The question is what are gases X and Y?
7TZXu6s0KA0-009|We're looking at the Maxwell Boltzmann distribution plots for two different gases at two different temperatures.
7TZXu6s0KA0-012|The temperature is doubling, but that wouldn't give us a factor of 2.
7TZXu6s0KA0-013|That would give us a factor of square root 2 in the rms velocity.
X5KLAdXoDxc-000|Let's look at the effect of temperature on a chemical reaction.
X5KLAdXoDxc-001|Now, we know intuitively if you raise the temperature, chemical reactions proceed more rapidly.
X5KLAdXoDxc-002|The reason is that at higher temperature, there are more molecules that have energies in excess of the activation energy for the chemical reaction.
X5KLAdXoDxc-003|So they can get over the barrier that separates products and reactants.
X5KLAdXoDxc-006|Now, that doesn't mean all those molecules will react.
X5KLAdXoDxc-007|They still have to collide, and collide with the appropriate orientation.
X5KLAdXoDxc-010|So now I have a higher percentage of molecules with energies in excess of the activation energy.
X5KLAdXoDxc-011|So why do reactions go faster with the increasing temperature?
X5KLAdXoDxc-012|More molecules, more particles with higher energies.
X5KLAdXoDxc-013|Now if you think about kinetics, our kinetic rate laws look like the rate is the rate constant times some function of the concentrations.
X5KLAdXoDxc-014|Now, the concentrations are not a function of temperature.
X5KLAdXoDxc-015|So for rates to increase with temperature, our rate constants must be a function of temperature.
X5KLAdXoDxc-018|So rate constants are a function of temperature.
X5KLAdXoDxc-023|k, the rate constant for the chemical reaction, is a function of temperature.
vhlYDz16s0w-000|Let's look at some examples of oxidation reduction reactions.
vhlYDz16s0w-001|Here's a common one-- the rusting of iron.
vhlYDz16s0w-002|Now, it's interesting to note, simply putting a nail made of iron in pure water, that won't rust.
vhlYDz16s0w-003|You actually need the presence of dissolved oxygen.
vhlYDz16s0w-004|So here's the actual reaction.
vhlYDz16s0w-007|Iron is in its 0 oxidation state, it's oxidation number increases.
vhlYDz16s0w-008|That's an oxidation to plus 2.
vhlYDz16s0w-009|So to help us balance this, let's go to our table of standard production potentials.
vhlYDz16s0w-012|We can use those to help balance this reaction, so let's do that.
vhlYDz16s0w-016|And I'll change the sign of my reduction potential.
vhlYDz16s0w-021|The cell potential for the oxidation of iron is 0.81 volts.
vhlYDz16s0w-023|These are farther down the table.
vhlYDz16s0w-026|They'll protect the iron from oxidation.
vhlYDz16s0w-027|And you know this.
NLGrUdD5uDw-000|To describe molecules quantum mechanically, we can take the atomic orbitals from the atoms and recombine them to form molecular orbitals.
NLGrUdD5uDw-001|And that works well for diatomics.
NLGrUdD5uDw-002|We can take s and p orbitals and overlap them and form molecular orbitals.
NLGrUdD5uDw-003|The s and p orbitals have the appropriate arrangement and symmetry for a diatomic molecule.
NLGrUdD5uDw-004|But what if we get more complicated?
NLGrUdD5uDw-005|Even if you go to a very simple triatomic or if you go to something with steric number 2, how do we accommodate steric number 2?
NLGrUdD5uDw-006|How do we have a carbon bond to something on this side and something on this side?
NLGrUdD5uDw-007|If I use my p orbital over here to make a molecular orbital with this, I don't have a p orbital available to make a molecular orbital with this.
NLGrUdD5uDw-008|I can't reuse my orbitals to make more bonds.
NLGrUdD5uDw-009|So I have to have bonds and orbitals that have the right orientation.
NLGrUdD5uDw-010|The atomic orbitals forming bond angle 180 can't quite do that.
NLGrUdD5uDw-011|So forming acetylene isn't really possible.
NLGrUdD5uDw-012|And it gets a little more complicated when I go to steric number 3.
NLGrUdD5uDw-013|So if I have a bond angle of 120 degrees, there aren't even atomic orbitals that have 120 degree orientation to each other.
NLGrUdD5uDw-017|So what I need are a set of atomic orbitals where I can combine them and get the various angles for the various steric numbers.
NLGrUdD5uDw-018|And that's what we'll look at next.
aLwZdRbk8_4-000|Let's look at the molecular orbital description of several atoms going across the periodic table.
aLwZdRbk8_4-001|We'll go from boron to neon and form the diatomic molecules, and look at how they look in a molecular orbital description.
aLwZdRbk8_4-002|When you start with boron, WE'LL take boron and we'll take the p orbital contributions only to the molecular orbitals.
aLwZdRbk8_4-004|So those don't contribute to the bond order.
aLwZdRbk8_4-005|Sigma s and sigma s star cancel each other out.
aLwZdRbk8_4-006|So we'll just look at the p contributions to the molecular orbitals.
aLwZdRbk8_4-008|It has a 2p electron, we can find that 2p electron, one from each boron and put it into the molecular orbitals.
aLwZdRbk8_4-009|So two electrons in the molecular orbitals.
aLwZdRbk8_4-010|Here's boron, the lowest energy molecular orbitals are the pi bonding orbitals.
aLwZdRbk8_4-012|Two bonding electrons, no anti-bonding electrons, so 2 divided by 2 is 1.
aLwZdRbk8_4-013|We can do the same thing with carbon.
aLwZdRbk8_4-015|So I have 1p electron, or 2p electrons from each carbon for a total of four going into my molecular orbitals.
aLwZdRbk8_4-016|And when I put in four electrons to my molecular orbitals I get a diamagnetic carbon molecule with a double bond.
aLwZdRbk8_4-017|Four bonding electrons divided by 2, bond order two.
aLwZdRbk8_4-019|Oxygen we would predict would be paramagnetic from this.
aLwZdRbk8_4-020|Now as we go across the periodic table, the lower energy orbitals are going to switch here.
aLwZdRbk8_4-021|The sigma bonding orbital becomes the lower energy as I go across the periodic table.
aLwZdRbk8_4-022|And you might predict that these molecular orbitals are formed from linear combinations of atomic orbitals.
aLwZdRbk8_4-023|And you have to do the mathematical calculation and calculate the energies, and you know the orbitals change as you go across the periodic table.
aLwZdRbk8_4-024|Atomic radii decrease.
aLwZdRbk8_4-025|So the character of the individual atomic orbital changes, and the character of the molecular orbitals that result changes.
aLwZdRbk8_4-026|When we give these orbital configurations, and it's necessary for you to know them, we'll make it very clear which orbital designation you should use.
aLwZdRbk8_4-027|So oxygen has two unpaired electrons, oxygen is paramagnetic.
aLwZdRbk8_4-028|We can go ahead to fluorine for the Lewis electron dot structure for fluorine, we'd predict a single bond.
aLwZdRbk8_4-029|Does the molecular orbital theory predict that?
aLwZdRbk8_4-033|So what we have is our molecular orbital theory predicting neon would not form a bond.
aLwZdRbk8_4-034|It would have a bond order zero.
wEwYuEk6dzs-000|Let's do a calculation involving both the ideal gas law and the van der Waals expression for gases and compare the two.
wEwYuEk6dzs-004|Well these last two we can answer right away.
wEwYuEk6dzs-005|Which guess has more collisions per second with the walls?
wEwYuEk6dzs-006|Well that's the gas that's moving at the greater average velocity.
wEwYuEk6dzs-007|And the greater average velocity goes as the temperature-- but they're both at the same temperature-- or the mass.
wEwYuEk6dzs-008|The lower the mass, the higher the average velocity.
wEwYuEk6dzs-009|So the one with the higher rms velocity, in this case, it's helium.
wEwYuEk6dzs-010|So helium will have more collisions with the wall.
wEwYuEk6dzs-011|Remember, they both can have the same pressure.
wEwYuEk6dzs-012|Just because helium is hitting the wall more often to impart its momentum, carbon dioxide hits with a bigger punch.
wEwYuEk6dzs-013|So they both exert the same pressure on the wall.
wEwYuEk6dzs-014|Now, what about the greater attractions?
wEwYuEk6dzs-015|Carbon dioxide will have the greater intermolecular attractions.
wEwYuEk6dzs-016|And we would see that reflected in the size of the A parameter in the van der Waals expression.
wEwYuEk6dzs-017|So we can go directly to that.
wEwYuEk6dzs-019|Now, they could be mixed or they could be in separate flasks, it doesn't matter.
wEwYuEk6dzs-020|If they're mixed, we'll calculate a partial pressure for each.
wEwYuEk6dzs-021|If they're separate, the partial pressure and the total pressure will be the same.
wEwYuEk6dzs-022|If they're mixed, the total pressure will be the sum of both pressures.
wEwYuEk6dzs-023|But either way, we can calculate the pressure for helium and carbon dioxide.
wEwYuEk6dzs-024|So first we'll do it for a real gas situation.
wEwYuEk6dzs-025|So we'll assume they behave like ideal gases.
wEwYuEk6dzs-030|The ideal gas law doesn't include the nature of the particle.
wEwYuEk6dzs-031|So both gases would have a pressure 1.63 atmospheres at 15 liters.
wEwYuEk6dzs-033|So the ideal gas calculation is easy because we don't have to include which gas we're talking about.
RI6HdlWwDtw-000|Let's do a calculation where we look at the standard state free energy difference for the combustion of glucose.
RI6HdlWwDtw-001|And we'll also look at what temperature range is that reaction spontaneous.
RI6HdlWwDtw-002|Now, it's the standard state free energy difference.
RI6HdlWwDtw-004|How can I calculate that?
RI6HdlWwDtw-006|So here it is.
RI6HdlWwDtw-013|What temperature range is this reaction spontaneous for?
RI6HdlWwDtw-014|Well, here's the chemical reaction in the free energy difference.
RI6HdlWwDtw-015|Is it spontaneous over a broad temperature range?
RI6HdlWwDtw-016|In order to do that, we need to know the enthalpy and the entropy.
RI6HdlWwDtw-017|Those are relatively independent of temperature.
RI6HdlWwDtw-018|And they tell us how delta G varies with temperature.
RI6HdlWwDtw-019|So delta S for this reaction-- I think we can intuitively say, without doing a calculation-- delta S is greater than zero.
RI6HdlWwDtw-020|We're producing a liquid and a gas from a solid and this gas.
RI6HdlWwDtw-022|The enthalpy-- you may know.
RI6HdlWwDtw-023|You've probably seen this chemical reaction, glucose burning.
RI6HdlWwDtw-024|It is exothermic.
RI6HdlWwDtw-025|But we don't have to use intuition for either one.
RI6HdlWwDtw-033|As temperature increases, delta S is positive.
RI6HdlWwDtw-034|That means the slope is negative.
njRitLAU3Ik-007|Now, four electrons from the o orbitals will fill these molecular orbitals.
njRitLAU3Ik-008|So we'll put those in-- one, two, three, four.
njRitLAU3Ik-010|The Lowest Unoccupied Molecular Orbital or LUMO is a pi-star or antibonding orbital.
njRitLAU3Ik-011|And it's common the transitions in these conjugated systems are a pi to pi-star transition.
njRitLAU3Ik-016|They go from a pi to pi-star.
njRitLAU3Ik-017|So the bond order here is reduced.
njRitLAU3Ik-019|Those are two names for the same compound.
njRitLAU3Ik-020|This compound that's in your eye acts as a light receptor.
njRitLAU3Ik-021|And when this transition occurs, this isomerism, that triggers a chain of chemical reactions that allow you to detect light in your eye.
njRitLAU3Ik-022|This is actually the light recepting molecule in your eye.
njRitLAU3Ik-023|So a pi to pi-star transition, very important in human vision.
9z0sMWVIm7A-000|Let's balance a redox reaction.
9z0sMWVIm7A-001|Here's a redox reaction here.
9z0sMWVIm7A-002|And when you balance redox reactions, you should follow a series of steps, and they're outlined in many textbooks.
9z0sMWVIm7A-003|I'll go through them right now.
9z0sMWVIm7A-007|Well, let's go through and see which is which.
9z0sMWVIm7A-008|So here's chromium.
9z0sMWVIm7A-010|And that turns out to be plus 6.
9z0sMWVIm7A-011|And the oxidation number here is obviously plus 3, so chromium is being reduced.
9z0sMWVIm7A-012|It's oxidation number is reduced.
9z0sMWVIm7A-013|So hopefully that means an oxidation is occurring over here.
9z0sMWVIm7A-014|And if we analyze this using the same rules, oxygen is always minus 2, hydrogen is always a plus 1.
9z0sMWVIm7A-019|So the next step, balance the elements in your two skeletal half cells.
9z0sMWVIm7A-020|Well, in the reduction, the 2 chromium here, only 1 there, so I need a 2.
9z0sMWVIm7A-021|And in fact, I'm just going to go through and apply the steps to the chromate first, and then come back and do it for the alcohol and aldehyde.
9z0sMWVIm7A-024|Now I have to balance the hydrogens by adding H plus.
9z0sMWVIm7A-030|So six pluses here and 2 times plus 3, 6 pluses there.
9z0sMWVIm7A-031|Now, let's do the same thing for the alcohol to aldehyde, balance the elements.
9z0sMWVIm7A-032|All the elements are balanced here, so nothing required there.
9z0sMWVIm7A-033|Balance the oxygen with water.
9z0sMWVIm7A-034|I have 1 oxygen on both sides, so nothing required there.
9z0sMWVIm7A-035|Balance the hydrogens by adding H plus.
9z0sMWVIm7A-039|Now I can bring my half reactions, now that they're balanced back together, and balance the electrons in both half cells.
9z0sMWVIm7A-040|So this half cell, I noticed 2 electrons and I noticed that's an oxidation.
9z0sMWVIm7A-041|6 electrons here and that's a reduction, that's good.
9z0sMWVIm7A-042|That's what I thought was happening and my procedure has confirmed that.
9z0sMWVIm7A-043|So, let's balance them out.
9z0sMWVIm7A-046|I'll add the two reactions together with my factor of 3.
9z0sMWVIm7A-050|And you can see now if you count up all the hydrogens, all the oxygens, all the chromium and carbons on this side, they exactly match up with that side.
9z0sMWVIm7A-052|That's how we balance redox reactions.
11QPT4cbBO8-000|Chemistry is the study of atoms, molecules, and their interactions.
11QPT4cbBO8-001|Now, virtually every process that happens in the universe involves molecules, atoms, and their interchange.
11QPT4cbBO8-005|So this reaction, the formation of water from hydrogen and oxygen, is favorable.
11QPT4cbBO8-006|In fact, when that reaction occurs, energy is released.
11QPT4cbBO8-007|It can be very dramatic.
11QPT4cbBO8-008|So in a sense, the products, water, are lower energy than the reactants.
11QPT4cbBO8-009|And we can plot that on a relative scale.
11QPT4cbBO8-010|And we'll do this a lot in chemistry.
11QPT4cbBO8-011|We'll plot energy on a vertical scale and the progress of a reaction along the horizontal scale.
11QPT4cbBO8-013|Now, in a mechanical example, you'd say, well, if they were a ball on the top of this hill, it would naturally roll down to the bottom of the hill.
11QPT4cbBO8-014|That's the natural way of things.
11QPT4cbBO8-015|And it's the same for this chemical reaction.
11QPT4cbBO8-016|The natural way is for hydrogen and oxygen to form water.
11QPT4cbBO8-017|There's a natural favorable direction of this chemical reaction.
11QPT4cbBO8-018|But there's one more interesting part.
11QPT4cbBO8-019|And that is, if I take oxygen and hydrogen, and I mix them together in a balloon, I can wait quite a while and never find water.
11QPT4cbBO8-020|I can come back and look at that balloon the next day, the next week, six months later, and it's still going to be a hydrogen oxygen balloon.
11QPT4cbBO8-021|You won't find a balloon full of water.
11QPT4cbBO8-022|And why is that?
11QPT4cbBO8-024|That pulling and stretching and rearranging of bonds requires energy.
11QPT4cbBO8-025|So there is a high energy intermediate state between the reactants and products.
11QPT4cbBO8-027|And you might say, well, if there's that high road barrier, why do they ever react?
11QPT4cbBO8-028|Well, let's go back to the mechanical example.
11QPT4cbBO8-029|If I had a ball here, could I get it to the other side of that hill?
11QPT4cbBO8-031|One way I could get it was to flick it-- give it some kinetic energy, and have it roll up over the hill.
11QPT4cbBO8-033|So it's possible in the mechanical example.
11QPT4cbBO8-034|Is it possible in the chemical example?
11QPT4cbBO8-035|Well, it is.
11QPT4cbBO8-036|I can't go in and flick individual atoms or molecules of hydrogen and oxygen, but I can get them to move faster, give them kinetic energy.
11QPT4cbBO8-037|I can heat them up.
11QPT4cbBO8-038|And throughout this course, we'll associate an increase in temperature with more molecular motion.
11QPT4cbBO8-039|So I can get reactants to have enough energy that some of them will react and go over this hill.
11QPT4cbBO8-040|And then it can become cooperative.
11QPT4cbBO8-041|They'll release some energy.
11QPT4cbBO8-042|they'll excite others in the mixture, and they'll all start to fall over that hill.
11QPT4cbBO8-049|So both those would be interesting.
11QPT4cbBO8-050|And in fact, a lot of these chemical reactions release energy, and they're fun to watch.
11QPT4cbBO8-051|And we'll watch a lot of chemical reactions through this series.
11QPT4cbBO8-052|The hydrogen oxygen reaction is one that I find most interesting, and I think you will, too.
XIJet3gIKRc-000|Let's do a calculation involving a weak acid, its pKa, and determining the relative concentrations of the acid and base form.
XIJet3gIKRc-001|So aspirin, acetylsalicylic acid has Ka, 3.2 times 10 to the minus 4.
XIJet3gIKRc-002|A typical weak acid.
XIJet3gIKRc-003|Now I can put it in solution and then adjust the pH to 4.13.
XIJet3gIKRc-008|We'll just tell you what the acidic proton is unless it's very obvious.
XIJet3gIKRc-009|So here's the acidic proton.
XIJet3gIKRc-010|I know the acid dissociation constant for the reaction where this proton leaves the molecule is 3.2 times 10 to the minus 4.
XIJet3gIKRc-013|I'm at a pH very close to that, really, within half a unit.
XIJet3gIKRc-017|About equal concentrations.
XIJet3gIKRc-018|I'm a little bit above, I'm a little bit to the basic side, of the pKa so that means the basic form will slightly predominate.
XIJet3gIKRc-019|Not a factor of 10.
XIJet3gIKRc-021|And we're not one full unit.
XIJet3gIKRc-022|We'd have to go up to pH 4.49 to get a tenfold difference.
XIJet3gIKRc-023|So we expect something less than tenfold difference.
XIJet3gIKRc-024|Let's actually do the calculation.
XIJet3gIKRc-025|So here's the acid and the base form, pKa 3.49.
XIJet3gIKRc-027|In this case, we know these two, so we can solve for this ratio.
XIJet3gIKRc-028|So let's just do that.
XIJet3gIKRc-033|So the ratio of base form to acid form is 4.36.
XIJet3gIKRc-034|There's about four times as much base form as there is acid form.
8y5KX4kzt0A-000|Let's talk about the isothermal expansion and compression for an ideal gas.
8y5KX4kzt0A-001|Isothermal means the temperature doesn't change.
8y5KX4kzt0A-002|And if the temperature doesn't change for an ideal gas, the energy and the enthalpy don't change.
8y5KX4kzt0A-003|Remember, the energy of an ideal gas, 3/2 nRT.
8y5KX4kzt0A-004|So isothermal processes delta-E and delta-H are zero.
8y5KX4kzt0A-005|So how can heat flow or work be done if the energy doesn't change?
8y5KX4kzt0A-006|Well, the work and heat just have to balance each other with opposite sign.
8y5KX4kzt0A-007|If you do a joule of work on the system, the system has to lose a joule of heat, joule for joule.
8y5KX4kzt0A-008|Do a joule of work, lose a joule of heat.
8y5KX4kzt0A-009|Do a joule of work, lose a joule of heat.
8y5KX4kzt0A-010|If I expand, the system does work.
8y5KX4kzt0A-011|So I do a joule of work, that would-- well, that's some of my energy, I need more energy to replace the energy I just used to do that work.
8y5KX4kzt0A-012|Well, absorb a joule of heat.
8y5KX4kzt0A-013|Do a joule of work, absorb a joule of heat.
8y5KX4kzt0A-020|That's expansion and contraction isothermally for an ideal gas.
VcGOL6Uthmw-000|Let's look at a couple of titrations and see if we can predict the equivalence point, or the end point, pH.
VcGOL6Uthmw-004|So two acids and a base.
VcGOL6Uthmw-005|We'll titrate them appropriately.
VcGOL6Uthmw-006|If they're acids, we'll titrate them with a base.
VcGOL6Uthmw-007|If it's a base, we'll titrate it with an acid.
VcGOL6Uthmw-008|The question is at equivalence point, which has a pH less than 7?
VcGOL6Uthmw-018|We're looking at the titration of some weak acid and base solutions.
VcGOL6Uthmw-019|Now, when you titrate a weak acid or a weak base with a strong base or a strong acid, the titration curve looks something like this.
VcGOL6Uthmw-020|We have the acid being converted to its conjugate base.
VcGOL6Uthmw-021|Now, if this were a base, it would be a base being converted to its conjugate acid.
VcGOL6Uthmw-024|If it were the weak base solution, you have titrating a weak base.
VcGOL6Uthmw-025|The pH is dropping.
VcGOL6Uthmw-026|But again, at equivalence point, you've added 1 mole of strong acid for every mole of weak base.
VcGOL6Uthmw-027|You convert the weak base into its conjugate acid.
VcGOL6Uthmw-028|That's what titration is.
VcGOL6Uthmw-029|It's converting the weak acid into its conjugate base, or the weak base into its conjugate acid.
VcGOL6Uthmw-030|So let's look at how that looks.
VcGOL6Uthmw-031|We have three possible species.
VcGOL6Uthmw-032|HCl, the first one.
VcGOL6Uthmw-033|When I take HCl to equivalence point and convert it all into its conjugate base, its conjugate base Cl minus, the chloride ion.
VcGOL6Uthmw-034|Chloride ion is a horrible base.
VcGOL6Uthmw-035|HCl, very strong acid, means its conjugate base, very weak base.
VcGOL6Uthmw-036|In fact, this is not basic at all.
VcGOL6Uthmw-037|You know a solution of sodium chloride, the chloride solution will be neutral.
VcGOL6Uthmw-038|So this will be around pH 7.
VcGOL6Uthmw-039|Here's the weak acid, formic acid.
VcGOL6Uthmw-041|For our third choice here, ammonia, that's a weak base.
VcGOL6Uthmw-042|I would titrate that with a strong acid and convert it into its conjugate acid.
VcGOL6Uthmw-043|So at equilibrium, I'd have a solution of a weak acid.
VcGOL6Uthmw-044|So this solution is slightly acidic.
VcGOL6Uthmw-048|So weak acids are converted into their conjugate bases, making the equivalence point pH basic.
VcGOL6Uthmw-049|Weak bases are converted into their conjugate acids, making the equivalence point slightly acidic.
VcGOL6Uthmw-050|In this case, ammonia has an acidic equivalence point.
Ibl75XCrHKQ-000|Here's a table of standard enthalpies of formation.
Ibl75XCrHKQ-001|Let's look at a few of them.
Ibl75XCrHKQ-004|Now this is common for energy calculations.
Ibl75XCrHKQ-005|For instance, if you're doing a gravitational energy, potential energy calculation, you could set the zero in gravitational potential to the bottom of a hill.
Ibl75XCrHKQ-006|And the top of the hill would be measured relative to that.
Ibl75XCrHKQ-007|Now this bottom of the hill is not really zero in gravitational potential.
Ibl75XCrHKQ-009|That's what we do in chemistry when we set the zero in standard enthalpies of formation as the elements in their standard state.
Ibl75XCrHKQ-010|It gives us a standard of comparison for all the other compounds.
Ibl75XCrHKQ-011|Now let's look at a few compounds and a few things formed from elements in their standard state.
Ibl75XCrHKQ-012|Here's hydrogen atoms in the gas phase.
Ibl75XCrHKQ-013|Now, hydrogen atoms in the gas phase is not the standard state of hydrogen.
Ibl75XCrHKQ-014|The standard state of hydrogen is how you would find hydrogen naturally at 1 atmosphere of pressure and 25 degrees C, and that's diatomic hydrogen gas.
Ibl75XCrHKQ-015|To make atoms out of diatomic hydrogen gas, you have to break hydrogen hydrogen bonds.
Ibl75XCrHKQ-016|Breaking bonds always requires energy.
Ibl75XCrHKQ-017|So there's an increase in energy-- that's an endothermic reaction to go from the molecules to the atoms.
Ibl75XCrHKQ-019|Now oxygen the same-- oxygen atoms will be formed from diatomic oxygen molecules.
Ibl75XCrHKQ-020|Here's carbon gas in its standard state.
Ibl75XCrHKQ-021|The standard state of carbon would be carbon graphite solid.
Ibl75XCrHKQ-022|So this would be atomizing solid graphite into carbon atoms in the gas phase.
Ibl75XCrHKQ-023|Of course, that requires energy to break all those intermolecular carbon bonds and atomize solid carbon.
Ibl75XCrHKQ-024|Carbon dioxide and these other compounds are lower in energy than the elements in their standard state.
Ibl75XCrHKQ-025|So when you form carbon dioxide from carbon solid in its standard state and oxygen gas in its standard state, you go down enthalpy hill.
Ibl75XCrHKQ-026|So these compounds are enthalpically more stable than the elements in their standard state.
Ibl75XCrHKQ-037|And that's why we tabulate standard enthalpies of formation.
O9fks01bccA-000|Let's look at some bond angles in lactic acid.
O9fks01bccA-001|Here's lactic acid and the carbon, oxygen, hydrogen bond angle is of interest here.
O9fks01bccA-009|We're looking at bond angles in lactic acid.
O9fks01bccA-012|But remember, the Lewis dot structure isn't predictive about bond angles until you use it to get the steric numbers.
O9fks01bccA-013|So the steric number on oxygen here is four-- one carbon, one hydrogen, and two lone pairs.
O9fks01bccA-014|Oxygen has to accommodate four things, a steric number of four, so that oxygen will sp3 hybridize.
O9fks01bccA-015|Four orbitals, sp3 equivalent orbitals.
O9fks01bccA-016|They have a tetrahedral configuration.
O9fks01bccA-017|The bond angle nominally will be 109 degrees.
nUxUTyL2raM-000|Let's look at a reaction between zinc metal and sulfuric acid.
nUxUTyL2raM-001|What happens when I mix the two?
nUxUTyL2raM-009|We're talking about mixing zinc metal and sulfuric acid.
nUxUTyL2raM-010|So let's look at the half cell reactions.
nUxUTyL2raM-014|So, indeed, that means the zinc ion reaction, reduction potential, will run in reverse.
nUxUTyL2raM-015|And when I add these two together, let's just do this.
nUxUTyL2raM-016|I reverse this one, so let's reverse the potential, so I can add directly down, as I add directly down here.
nUxUTyL2raM-017|It's the same thing as saying I'm going to subtract a negative 0.76 volts.
nUxUTyL2raM-021|In this chemical reaction, the favored state is zinc ions and hydrogen gas.
nUxUTyL2raM-022|So let's watch that happen.
nUxUTyL2raM-026|The products are hydrogen gas, which is bubbling off, and zinc ions forming in solution.
nUxUTyL2raM-029|In this case, the answers are A, zinc ions, and B, hydrogen gas.
edezISdNZj0-001|In order to do that, you need to know the bond enthalpies of all the reactant bonds and all the bond enthalpies of the product bonds.
edezISdNZj0-002|So you need a table that looks like this.
edezISdNZj0-003|Here, I've calculated average bond enthalpies.
edezISdNZj0-005|It's an average over a lot of molecules.
edezISdNZj0-006|So it's the average carbon-carbon bond, say, in ethane, butane, benzene, over a wide variety of molecules.
edezISdNZj0-007|Same thing for the double bond and the triple bond.
edezISdNZj0-008|Some of these are exact, of course.
edezISdNZj0-009|Hydrogen-hydrogen, there's only one kind of hydrogen-hydrogen bond.
edezISdNZj0-010|So 436 kilojoules is the exact bond enthalpy for hydrogen-hydrogen. But carbon-oxygen, that's an average.
edezISdNZj0-011|Now, there's a couple of interesting things to notice on this table.
edezISdNZj0-015|So we can't make direct correlations between the bond strength and the number of bonds from our average bond enthalpy tables.
edezISdNZj0-016|Now, we've calculated reaction enthalpies in a different way.
edezISdNZj0-017|We also said, take the enthalpies of formation of all the reactants and subtract away the enthalpies of formation of all the products.
edezISdNZj0-019|So what's the difference between those two methods using the standard enthalpies of formation or the standard bond enthalpies?
edezISdNZj0-020|Well, think about this.
edezISdNZj0-022|So you need an incredibly extensive table of enthalpies of formation.
edezISdNZj0-025|So that's the power of this method.
edezISdNZj0-026|But you do give up something.
edezISdNZj0-027|Remember, these are average bond enthalpies.
edezISdNZj0-028|So you're making an estimate of the enthalpy of reaction.
edezISdNZj0-029|If you use the standard enthalpies of formation, that would be exact.
edezISdNZj0-030|So it's a trade-off between a large table and a small table-- exactness and an estimate.
edezISdNZj0-031|Now, notice these two tables are related.
edezISdNZj0-032|So if you look, say, what's the standard enthalpy of formation of hydrogen atoms?
edezISdNZj0-033|That's 218 kilojoules per mole of hydrogen atoms.
edezISdNZj0-034|If you look at breaking the hydrogen-hydrogen bond, that says 436 kilojoules.
edezISdNZj0-035|And that makes sense.
edezISdNZj0-036|That's twice my enthalpy of formation because when I break this bond, I add 436 kilojoules.
edezISdNZj0-037|I get two moles of hydrogen atoms, so two moles, twice the standard enthalpy of formation of hydrogen.
edezISdNZj0-038|So I find a relationship between the tables, their strengths and weaknesses of each method of calculating enthalpies for chemical reactions.
edezISdNZj0-039|The strength here is, let me remember just a small number of things and be able to calculate for a wide variety of chemical reaction.
Zwt_x-8hM54-000|Let's look at the titration curve of a weak base with a strong acid.
Zwt_x-8hM54-002|The formation of water is favored.
Zwt_x-8hM54-003|So this reaction lies far towards products.
Zwt_x-8hM54-004|And basically, every mole of acid you have will consume a mole of your weak base.
Zwt_x-8hM54-006|So let's look at the various points along this curve.
Zwt_x-8hM54-013|When I get to point, the equivalence point here, now I've converted all my base to the conjugate acid.
Zwt_x-8hM54-014|I've added a mole of strong acid for every mole of base that I originally had, converting it all to its conjugate base.
Zwt_x-8hM54-015|So you have a solution here of a conjugate acid.
Zwt_x-8hM54-016|You can calculate the pH of a solution of a weak acid.
Zwt_x-8hM54-017|As you can continue out, the pH will start to be dominated by the fact that you're just adding strong acid.
Zwt_x-8hM54-018|So these are the regions of the titration curve for a weak base by a strong acid.
yoq9Vx6FvSE-000|Among the most common equilibria that we'll look, and the most important, are acid base equilibria.
yoq9Vx6FvSE-001|Acid base equilibria involve transferring a proton from one species to another.
yoq9Vx6FvSE-003|Now I'm saying proton, by that I mean when this hydrogen leaves the HA compound, it leaves its electron behind.
yoq9Vx6FvSE-004|So it's only the nucleus of the hydrogen that leaves.
yoq9Vx6FvSE-005|And the nucleus of a hydrogen atom is a proton.
yoq9Vx6FvSE-006|So we call this proton transfer.
yoq9Vx6FvSE-011|Because this is an equilibrium.
yoq9Vx6FvSE-013|So that'll be our definition of acid and base.
yoq9Vx6FvSE-017|So I've abbreviated significantly here.
yoq9Vx6FvSE-018|And of those four hydrogens only one is donated as a proton.
yoq9Vx6FvSE-019|One is the acidic proton on acetic acid.
yoq9Vx6FvSE-020|H3O plus, that's water with a proton added.
yoq9Vx6FvSE-021|So it has an extra proton, and it will very readily donate that one.
yoq9Vx6FvSE-025|So water can act as an acid or a base.
yoq9Vx6FvSE-027|So here would be water donating a proton, you'd have OH minus leftover.
yoq9Vx6FvSE-031|NH3, it can accept the proton, act as a base, and OH minus, as we've already discussed, the strongest of these three bases.
yoq9Vx6FvSE-032|The most able to take a proton.
yoq9Vx6FvSE-035|So our definition, an acid gives up a proton, a base accepts a proton.
AHUBz-4aAbE-000|Let's look at some more orbital shapes and orientation.
AHUBz-4aAbE-008|We're talking about orbitals in terms of their shape, and the number of nodes, and their orientation.
AHUBz-4aAbE-009|Orbitals are described by three quantum numbers, n, l, and m sub l.
AHUBz-4aAbE-010|n gives you the overall size and energy, l gives you the shape, and m sub l an orientation.
AHUBz-4aAbE-011|Now what we want to know for this case is where and how are the nodes oriented.
AHUBz-4aAbE-012|Well, let's look at nodes.
AHUBz-4aAbE-013|Remember, for angular nodes, they go as the value of l.
AHUBz-4aAbE-014|So if l is 0, there are 0 angular nodes.
AHUBz-4aAbE-015|An s orbital, l equals 0, has 0 angular nodes.
AHUBz-4aAbE-018|If you understand that there are angular nodes slicing up space, that pretty much tells you about the shape and orientation of the orbital.
AHUBz-4aAbE-020|Or 2, 3 minus 1, or 1.
AHUBz-4aAbE-021|So there's three nodes here, two nodes here, one node here.
AHUBz-4aAbE-025|When l is 2, that's a d, there are two angular nodes.
AHUBz-4aAbE-026|When l is 3, that's an f orbital, there are three angular nodes.
AHUBz-4aAbE-027|So now we can easily tell what's going on, remember there's three nodes in the 4f, but they're all angular.
AHUBz-4aAbE-028|So there are no radial nodes in a 4f orbital.
AHUBz-4aAbE-029|The 3d, there's two total nodes, n minus 1, but they're both angular.
AHUBz-4aAbE-030|Again, no radial nodes in the 3d.
AHUBz-4aAbE-031|The 2s, there's one node, s equals 0, so the value of l is 0.
AHUBz-4aAbE-032|So there are zero angular nodes, so that one node must be a radial node.
AHUBz-4aAbE-033|So no radial nodes, no radial nodes, 1 radial node.
9JM7zZk1j_g-000|Let's look at the isothermal compression of a gas.
9JM7zZk1j_g-001|I'll take a gas, compress it, and I'd like to know-- what is the heat flow?
9JM7zZk1j_g-010|We're talking about the isothermal compression of an ideal gas.
9JM7zZk1j_g-011|Now isothermal is the key word because if the temperature doesn't change for an ideal gas, its energy didn't change.
9JM7zZk1j_g-012|So no energy change in this system.
9JM7zZk1j_g-013|Now it was a compression, so work is being done on the system.
9JM7zZk1j_g-014|Delta-V will be negative, but we have a negative sign minus the external delta-V-- positive work.
9JM7zZk1j_g-019|Now let's think about this practically.
9JM7zZk1j_g-020|I compress the gas, I'm doing work on the gas.
9JM7zZk1j_g-021|I'm giving energy to the gas.
9JM7zZk1j_g-022|And you say, wait, but if I give energy to the gas, its temperature has to change.
9JM7zZk1j_g-023|How can I give it energy without its temperature changing?
9JM7zZk1j_g-024|Well, that energy has to flow out at the same time I give it.
9JM7zZk1j_g-025|So how can that energy flow out?
9JM7zZk1j_g-026|Well, I'll have work done on the system, but every joule of work will flow out as a joule of heat.
9JM7zZk1j_g-027|And we call that negative heat or an exothermic process when heat leaves the system.
9JM7zZk1j_g-028|Practically, it means you probably do this process very slowly.
9JM7zZk1j_g-029|You compress it a little bit, wait for some heat to flow out.
9JM7zZk1j_g-030|You compress a little more, wait for heat to flow out.
9JM7zZk1j_g-031|Or you compress it and wait a while until all the heat flows out.
9JM7zZk1j_g-032|In general, you can do work quickly on a system, but heat flow always takes time.
9JM7zZk1j_g-033|So when you see isothermal, you often think slow, because I have to wait for the system to come back to the same temperature or become isothermal.
9JM7zZk1j_g-034|That's an isothermal compression of an ideal gas.
jzHAfhGPQxg-003|Well, this chemical reaction releases energy, delta H is negative it's exothermic.
jzHAfhGPQxg-004|Delta acid is positive and you might have guessed that because there's a gas released from a solid and a liquid.
jzHAfhGPQxg-005|So that's going to have many more microstate.
jzHAfhGPQxg-006|You're going to have a dispersion of energy when a gas is created.
jzHAfhGPQxg-007|So delta H negative, delta S positive.
jzHAfhGPQxg-008|This reaction will have a negative delta G for all temperatures.
jzHAfhGPQxg-013|We've actually seen this chemical reaction, but in reverse.
jzHAfhGPQxg-014|We saw peroxide going to water and oxygen.
jzHAfhGPQxg-015|And we know energy is released in this direction.
jzHAfhGPQxg-016|So it has to be absorbed in this direction.
jzHAfhGPQxg-017|Delta S, in this case, is negative.
jzHAfhGPQxg-018|Again you might guess that.
jzHAfhGPQxg-019|Gases forming liquids.
jzHAfhGPQxg-020|So this reaction in this direction is never favorable, at any temperature.
jzHAfhGPQxg-024|You're forming a liquid from a solid, so you'd expect an increase in entropy.
jzHAfhGPQxg-028|This one it's a little less clear, everything's a solid.
jzHAfhGPQxg-029|So entropy wise it's hard to predict.
jzHAfhGPQxg-030|Entropy wise this is an exothermic chemical reaction, and entropically there there's a decrease in entropy as you go.
jzHAfhGPQxg-031|So this reaction is favored at low temperatures.
jzHAfhGPQxg-032|So we need, since we have this negative delta S contribution, we need T to be small.
jzHAfhGPQxg-033|So that the overall and felt that contribution will have a contribution to delta G that is negative.
jzHAfhGPQxg-034|So this reaction favored at low temperature, unfavorable at high temperature.
pxxhXa3TZAU-001|Now, the change in concentration over time is the slope of the concentration versus time curve.
pxxhXa3TZAU-002|And we could start at any initial concentrations and let this reaction go.
pxxhXa3TZAU-011|We're talking about writing an initial rate for a chemical reaction, and we can write initial rates in terms of changes in concentration over a change in time.
pxxhXa3TZAU-013|That is, two hydrogen ions will be consumed when a zinc ion is produced.
pxxhXa3TZAU-014|So two things will happen.
pxxhXa3TZAU-015|The rates, if I cast it in terms of the zinc ion, should have the opposite sign, and there should be a factor of two.
pxxhXa3TZAU-016|And that is, of course, C here.
pxxhXa3TZAU-018|In this case, the correct answer is C.
7UFQPsGBreg-000|Let's continue writing Lewis electron dot structures.
7UFQPsGBreg-001|The cyanide ion, CN minus, is written down partially here.
7UFQPsGBreg-002|Or is it partial?
7UFQPsGBreg-003|Are there no bonds required to complete it?
7UFQPsGBreg-004|One bond required to complete it?
7UFQPsGBreg-012|We're writing the Lewis electron structure for CN minus, the cyanide ion.
7UFQPsGBreg-013|First, we count up the valence electrons.
7UFQPsGBreg-014|Carbon contributes 4, nitrogen contributes 5, and the ion has a negative charge, so there must be an extra electron.
7UFQPsGBreg-015|That contributes 1.
7UFQPsGBreg-016|So 10 total electrons.
7UFQPsGBreg-017|That's actually the same number of electrons as the N2 molecule.
7UFQPsGBreg-018|So CN minus requires two more bonds to get 10 total electrons.
7UFQPsGBreg-019|I have 2, 4, 6, 8, 10 electrons in the cyanide ion.
7UFQPsGBreg-021|CN minus, isoelectronic with N2, is a triple bonded structure.
K6LVoYlzJ08-000|An acid is a compound that donates a proton.
K6LVoYlzJ08-001|A base is a compound that accepts a proton.
K6LVoYlzJ08-002|In water solutions, the acid donates its proton to water.
K6LVoYlzJ08-011|So this Ka has the special label, but it behaves like a normal equilibrium constant.
K6LVoYlzJ08-012|In particular, if it's large, then the equilibrium lies towards the products.
K6LVoYlzJ08-013|A lot of H3O+ is produced, a lot of the HA, a high percentage, dissociates for strong acids.
K6LVoYlzJ08-014|For weak acids, a smaller percentage dissociates.
K6LVoYlzJ08-019|By accepting a proton, it forms OH- in water solution.
K6LVoYlzJ08-020|So the presence of OH- is indicative of a base in water.
K6LVoYlzJ08-024|Again, water, a pure liquid, doesn't appear in the expression.
K6LVoYlzJ08-025|Again, the size of the K determines the strength of the base.
K6LVoYlzJ08-027|So we have Ka and Kb, acid and base dissociation constants.
vC2v1n-bZQU-000|Equilibrium is the condition where macroscopically, a reaction appears to have stopped-- the macroscopic concentrations or pressures aren't changing.
vC2v1n-bZQU-001|But microscopically, there's a dynamic equilibrium.
vC2v1n-bZQU-002|Products are changing into reactants and reactants are changing into products.
vC2v1n-bZQU-003|You can approach equilibrium in a variety of ways.
vC2v1n-bZQU-010|So this equilibrium constant K would be large, it would favor the products.
vC2v1n-bZQU-013|Now, how can I approach this equilibrium state?
vC2v1n-bZQU-018|I can demonstrate that.
vC2v1n-bZQU-020|So let's see that happening.
vC2v1n-bZQU-022|So equilibrium reached in a step.
vC2v1n-bZQU-023|We can also reach equilibrium in an oscillatory fashion.
vC2v1n-bZQU-025|So let's see that chemical reaction occur.
vC2v1n-bZQU-026|So here's a chemical reaction, turns to yellow forming, but then to blue as it goes by the products.
vC2v1n-bZQU-031|There's an overall envelope of reaching equilibrium, but it happens in an oscillatory fashion.
vC2v1n-bZQU-032|There's another instance you could imagine, have an approach to equilibrium that's chaotic.
vC2v1n-bZQU-034|All reactions if you wait long enough will approach the equilibrium.
vC2v1n-bZQU-035|The thermodynamically favored state is the favored state at long times.
vC2v1n-bZQU-036|So we expect favored states to occur if we wait long enough.
vC2v1n-bZQU-037|So several ways to approach equilibrium.
vC2v1n-bZQU-038|Either way, they're dynamic.
TUBGGXVJj3Q-000|Let's look at adding a strong base solution to a strong acid solution.
TUBGGXVJj3Q-002|Now, strong acids and strong bases totally dissociate in water-- that's the definition of strong acid and strong base.
TUBGGXVJj3Q-003|So HCl completely dissociates into H3O+ and Cl-.
TUBGGXVJj3Q-004|We can look at in steps, though.
TUBGGXVJj3Q-017|I could, but it doesn't make much sense, the H3O+ plus concentration of 0.2 plus to the minus 7, 10 to the minus 7 is tiny by comparison, right?
TUBGGXVJj3Q-018|This is 0.2000001.
TUBGGXVJj3Q-019|So that's essentially 0.2.
TUBGGXVJj3Q-021|Now, we want to add 10 mils of sodium hydroxide at one molar.
TUBGGXVJj3Q-022|How many moles of OH- is that?
TUBGGXVJj3Q-026|If I add 0.01 moles of sodium hydroxide, I'm actually adding 0.01 moles of hydroxide ion.
TUBGGXVJj3Q-027|So let's look at that.
TUBGGXVJj3Q-033|Do I need to do an equilibrium calculation here?
TUBGGXVJj3Q-034|No I don't, because the K for this reaction, the reverse reaction is K W, So 1 over a K W is the K for this reaction.
TUBGGXVJj3Q-035|K W is times 10 to the minus 14, so the K for this reaction 10 to the plus 14.
TUBGGXVJj3Q-036|So this goes completely to the water.
TUBGGXVJj3Q-037|So every mole of OH- finds a mole of H3O+ and forms water.
TUBGGXVJj3Q-038|One for one.
TUBGGXVJj3Q-048|And that 0.19 moles is in about a liter.
TUBGGXVJj3Q-049|It's in a liter plus the 10 mils of any OH we added.
TUBGGXVJj3Q-050|About a liter.
TUBGGXVJj3Q-051|So it's about 0.19 molar.
TUBGGXVJj3Q-052|So the H3O+ concentration now will be about 0.19 molar.
TUBGGXVJj3Q-053|Minus log of 0.19 molar, 0.72.
TUBGGXVJj3Q-054|So the pH has changed from 0.7 to 0.72.
TUBGGXVJj3Q-055|Not very much.
TUBGGXVJj3Q-061|Now, I can look at adding acid and base.
TUBGGXVJj3Q-062|Here I have an acid solution and a base solution.
TUBGGXVJj3Q-063|So this is concentrated, this is not as concentrated.
TUBGGXVJj3Q-064|As I add base to acid, we can watch the pH change by the color of the solution.
TUBGGXVJj3Q-066|So let's watch this acid-base titration.
TUBGGXVJj3Q-067|Adding a strong base to a strong acid solution, let's watch that happen.
TUBGGXVJj3Q-068|PH changes slowly at first, and then rapidly to blue-- the base solution.
TUBGGXVJj3Q-069|So this is well down here on the titration curve.
TUBGGXVJj3Q-070|I've gone past the equivalence point to the point where the base is determining the pH.
TUBGGXVJj3Q-071|A basic solution here maybe around pH 10 to 13-- a blue color for my indicator.
TUBGGXVJj3Q-072|So there's adding a strong base to a strong acid solution.
vApxjo9sAmI-000|A rubber band is a system that could do work or have work done on it.
vApxjo9sAmI-001|So let's take a rubber band and stretch it, hold it there so it comes to thermal equilibrium, and then suddenly release it.
vApxjo9sAmI-007|We're talking about stretching and releasing a rubber band, or any piece of rubber.
vApxjo9sAmI-008|I have a rubber glove here.
vApxjo9sAmI-009|I can stretch it and hold it, let it come to thermal equilibrium.
vApxjo9sAmI-010|I can even check that against my lip.
vApxjo9sAmI-011|Yeah, that's room temperature.
vApxjo9sAmI-012|And then suddenly release it and touch it to my lip again.
vApxjo9sAmI-013|What happened to the temperature change?
vApxjo9sAmI-014|Well you can do that experiment, I'll do the math.
vApxjo9sAmI-015|It's an adiabatic change, and we arranged for it to be adiabatic simply by doing it quickly.
vApxjo9sAmI-020|The rubber has to pull my hands back together.
vApxjo9sAmI-021|So a system that does work uses its internal energy to do the work.
vApxjo9sAmI-022|So the energy must go down.
vApxjo9sAmI-024|The energy change and the temperature change are locked for an ideal gas.
vApxjo9sAmI-025|That is, for an ideal gas, the energy and the temperature change are essentially the same parameter.
vApxjo9sAmI-026|E is 3/2 nRT.
vApxjo9sAmI-027|For other systems like rubber, it's almost the same.
vApxjo9sAmI-028|That is, energy changes and temperature changes are very often correlated, and in this case, they are.
TZuRh_epSyI-000|Let's talk some more about bonding.
TZuRh_epSyI-001|When two elements come together to bond, they can share electrons to fill their octets.
TZuRh_epSyI-002|And when we write down elements and their electrons as dots to form bonds, we're writing down what's known as Lewis electron dot structures.
TZuRh_epSyI-003|It was Gilbert Lewis who first proposed that eight electrons form a stable octet and the basis of much of bonding.
TZuRh_epSyI-004|So let's look at a few examples.
TZuRh_epSyI-005|Here's carbon bonded to four hydrogens.
TZuRh_epSyI-006|And why does that occur?
TZuRh_epSyI-007|Well, carbon has four valence electrons.
TZuRh_epSyI-008|Hydrogen has one.
TZuRh_epSyI-009|So if a carbon wants to fill its octet, it needs four more electrons.
TZuRh_epSyI-010|Each hydrogen has one.
TZuRh_epSyI-011|So if it bonds to four hydrogens it can get to its octet.
TZuRh_epSyI-012|Carbon says I'll share one with you.
TZuRh_epSyI-013|I'll share one with you.
TZuRh_epSyI-014|I'll share one with you and one with you.
TZuRh_epSyI-015|And what I'll end up with is a carbon with an octet.
TZuRh_epSyI-016|And here, instead of the dots, I've drawn a single line to indicate the pair of electrons.
TZuRh_epSyI-017|So carbon has two, four, six, eight electrons around it.
TZuRh_epSyI-018|Each hydrogen has a pair of electrons.
TZuRh_epSyI-019|And a pair is sufficient for hydrogen because hydrogen only goes up to two.
TZuRh_epSyI-020|It fills up principal quantum level one.
TZuRh_epSyI-021|And the maximum number of electrons there is two.
TZuRh_epSyI-022|Now, there's some other examples.
TZuRh_epSyI-025|Well, six on one and six on another is a total of 12.
TZuRh_epSyI-026|How do these oxygens share 12 electrons so that each octet is fulfilled?
TZuRh_epSyI-027|Well, here's the best way to do it.
TZuRh_epSyI-028|I write the oxygens with a double bond.
TZuRh_epSyI-029|Each of these lines represent two electrons that the oxygens are sharing.
TZuRh_epSyI-030|So if I count up the electrons-- two, four, six, eight, 10, 12, that's the 12 valence electrons that I can use for bonding.
TZuRh_epSyI-031|How did I share them?
TZuRh_epSyI-032|Well, this oxygen can count two, four, six, eight in its octet.
TZuRh_epSyI-033|And this oxygen can count two, four, six, eight in its octet.
TZuRh_epSyI-034|The shared electrons can be counted on both atoms.
TZuRh_epSyI-035|That's how nitrogen bonds as well.
TZuRh_epSyI-036|Nitrogen, the molecule, is a 10 electron system.
TZuRh_epSyI-037|Each nitrogen has five valence electrons.
TZuRh_epSyI-039|We write the Lewis electron dot structure with a triple bond.
TZuRh_epSyI-040|Here it is.
TZuRh_epSyI-041|Nitrogen each sharing six electrons.
TZuRh_epSyI-042|The total count is 10- two, four, six, eight, 10.
TZuRh_epSyI-043|Those are all the valence electrons.
TZuRh_epSyI-044|For each nitrogen it's two, four, six, eight, a stable octet.
TZuRh_epSyI-045|Two, four, six, eight on the other nitrogen.
TZuRh_epSyI-046|Now the octet rule is followed a lot of the time, but there are exceptions.
TZuRh_epSyI-048|The boron trifluoride molecule is relatively stable, but boron only has six electrons around it.
TZuRh_epSyI-049|But because it's not a stable octet, it turns out boron trifluoride as a molecule is very reactive.
TZuRh_epSyI-050|Boron is looking to fill its octet.
TZuRh_epSyI-051|In fact, boron trifluoride will react with ammonia.
TZuRh_epSyI-053|When you mix boron trifluoride and ammonia, immediately they form a compound between the nitrogen and boron.
TZuRh_epSyI-054|And a new molecule is formed.
TZuRh_epSyI-055|And we'll see that in the demo lab.
TZuRh_epSyI-056|Here's xenon with four florines bonded to it and two lone pairs.
TZuRh_epSyI-057|That's two, four, six, eight 10, 12, 12 electrons around xenon.
TZuRh_epSyI-058|That's an expanded octet, more than eight.
TZuRh_epSyI-059|Elements that have larger nuclei have higher principle quantum levels they can access.
TZuRh_epSyI-060|So xenon can go to d orbitals in higher principle quantum levels.
TZuRh_epSyI-061|And it's not restricted to an octet it can expand its octet using extra orbitals for bonding.
TZuRh_epSyI-062|The same thing with SF6.
TZuRh_epSyI-064|So as you go below principle quantum level three, you get extra orbitals.
TZuRh_epSyI-065|You don't have to satisfy yourself with just your s and p orbitals in an octet.
TZuRh_epSyI-067|And that's stable.
TZuRh_epSyI-068|But expanded octets are also explained by quantum mechanics.
TZuRh_epSyI-069|It says as you go to higher principle quantum levels, you have more orbitals that you can use for bonding.
DVeicfdiyQk-003|Here, we're not at that standard-state conditions.
DVeicfdiyQk-004|We're at a different set of conditions.
DVeicfdiyQk-005|We want to determine if that set of conditions is at equilibrium or will the reaction proceed towards products or reactants to reach equilibrium?
DVeicfdiyQk-006|So we know the equilibrium constant is 41.
DVeicfdiyQk-011|I can plug in the conditions I've been given.
DVeicfdiyQk-015|So Q is less than K. That means-- less than k, that means the denominator must be too big.
DVeicfdiyQk-016|The reactants are too big.
DVeicfdiyQk-017|So I'm going to proceed toward products to get Q to equal k.
DVeicfdiyQk-019|Now, I should anticipate that that's going to be negative.
DVeicfdiyQk-020|Because I've already said the reaction is going to go towards products.
DVeicfdiyQk-021|That means it has to proceed further downhill.
DVeicfdiyQk-022|The reactant free energies are still too high compared to the product free energies.
DVeicfdiyQk-035|So everything rings true.
DVeicfdiyQk-036|I can calculate reaction quotients.
nEniUpvACdY-000|Let's a get a chemical reaction and some data and see if we can determine the rate orders.
nEniUpvACdY-001|So, what's the order with respect to the partial pressure of bromine gas in this chemical reaction?
nEniUpvACdY-002|Hydrogen gas plus bromine gas makes hydrogen bromide gas.
nEniUpvACdY-003|I have some data here, some possible rate orders.
nEniUpvACdY-010|We're looking at a chemical reaction and some data and trying to determine rate orders.
nEniUpvACdY-011|So our chemical reaction hydrogen gas bromine gas goes to hydrogen bromide, and some data for initial partial pressures and rates.
nEniUpvACdY-015|When I quadruple the concentration of bromine, holding everything else constant, the rate doubles.
nEniUpvACdY-016|So if I'm going to write down a rate law, it sounds like the rate should be proportional to Br2 to the 1/2 power, the square root.
nEniUpvACdY-017|Quadruple the concentration gives you a doubling of the rate.
nEniUpvACdY-018|Well, as long as we're here, what about hydrogen?
nEniUpvACdY-019|Can we figure that out too, the power with respect to hydrogen or the order with respect to hydrogen?
nEniUpvACdY-021|And when I double the H2 partial pressure, I double the rate holding the Br concentration constant.
nEniUpvACdY-022|So that's constant, doesn't matter, folded into the rate constant.
nEniUpvACdY-023|Doubling this doubles that, so that should be a power of one.
nEniUpvACdY-024|So what it looks like is our overall power is 1 and 1/2, or 3/2, the overall order of the rate is 3/2.
nEniUpvACdY-025|We just asked for the order with respect to Br2.
0JyN9PH2oNg-000|Water can act as an acid or a base.
0JyN9PH2oNg-001|Water can react with itself, one molecule acting like an acid, another acting like a base, to form H3O+.
0JyN9PH2oNg-002|When this water molecule donates a proton to this water molecule, I'll form an H3O+ and leave behind an OH-.
0JyN9PH2oNg-007|So the equilibrium expression for the autodissociation of water is the product of H3O+ and OH- concentrations.
0JyN9PH2oNg-008|Now, this equilibrium is always true in water solution.
0JyN9PH2oNg-009|Remember regardless of what other equilibria are occurring, all equilibria work together in a solution to satisfy themselves.
0JyN9PH2oNg-010|So in water, this is always true.
0JyN9PH2oNg-014|Acid-base forms of water have a very small K.
0JyN9PH2oNg-015|So this reaction very much favors the water.
0JyN9PH2oNg-020|So pure liquid water in equilibrium, the H3O+ and OH- concentrations are equal.
0JyN9PH2oNg-023|And recall, when you take the log, you're taking the exponent, log base 10, of 10 to the exponent is just the exponent.
0JyN9PH2oNg-024|So minus log of 10 to the minus 7 is 7.
0JyN9PH2oNg-025|So the pH of pure liquid water is 7.
0JyN9PH2oNg-026|The pOH of pure liquid water is 7 due to the autoionization and autodissociation of water.
f8b4i8DtA1Y-000|An interesting application of the ideal gas law is swimming and diving.
f8b4i8DtA1Y-001|When you dive beneath the surface of water, your body experiences an increase in pressure.
f8b4i8DtA1Y-002|That's because the water has mass, and that mass, pushing down on your body, increases pressure.
f8b4i8DtA1Y-003|The increase in pressure goes, for every 10 meters in depth, you get an additional 1 atmosphere of pressure.
f8b4i8DtA1Y-004|So if you're at depth and you hold your breath, with a fixed amount of air in your lungs, and you ascend, that can actually be dangerous, as that volume expands.
f8b4i8DtA1Y-005|Which is the most dangerous, is the question I have for you?
f8b4i8DtA1Y-015|We're talking about diving, and when you dive beneath the surface of water, the pressure changes one atmosphere for every 10 meters in change of depth.
f8b4i8DtA1Y-016|So how can I calculate the volume in my lungs, how that will change?
f8b4i8DtA1Y-017|If I hold my breath and fix the amount of air, then the volume will expand or contract with the pressure changes.
f8b4i8DtA1Y-018|As I ascend, the volume wants to increase, because the pressure decreases.
f8b4i8DtA1Y-019|And if I hold my breath, that increase in volume can actually damage my lung tissues and my chest cavity.
f8b4i8DtA1Y-022|So P2 times V2 is P1 times V1.
f8b4i8DtA1Y-025|Let's do that for each ascent.
f8b4i8DtA1Y-030|So this is a change of 2-- a factor of 2 in volume.
f8b4i8DtA1Y-031|The volume will double if I do this ascent.
f8b4i8DtA1Y-032|These two are a factor of 1 and 1/2.
f8b4i8DtA1Y-033|So it's interesting that the greatest change in volume occurs nearest the surface.
f8b4i8DtA1Y-034|And since that's where we do most of our swimming, that's where we have to be most careful.
f8b4i8DtA1Y-035|So the greatest danger in holding your breath, you get the greatest volume change due to pressure, is from 10 meters to the surface.
wihnb2KZ1Bs-001|For instance, here's two entries carbon-12 and naturally occurring carbon.
wihnb2KZ1Bs-002|They have different relative masses.
wihnb2KZ1Bs-003|That's because I could take a sample of pure carbon-12, where every atom is carbon-12, and that would have mass 12.
wihnb2KZ1Bs-004|But in naturally occurring carbon, 1 out of every 100 atoms is a carbon-13, it's slightly more massive.
wihnb2KZ1Bs-005|So if I took this piece of carbon and I started picking out atoms, 1 in 100 would be that carbon-13 with a slightly more massive nucleus.
wihnb2KZ1Bs-006|That's what gives this naturally occurring carbon that slightly higher relative mass.
wihnb2KZ1Bs-007|It's the mass weighted average of all the isotopes.
wihnb2KZ1Bs-008|And you can see many elements have a potpourri of different isotopes.
wihnb2KZ1Bs-009|They don't have integer molecular masses and atomic masses, because of the presence of all the different isotopes.
wihnb2KZ1Bs-010|Now for the most part, isotopes are chemically similar, that is they do the same chemical reactions, but of course they have different masses.
wihnb2KZ1Bs-011|A lot of times the mass difference is very small.
wihnb2KZ1Bs-012|For hydrogen, there's hydrogen with mass 1 and deuterium with mass 2.
wihnb2KZ1Bs-013|That's a factor of 2 in mass-- that's very large.
wihnb2KZ1Bs-014|But for carbon already, carbon-12 versus carbon-13, that's a 1% difference in mass.
wihnb2KZ1Bs-015|That's relatively small.
wihnb2KZ1Bs-017|That's a very small percentage difference in mass.
wihnb2KZ1Bs-018|Now you can separate pure isotopes and this is very common.
wihnb2KZ1Bs-019|For instance, uranium is a good example because we take uranium-238 and we separate out the other isotopes and we make pure uranium-238.
wihnb2KZ1Bs-020|It's called depleted uranium because the radioactive fissible nuclei are removed.
wihnb2KZ1Bs-022|Uranium-235 is the fissible nucleus.
wihnb2KZ1Bs-023|It's used in nuclear reactors.
wihnb2KZ1Bs-024|So we have different properties of the nuclei that are different isotopes.
wihnb2KZ1Bs-025|Chemically they're very similar, but sometimes their mass gives them different physical properties that we can use.
wihnb2KZ1Bs-026|In general, we catalog them based on the potpourri, the mass weighted average of all the isotopes in the mixture.
wihnb2KZ1Bs-027|And that's how we achieve relative atomic and molecular masses.
JOjrtRHsozI-000|Let's look at some naturally-occurring polychromic acids, amino acids.
JOjrtRHsozI-001|Amino acids are found in your body.
JOjrtRHsozI-002|They're the building blocks of the proteins in your body.
JOjrtRHsozI-003|They're called amino acids because each one has an amino group and an acid group.
JOjrtRHsozI-004|Here's the amino group and the carboxylic COOH group.
JOjrtRHsozI-005|Now, I've drawn them at pH 7.
JOjrtRHsozI-006|And I draw them at pH 7, that means this carboxylic acid group with pKa 2 will be in its basic form.
JOjrtRHsozI-011|Here, tyrosine, as well, has a side group.
JOjrtRHsozI-016|And in this case, that will be a neutral charge.
JOjrtRHsozI-017|Notice, when I protonate the amino group here at the end of lysine, that attains a positive charge.
JOjrtRHsozI-022|That's going to be in its basic form at pH 7.
JOjrtRHsozI-024|Here's histidine.
JOjrtRHsozI-028|And notice, I'm very close here.
JOjrtRHsozI-029|There's just a factor of one pH unit.
JOjrtRHsozI-030|So you know there's a factor of 10.
JOjrtRHsozI-031|There's 10 times as much of the unprotonated basic form as the protonated form.
JOjrtRHsozI-033|Now, that's very important, because the charge on amino acids, especially their side groups, affects the structure of proteins.
JOjrtRHsozI-034|Proteins are molecules in your body that do all the catalysis in your body.
JOjrtRHsozI-035|They help your chemical reactions go.
JOjrtRHsozI-036|And proteins are long chains of amino acids, and one of the reasons proteins can operate is their three-dimensional structure.
JOjrtRHsozI-038|One of the things that holds those structures together are charges on the side groups.
JOjrtRHsozI-039|You have a positively-charged side group here, a negatively-charged side group here.
JOjrtRHsozI-040|Those are attracted coulombically, and can anchor the three-dimensional structure, hold it together by that coulombic attraction.
JOjrtRHsozI-042|If the molecule falls apart, it can't perform its catalytic operation anymore.
JOjrtRHsozI-043|And I can actually demonstrate that for you.
JOjrtRHsozI-044|There's enzymes, one of them peroxidase, which are very common in nature.
JOjrtRHsozI-045|In fact, in turnips.
JOjrtRHsozI-046|I have some turnip here.
JOjrtRHsozI-048|That is, it will break down hydrogen peroxide into water and oxygen. So I'm going to do that.
JOjrtRHsozI-049|I'm going to bring in a blender here-- and I've already ground up some turnip in advance here, kind of in the Julia Child child style.
JOjrtRHsozI-050|So I've got a slurry already.
JOjrtRHsozI-051|I'm going to make a little more here.
JOjrtRHsozI-053|To one of the beakers, I'm going to add HCL, a strong acid.
JOjrtRHsozI-054|So let me get my safety glasses on here.
JOjrtRHsozI-055|I'll add a strong acid.
JOjrtRHsozI-058|So if I change the PH, I change the protonated state, and I should affect the activity of the peroxidase.
JOjrtRHsozI-059|That molecule should denature.
JOjrtRHsozI-060|So what I'm going to do now is add hydrogen peroxide to each flask, and we should see peroxidase activity, bubbling oxygen here.
wXt_tUPtAe8-000|Let's look at some electronic configurations.
wXt_tUPtAe8-001|Fluorine minus, sodium, or sodium plus.
wXt_tUPtAe8-002|Which of those three can have the electronic configuration helium 2s2 2p5 3s1?
wXt_tUPtAe8-011|Sodium plus has one fewer electrons, so it's a 10-electron system.
wXt_tUPtAe8-013|Fluorine minus, same situation.
wXt_tUPtAe8-014|It's a 10-electron species, and it's in a slightly excited state.
wXt_tUPtAe8-015|The 2p electron promoted to the 3s.
wXt_tUPtAe8-016|So what we have are two 10-electron species.
wXt_tUPtAe8-017|Every 10-electron species that's slightly excited-- that is, has a 3s1 electron-- will have this electronic configuration.
wXt_tUPtAe8-018|Electronic configurations are not unique.
wXt_tUPtAe8-019|If you have to put 10 electrons around an atom, you have to put them in a certain way.
wXt_tUPtAe8-020|It's the number of protons that determine what the atom actually is, whether it's a sodium or fluorine.
wXt_tUPtAe8-021|The number of electrons is completely independent.
wXt_tUPtAe8-022|I could have too many electrons, I could have a negative ion, I can remove electrons and have the positive ion.
wXt_tUPtAe8-023|So almost any element can have almost any electronic configuration.
wXt_tUPtAe8-024|So you have to know, to determine the species, the number of protons in the nucleus.
wXt_tUPtAe8-025|We don't know that here.
k3ISd1oHxKA-000|If I have a particle that has wavelike properties and is trapped in a box, we understand that it has different energy states.
k3ISd1oHxKA-001|And how do I make a transition between one energy state and the next energy state?
k3ISd1oHxKA-002|Well, I absorb energy.
k3ISd1oHxKA-006|Now, I haven't drawn A and B to scale.
k3ISd1oHxKA-007|The question is, which transition is bigger, has the larger energy transition?
k3ISd1oHxKA-008|Is A bigger than B?
k3ISd1oHxKA-009|Are they about the same, or is A smaller than B?
k3ISd1oHxKA-013|We're looking at identical particles trapped in two different boxes, one box twice the size of the other box.
k3ISd1oHxKA-015|Well, you could calculate the energies based on this expression.
k3ISd1oHxKA-016|But you could also just look at the energy levels of both boxes.
k3ISd1oHxKA-017|Here's n equals 1, for instance, for the smaller box, and n equals 2 for the smaller box.
k3ISd1oHxKA-018|n equals 1 for the larger box here, and n equals 2 for the larger box.
k3ISd1oHxKA-019|And here's where you notice something interesting.
k3ISd1oHxKA-020|If n equals 2 and you put that into the expression, you get a 2 here and a 2L here.
k3ISd1oHxKA-022|So you get cancellation in the 2 case for both the larger box and n equal 1 case for the smaller box.
k3ISd1oHxKA-023|Those energy levels turn out to be the same.
k3ISd1oHxKA-026|So the correct answer here, A transition smaller than B transition.
SKjxtaGYxjk-000|Let's look at the periodic table in a slightly different way.
SKjxtaGYxjk-001|What I've done here is I've written out the periodic table and I've represented by dots the number of electrons in the outermost shell.
SKjxtaGYxjk-006|And if you have the same number of valence electrons, you'll have the same properties.
SKjxtaGYxjk-007|You'll react the same.
SKjxtaGYxjk-008|You'll bond the same to other atoms.
SKjxtaGYxjk-009|And that's why, in the periodic table, elements with similar properties line up because they have similar numbers of valence electrons.
SKjxtaGYxjk-010|So our valence electrons go from one to eight along principal quantum level two and from one to eight along principal quantum level three.
SKjxtaGYxjk-014|Neon and argon have a closed shell, stable octet of electrons.
SKjxtaGYxjk-015|They're not as reactive as their counterparts in the corresponding rows.
SKjxtaGYxjk-016|Same thing with helium, it's a closed shell of two electrons.
SKjxtaGYxjk-017|That's the maximum number of electrons in principle quantum level one.
SKjxtaGYxjk-018|And helium is also not very reactive.
SKjxtaGYxjk-019|But what we're going to look at now is how these valence electrons affect the bonding and the reactions of the elements in the periodic table.
SKjxtaGYxjk-020|In fact, reactions and bonding are the subject of pretty much the rest of chem one.
WVLx-q-hZGk-000|Let's look at the titration of a strong acid with a strong base, our strong acid, HCl, and our strong base, NaOH.
WVLx-q-hZGk-001|HCl, the strong acid, totally dissociates in water.
WVLx-q-hZGk-002|In fact, that's the definition of strong acid-- totally dissociates in water.
WVLx-q-hZGk-003|NaOH totally dissociates in water.
WVLx-q-hZGk-004|That's a strong base.
WVLx-q-hZGk-005|If I titrate an HCl solution with 0.1 molar NaOH, I get this titration curve.
WVLx-q-hZGk-006|The question is, how does the titration curve change if I titrate with 0.2 molar NaOH?
WVLx-q-hZGk-014|We're talking about titrating HCl, a strong acid, with two different strong base solutions.
WVLx-q-hZGk-015|One, 0.1 molar, and then doing it again with 0.2 molar strong base.
WVLx-q-hZGk-016|The question is, how does that second-- 0.2 molar-- pH curve look?
WVLx-q-hZGk-017|Well, the initial HCl concentration is the same, so I'm titrating the same acid solution.
WVLx-q-hZGk-018|So the starting point is the same.
WVLx-q-hZGk-019|So A is out already.
WVLx-q-hZGk-024|So the new pH curve will look like this, an equivalence point twice as early in half the volume.
WVLx-q-hZGk-025|And that's selection C. The correct answer here is C.
4nVIxjOAAGY-002|Now, that's what I do for steric number two.
4nVIxjOAAGY-005|How about steric number three?
4nVIxjOAAGY-006|When I've steric number three, I need to accommodate bond angles of 120 degrees.
4nVIxjOAAGY-011|So hybridisation allows us to accommodate the geometry for various steric numbers.
ZwsWjelzqDA-000|Particles, especially tiny ones, have a wavelike property that we can resolve.
ZwsWjelzqDA-001|And we saw that with electrons going through a crystal diffracting.
ZwsWjelzqDA-003|It actually gets a little spookier than that.
ZwsWjelzqDA-006|It's a very, very, very spooky property of matter.
ZwsWjelzqDA-007|And from here on in chemistry we're going to talk about those very spooky quantum properties of matter.
ZwsWjelzqDA-008|Let's start by looking at a classic experiment or a classic calculation, a particle in a box.
ZwsWjelzqDA-009|You can imagine taking a particle that has a wavelike property and trapping it in a small region of space.
ZwsWjelzqDA-012|I'll say a box of length L, and the particle has to be between here and here.
ZwsWjelzqDA-013|It'll have a wavelike property, so I'll draw a wavelike expression on this box.
ZwsWjelzqDA-014|Now, this wavelike expression I'll call the wave function of that particle.
ZwsWjelzqDA-015|The wave function of that particle will give the designation psi, in this case psi of x because this is the x dimension in space.
ZwsWjelzqDA-016|Now, we've already said that the intensity of the wave squared is going to give us indication of the probability of finding the particle.
ZwsWjelzqDA-017|So the probability size squared gives us the probability of finding the particle.
ZwsWjelzqDA-018|Clearly in this case, the probability of finding the particle right in the middle of the box would be very high if this particle behaves like a wave.
ZwsWjelzqDA-019|Well, how do I come up with these wave functions?
ZwsWjelzqDA-020|Well, I can actually do some mathematics.
ZwsWjelzqDA-022|And when I do those things, I can solve the differential equation for the particle stuck in this tiny little box.
ZwsWjelzqDA-023|It's behaving like a wave.
ZwsWjelzqDA-024|There are those wave functions psi.
ZwsWjelzqDA-025|These are how the wave functions psi must interact with each other if it's a particle to be stuck in this box.
ZwsWjelzqDA-026|Now, we won't go through all the mathematics, but you can solve this.
ZwsWjelzqDA-027|And when you solve this equation, you get an expression for psi.
ZwsWjelzqDA-028|It's not that hard to understand.
ZwsWjelzqDA-029|And it looks like a wave function.
ZwsWjelzqDA-033|There's not just one solution to this expression.
ZwsWjelzqDA-034|There's multiple solutions.
ZwsWjelzqDA-035|All the integers n work and give you a solution to this equation.
ZwsWjelzqDA-036|So that means there's not one wave function that works.
ZwsWjelzqDA-037|There are several wave functions that work.
ZwsWjelzqDA-038|You can have a ground state, the smallest value of n, but you can have what we call excited states.
ZwsWjelzqDA-039|Higher values of n also give wave functions, and you can tell what happens.
ZwsWjelzqDA-043|We can also say, well, what's the energy of that particle?
ZwsWjelzqDA-044|We can calculate the energy using our expression for the wave function and the probability squared.
ZwsWjelzqDA-045|And we can say, well, I'll rank the energy.
ZwsWjelzqDA-046|These particles increase in energy as n increases.
ZwsWjelzqDA-048|The length of the box, the quantum number, Planck's constant are all in there.
ZwsWjelzqDA-049|So I can then plot out, well, the lowest energy state n equal 1.
ZwsWjelzqDA-050|We have no n equal 0 state.
ZwsWjelzqDA-051|N equal 1 is where we start our counting.
ZwsWjelzqDA-052|N equal 2, higher energy state.
ZwsWjelzqDA-053|You'll see n goes in as the square, so higher n, higher energy state.
ZwsWjelzqDA-054|N equal 3.
ZwsWjelzqDA-056|Zero probability are the places where the wave function goes to 0.
ZwsWjelzqDA-057|After you square 0, you still get 0.
ZwsWjelzqDA-058|So the probability of finding the particle at these crossings is zero.
ZwsWjelzqDA-059|And that's a strange characteristic of particles that behave like waves.
ZwsWjelzqDA-060|There's portions of the box where the particle is forbidden to be.
ZwsWjelzqDA-061|So this is interesting.
ZwsWjelzqDA-062|I'll have two nodes, 1, 2.
ZwsWjelzqDA-063|Node a, area where the wave function goes to 0 in the high energy state, one node here, and zero nodes here.
ZwsWjelzqDA-064|When the number of nodes increase, the energy state increases.
ZwsWjelzqDA-065|That's a higher energy situation.
ZwsWjelzqDA-066|So here's what I have.
ZwsWjelzqDA-067|I have a wave.
ZwsWjelzqDA-069|Only certain wavelengths can exist.
ZwsWjelzqDA-070|So half a wavelength, one full wavelength, 1 and 1/2 wavelengths.
ZwsWjelzqDA-072|So if I fix the ends of the box, I put boundaries on a wave.
ZwsWjelzqDA-073|I naturally get what I call quantization.
ZwsWjelzqDA-074|Not every wave can fit in these boxes.
ZwsWjelzqDA-075|Only certain waves can fit in these boxes.
ZwsWjelzqDA-076|And there's a gap.
ZwsWjelzqDA-077|I go from here, n equal 1, all the way up to here, and I skip all those energies in between.
ZwsWjelzqDA-078|I can only have this energy state for the box or this energy state for the box and no energy states in between.
ZwsWjelzqDA-079|The energy is quantized.
ZwsWjelzqDA-080|And waves naturally do this.
ZwsWjelzqDA-081|It's not unusual to see it.
ZwsWjelzqDA-082|We can demonstrate it with audio waves.
ZwsWjelzqDA-083|Audio, sound, is a wavelike property, and here's a tube of fixed length.
ZwsWjelzqDA-084|So if I want waves to exist on this tube, only certain wavelengths will fit.
ZwsWjelzqDA-085|I'll be able to fit a full wave on this, a wave on this, a wave and a half on this.
ZwsWjelzqDA-086|So only certain sounds will fit in this tube.
ZwsWjelzqDA-087|That's interesting characteristic.
ZwsWjelzqDA-088|I can demonstrate a couple of the sounds that fit in this tube.
ZwsWjelzqDA-089|We won't hear when we figure the sounds that fit in this tube.
ZwsWjelzqDA-090|We won't hear a continuous sound.
ZwsWjelzqDA-091|We won't hear [MAKING SOUND EFFECT] continuous wavelengths.
ZwsWjelzqDA-092|We'll have [MAKING SOUND EFFECT] and [MAKING SOUND EFFECT],, two individual wavelengths that fit in this box.
ZwsWjelzqDA-093|Let's actually demonstrate that.
ZwsWjelzqDA-094|I'll try to spin this, and you guys can listen.
ZwsWjelzqDA-095|There is one wave that fits.
ZwsWjelzqDA-097|And there's that low frequency, long wavelength, two energy levels.
ZwsWjelzqDA-098|And you can see energy.
ZwsWjelzqDA-100|I wonder if I get a higher one.
ZwsWjelzqDA-102|That's a beautiful acoustic example of waves and quantization.
ZwsWjelzqDA-103|All you need to get quantization is take a wave and force it to exist in a certain area of space.
ZwsWjelzqDA-104|If you fix the ends of a wave, you get quantization.
ZwsWjelzqDA-106|And when waves are bounded, you get quantization.
ZwsWjelzqDA-107|You get a particle in a box.
sY-0uDLNYmk-000|Let's look at the bromination of diene.
sY-0uDLNYmk-002|Now, as these reactions go, there's an equilibrium here, so these two products can interconvert.
sY-0uDLNYmk-005|Notice that the 1-bromo is the same throughout.
sY-0uDLNYmk-012|We're looking at the bromination of a diene.
sY-0uDLNYmk-013|It can be brominated in the 1 position or the 3 position, and we have some data about the chemical reaction.
sY-0uDLNYmk-014|So as the chemical reaction occurs initially, the 1-bromo is favored, so it's kinetically favored.
sY-0uDLNYmk-015|It has a short pathway or a low activation energy to forming relative to the 3-bromo.
sY-0uDLNYmk-016|If you look at the reaction over time though, as I go out to four hours, what's happening is the 3-bromo is building up in concentration.
sY-0uDLNYmk-017|So over time, the 3-bromo is more stable, so it's more thermodynamically stable.
sY-0uDLNYmk-020|The thermodynamically favored means there should be a larger difference in energy between the reactants and the products.
sY-0uDLNYmk-022|So in this case, the correct answer is B.
xcJkUZ4fzEE-000|Let's do a calculation involving isotopes.
xcJkUZ4fzEE-003|Here we're going to take a sample of pure carbon and look at the mass spectrum.
xcJkUZ4fzEE-004|Now remember, pure carbon means we take a sample of pure carbon from the ground and we run it through the mass spectrometer.
xcJkUZ4fzEE-005|The question is, what is this peak at mass 13?
xcJkUZ4fzEE-006|What's the molar mass of the sample?
xcJkUZ4fzEE-007|And how many protons and neutrons are in carbon-12?
xcJkUZ4fzEE-008|We can go through these initially.
xcJkUZ4fzEE-009|The sample is naturally occurring carbon.
xcJkUZ4fzEE-010|So it's 1% carbon-13 and 99% carbon-12.
xcJkUZ4fzEE-011|And that's reflected in the mass spectrum, a peak about 100 times as big at 12 as you have at 13.
xcJkUZ4fzEE-012|So that explains the mass spectrum, the naturally occurring isotope carbon-13.
xcJkUZ4fzEE-013|Now, the molar mass is the weighted average of those two isotopes.
xcJkUZ4fzEE-014|The ratio, in every 100 atoms, there are 99 carbon 12's and one carbon-13.
xcJkUZ4fzEE-024|And it determines the identity of the atom.
xcJkUZ4fzEE-025|Remember, if you know the number of protons, you know the identity of the element-- not the mass, the number of protons.
xcJkUZ4fzEE-026|It has mass 12.
xcJkUZ4fzEE-027|So those other 6 mass units, 6 for the proton, mass 12, there's 6 more mass units.
xcJkUZ4fzEE-028|Those must be the neutrons.
xcJkUZ4fzEE-029|So there's 6 neutrons in carbon-12 nucleus.
8qqpmXD8cjU-000|Let's look at some weak acids and weak bases in solution and see if we can predict which will have the highest pH.
8qqpmXD8cjU-001|So here I have three solutions.
8qqpmXD8cjU-002|They'll all be at 0.1 molar.
8qqpmXD8cjU-003|Which has the highest pH?
8qqpmXD8cjU-012|A weak base, weak base equilibrium constant-- 10 to the minus 8.
8qqpmXD8cjU-013|So this reacts as a weak base.
8qqpmXD8cjU-014|This will be a slightly basic solution.
8qqpmXD8cjU-015|Another of ours is sodium acetate.
8qqpmXD8cjU-019|Here's the acetate ion reacting with water as a base.
8qqpmXD8cjU-020|We had a Kb 10 to the minus 10.
8qqpmXD8cjU-021|So a Kb 10 to the minus 10, this base considerably stronger than this base.
8qqpmXD8cjU-022|The last one we have, NH4Cl, that forms NH4 plus ions in solution-- and Cl minus.
8qqpmXD8cjU-023|And we've seen that.
8qqpmXD8cjU-025|So this last one is an acid.
8qqpmXD8cjU-030|And of these two, NH3 is the stronger base.
8qqpmXD8cjU-031|So the highest pH, the highest OH minus concentration, comes from the strongest base.
8qqpmXD8cjU-032|And in this case that is NH3.
Xt-yom79grM-000|Let's look at the equilibrium between liquid water and gaseous water at 25 degrees C.
Xt-yom79grM-001|And what can we say about the standard state free energy difference and K for that physical process?
Xt-yom79grM-014|That's where water comes up to the boiling point.
Xt-yom79grM-015|1 atmosphere of pressure is the pressure that defines the boiling point.
Xt-yom79grM-016|So, I have 1 atmosphere of pressure.
Xt-yom79grM-017|That's too high for 25 degrees C.
Xt-yom79grM-018|So you would guess that this reaction in the standard state must favor the reactants.
Xt-yom79grM-019|The pressure is too high.
Xt-yom79grM-020|1 atmosphere is too high a pressure for 25 degrees C.
Xt-yom79grM-021|So, we're going to favor the reactant.
Xt-yom79grM-022|That means delta G will be positive.
Xt-yom79grM-023|Delta G positive in the standard means K has to be less than 1.
Xt-yom79grM-024|Those two things go together by our definition of delta G standard is minus RTLNK.
Xt-yom79grM-025|So, I can make the correlation that delta G is bigger than zero, and K is less than 1.
Xt-yom79grM-027|And it's equal to the equilibrium constant.
Xt-yom79grM-028|So, I could have started from that direction and said, the equilibrium constant for this reaction at 25 degrees C is less than 1.
Xt-yom79grM-029|That means the standard state free energy difference must be greater than zero.
Xt-yom79grM-031|And that's what I'm looking at.
Xt-yom79grM-032|So, in either case, standard state free energy, greater than zero.
QL6TFnaXvMg-000|Let's look at a can of a carbonated beverage that's been, say, left in the back of a car and it's warmed up.
QL6TFnaXvMg-001|A warm day, carbonated beverage expands.
QL6TFnaXvMg-002|So you'll notice the can is larger, about ready to pop.
QL6TFnaXvMg-003|What does that mean about the dissolution of carbon dioxide?
QL6TFnaXvMg-011|We're talking about carbon dioxide gas dissolved in a carbonated beverage.
QL6TFnaXvMg-012|As the beverage warms, the can expands.
QL6TFnaXvMg-013|Now, the can expands because carbon dioxide is coming out of solution.
QL6TFnaXvMg-014|You might have thought, well, the can expands because there's a little gas in the can and that warms up and that expands.
QL6TFnaXvMg-015|But the amount of gas in the can is very small, so it doesn't account for all the expansion of the can.
QL6TFnaXvMg-016|The big factor accounting for the expansion of the can is carbon dioxide in the solution coming out and creating a larger volume of carbon dioxide.
QL6TFnaXvMg-017|And why would that happen?
QL6TFnaXvMg-018|Well, as you warm it, if this is an exothermic reaction, heat would be a product.
QL6TFnaXvMg-019|Then warming it would cause the reaction to shift back towards reactants-- the carbon dioxide gas.
pKQgwIPlNL4-000|Let's look at the equilibrium in the bloodstream between hemoglobin, carbon monoxide, and oxygen.
pKQgwIPlNL4-001|Now, carbon monoxide bonds strongly to hemoglobin.
pKQgwIPlNL4-002|And that's why carbon monoxide is poison.
pKQgwIPlNL4-003|It can displace the oxygen from your blood and replace it with carbon monoxide.
pKQgwIPlNL4-004|That ties up that oxygen-carrying molecule.
pKQgwIPlNL4-005|And even though you could breathe in and out, you can't bind any of the oxygen you're breathing in.
pKQgwIPlNL4-006|So you can actually suffocate while you breathe because carbon monoxide has bound the hemoglobin.
pKQgwIPlNL4-007|The question I have for you is, how could you reverse that?
pKQgwIPlNL4-008|What's the best procedure?
pKQgwIPlNL4-016|We're looking at the reaction between hemoglobin oxygenated and carbon monoxide.
pKQgwIPlNL4-017|Carbon monoxide strongly binds the hemoglobin and displaces oxygen. What we're looking for is a strategy to reverse that.
pKQgwIPlNL4-018|So our options are increase the partial pressure of oxygen.
pKQgwIPlNL4-019|And if we increase the partial pressure of oxygen, that's a product.
pKQgwIPlNL4-020|Increasing something on the product side will tend to shift the reaction back towards reactants.
pKQgwIPlNL4-021|Increasing a product will automatically make Q larger than K. You're increasing the numerator.
pKQgwIPlNL4-022|If you increase the numerator, what do you need to do?
pKQgwIPlNL4-023|Shift back towards reactants, increase the size of the denominator, decrease the size of the numerator, a shift back towards reactants.
pKQgwIPlNL4-024|We could also just look at it as you put something on this side, the reaction we want to shift back towards this side.
pKQgwIPlNL4-027|You will put an oxygen mask, have them breathe pure oxygen.
pKQgwIPlNL4-028|That increases the partial pressure of oxygen and drives the carbon monoxide out of the system.
pKQgwIPlNL4-029|So here, the correct answer, increase the partial pressure of oxygen.
V_0Bnc_URos-000|What causes condensation in a real gas?
V_0Bnc_URos-001|Well, in an ideal gas, the particles don't interact.
V_0Bnc_URos-002|There's no attraction or repulsion energy between them.
V_0Bnc_URos-003|In fact, if you plotted their interaction energy versus distance, it would be very boring.
V_0Bnc_URos-004|There's no interaction energy, regardless of the distance between the ideal gas particles.
V_0Bnc_URos-005|For real gas particles, there is an interaction energy.
V_0Bnc_URos-009|Large distances, the interaction between the particles, isn't very important.
V_0Bnc_URos-010|As the particles get closer, that interaction energy is more important.
V_0Bnc_URos-011|And you'll see, the plot goes to a stabilization energy as the particles get closer.
V_0Bnc_URos-012|As the particles get very close, then that interaction energy becomes unfavorable.
V_0Bnc_URos-013|That is, there's an actual repulsion.
V_0Bnc_URos-014|You can't push the particles on top of each other.
V_0Bnc_URos-017|And this attraction energy, if that is higher than the kinetic energy of the particles, then the particles can actually associate.
V_0Bnc_URos-018|That is, for low attraction energies, the kinetic energy is larger than the attraction energy, and the kinetic energy of the particles overwhelms the attraction energy.
V_0Bnc_URos-019|But as the kinetic energy goes down-- and how do I lower kinetic energy?
V_0Bnc_URos-020|Lower the temperature.
V_0Bnc_URos-021|As I lower the kinetic energy, then the attraction energy becomes a more important factor, and the particles associate.
V_0Bnc_URos-022|That's the nature of condensation.
o6fk49PNN2s-000|The natural direction of a process is determined by the entropy change in the universe, but it's difficult to track the entropy of the system in the surroundings.
o6fk49PNN2s-001|So we'd rather have an expression that involves just the system.
o6fk49PNN2s-005|Now, this entropy of the surroundings.
o6fk49PNN2s-006|What's happening in this reaction?
o6fk49PNN2s-007|Well, water liquid is changing into water gas.
o6fk49PNN2s-008|In order for that to happen, heat needs to flow from the surroundings into the system.
o6fk49PNN2s-009|How much heat?
o6fk49PNN2s-010|Well, the enthalpy of vaporization of water.
o6fk49PNN2s-011|So I can write this entropy change of the surroundings in terms of the enthalpy change in the system.
o6fk49PNN2s-012|And I'm going to give it a negative sign because it's the surroundings that's sending heat into the system, so the surroundings is losing heat.
o6fk49PNN2s-018|So when minus delta S is less than 0, the forward direction is favored by the universe.
o6fk49PNN2s-021|And when the Gibbs free energy of the system is less than or equal to 0, the forward process is favored.
o6fk49PNN2s-022|When it's equal to 0 exactly, you're in equilibrium.
LOSpsszfwWU-000|Let's look at the effect of temperature on a chemical reaction from a kinetic standpoint.
LOSpsszfwWU-001|So for the following chemical reaction, the forward rate constant is larger than the reversed rate constant-- this chemical reaction here.
LOSpsszfwWU-002|What does that mean about the equilibrium constant for an increase in temperature?
LOSpsszfwWU-011|We're looking at the effect of temperature on an equilibrium constant from a kinetic standpoint.
LOSpsszfwWU-012|So the equilibrium constant is the ratio of K forward to K reverse, the rate constants.
LOSpsszfwWU-013|So we should be able to predict the temperature effect on that equilibrium constant.
LOSpsszfwWU-015|This doesn't give us the whole rate.
LOSpsszfwWU-017|But, what this ratio tells us is that the rate constant for the forward is larger.
LOSpsszfwWU-018|That means the activation energy for the forward is smaller.
LOSpsszfwWU-019|There is a lower barrier, that's what makes the forward reaction intrinsically faster.
LOSpsszfwWU-020|The reverse activation energy is slightly larger.
LOSpsszfwWU-024|The larger the activation energy, the more sensitive the rate constant is to a change in temperature.
LOSpsszfwWU-025|So for a large activation energy, K changes a lot with a small change in temperature.
LOSpsszfwWU-029|This one will change more dramatically.
LOSpsszfwWU-030|This will get larger with temperature faster.
LOSpsszfwWU-031|That will make the overall equilibrium constant get smaller with an increase in temperature.
LOSpsszfwWU-032|So for this, the equilibrium constant decreases with temperature.
LOSpsszfwWU-033|Now there's another way you could have looked at that.
LOSpsszfwWU-035|And if it's exothermic, heat is a product.
LOSpsszfwWU-036|I increase the temperature, that favors the reactants and leads to a decrease in equilibrium constant.
LOSpsszfwWU-037|Two arguments that lead to the same conclusion for the equilibrium constant and temperature.
ESX5q4FtkAo-000|For a sample of gases, even if that's a mixture of gases, the gases have the same pressure and the same kinetic energy, as long as they're at the same temperature.
ESX5q4FtkAo-001|Now the gases could be very different.
ESX5q4FtkAo-003|Now they do it differently.
ESX5q4FtkAo-004|That is bromine comes in with large, massive shots to the walls to exert pressure, where hydrogen comes in with lighter shot, but more rapidly.
ESX5q4FtkAo-005|The property that differentiates the gases is their velocity.
ESX5q4FtkAo-006|They have different velocities based on their masses.
ESX5q4FtkAo-007|So the velocity is the distinguishing characteristic, and it would be nice if we could solve for that velocity.
ESX5q4FtkAo-008|Our expression for the kinetic energy has the velocity squared, the mean velocity squared, in it.
ESX5q4FtkAo-009|So we can solve for the mean velocity squared, take the square root, and we'll have a quantity called the root mean squared velocity.
ESX5q4FtkAo-010|Now it's not exactly the mean velocity, but it's very close, and it depends on fundamental properties like the temperature and the nature of the gas.
ESX5q4FtkAo-011|So we can easily solve for that quantity.
ESX5q4FtkAo-016|Here we'll express R in joules, so our velocities come out in meters per second.
ESX5q4FtkAo-018|So I'll take the square root of these three quantities, and I'll find a very simple expression for an important distinguishing property for the gases-- their actual velocity.
ESX5q4FtkAo-019|The root mean squared velocity for a mole of gas is 3RT over the molar mass, square root.
VY3yL2nDHuQ-000|We all know glass is transparent, so let's look at absorption over a broader range of the electromagnetic spectrum.
VY3yL2nDHuQ-001|Transparent glass should have an absorption spectrum that looks like either 1, 2, or 3.
VY3yL2nDHuQ-008|The absorption spectrum of transparent glass is what we're talking about.
VY3yL2nDHuQ-009|And we're talking about a rather broad region of the electromagnetic spectrum from ultraviolet wavelengths down to infrared wavelengths.
VY3yL2nDHuQ-010|And I've overlaid the common visible colored wavelengths in the center.
VY3yL2nDHuQ-013|But in order for the glass to be transparent, all visible wavelengths must pass through.
VY3yL2nDHuQ-014|Transparent glass will past white light directly through it.
VY3yL2nDHuQ-015|You can see all colors through transparent glass, so all colors must pass through.
VY3yL2nDHuQ-016|So we have to choose one that has no absorption spectra in the visible range.
VY3yL2nDHuQ-017|That is option 1.
VY3yL2nDHuQ-018|So the correct answer here is A.
p26UwTYXXF0-000|Let's calculate the heat involved for a physical process.
p26UwTYXXF0-001|How much energy is required to heat 250 grams of water from 50 degrees C to 110 degrees C?
p26UwTYXXF0-002|Now, the way you do this, is to break it up into distinct energy changes.
p26UwTYXXF0-008|The sum of those three enthalpies will give me the enthalpy for the total process.
p26UwTYXXF0-009|So we can calculate each one of them.
YttroH0ftiA-001|And the energy amount is h times nu, Planck's constant times the frequency of the light.
YttroH0ftiA-002|Now a particle has momentum.
YttroH0ftiA-003|And we've seen the momentum of the particle, that photon, can cause an electron to be ejected from a metal.
YttroH0ftiA-004|We saw in the photoelectric effect a incoming photon ejecting electrons from a metal.
YttroH0ftiA-005|So how does that particle nature and wave nature reconcile themselves?
YttroH0ftiA-006|Well, let's talk about that.
YttroH0ftiA-007|The light wave particle duality.
YttroH0ftiA-009|Now the wave particle, the light particle, that we call a photon has a momentum.
YttroH0ftiA-010|We've seen it can transfer momentum from the photon to the electron, but the momentum is we often associate with mass.
YttroH0ftiA-011|But the photon has no mass.
YttroH0ftiA-012|The photon is a particle and it massless, moving at the speed of light.
YttroH0ftiA-014|Now we have two expressions for the energy, the energy of the photon and the relativistic energy, m c squared.
YttroH0ftiA-016|So the momentum is m times c times another c gives you m c squared.
YttroH0ftiA-020|Waves and particles acting the way they choose, sometimes light will behave like a wave, sometimes it behaves like a particle.
YttroH0ftiA-021|There's a duality between them expressed by this beautiful relationship between the wavelength and the momentum.
YttroH0ftiA-022|Planck's constant, again, an extremely small number, is the proportionality constant between the momentum and the wavelength.
YttroH0ftiA-023|So waves, particles, light, we have to think of them all at the same time.
JnNVdsrkKV8-000|Let's do some calculations with weak acids and weak bases.
JnNVdsrkKV8-001|We're going to talk about the salt of hydrocyanic acid and formic acid.
JnNVdsrkKV8-002|Now, the salt is the compound that's formed when you react the acid with a strong base.
JnNVdsrkKV8-003|You form the salt. You take the counter ion.
JnNVdsrkKV8-005|The conjugate ions from the base and the acid.
JnNVdsrkKV8-006|Sodium coming from the strong base, and cyanide coming from the acid.
JnNVdsrkKV8-007|So the salt of acids contains the weak base.
JnNVdsrkKV8-008|If I did that with formate, I'd take formic acid and react it with sodium hydroxide.
JnNVdsrkKV8-009|I'd form sodium formate.
JnNVdsrkKV8-010|And sodium formate contains the formate ion.
JnNVdsrkKV8-011|That's the conjugate base of formic acid.
JnNVdsrkKV8-013|So oh actually I've said, which is less basic?
JnNVdsrkKV8-014|So which has the lower pH?
JnNVdsrkKV8-024|So let's do that.
JnNVdsrkKV8-028|So the formate ion is the weaker base.
JnNVdsrkKV8-029|It will be less basic in solution.
JnNVdsrkKV8-030|It will have the lower pH.
JnNVdsrkKV8-031|So let's calculate that.
JnNVdsrkKV8-037|So formate ion plus water forms OH minus, the base, and of course the conjugate acid of the formate ion.
JnNVdsrkKV8-038|So by forming this base, that's what raises the pH above 7 when you put sodium formate, a salt, in water.
JnNVdsrkKV8-039|How much base is formed?
JnNVdsrkKV8-040|Well, we can do the calculations.
JnNVdsrkKV8-041|We can say, well, I took 0.1 Molar of this ion from the salt and put it in water.
JnNVdsrkKV8-042|Initially, there was no OH minus.
JnNVdsrkKV8-060|And you see it's 2.37 times 10 to the minus 6.
JnNVdsrkKV8-061|And indeed, that's much smaller than 10 to the minus 1.
JnNVdsrkKV8-062|Our original concentration, 10 to the minus 1, the amount that associates, about 10 to the minus 6.
JnNVdsrkKV8-063|So it's indeed true that x is small compared to 0.1.
JnNVdsrkKV8-064|I have the OH minus concentration.
JnNVdsrkKV8-065|And in water, H3O plus times OH minus is always 10 to the minus 14.
JnNVdsrkKV8-069|Now, I solved for H3O plus because that's how we get the pH.
JnNVdsrkKV8-070|We take minus log of H3O plus.
JnNVdsrkKV8-071|I could have taken minus log of OH minus.
JnNVdsrkKV8-072|That would be the pOH, and then subtracted that from 14.
JnNVdsrkKV8-073|That's the same mathematics basically.
JnNVdsrkKV8-074|Or say the pH is minus log of H3O plus and 8.37.
JnNVdsrkKV8-075|So here, we have a salt of a weak acid, formate ion, added to water.
JnNVdsrkKV8-076|It makes the water slightly basic.
rvH5eufKoe8-009|We're comparing two acids, HF, a weak acid, and HBr, a strong acid.
rvH5eufKoe8-010|We know the relative bond enthalpies go like this.
rvH5eufKoe8-014|Now, that's really the indicator of what we're looking for here because, if K is less than 1, then the standard state free energy difference must be positive somewhere.
rvH5eufKoe8-015|It must have some positive values.