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yHIDWAYbYBs-015|This second balloon has only oxygen. |
yHIDWAYbYBs-016|Now, many people think that oxygen is a very dangerous gas, very explosive. |
yHIDWAYbYBs-017|But it turns out oxygen is an explosive gas in the presence of other molecules to react with. |
yHIDWAYbYBs-018|Oxygen by itself is not an explosive gas. |
yHIDWAYbYBs-019|It will react rapidly with other molecules. |
yHIDWAYbYBs-020|In this case, we haven't given it any other molecules, except maybe the material in the balloon. |
yHIDWAYbYBs-021|But in general, this is going to give almost no explosion. |
yHIDWAYbYBs-022|The third case, I have hydrogen and oxygen intimately mixed together. |
yHIDWAYbYBs-023|Now in this case when the flame comes through the balloon, hydrogen and oxygen are primed to react. |
yHIDWAYbYBs-024|They're very close to each other. |
yHIDWAYbYBs-025|A hydrogen doesn't have to go very far to find an oxygen. |
yHIDWAYbYBs-028|Now Chemical Quizzes that we're going to do-- ChemQuizzes-- throughout this series are going to require you to think outside the box or outside the balloon. |
yHIDWAYbYBs-029|This one had a lot of chemical principles, especially for early on in the course, but I wanted to get you a feel for what you have to think about to do ChemQuizzes. |
yHIDWAYbYBs-030|You might not have got this one because of the complicated nature of the quick reaction versus the slow reaction, but keep trying. |
yHIDWAYbYBs-031|We're going to give you more principles and more tools to be able to perform on these ChemQuizzes. |
eqHY6DH8QRI-000|For an ideal gas, the product of the pressure and the volume is a constant for a constant temperature. |
eqHY6DH8QRI-001|If you change the temperature, the value of the constant changes, but the product of pressure and volume is still a constant. |
eqHY6DH8QRI-002|Let's look at how changing the temperature affects our curves. |
eqHY6DH8QRI-007|Now the product of pressure and volume being a constant, means pressure and volume are inversely proportional. |
eqHY6DH8QRI-008|As I increase pressure, volume decreases. |
eqHY6DH8QRI-009|Decrease pressure volume increases. |
eqHY6DH8QRI-010|So their product is always a constant. |
eqHY6DH8QRI-011|It turns out, that's how your lungs work. |
eqHY6DH8QRI-012|In order for your lungs to draw air in, your chest cavity expands higher value, that means lower pressure. |
eqHY6DH8QRI-013|And high pressure air from the atmosphere can rush into your lungs. |
eqHY6DH8QRI-016|There's a model chest cavity. |
eqHY6DH8QRI-017|The model chest cavity has that muscle in your chest, the diaphragm. |
eqHY6DH8QRI-018|And when that contracts, the volume of your chest cavity increases. |
eqHY6DH8QRI-019|Higher volume, lower pressure, air rushes in. |
eqHY6DH8QRI-022|Decrease volume, higher pressure, air rushes out. |
eqHY6DH8QRI-025|Taking advantage of the relationship p times v is the constant. |
9mTP53W2_RY-000|Glucose is an energy source for your body. |
9mTP53W2_RY-001|That is glucose is oxidized to release energy. |
9mTP53W2_RY-002|Now, that oxidation you could do on the benchtop. |
9mTP53W2_RY-003|You could burn glucose in oxygen and it would release energy. |
9mTP53W2_RY-004|In your body, it's done via enzymes, and the reaction occurs at body temperatures. |
9mTP53W2_RY-006|So if the initial and final states are the same, we know that free energy and enthalpy and entropy changes are the same. |
9mTP53W2_RY-007|They're a state function. |
9mTP53W2_RY-008|So the same amount of energy that you release on the benchtop is available to your body when you oxidize glucose in your body. |
9mTP53W2_RY-013|So rather than drawing it as a ring, I just bent this bond all the way around to show that this carbon is attached to that carbon. |
9mTP53W2_RY-014|Here's oxygen, carbon dioxide, and water. |
9mTP53W2_RY-015|And we know how to assign oxidation numbers from Lewis structures. |
9mTP53W2_RY-019|So oxygen gets these two. |
9mTP53W2_RY-020|Carbon and carbon here, that's the same elements. |
9mTP53W2_RY-021|So I'll share these electrons and assign one to this carbon. |
9mTP53W2_RY-022|And carbon is slightly more electronegative than hydrogen, so I'll assign both of them to the carbon here. |
9mTP53W2_RY-023|So I'm assigning one, two, three, four electrons to the carbon based on relative electronegativities. |
9mTP53W2_RY-024|And carbon normally has four electrons around it in its valence shell. |
9mTP53W2_RY-025|So it has oxidation number 0. |
9mTP53W2_RY-027|Then I see that it has a plus 4 oxidation number here. |
9mTP53W2_RY-028|So it goes from 0 to plus 4, that's an increase in oxidation number-- formerly, an oxidation. |
9mTP53W2_RY-029|So something has to be reduced. |
9mTP53W2_RY-030|If there's an oxidation, there must be a reduction as well. |
9mTP53W2_RY-031|Here's elemental oxygen, where the oxidation number is 0. |
9mTP53W2_RY-032|And what about the oxygen over here? |
9mTP53W2_RY-033|The oxygen in water, I assign all of the electrons to the oxygen. It's the more electronegative element. |
9mTP53W2_RY-034|So it has 1, 2, 3, 4, 5, 6, 7, 8 electrons around it. |
9mTP53W2_RY-035|The atomic oxygen would have 6, so it has two more than it normally has as an atom. |
9mTP53W2_RY-036|So that's an oxidation number of minus 2. |
9mTP53W2_RY-037|Indeed, that's a reduction. |
9mTP53W2_RY-038|I go from 0 to minus 2. |
9mTP53W2_RY-039|A reduction in oxidation number is a reduction. |
g0nvNTN4XHc-000|Let's look at the valence electrons of some atoms and see how they participate in bonding. |
g0nvNTN4XHc-001|Now, we've talked about several kinds of bonding-- ionic bonding and covalent bonding. |
g0nvNTN4XHc-002|Let's look at ionic bonding first. |
g0nvNTN4XHc-003|If you look at lithium and fluorine, sodium and chlorine and their valence electrons. |
g0nvNTN4XHc-004|You'll notice that fluorine and chlorine has seven valence electrons. |
g0nvNTN4XHc-005|If they could pick up one more, they'd get to eight and that stable octet. |
g0nvNTN4XHc-006|Sodium and lithium, should they lose one, then they'll have a closed shell underneath that primary shell. |
g0nvNTN4XHc-007|So lithium and sodium are kind of primed to give up an electron and go to the closed shell underneath. |
g0nvNTN4XHc-008|Fluorine and chlorine are primed to accept electrons and close their existing shells at that stable octet. |
g0nvNTN4XHc-009|So one way that you can describe the bonding is an electron transfer. |
g0nvNTN4XHc-010|One element gives up an electron to another element. |
g0nvNTN4XHc-011|You form a positive ion and a negative ion. |
g0nvNTN4XHc-012|And then Coulombic attraction, the attraction between plus and minus charges, holds those elements together in a bond. |
g0nvNTN4XHc-013|So for lithium and fluorine, an electron goes from the lithium to the fluorine. |
g0nvNTN4XHc-014|You get a Li plus and F minus holding each other together. |
g0nvNTN4XHc-016|And this can happen with multiple atoms as well. |
g0nvNTN4XHc-018|It can give up one to each of two chlorines. |
g0nvNTN4XHc-019|It becomes plus 2 charge. |
g0nvNTN4XHc-020|Each chlorine is negative 1 charge. |
g0nvNTN4XHc-021|And all three of those are held together by Coulombic interaction. |
g0nvNTN4XHc-022|So ionic bonding is the transfer of electrons from one species to another species. |
g0nvNTN4XHc-023|And then the plus-minus interaction holds those two together in a bond. |
-WxXRDp-IuU-000|Let's look at the sublimation of two different solids, iodine and carbon dioxide. |
-WxXRDp-IuU-002|Would it look like A, B, or C? |
-WxXRDp-IuU-003|We can start these two subliming. |
-WxXRDp-IuU-012|We're talking about the sublimation of iodine and carbon dioxide. |
-WxXRDp-IuU-015|So with that information, can we determine how the plot of lnK versus 1/T look relative to each other? |
-WxXRDp-IuU-016|Well, lnK versus 1/T, the slope is determined by the enthalpy, the intercept by the entropy. |
-WxXRDp-IuU-017|So the entropy of sublimation probably about the same. |
-WxXRDp-IuU-019|So we expect the entropy of sublimation is determined mainly by the fact that it's a phase change from solid to gas. |
-WxXRDp-IuU-020|Entropy of sublimation about the same means the intercept of these two plots should be about the same. |
-WxXRDp-IuU-023|That sublimes at a very low temperature. |
-WxXRDp-IuU-024|And the bonds of carbon dioxide, not held together in the solid as strongly as iodine. |
-WxXRDp-IuU-025|Because we see the iodine at room temperature and we don't see the solid carbon dioxide at room temperature. |
-WxXRDp-IuU-026|So we expect the enthalpy of sublimation to be lower. |
-WxXRDp-IuU-027|The slope determines the enthalpy. |
-WxXRDp-IuU-028|Enthalpy is determined by the slope for lnK versus 1/T. |
-WxXRDp-IuU-029|So we expect a less dramatic slope for the carbon dioxide, a lower enthalpy of sublimation, so a less negative slope. |
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