ceaglex/VisualTTS_Chem · Datasets at Hugging Face

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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.

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.