ceaglex/VisualTTS_Chem · Datasets at Hugging Face

text stringlengths 19 206
ksZoOde-7zY-026\|So what you're doing is saying I have three p electrons in the nitrogen state.
ksZoOde-7zY-027\|When I go to oxygen, I add a paired electron.
ksZoOde-7zY-028\|That's a little bit tough.
ksZoOde-7zY-029\|Pairing electrons, there's a little bit of repulsion energy.
ksZoOde-7zY-031\|So oxygen, slightly easier to ionize than its counterparts.
ksZoOde-7zY-033\|And of course, then sodium, look how similar sodium is to lithium.
ksZoOde-7zY-043\|And we understood that in terms of the fine structure of the electronic configurations.
ksZoOde-7zY-044\|A stable half-filled shell or the pairing energy affecting ionization energies.
ksZoOde-7zY-045\|So this trend is one that we can look at in detail and with the quantum mechanics of a very good understanding of the trends.
e7Ofj4bDdqI-000\|Let's look at several molecules and see if we can predict their relative boiling or condensation points.
e7Ofj4bDdqI-001\|Here's three molecules-- H2O, H2S, or NH3.
e7Ofj4bDdqI-011\|Now, all these particles have the three classic intermolecular interactions.
e7Ofj4bDdqI-012\|Dispersion forces, they have dipole-dipole interactions, and hydrogen bonding.
e7Ofj4bDdqI-013\|Whenever hydrogen bonding is involved though, that's the dominant force.
e7Ofj4bDdqI-014\|It's orders of magnitude greater than the other two forces.
e7Ofj4bDdqI-015\|So what we have to determine is, which of these has the strongest hydrogen bonding interaction?
e7Ofj4bDdqI-016\|The strongest hydrogen bonding interaction will come from the greatest dipole.
e7Ofj4bDdqI-017\|So is the OH bond, the SH bond, or the NH bond the greatest dipole moment?
e7Ofj4bDdqI-018\|Well, we determine that by the difference in electronegativity.
e7Ofj4bDdqI-019\|You know O is more electronegative than S is more electron give than N, nitrogen.
e7Ofj4bDdqI-023\|Sulphur slightly smaller, and the NH bond slightly smaller yet.
e7Ofj4bDdqI-024\|So the greatest dipole-dipole gives the greatest hydrogen bonding interaction, and in that case, that's water.
e7Ofj4bDdqI-025\|So among these three, the highest boiling point is the one with the strongest intermolecular interaction.
e7Ofj4bDdqI-026\|Remember, the intermolecular interaction has to be overcome by kinetic energy to boil.
e7Ofj4bDdqI-027\|Higher kinetic energy, higher temperature.
e7Ofj4bDdqI-028\|So more intermolecular interactions that are very strong means higher temperature to boil.
e7Ofj4bDdqI-029\|So in this case, water is the largest intermolecular interaction, largest hydrogen bonding interaction, and that's the correct answer.
e7Ofj4bDdqI-032\|So we have the hydrogen bonding interactions reflected in the boiling points and the size of the van der Waals A parameters.
QoHZH6OkzQs-000\|Let's look at molecular oxygen.
QoHZH6OkzQs-001\|Molecular oxygen is paramagnetic, but it can undergo a transition where it becomes diamagnetic.
QoHZH6OkzQs-002\|The question I have for you is, which process involving a photon makes paramagnetic oxygen into diamagnetic oxygen?
QoHZH6OkzQs-011\|Now these would be effectively paired and this would be a non-magnetic species, diamagnetic.
QoHZH6OkzQs-012\|An ionization event won't allow you to go to a diamagnetic species.
QoHZH6OkzQs-013\|If you remove an electron from a paramagnetic oxygen, you still have an unpaired electron and it's still paramagnetic.
QoHZH6OkzQs-015\|Well I think you remember from filling electrons into orbitals, we put them in spin parallel because anti-parallel is the higher energy state.
QoHZH6OkzQs-016\|Making those magnets line up anti-parallel is a higher energy interaction.
QoHZH6OkzQs-018\|Now it's an absorption in the red region of the visible spectrum, a low energy red photon.
QoHZH6OkzQs-019\|And oxygen, liquid oxygen will absorb red photons and it will appear blue.
QoHZH6OkzQs-020\|Remember, things that absorb in the red will transmit to blues.
QoHZH6OkzQs-021\|So liquid oxygen appears blue.
QoHZH6OkzQs-022\|We can see that in the demonstration lab.
JXHl3rYqlHU-000\|For an ideal gas, the product of the pressure and volume is constant for a fixed temperature.
JXHl3rYqlHU-001\|What happens when we change the temperature?
JXHl3rYqlHU-010\|Now, on this scale, as I go to zero in volume, that will represent an absolute zero, because I can't go negative in volume.
JXHl3rYqlHU-011\|There's no such thing as negative volume.
JXHl3rYqlHU-012\|So as I get to zero in volume, I reach a logical zero in temperature.
JXHl3rYqlHU-013\|And we call that absolute zero because we can't measure a temperature on this scale below that temperature.
JXHl3rYqlHU-014\|The Celsius temperature scale has a zero set at the freezing point of water.
JXHl3rYqlHU-015\|And I can measure this zero on this temperature scale on the Celsius scale.
JXHl3rYqlHU-016\|It turns out to be minus 273 Celsius.
JXHl3rYqlHU-017\|So minus 273 Celsius is an absolute zero in temperature.
JXHl3rYqlHU-020\|Our standard temperature and pressure will be 1 atmosphere of pressure and 0 Celsius, or 273 Kelvin.
JXHl3rYqlHU-021\|The Kelvin unit, the Kelvin degree, and the Celsius degree are the same size.
JXHl3rYqlHU-022\|The 0 is offset by 273 degrees.
JXHl3rYqlHU-023\|Now, let's look at our temperature-volume relationship in a demonstration.
JXHl3rYqlHU-024\|I can take a volume of gas and expose it to very low temperature.
JXHl3rYqlHU-025\|Here's liquid nitrogen. This is at about 70 on the absolute temperature scale, 70 Kelvin.
JXHl3rYqlHU-026\|As I expose a volume of gas to that very low temperature, the volume decreases.
JXHl3rYqlHU-027\|And we understand that this volume is decreasing linearly.
JXHl3rYqlHU-028\|As the temperature decreases, the volume decreases in a corresponding linear fashion.
JXHl3rYqlHU-029\|I'm going to go down to very near, within 70 degrees of that absolute zero temperature.
JXHl3rYqlHU-030\|That's where this liquid nitrogen is.
JXHl3rYqlHU-031\|The volume of the balloon has gone very, very, very, very low.
JXHl3rYqlHU-032\|Now, let's remove the balloon from liquid nitrogen and let it expand.
JXHl3rYqlHU-033\|As the temperature increases, the volume of the gas increases.
JXHl3rYqlHU-034\|The volume of the gas increases linearly as the temperature increases.
JXHl3rYqlHU-035\|And you can see this volume comes back to near its original volume.
JXHl3rYqlHU-036\|Linear relationship between temperature and volume.
JXHl3rYqlHU-042\|Let's say I take a 1 mole sample of gas, and I look at just half of the mole, half of the gas, half a mole.
JXHl3rYqlHU-043\|That half a mole of gas, those particles, they behave as if they have the whole volume to themselves.
JXHl3rYqlHU-044\|And that's how gas particles always behave.
JXHl3rYqlHU-045\|Gas particles don't interact with each other in the ideal gas sense.
JXHl3rYqlHU-046\|So they behave like they have the whole volume.
JXHl3rYqlHU-047\|Half a mole of a sample will exert half the pressure.
JXHl3rYqlHU-048\|It'll occupy the whole volume, but exert half the pressure.
JXHl3rYqlHU-049\|1/3 third of the particles will exert 1/3 of the pressure.
JXHl3rYqlHU-050\|So we can look at fractional pressures and partial pressures in terms of partial number of particles.
fysCY3d9zAU-000\|Equilibrium is the condition where products and reactants freely interchange.
fysCY3d9zAU-001\|The free energy difference between the products and the reactants is 0, so there's no energy penalty to switch back and forth.
fysCY3d9zAU-002\|And they do so freely.
fysCY3d9zAU-003\|Now at equilibrium, the products or the reactants may be favored.
fysCY3d9zAU-005\|So those two things are correlated.
fysCY3d9zAU-007\|So if I looked at this in the standard state, I'd have 1 molar d and c, 1 molar a and b.
fysCY3d9zAU-008\|And the free energy difference between those would be the standard free energy difference.
fysCY3d9zAU-009\|Now that's not the equilibrium situation.
fysCY3d9zAU-010\|Obviously, if I have 1 molar of everything, that's an arbitrary standard condition that I've chosen.
fysCY3d9zAU-013\|K greater than 1 indicates products will be favored.
fysCY3d9zAU-014\|K less than 1 indicates reactants will be favored.
fysCY3d9zAU-019\|Products over reactants, if the product is larger than reactants, then that's greater than 1 in the ratio.
fysCY3d9zAU-020\|And that's a K greater than 1.
fysCY3d9zAU-021\|Products are favored.
fysCY3d9zAU-022\|So if both of these favor products, spontaneous is the label we give them.
fysCY3d9zAU-023\|Now notice that that label, spontaneous, means products are favored at equilibrium.
fysCY3d9zAU-024\|It doesn't have anything to do with the rate of the reaction or how fast we get there.
fysCY3d9zAU-025\|It just means that once we get to equilibrium, the products are favored.
fysCY3d9zAU-027\|Those are called not spontaneous reactions or non spontaneous.
fysCY3d9zAU-028\|Now just because a reaction is non spontaneous does not mean it doesn't form any products.
fysCY3d9zAU-029\|It doesn't mean the reaction doesn't go.
fysCY3d9zAU-030\|It just means when I get to equilibrium, it's the reactants that are favored.
fysCY3d9zAU-031\|There will be some products, but the reactants are favored.

ksZoOde-7zY-026|So what you're doing is saying I have three p electrons in the nitrogen state.

ksZoOde-7zY-027|When I go to oxygen, I add a paired electron.

ksZoOde-7zY-028|That's a little bit tough.

ksZoOde-7zY-029|Pairing electrons, there's a little bit of repulsion energy.

ksZoOde-7zY-031|So oxygen, slightly easier to ionize than its counterparts.

ksZoOde-7zY-033|And of course, then sodium, look how similar sodium is to lithium.

ksZoOde-7zY-043|And we understood that in terms of the fine structure of the electronic configurations.

ksZoOde-7zY-044|A stable half-filled shell or the pairing energy affecting ionization energies.

ksZoOde-7zY-045|So this trend is one that we can look at in detail and with the quantum mechanics of a very good understanding of the trends.

e7Ofj4bDdqI-000|Let's look at several molecules and see if we can predict their relative boiling or condensation points.

e7Ofj4bDdqI-001|Here's three molecules-- H2O, H2S, or NH3.

e7Ofj4bDdqI-011|Now, all these particles have the three classic intermolecular interactions.

e7Ofj4bDdqI-012|Dispersion forces, they have dipole-dipole interactions, and hydrogen bonding.

e7Ofj4bDdqI-013|Whenever hydrogen bonding is involved though, that's the dominant force.

e7Ofj4bDdqI-014|It's orders of magnitude greater than the other two forces.

e7Ofj4bDdqI-015|So what we have to determine is, which of these has the strongest hydrogen bonding interaction?

e7Ofj4bDdqI-016|The strongest hydrogen bonding interaction will come from the greatest dipole.

e7Ofj4bDdqI-017|So is the OH bond, the SH bond, or the NH bond the greatest dipole moment?

e7Ofj4bDdqI-018|Well, we determine that by the difference in electronegativity.

e7Ofj4bDdqI-019|You know O is more electronegative than S is more electron give than N, nitrogen.

e7Ofj4bDdqI-023|Sulphur slightly smaller, and the NH bond slightly smaller yet.

e7Ofj4bDdqI-024|So the greatest dipole-dipole gives the greatest hydrogen bonding interaction, and in that case, that's water.

e7Ofj4bDdqI-025|So among these three, the highest boiling point is the one with the strongest intermolecular interaction.

e7Ofj4bDdqI-026|Remember, the intermolecular interaction has to be overcome by kinetic energy to boil.

e7Ofj4bDdqI-027|Higher kinetic energy, higher temperature.

e7Ofj4bDdqI-028|So more intermolecular interactions that are very strong means higher temperature to boil.

e7Ofj4bDdqI-029|So in this case, water is the largest intermolecular interaction, largest hydrogen bonding interaction, and that's the correct answer.

e7Ofj4bDdqI-032|So we have the hydrogen bonding interactions reflected in the boiling points and the size of the van der Waals A parameters.

QoHZH6OkzQs-000|Let's look at molecular oxygen.

QoHZH6OkzQs-001|Molecular oxygen is paramagnetic, but it can undergo a transition where it becomes diamagnetic.

QoHZH6OkzQs-002|The question I have for you is, which process involving a photon makes paramagnetic oxygen into diamagnetic oxygen?

QoHZH6OkzQs-011|Now these would be effectively paired and this would be a non-magnetic species, diamagnetic.

QoHZH6OkzQs-012|An ionization event won't allow you to go to a diamagnetic species.

QoHZH6OkzQs-013|If you remove an electron from a paramagnetic oxygen, you still have an unpaired electron and it's still paramagnetic.

QoHZH6OkzQs-015|Well I think you remember from filling electrons into orbitals, we put them in spin parallel because anti-parallel is the higher energy state.

QoHZH6OkzQs-016|Making those magnets line up anti-parallel is a higher energy interaction.

QoHZH6OkzQs-018|Now it's an absorption in the red region of the visible spectrum, a low energy red photon.

QoHZH6OkzQs-019|And oxygen, liquid oxygen will absorb red photons and it will appear blue.

QoHZH6OkzQs-020|Remember, things that absorb in the red will transmit to blues.

QoHZH6OkzQs-021|So liquid oxygen appears blue.

QoHZH6OkzQs-022|We can see that in the demonstration lab.

JXHl3rYqlHU-000|For an ideal gas, the product of the pressure and volume is constant for a fixed temperature.

JXHl3rYqlHU-001|What happens when we change the temperature?

JXHl3rYqlHU-010|Now, on this scale, as I go to zero in volume, that will represent an absolute zero, because I can't go negative in volume.

JXHl3rYqlHU-011|There's no such thing as negative volume.

JXHl3rYqlHU-012|So as I get to zero in volume, I reach a logical zero in temperature.

JXHl3rYqlHU-013|And we call that absolute zero because we can't measure a temperature on this scale below that temperature.

JXHl3rYqlHU-014|The Celsius temperature scale has a zero set at the freezing point of water.

JXHl3rYqlHU-015|And I can measure this zero on this temperature scale on the Celsius scale.

JXHl3rYqlHU-016|It turns out to be minus 273 Celsius.

JXHl3rYqlHU-017|So minus 273 Celsius is an absolute zero in temperature.

JXHl3rYqlHU-020|Our standard temperature and pressure will be 1 atmosphere of pressure and 0 Celsius, or 273 Kelvin.

JXHl3rYqlHU-021|The Kelvin unit, the Kelvin degree, and the Celsius degree are the same size.

JXHl3rYqlHU-022|The 0 is offset by 273 degrees.

JXHl3rYqlHU-023|Now, let's look at our temperature-volume relationship in a demonstration.

JXHl3rYqlHU-024|I can take a volume of gas and expose it to very low temperature.

JXHl3rYqlHU-025|Here's liquid nitrogen. This is at about 70 on the absolute temperature scale, 70 Kelvin.

JXHl3rYqlHU-026|As I expose a volume of gas to that very low temperature, the volume decreases.

JXHl3rYqlHU-027|And we understand that this volume is decreasing linearly.

JXHl3rYqlHU-028|As the temperature decreases, the volume decreases in a corresponding linear fashion.

JXHl3rYqlHU-029|I'm going to go down to very near, within 70 degrees of that absolute zero temperature.

JXHl3rYqlHU-030|That's where this liquid nitrogen is.

JXHl3rYqlHU-031|The volume of the balloon has gone very, very, very, very low.

JXHl3rYqlHU-032|Now, let's remove the balloon from liquid nitrogen and let it expand.

JXHl3rYqlHU-033|As the temperature increases, the volume of the gas increases.

JXHl3rYqlHU-034|The volume of the gas increases linearly as the temperature increases.

JXHl3rYqlHU-035|And you can see this volume comes back to near its original volume.

JXHl3rYqlHU-036|Linear relationship between temperature and volume.

JXHl3rYqlHU-042|Let's say I take a 1 mole sample of gas, and I look at just half of the mole, half of the gas, half a mole.

JXHl3rYqlHU-043|That half a mole of gas, those particles, they behave as if they have the whole volume to themselves.

JXHl3rYqlHU-044|And that's how gas particles always behave.

JXHl3rYqlHU-045|Gas particles don't interact with each other in the ideal gas sense.

JXHl3rYqlHU-046|So they behave like they have the whole volume.

JXHl3rYqlHU-047|Half a mole of a sample will exert half the pressure.

JXHl3rYqlHU-048|It'll occupy the whole volume, but exert half the pressure.

JXHl3rYqlHU-049|1/3 third of the particles will exert 1/3 of the pressure.

JXHl3rYqlHU-050|So we can look at fractional pressures and partial pressures in terms of partial number of particles.

fysCY3d9zAU-000|Equilibrium is the condition where products and reactants freely interchange.

fysCY3d9zAU-001|The free energy difference between the products and the reactants is 0, so there's no energy penalty to switch back and forth.

fysCY3d9zAU-002|And they do so freely.

fysCY3d9zAU-003|Now at equilibrium, the products or the reactants may be favored.

fysCY3d9zAU-005|So those two things are correlated.

fysCY3d9zAU-007|So if I looked at this in the standard state, I'd have 1 molar d and c, 1 molar a and b.

fysCY3d9zAU-008|And the free energy difference between those would be the standard free energy difference.

fysCY3d9zAU-009|Now that's not the equilibrium situation.

fysCY3d9zAU-010|Obviously, if I have 1 molar of everything, that's an arbitrary standard condition that I've chosen.

fysCY3d9zAU-013|K greater than 1 indicates products will be favored.

fysCY3d9zAU-014|K less than 1 indicates reactants will be favored.

fysCY3d9zAU-019|Products over reactants, if the product is larger than reactants, then that's greater than 1 in the ratio.

fysCY3d9zAU-020|And that's a K greater than 1.

fysCY3d9zAU-021|Products are favored.

fysCY3d9zAU-022|So if both of these favor products, spontaneous is the label we give them.

fysCY3d9zAU-023|Now notice that that label, spontaneous, means products are favored at equilibrium.

fysCY3d9zAU-024|It doesn't have anything to do with the rate of the reaction or how fast we get there.

fysCY3d9zAU-025|It just means that once we get to equilibrium, the products are favored.

fysCY3d9zAU-027|Those are called not spontaneous reactions or non spontaneous.

fysCY3d9zAU-028|Now just because a reaction is non spontaneous does not mean it doesn't form any products.

fysCY3d9zAU-029|It doesn't mean the reaction doesn't go.

fysCY3d9zAU-030|It just means when I get to equilibrium, it's the reactants that are favored.

fysCY3d9zAU-031|There will be some products, but the reactants are favored.