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pxxhXa3TZAU-015|The rates, if I cast it in terms of the zinc ion, should have the opposite sign, and there should be a factor of two. |
pxxhXa3TZAU-016|And that is, of course, C here. |
pxxhXa3TZAU-018|In this case, the correct answer is C. |
7UFQPsGBreg-000|Let's continue writing Lewis electron dot structures. |
7UFQPsGBreg-001|The cyanide ion, CN minus, is written down partially here. |
7UFQPsGBreg-002|Or is it partial? |
7UFQPsGBreg-003|Are there no bonds required to complete it? |
7UFQPsGBreg-004|One bond required to complete it? |
7UFQPsGBreg-012|We're writing the Lewis electron structure for CN minus, the cyanide ion. |
7UFQPsGBreg-013|First, we count up the valence electrons. |
7UFQPsGBreg-014|Carbon contributes 4, nitrogen contributes 5, and the ion has a negative charge, so there must be an extra electron. |
7UFQPsGBreg-015|That contributes 1. |
7UFQPsGBreg-016|So 10 total electrons. |
7UFQPsGBreg-017|That's actually the same number of electrons as the N2 molecule. |
7UFQPsGBreg-018|So CN minus requires two more bonds to get 10 total electrons. |
7UFQPsGBreg-019|I have 2, 4, 6, 8, 10 electrons in the cyanide ion. |
7UFQPsGBreg-021|CN minus, isoelectronic with N2, is a triple bonded structure. |
K6LVoYlzJ08-000|An acid is a compound that donates a proton. |
K6LVoYlzJ08-001|A base is a compound that accepts a proton. |
K6LVoYlzJ08-002|In water solutions, the acid donates its proton to water. |
K6LVoYlzJ08-011|So this Ka has the special label, but it behaves like a normal equilibrium constant. |
K6LVoYlzJ08-012|In particular, if it's large, then the equilibrium lies towards the products. |
K6LVoYlzJ08-013|A lot of H3O+ is produced, a lot of the HA, a high percentage, dissociates for strong acids. |
K6LVoYlzJ08-014|For weak acids, a smaller percentage dissociates. |
K6LVoYlzJ08-019|By accepting a proton, it forms OH- in water solution. |
K6LVoYlzJ08-020|So the presence of OH- is indicative of a base in water. |
K6LVoYlzJ08-024|Again, water, a pure liquid, doesn't appear in the expression. |
K6LVoYlzJ08-025|Again, the size of the K determines the strength of the base. |
K6LVoYlzJ08-027|So we have Ka and Kb, acid and base dissociation constants. |
vC2v1n-bZQU-000|Equilibrium is the condition where macroscopically, a reaction appears to have stopped-- the macroscopic concentrations or pressures aren't changing. |
vC2v1n-bZQU-001|But microscopically, there's a dynamic equilibrium. |
vC2v1n-bZQU-002|Products are changing into reactants and reactants are changing into products. |
vC2v1n-bZQU-003|You can approach equilibrium in a variety of ways. |
vC2v1n-bZQU-010|So this equilibrium constant K would be large, it would favor the products. |
vC2v1n-bZQU-013|Now, how can I approach this equilibrium state? |
vC2v1n-bZQU-018|I can demonstrate that. |
vC2v1n-bZQU-020|So let's see that happening. |
vC2v1n-bZQU-022|So equilibrium reached in a step. |
vC2v1n-bZQU-023|We can also reach equilibrium in an oscillatory fashion. |
vC2v1n-bZQU-025|So let's see that chemical reaction occur. |
vC2v1n-bZQU-026|So here's a chemical reaction, turns to yellow forming, but then to blue as it goes by the products. |
vC2v1n-bZQU-031|There's an overall envelope of reaching equilibrium, but it happens in an oscillatory fashion. |
vC2v1n-bZQU-032|There's another instance you could imagine, have an approach to equilibrium that's chaotic. |
vC2v1n-bZQU-034|All reactions if you wait long enough will approach the equilibrium. |
vC2v1n-bZQU-035|The thermodynamically favored state is the favored state at long times. |
vC2v1n-bZQU-036|So we expect favored states to occur if we wait long enough. |
vC2v1n-bZQU-037|So several ways to approach equilibrium. |
vC2v1n-bZQU-038|Either way, they're dynamic. |
TUBGGXVJj3Q-000|Let's look at adding a strong base solution to a strong acid solution. |
TUBGGXVJj3Q-002|Now, strong acids and strong bases totally dissociate in water-- that's the definition of strong acid and strong base. |
TUBGGXVJj3Q-003|So HCl completely dissociates into H3O+ and Cl-. |
TUBGGXVJj3Q-004|We can look at in steps, though. |
TUBGGXVJj3Q-017|I could, but it doesn't make much sense, the H3O+ plus concentration of 0.2 plus to the minus 7, 10 to the minus 7 is tiny by comparison, right? |
TUBGGXVJj3Q-018|This is 0.2000001. |
TUBGGXVJj3Q-019|So that's essentially 0.2. |
TUBGGXVJj3Q-021|Now, we want to add 10 mils of sodium hydroxide at one molar. |
TUBGGXVJj3Q-022|How many moles of OH- is that? |
TUBGGXVJj3Q-026|If I add 0.01 moles of sodium hydroxide, I'm actually adding 0.01 moles of hydroxide ion. |
TUBGGXVJj3Q-027|So let's look at that. |
TUBGGXVJj3Q-033|Do I need to do an equilibrium calculation here? |
TUBGGXVJj3Q-034|No I don't, because the K for this reaction, the reverse reaction is K W, So 1 over a K W is the K for this reaction. |
TUBGGXVJj3Q-035|K W is times 10 to the minus 14, so the K for this reaction 10 to the plus 14. |
TUBGGXVJj3Q-036|So this goes completely to the water. |
TUBGGXVJj3Q-037|So every mole of OH- finds a mole of H3O+ and forms water. |
TUBGGXVJj3Q-038|One for one. |
TUBGGXVJj3Q-048|And that 0.19 moles is in about a liter. |
TUBGGXVJj3Q-049|It's in a liter plus the 10 mils of any OH we added. |
TUBGGXVJj3Q-050|About a liter. |
TUBGGXVJj3Q-051|So it's about 0.19 molar. |
TUBGGXVJj3Q-052|So the H3O+ concentration now will be about 0.19 molar. |
TUBGGXVJj3Q-053|Minus log of 0.19 molar, 0.72. |
TUBGGXVJj3Q-054|So the pH has changed from 0.7 to 0.72. |
TUBGGXVJj3Q-055|Not very much. |
TUBGGXVJj3Q-061|Now, I can look at adding acid and base. |
TUBGGXVJj3Q-062|Here I have an acid solution and a base solution. |
TUBGGXVJj3Q-063|So this is concentrated, this is not as concentrated. |
TUBGGXVJj3Q-064|As I add base to acid, we can watch the pH change by the color of the solution. |
TUBGGXVJj3Q-066|So let's watch this acid-base titration. |
TUBGGXVJj3Q-067|Adding a strong base to a strong acid solution, let's watch that happen. |
TUBGGXVJj3Q-068|PH changes slowly at first, and then rapidly to blue-- the base solution. |
TUBGGXVJj3Q-069|So this is well down here on the titration curve. |
TUBGGXVJj3Q-070|I've gone past the equivalence point to the point where the base is determining the pH. |
TUBGGXVJj3Q-071|A basic solution here maybe around pH 10 to 13-- a blue color for my indicator. |
TUBGGXVJj3Q-072|So there's adding a strong base to a strong acid solution. |
vApxjo9sAmI-000|A rubber band is a system that could do work or have work done on it. |
vApxjo9sAmI-001|So let's take a rubber band and stretch it, hold it there so it comes to thermal equilibrium, and then suddenly release it. |
vApxjo9sAmI-007|We're talking about stretching and releasing a rubber band, or any piece of rubber. |
vApxjo9sAmI-008|I have a rubber glove here. |
vApxjo9sAmI-009|I can stretch it and hold it, let it come to thermal equilibrium. |
vApxjo9sAmI-010|I can even check that against my lip. |
vApxjo9sAmI-011|Yeah, that's room temperature. |
vApxjo9sAmI-012|And then suddenly release it and touch it to my lip again. |
vApxjo9sAmI-013|What happened to the temperature change? |
vApxjo9sAmI-014|Well you can do that experiment, I'll do the math. |
vApxjo9sAmI-015|It's an adiabatic change, and we arranged for it to be adiabatic simply by doing it quickly. |
vApxjo9sAmI-020|The rubber has to pull my hands back together. |
vApxjo9sAmI-021|So a system that does work uses its internal energy to do the work. |
vApxjo9sAmI-022|So the energy must go down. |
vApxjo9sAmI-024|The energy change and the temperature change are locked for an ideal gas. |
vApxjo9sAmI-025|That is, for an ideal gas, the energy and the temperature change are essentially the same parameter. |
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