ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
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.

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.