ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
-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.

-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.