ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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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.

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.