WEBVTT chemistry/ap-chemistry/hovasapian
00:00:00.000 --> 00:00:05.000
Hello, and welcome back to Educator.com; welcome back to AP Chemistry.
00:00:05.000 --> 00:00:11.000
We are going to continue our discussion of equilibrium, and today we are going to introduce something called the reaction quotient.
00:00:11.000 --> 00:00:21.000
As you will see in a minute, the reaction quotient is just like the equilibrium expression, except it's used at any given time during the reaction.
00:00:21.000 --> 00:00:27.000
It actually will tell us in which direction the reaction must go in order to reach equilibrium.
00:00:27.000 --> 00:00:32.000
Sometimes it's too far to the left; it wants to go to the right; sometimes the reaction is too far to the right; it wants to go to the left.
00:00:32.000 --> 00:00:35.000
Sometimes, it is exactly at equilibrium.
00:00:35.000 --> 00:00:45.000
So, this reaction quotient is going to be one of the fundamental things that we use, and we will actually see it over and over again as we continue to decide where our reaction is and in what direction it needs to proceed.
00:00:45.000 --> 00:00:55.000
And, as you will see when we start doing the problems, it is going to describe the mathematics--what are we adding? What are we subtracting?--things like that.
00:00:55.000 --> 00:00:59.000
So, anyway, let's go ahead and get started.
00:00:59.000 --> 00:01:05.000
OK, so let's do a quick review; so let's do this in blue.
00:01:05.000 --> 00:01:19.000
We'll take our equation that we have been dealing with, which is nitrogen gas, plus 3 moles of hydrogen gas, in equilibrium to form 2 moles of NH₃ gas--ammonia.
00:01:19.000 --> 00:01:38.000
OK, now we said that the equilibrium expression, K...we are no longer writing K < font size="-6" > eq < /font > , remember--if we just do K, that will mean that we are talking about moles per liter, as opposed to K < font size="-6" > P < /font > , which means we are definitely dealing in partial pressures, either in torr or atmosphere, as long as the units are consistent.
00:01:38.000 --> 00:02:01.000
OK, so K is products over reactants, so it is going to be the concentration of NH₃ squared, over the concentration of N₂ (and again, I am using parentheses for concentration instead of the standard brackets, simply to make this a little cleaner, because my brackets tend to be a bit messy) and H₂³.
00:02:01.000 --> 00:02:17.000
So, this is our equilibrium expression; and now, just as a review for what this equilibrium expression tells us: depending on what this number is--if it's a really, really big number, well, notice: if it's a big number, that means the numerator is a lot bigger than the denominator.
00:02:17.000 --> 00:02:22.000
That means that there is more product than there is reactant.
00:02:22.000 --> 00:02:31.000
That means that the equation favors the products; that means that, at equilibrium, most of what is in that flask is going to be products.
00:02:31.000 --> 00:02:33.000
We say it is really far to the right.
00:02:33.000 --> 00:02:42.000
If this is a really small number, this ratio, that means that the numerator is small and the denominator is big.
00:02:42.000 --> 00:02:53.000
That means that the reactants actually are favored in the equilibrium; so at equilibrium, you are going to find most of the stuff on the left-hand side; it's going to be mostly nitrogen and hydrogen, instead of the ammonia.
00:02:53.000 --> 00:03:01.000
That is what this equilibrium constant is a measure of; it is a measure of the extent to which a reaction moves forward or doesn't move forward.
00:03:01.000 --> 00:03:05.000
That is all it is; it is just a numerical value expressing that.
00:03:05.000 --> 00:03:17.000
So, we can speak about it qualitatively--"It's far to the right; it's far to the left"--or we can be precise and quantitative--"It is far to the right, because the K < font size="-6" > eq < /font > is 500," or "The K < font size="-6" > eq < /font > is .001."
00:03:17.000 --> 00:03:22.000
This tells us specifically how far to the left or right; so that is all that that is.
00:03:22.000 --> 00:03:36.000
OK, now, as we said, there is also something called the reaction quotient, which we can use to tell us in which direction a reaction must go in order to reach equilibrium--in other words, where it is at that given moment.
00:03:36.000 --> 00:03:49.000
OK, so let's go ahead and define our Q; this is called the reaction quotient.
00:03:49.000 --> 00:04:01.000
It is the same expression as the equilibrium constant; so, in other words, it's the same thing as this (products over reactants, raised to their stoichiometric coefficients).
00:04:01.000 --> 00:04:35.000
It's the same expression as K < font size="-6" > eq < /font > , but concentrations/pressures are taken at any given moment.
00:04:35.000 --> 00:04:43.000
So you know that this constant--equilibrium constant--it is the ratio of these concentrations, once the system has come to equilibrium.
00:04:43.000 --> 00:05:04.000
Well, we can measure the concentration any time we want (of the NH₃, the N₂, the H₂, or the products, or the reactants), and we can put those into this expression, and we can see how far away from the equilibrium constant it is, and that will tell us whether it is too far to the left or too far to the right.
00:05:04.000 --> 00:05:30.000
That is all we are doing here; so, for a general reaction, aA + bB going to cC + dD, we have that the reaction quotient, Q, is equal to the concentration of C raised to the c power, the concentration of D raised to the d power, over the concentration of A raised to the a power, the concentration of B raised to the b power.
00:05:30.000 --> 00:05:37.000
It is exactly the same as the equilibrium expression, except these concentrations are at any given moment.
00:05:37.000 --> 00:05:46.000
That is all; now, here are the criteria by which we decide where a reaction is--how far from equilibrium.
00:05:46.000 --> 00:06:21.000
If Q is bigger than K, then the reaction will proceed (let's say...yes, you know, I think "proceed" is a good word; I wanted to use "shift," but I think "proceed" is better, or "move") to the left to reach equilibrium.
00:06:21.000 --> 00:06:36.000
In other words, if Q is bigger than K, that means this is bigger than that, that means it has gone too far to the right; or it has not gone too far to the right--it is too far to the right.
00:06:36.000 --> 00:06:48.000
In order for it to reach its equilibrium point (which, as we said, is a fingerprint for that particular reaction at a given temperature), it needs to move to the left; that means it needs to decompose product to form more reactant.
00:06:48.000 --> 00:06:53.000
It needs to decompose this to form more this; it needs to move to the left--that is what that means.
00:06:53.000 --> 00:07:18.000
So now, if Q is less than K, well, it's just the opposite: then, the reaction will proceed to the right to reach equilibrium.
00:07:18.000 --> 00:07:32.000
At any given moment, if we take a bunch of concentrations of products and reactants and we stick it into this expression, we solve it, and it ends up being less than the K, that means there is too much of this and it needs to move in this direction to reach equilibrium.
00:07:32.000 --> 00:07:36.000
That means this denominator is too big; Q is too small.
00:07:36.000 --> 00:07:41.000
It needs to go this way, so it's going to proceed to the right to reach equilibrium.
00:07:41.000 --> 00:07:47.000
So, in this case, reactants are depleting; products are forming.
00:07:47.000 --> 00:08:07.000
And of course, last but not least, if Q equals K, well, you know the answer to this one: then the reaction is at equilibrium (here we go again with the stray lines; OK, that is nice).
00:08:07.000 --> 00:08:16.000
That is it--nice and simple; a mathematical way to see where a reaction is and to see where a reaction is going.
00:08:16.000 --> 00:08:30.000
OK, let's go ahead and do an example.
00:08:30.000 --> 00:09:21.000
Example 1 (let's see): For the equation H₂O gas (you know, that's OK...well, I don't know; I want to sort of skip writing the gas, but I guess it's pretty important) + Cl₂O gas (oops, can't have a double arrow going--I need it to go that way and that way) forms 2 HOCl gas at 25 degrees Celsius, the equilibrium constant equals 0.0900.
00:09:21.000 --> 00:09:46.000
OK, so: for the reaction H₂O gas + dichlorine monoxide, that goes to 2 HOCl gas (hydrogen hypochloride gas, OK?--this is not hypochlorous acid; it's in the gaseous state, so it's not aqueous, so it's not the acid--this is the hydrogen hypochloride), the equilibrium constant for this reaction at 25 degrees Celsius is .0900.
00:09:46.000 --> 00:09:50.000
Notice, there is no unit here; we will get to that in just a minute.
00:09:50.000 --> 00:10:37.000
Here is what we want to do: For the following concentrations, determine the direction (I don't want to run out of room over there) the reaction must go in, in order to reach equilibrium.
00:10:37.000 --> 00:10:45.000
OK, so: For the following concentrations, determine the direction the reaction must go in order to reach equilibrium: a standard reaction quotient problem.
00:10:45.000 --> 00:10:51.000
It is going to be basic; we are going to do it for most equilibrium problems, because we want to know where equilibrium is.
00:10:51.000 --> 00:11:20.000
OK, so the first one: we have: A partial pressure of H₂ O is equal to 200 torr, and the partial pressure of Cl₂O is going to equal 49.8 torr, and the P of HOCl is equal to 21.0 torricelli.
00:11:20.000 --> 00:11:34.000
Now, our Q is equal to the partial pressure of HOCl squared, over the partial pressure of H₂O times the partial pressure of Cl₂O.
00:11:34.000 --> 00:11:55.000
Well, that equals 21.0 squared (that is the HOCl), divided by the partial pressure of H₂O, which is 200, and it is 49.8; there we go.
00:11:55.000 --> 00:12:05.000
Now, in this particular case, this is torr and this is torr squared; so the unit on top is going to be torr squared; this is torr; this is torr; it's going to be torr squared, so it's going to end up without a unit.
00:12:05.000 --> 00:12:09.000
That is why this equilibrium constant doesn't have a unit.
00:12:09.000 --> 00:12:13.000
It is because the torricelli, torricelli cancels with torricelli, torricelli, down at the bottom.
00:12:13.000 --> 00:12:46.000
So, now, we get that Q is equal to 0.0443; so Q is equal to .0443; K is equal to .0900; clearly, Q is less than K, which implies that the reaction will move to the right to reach equilibrium.
00:12:46.000 --> 00:12:54.000
In other words, reactant will deplete; product will form; it will move to the right to reach equilibrium.
00:12:54.000 --> 00:13:00.000
It hasn't reached equilibrium yet; it's still moving to the right to reach eq.
00:13:00.000 --> 00:13:06.000
That is it; that is all you are using the reaction quotient for--to tell you which direction it is going in.
00:13:06.000 --> 00:13:16.000
OK, all right, let's see: so, let's do another one--another set of conditions.
00:13:16.000 --> 00:13:48.000
This time, we will do: A 3.0-liter flask contains 0.25 moles of HOCl, 0.0100 mol of Cl₂O, and 0.56 mol of H₂O.
00:13:48.000 --> 00:13:52.000
OK, notice: they give us moles, and they give us the actual volume of the flask.
00:13:52.000 --> 00:14:01.000
Well, again, when we deal with these reaction quotients and equilibrium expressions, they have to be in moles per liter--in concentrations.
00:14:01.000 --> 00:14:06.000
Let's just take each one and find the concentrations before we put them into our reaction quotient.
00:14:06.000 --> 00:14:35.000
So, our concentration of HOCl is equal to 0.25 mol, divided by 3.0 liters; that is going to equal 0.0833 Molar (that m with a line over it means molarity; it's an older expression; you are accustomed to seeing it: capital M).
00:14:35.000 --> 00:14:55.000
The Cl₂O: that is equal to 0.0100 moles, again divided by 3 liters, because it is in the flask; so, its concentration is 0.00333 Molar.
00:14:55.000 --> 00:15:13.000
And finally, our H₂O concentration is equal to 0.56 mol, divided by 3.0 liter; and this concentration is 0.1867 Molar.
00:15:13.000 --> 00:15:22.000
Now that we have the molarities, we can put them into our reaction quotient; the reaction quotient is going to be exactly the same thing.
00:15:22.000 --> 00:16:08.000
Now, we have: Q is equal to the concentration, as we said, of HOCl squared, over the concentration of H₂O, times the concentration of Cl₂O, which is equal to (drop down a little bit here) 0.0833 squared, times 0.1867, times 0.0033; and we get 11.16.
00:16:08.000 --> 00:16:32.000
Now, Q is 11.16; we said that K was equal to 0.0900; so, clearly, Q is much larger than K, which implies that the reaction will move to the right to reach equilibrium.
00:16:32.000 --> 00:16:38.000
In other words, it is still moving--I'm sorry; not to the right--to the left!
00:16:38.000 --> 00:16:44.000
The Q is bigger; yes, Q is bigger--it is going to move to the left.
00:16:44.000 --> 00:16:59.000
Sorry about that; that means that there is too much product at this temperature, given the equilibrium, which is a fingerprint for that reaction; so, there is too much product; the product needs to decompose to form reactant.
00:16:59.000 --> 00:17:08.000
It is moving to the left to reach equilibrium.
00:17:08.000 --> 00:17:20.000
That is it; OK, now, again, notice that there are no units for that; there are no units for this particular equilibrium constant.
00:17:20.000 --> 00:17:30.000
In fact, notice in the question: the question actually said, "Equilibrium constant equals"--it didn't say "K equals" or "K < font size="-6" > P < /font > equals."
00:17:30.000 --> 00:17:49.000
If you are given K < font size="-6" > eq < /font > , K, or K < font size="-6" > P < /font > , it will specifically mean that we are talking about pressures, or we are talking about moles per liter; but in this particular case, because there is no unit, that actually means that the K < font size="-6" > P < /font > and the K are the same.
00:17:49.000 --> 00:18:05.000
You remember the definition of the relationship between K < font size="-6" > P < /font > and K; well, the fact that there is no unit means that there is no...if you look at the equation, Δn=0, so that RT that we had is 1.
00:18:05.000 --> 00:18:23.000
So, when it says the "equilibrium constant," but it doesn't specifically specify whether it is a K or a K < font size="-6" > P < /font > , well, it's the same thing--they are actually equal to each other, which is why we use the same number, .0900, with molarity and with the one that we just did, which was done in terms of pressure.
00:18:23.000 --> 00:18:32.000
In both cases, we use the .0900; that comes from the fact that there is no unit that tells us that the K and the K < font size="-6" > P < /font > are the same.
00:18:32.000 --> 00:18:52.000
These are the little things that you have to watch out for; in other circumstances, when you do have a unit, you have to watch out; you have to actually (if you are dealing with molarity and you are given the K < font size="-6" > P < /font > , you have to) either convert the K < font size="-6" > P < /font > to a K, or you have to convert the molarities to pressures, if you can, depending on what the problem is asking.
00:18:52.000 --> 00:18:59.000
Again, these are the sort of things; there is a lot that is going to be going on--there is a lot that you have to watch out for.
00:18:59.000 --> 00:19:05.000
It isn't just "plug and play"--you don't just put numbers into an equation and hope things will fall out.
00:19:05.000 --> 00:19:15.000
You have to understand what is happening; this is real science, and real science means conversions, units, and strange things.
00:19:15.000 --> 00:19:20.000
OK, so let's do another example.
00:19:20.000 --> 00:19:25.000
Here is where we begin to actually explore some of the diversity of these equilibrium problems.
00:19:25.000 --> 00:19:30.000
What we are going to do is: most of our learning is actually going to come through the problems themselves.
00:19:30.000 --> 00:19:51.000
That is why we are going to do a fair number of these equilibrium problems; it's very, very important that you have a reasonably solid understanding of how to handle these things, because it's going to be the bread and butter of what you do for the rest of chemistry--certainly for the rest of the AP and the free response questions; the electrochemistry; acid-base; the thermodynamics.
00:19:51.000 --> 00:19:56.000
It's precisely this kind of reasoning, and equilibrium is fundamental to it all.
00:19:56.000 --> 00:19:58.000
That is why it comes before everything else does.
00:19:58.000 --> 00:20:03.000
Equilibrium is chemistry; it is that simple.
00:20:03.000 --> 00:20:11.000
OK, so let's do Example #2: let's do this one in red--how is that?
00:20:11.000 --> 00:20:21.000
OK, Example 2: The question is going to be a bit long, but...let's see.
00:20:21.000 --> 00:20:57.000
At a certain temperature, a 1.0-liter flask contains 0.298 mol of PCl₃ and 8.70x10^-3 mol of PCl₅.
00:20:57.000 --> 00:21:29.000
OK, now after the system comes to equilibrium (comes to eq), 2.00x10^-3 mol of Cl₂ gas was formed in the flask.
00:21:29.000 --> 00:21:50.000
Now, PCl₅ decomposes according to the following: PCl₅ decomposes into PCl₃ + Cl₂ gas.
00:21:50.000 --> 00:22:17.000
What we would like you to do is calculate the equilibrium concentrations of all species and the K < font size="-6" > eq < /font > .
00:22:17.000 --> 00:22:29.000
OK, so we would like you to calculate the equilibrium concentrations of all the species (in other words, the PCl₅, the PCl₃, and the Cl₂) and we would like you to tell us what the K < font size="-6" > eq < /font > is--what the K is.
00:22:29.000 --> 00:22:42.000
OK, so let's read this again: At a certain temperature, a 1-liter flask contains .298 moles of PCl₃, and 8.70x10^-3 moles of PCl₅.
00:22:42.000 --> 00:22:48.000
After the system comes to equilibrium, 2.0x10^-3 moles of Cl₂ is formed in the flask.
00:22:48.000 --> 00:22:58.000
Great! So, let's go ahead and start this off; so I'm going to go ahead and move to a new page so I can rewrite the equation.
00:22:58.000 --> 00:23:03.000
We are going to do our little ICE chart here: Initial, Change, Equilibrium.
00:23:03.000 --> 00:23:16.000
PCl₅ decomposes into PCl₃ plus Cl₂; our initial concentration, our change, and our equilibrium concentration--which is what we actually want here.
00:23:16.000 --> 00:23:26.000
It's telling me that PCl₅ was 8.7x10^-3 initially, right?
00:23:26.000 --> 00:23:40.000
8.70x10 to the negative...oh, let's do...yes, that's fine; OK; I can just do this down below.
00:23:40.000 --> 00:23:56.000
Notice: they give us (well, here; let me do this over to the side)--with the PCl₅, they gave us 8.70x10^-3 moles.
00:23:56.000 --> 00:24:03.000
Now, that is not a concentration--that is moles, not moles per liter, but they said we have a 1.0-liter flask.
00:24:03.000 --> 00:24:17.000
Again, this is one of the other things: we have to make sure to actually calculate the concentrations; so, in this case, it's going to be 8.70x10^-3 moles, over 1.0 liters.
00:24:17.000 --> 00:24:22.000
Well, because it's 1 liter--it's a 1-liter flask--the number of moles is equal to the molarity.
00:24:22.000 --> 00:24:25.000
So, I can just go ahead and put these numbers here.
00:24:25.000 --> 00:24:36.000
If this were not a 1-liter flask--if it were anything other than a 1-liter flask--I would have to actually calculate the initial concentration, and those are the values that I use in my ICE chart.
00:24:36.000 --> 00:24:42.000
OK, so in my ICE chart, I'm working with concentrations, not moles.
00:24:42.000 --> 00:24:56.000
OK, so we have: 8.70x10^-3 molarity, and we said the PCl₃ was 0.298, and there is no chlorine gas.
00:24:56.000 --> 00:25:06.000
Well, they said that, at equilibrium, there is 2.0x10^-3 moles per liter of chlorine gas.
00:25:06.000 --> 00:25:26.000
So, chlorine gas showed up; so the change was: well, for every mole of chlorine gas that shows up, a mole of PCl₃ shows up, and a mole of PCl₅ decomposes, because the ratio is 1:1, 1:1, 1:1.
00:25:26.000 --> 00:25:45.000
If 2.00x10^-3 moles shows up, that means here, also, 2.00x10^-3 moles shows up; here, it is -2.00x10^-3 moles.
00:25:45.000 --> 00:25:53.000
Again, we are using just basic intuition and what we know about the physical system to decide how the math works.
00:25:53.000 --> 00:25:57.000
This is what chemistry is all about--this is the single biggest problem with chemistry.
00:25:57.000 --> 00:26:11.000
There is nothing intuitively strange about any concept in chemistry--it's all very, very clear--it's all very, very basic as far as what is happening; there is nothing esoteric; there is nothing metaphysical going on.
00:26:11.000 --> 00:26:16.000
It is just that...how does one change the physical situation into the math?
00:26:16.000 --> 00:26:23.000
Well, this is how you do it; you have to know what is going on physically, and then the math should fall out; just trust your instincts.
00:26:23.000 --> 00:26:27.000
One mole of this shows up; well, the equation says one mole of this shows up.
00:26:27.000 --> 00:26:32.000
That means, if one mole of this shows up--that means one mole of this was used up; that is it.
00:26:32.000 --> 00:26:46.000
So now, we do 8.7x10^-3, minus 2.0x10^-3, and we end up with 6.7x10^-3, which I am going to express as a decimal; so, 0.067.
00:26:46.000 --> 00:26:59.000
And then, we have this one; when we add it together, we end up with 0.300; and here, we have 2.00x10^-3.
00:26:59.000 --> 00:27:07.000
It is probably not a good idea to mix the scientific notation and decimals, but you know what--actually, it is not that big of a deal--it is what science is all about.
00:27:07.000 --> 00:27:15.000
OK, so now, because we have the equilibrium concentrations, we have solved the first part of the problem.
00:27:15.000 --> 00:27:28.000
At equilibrium, we are going to have .067 Molar of the PCl₅, .300 Molar of the PCl₃, and 2.0x10^-3 Molar of the Cl₂.
00:27:28.000 --> 00:27:34.000
Well now, let's just go ahead and put it into our equilibrium expression, and find our K < font size="-6" > eq < /font > .
00:27:34.000 --> 00:27:36.000
That is the easy part.
00:27:36.000 --> 00:27:51.000
OK, so K < font size="-6" > eq < /font > is equal to...it is going to be...the Cl₂ concentration, times the PCl₃ concentration, divided by the PCl₅ concentration.
00:27:51.000 --> 00:28:08.000
That equals 2.00x10^-3, times 0.300, divided by 0.067.
00:28:08.000 --> 00:28:18.000
Yes...no; 6.7x10^-3...oops, I think I have my numbers wrong here; this is supposed to be...yes, I should have just left it as scientific notation.
00:28:18.000 --> 00:28:22.000
You know what, I'm just going to go ahead and leave it as scientific notation.
00:28:22.000 --> 00:28:33.000
I shouldn't mess with things; 6.7x10^-3, and we will write this as 6.7x10^-3, also.
00:28:33.000 --> 00:28:40.000
6.7x10^-3: I ended up forgetting a 0; it was .0067; OK.
00:28:40.000 --> 00:28:52.000
And then, when we solve this, we end up with...(let's see, what number did we get?) 8.96x10^-2 Molar.
00:28:52.000 --> 00:28:58.000
So, in this case, it does have a unit; this is a K < font size="-6" > eq < /font > --this is not a concentration.
00:28:58.000 --> 00:29:13.000
That is why I am not a big fan of units when it comes to equilibrium constants; as far as equilibrium constants are concerned, I think that units should only be used to decide about conversions.
00:29:13.000 --> 00:29:16.000
Other than that, I think they should be avoided; but you know what, we will just go ahead and leave it there.
00:29:16.000 --> 00:29:21.000
This is the K < font size="-6" > eq < /font > ; K < font size="-6" > eq < /font > equals 8.96x10^-2.
00:29:21.000 --> 00:29:25.000
That is actually a pretty small number; what does a small K < font size="-6" > eq < /font > mean?
00:29:25.000 --> 00:29:29.000
That means most of this reaction is over here, on the left; there are not a lot of products.
00:29:29.000 --> 00:29:33.000
Most of it is PCl₅; that is what that means.
00:29:33.000 --> 00:29:46.000
Don't let these numbers say that it is mostly this, because, remember: we started off with a certain amount of the PCl₅; we started off with a whole bunch of the PCl₃.
00:29:46.000 --> 00:29:55.000
So, .298--it only went up to .3; that means it only went up .002; that is not very far.
00:29:55.000 --> 00:30:11.000
So, don't let these equilibrium amounts fool you into thinking that the reaction went forward; it is this equilibrium constant which tells you the relationship between these three numbers under these conditions.
00:30:11.000 --> 00:30:16.000
But, this is a fingerprint; this reaction at this temperature will not go very far forward.
00:30:16.000 --> 00:30:24.000
Most of it is still PCl₅; that is what is going on--don't let this .3 fool you; it doesn't mean that it has formed that much.
00:30:24.000 --> 00:30:33.000
You already started at .298; you only formed .002 moles per liter of the PCl₃--not very much at all.
00:30:33.000 --> 00:30:35.000
It is confirmed by the K < font size="-6" > eq < /font > .
00:30:35.000 --> 00:30:42.000
OK, let's do another example; that is what we are here to do.
00:30:42.000 --> 00:30:45.000
Let's see what we have.
00:30:45.000 --> 00:30:56.000
Yes, OK; let me write it out, and then...well, you know what, I am going to actually start a new page for this one, because I would like to see part of the problem while we are reading; OK.
00:30:56.000 --> 00:31:41.000
This Example 3: Now, carbon monoxide reacts with steam (this is H₂O gas) to produce carbon dioxide and hydrogen at 700 Kelvin (we don't need a comma there).
00:31:41.000 --> 00:31:54.000
At 700 Kelvin, the eq constant (equilibrium constant) is 5.10.
00:31:54.000 --> 00:32:33.000
Calculate the eq concentrations of all species if 1.000 mol of each component (each component means each species) is mixed in a 1.0-liter flask.
00:32:33.000 --> 00:32:46.000
OK, our reaction is: they said: Carbon monoxide gas plus steam, which is H₂O gas, forms carbon dioxide gas, plus hydrogen gas.
00:32:46.000 --> 00:32:53.000
That is our reaction, and it is balanced; so it is 1:1:1:1--not a problem.
00:32:53.000 --> 00:32:58.000
OK, so let's read this: Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen.
00:32:58.000 --> 00:33:03.000
At 700 Kelvin, the equilibrium constant is 5.10.
00:33:03.000 --> 00:33:07.000
So, we have our (let me do this in blue now) 5.10.
00:33:07.000 --> 00:33:19.000
Calculate the equilibrium concentrations of all species if 1.0 mol of each of the components (that means 1, 1, 1, 1 mole of each) is mixed in a 1-liter flask.
00:33:19.000 --> 00:33:30.000
OK, now the first thing we need to do is: we are going to (and again, this is sort of something we are always going to do--this is the procedure) write the equation, which we have.
00:33:30.000 --> 00:33:39.000
We are going to write the equilibrium expression, and then we are going to check the reaction quotient to see where the reaction is at that moment, to see which direction it is actually going to be moving in.
00:33:39.000 --> 00:33:46.000
Let's write the equilibrium expression for this, which is going to be the same as the reaction quotient.
00:33:46.000 --> 00:33:57.000
It is the concentration of H₂, times the concentration of CO₂, divided by the concentration of CO, times the concentration of water.
00:33:57.000 --> 00:34:01.000
OK, and that is also equal to the reaction quotient.
00:34:01.000 --> 00:34:08.000
The first thing we want to do is calculate the reaction quotient to see which direction it is moving in.
00:34:08.000 --> 00:34:14.000
Now, they say they put in 1.000 mol of each component.
00:34:14.000 --> 00:34:28.000
Well, it is sitting in a 1.0-liter flask, so basically, I can work with moles: 1 mole, divided by 1 liter, is 1 mole per liter, so the concentration of each species is 1 mole per liter.
00:34:28.000 --> 00:34:31.000
Well, now let's calculate the Q.
00:34:31.000 --> 00:34:48.000
The Q equals...well, it is 1 Molar of the H₂, 1 Molar of the CO₂, divided by 1 Molar of the CO and 1 Molar of the H₂O; the Q equals 1.
00:34:48.000 --> 00:34:56.000
Now, the reaction quotient Q, which is equal to 1, is less than 5.10 (remember, they gave us the 5.10, which is equal to K).
00:34:56.000 --> 00:35:05.000
When Q is less than K, that means the reaction wants to move forward to produce more product, in order to reach equilibrium.
00:35:05.000 --> 00:35:13.000
That means it hasn't reached equilibrium yet; it is still moving forward--it is producing more product to reach equilibrium.
00:35:13.000 --> 00:35:24.000
That means carbon monoxide and H₂O gas are being depleted, and for each amount that these are depleted (because the ratio is 1:1), an equal amount of CO₂ and H₂ are being formed.
00:35:24.000 --> 00:35:28.000
Now, we can do the actual equilibrium part of this problem.
00:35:28.000 --> 00:35:33.000
So again, these "1 Molar"--that came from where we started at that moment.
00:35:33.000 --> 00:35:40.000
At any given moment, I stick in 1 mole of each in a 1-liter flask, and let me find out what this value is.
00:35:40.000 --> 00:35:46.000
It is 1; it is less than the equilibrium constant; that means it is going to move forward, to the right.
00:35:46.000 --> 00:36:13.000
OK, so now let's do our equilibrium part, and we do that by doing our ICE chart, so let me rewrite CO + H₂O (I tend to rewrite things a lot--sorry about that; I hope it's not a problem--I'm sure you have different ways of doing it yourself--as long as each of these is here...) goes to CO₂ + H₂.
00:36:13.000 --> 00:36:21.000
OK, we have an initial concentration; we have the change; and we have our equilibrium concentrations, which is what we are looking for.
00:36:21.000 --> 00:36:26.000
Our initial concentrations are 1.000, right?--that is how much we started with.
00:36:26.000 --> 00:36:29.000
We stuck each of those in a flask.
00:36:29.000 --> 00:36:36.000
Now, this reaction quotient tells us that the reaction is moving in that direction to reach equilibrium.
00:36:36.000 --> 00:36:42.000
It is moving in that direction; that means CO is disappearing.
00:36:42.000 --> 00:36:54.000
H₂O is also disappearing by an amount x--that means a certain amount is decomposing--a certain amount of CO is being lost, is being converted.
00:36:54.000 --> 00:37:06.000
Well, since it is moving to the right, and this is 1:1:1:1, that means this is +x; that means CO₂ is forming for every 1 mole of this that is disappearing.
00:37:06.000 --> 00:37:11.000
This is also +x; I hope that makes sense.
00:37:11.000 --> 00:37:22.000
Our equilibrium concentration (at equilibrium, once everything has stopped, a certain amount of CO has been used up, so our equilibrium concentration) is going to be 1.00-x.
00:37:22.000 --> 00:37:28.000
A certain amount of H₂O has been used up; that is 1.00-x.
00:37:28.000 --> 00:37:32.000
A certain amount of CO₂ has formed, so it is going to be 1+x.
00:37:32.000 --> 00:37:38.000
A certain amount of H₂ has formed: 1+x.
00:37:38.000 --> 00:37:54.000
These are our equilibrium (oh, wow, that is interesting; look at that--let's get these lines out of the way) expression, but notice: now, we have x, so we need to actually find x.
00:37:54.000 --> 00:38:06.000
Fortunately, we can do that: we can plug it into the equilibrium expression, right?--because the equilibrium expression is a measure of these concentrations at equilibrium; that is what these values are.
00:38:06.000 --> 00:38:11.000
We stick it in here; we know what the K < font size="-6" > eq < /font > is--it's 5.10; and we solve for x.
00:38:11.000 --> 00:38:14.000
That is it; it is just an algebra problem.
00:38:14.000 --> 00:38:49.000
So, let's go ahead and do that: so K, which is equal to 1.000+x, times 1.000+x (that is the CO₂ and H₂ concentrations), divided by the CO and H₂O concentrations, which at equilibrium are 1.000-x, 1.000-x, and we know that that equals 5.10.
00:38:49.000 --> 00:38:53.000
OK, so now let's just handle this algebraically.
00:38:53.000 --> 00:39:13.000
This is (1.000+x) squared, over 1.000 (oops, too many 0's) minus x, squared, equals 5.10.
00:39:13.000 --> 00:39:23.000
Now, I know that I said earlier (I think a lesson or two ago): when you are writing out the equilibrium expression, don't put the square--don't square it immediately--if two of the things are the same.
00:39:23.000 --> 00:39:33.000
That is different than what I am doing now; I wrote the expression as each species separately, so that I can see that I actually have four species in my equilibrium expression.
00:39:33.000 --> 00:39:35.000
Here, now, I am just dealing with the math.
00:39:35.000 --> 00:39:40.000
Once you have actually written it out, then you can go ahead and write it like this to deal with the math.
00:39:40.000 --> 00:39:44.000
Now, it's (1+x)² over (1-x)².
00:39:44.000 --> 00:39:49.000
That is fine; but when you initially write the expression, don't cut corners; write down everything.
00:39:49.000 --> 00:39:55.000
We want to know that the expression actually consists of four terms, not two terms, each squared.
00:39:55.000 --> 00:40:06.000
OK, and now we just...well, we have a square here and a square here, so we'll just go ahead and take the square root of both sides.
00:40:06.000 --> 00:40:40.000
We end up with 1+x over 1-x, equals 2.258, and then we multiply through to get 1.000+x=2.258-2.258 times x (I'm hoping that I am doing my math right here).
00:40:40.000 --> 00:40:58.000
We end up with 3.258x equals 1.258, and x is equal to 0.386 Molar.
00:40:58.000 --> 00:41:04.000
We found x; x is .386 Molar.
00:41:04.000 --> 00:41:11.000
Now, if you go back to your ICE chart, it didn't ask for what x was; it asked for the final concentration.
00:41:11.000 --> 00:41:16.000
Well, the final concentrations were the 1-x and the 1+x for those four species.
00:41:16.000 --> 00:41:41.000
So, the CO concentration (carbon monoxide concentration), which also equals the H₂O concentration, is equal to 1.000-x, 1.000-.386, equals 0.613 Molar.
00:41:41.000 --> 00:41:45.000
The carbon monoxide and the water are at .613 molarity.
00:41:45.000 --> 00:42:31.000
Now, the CO₂ concentration, which also happens to equal the H₂ concentration, is equal to (OK, let's see if we can clean this up a little bit; I'm not going to have these stray lines driving us crazy all day)...CO₂ (I just really need to learn to write slower; I know that that is what it is) equals H₂ concentration, equals 1.000+x, equals 1.000+0.386, =1.386 molarity.
00:42:31.000 --> 00:42:44.000
OK, and my friends, we have done it; we have taken a standard equilibrium problem, and here is how we have approached it.
00:42:44.000 --> 00:42:50.000
1) Write the equation; this is chemistry--chemistry always begins with some equation; don't just go into the math.
00:42:50.000 --> 00:42:56.000
Look at the equation--the equation gives you all of the information that you need--in fact, it will tell you everything you want.
00:42:56.000 --> 00:43:03.000
2) Write the equilibrium expression--write the K expression.
00:43:03.000 --> 00:43:09.000
Write it out explicitly--don't count on the fact that you will know how to do it later on when you are ready to plug things in.
00:43:09.000 --> 00:43:18.000
Write it out; don't cut corners; doing things quickly is not impressive--doing things correctly is impressive.
00:43:18.000 --> 00:43:34.000
3) Find Q, the reaction quotient; find Q to decide which direction the reaction is going in--to decide the reaction direction, if any.
00:43:34.000 --> 00:43:43.000
That is how we knew...in this problem that we just did, the reaction quotient was less than the equilibrium constant--so that means the reaction was actually moving forward.
00:43:43.000 --> 00:43:48.000
Well, that is how we knew that the reactants get the -x and the products get the +x.
00:43:48.000 --> 00:43:57.000
If it were the other way around and the reaction were moving to the left, that means the products would get the -x and the reactants would get the +x.
00:43:57.000 --> 00:44:02.000
You have to do this; you have to do the reaction quotient--very, very important.
00:44:02.000 --> 00:44:12.000
After the reaction quotient, well, you set up your ICE chart: Initial, Change, Equilibrium concentration.
00:44:12.000 --> 00:44:45.000
Once you get a value for the equilibrium concentration, you put the eq concentration expressions (in other words, the concentrations that you calculated there--the equations/expressions) into the K < font size="-6" > eq < /font > expression.
00:44:45.000 --> 00:44:49.000
Then, last but not least, you solve, depending on what they want.
00:44:49.000 --> 00:45:01.000
If they want x, you stop there; if they want equilibrium concentrations, you take the x value; you plug it back into the equilibrium expressions; and you add and subtract until you get your equilibrium expressions.
00:45:01.000 --> 00:45:05.000
If they want something else, they will tell you that they want something else.
00:45:05.000 --> 00:45:14.000
Again, this process is what you are always going to be doing; this is the algorithm, the general, broad-strokes algorithm.
00:45:14.000 --> 00:45:23.000
Write the equation; write the expression; find the Q; set up your ICE chart; put the equilibrium concentrations in; and then solve it.
00:45:23.000 --> 00:45:31.000
Within this are the different variations that make up the different number of problems, which seem to be infinite (I understand completely).
00:45:31.000 --> 00:45:34.000
That is all you are doing here.
00:45:34.000 --> 00:45:40.000
OK, thank you for joining us here at Educator.com for our continuation of AP Chemistry in equilibrium.
00:45:40.000 --> 00:45:51.000
In our next lesson, we are actually going to do more equilibrium problems, because again, this is a profoundly important concept; you have to be able to have a good, good, solid, intuitive understanding of what is going on.
00:45:51.000 --> 00:45:53.000
So, we will do some more practice.