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VisualTTS_Chem

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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.