WEBVTT chemistry/ap-chemistry/hovasapian
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Hello, and welcome back to Educator.com; welcome back to AP Chemistry.
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We are going to continue our discussion of equilibrium--in my opinion, absolutely the single most important concept, not just in chemistry, but in all of science.
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Equilibrium is the thing that all systems tend to--this notion of balance; they don't like being off balance (too much of this, too much of that).
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All systems tend toward equilibrium, and chemical systems are no different.
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Last lesson, we introduced the notion of an equilibrium expression, this K < font size="-6" > eq < /font > where, beginning with any sort of concentrations of a particular reaction, once the system has come to equilibrium, we measure those concentrations at equilibrium.
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We put it into the equilibrium expression, and it ends up being the constant.
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Different equilibrium conditions=different values; but the relationship among those values is a constant.
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Today, we are going to continue that discussion, and we are going to introduce a variation of the expression for reactions that involve a gas.
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As it turns out, when you have a reaction that involves a gas, you can either work in moles per liter (like we did last time--you are certainly welcome to do so by taking the number of moles in a flask and the volume of the flask, and dividing to get your concentration in moles per liter) or, as it turns out, pressure--you can work with pressures, because pressure is actually a measure of concentration.
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We will show you how, mathematically; it is actually very, very nice.
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Let's get started.
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OK, so now, let's just go ahead and write, "For equilibriums involving gases, the K can be expressed in terms of P, pressure."
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What that means is that I can either express my equilibrium expression with moles per liter concentrations or with pressure.
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I'm going to write the two expressions, and then we will see how they are actually related, because there is a relationship between the two, which allows us to go back and forth, depending on what the problem is asking.
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Because sometimes, it might be easier to deal with pressure--sometimes, easier with concentration; it just depends.
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So again, we are going to stick with our tried-and-true nitrogen, plus 3 moles of hydrogen gas, going to 2 moles of ammonia.
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We know that the actual K < font size="-6" > eq < /font > expression in concentrations is: the concentration of ammonia squared, over the concentration of nitrogen times the concentration of hydrogen cubed.
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Well, as it turns out, I can also express it this way: and now, I put a P down as a subscript to the equilibrium expression--P for pressure.
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K < font size="-6" > P < /font > --it's the same thing, and it equals the partial pressure of NH₃ (partial pressure means the pressure just of that gas in the container, because you have three gases in a container: one gas has one pressure, on has another pressure--we call those the partial pressures), squared (so again, everything is the same; it's products, over reactants, raised to the stoichiometric coefficients; the pressure of NH₃, squared...), because this is a 2, divided by the pressure of nitrogen gas, raised to a 1 power, because this stoichiometric coefficient is 1, times the partial pressure of H₂ cubed, because its stoichiometric coefficient is 3.
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So now, let's talk about how these two are related.
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Well, whenever we talk about gases, the one equation that we talk about is the PV=nRT expression.
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Let me just erase a little bit of this and give me a little more room.
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Pressure times volume equals (let me make my V a little more clear) the number of moles, times the gas constant, times the temperature in Kelvin.
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Let's rearrange this a little bit.
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I'm going to write P=nRT/V.
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Now, I'm going to take two of these and combine them: n/V times RT: just a little bit of mathematical manipulation--I can do this.
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I can take the denominator and put it with one of the numbers in the numerator.
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Now, notice what I have: what is n/V?--n is the number of moles, and V is volume in liters; so, as it turns out, P and n/V are related by a constant, RT.
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As it turns out, pressure is an alternative form of representing concentration in moles per liter.
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n/V is just moles per liter, so this says, if I have something which is a certain number of moles per a certain number of liters, if I multiply that by RT, I actually get the pressure.
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Therefore, instead of concentration, I can just put the particular pressure.
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That is all this is: in dealing with gases, it is often difficult to...not difficult to deal in concentrations; it tends to be easier, simply by nature of gases, by measuring pressure.
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That is all this is; so, whenever we have an equilibrium condition that involves some gases, we can use its pressure, because pressure is just an alternative form of concentration; that is it.
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Heads, tails--it's just another way of looking at concentration.
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Now, let's see what the actual relationship is between these two values.
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OK, so P is equal to n/V(RT); so P is equal to concentration times RT (we will just use C as concentration, instead of writing it as n/V) or, we can write C=P/RT: concentration is equal to pressure over RT.
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This is just standard mathematical manipulation.
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So now, this is concentration; we are going to rewrite the expression here, and we are going to put concentration back in to see how it is related here.
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We will do: K < font size="-6" > eq < /font > is equal to...it's pressure over RT, the concentration: so we get the partial pressure of NH₃, over RT, squared.
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So all I have done is: I have just taken this expression, put it into here, and divided by pressure of N₂/RT to the first power, times the pressure of H₂/RT (that is concentration squared, concentration raised to the first, concentration cubed, right?)--just basic math.
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And then, I have partial pressure of NH₃, squared, times 1/RT, squared, over the partial pressure of N₂ times 1/RT, times the partial pressure of H₂, cubed, times 1/RT, cubed--so far, so good.
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I have: partial pressure of NH₃, squared, over partial pressure of N₂, partial pressure of H₂, cubed, times 1/RT, squared, over 1/RT to the fourth power.
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All right, and now, I am going to have...this 2 is going to cancel that; I'm going to end up with (let me write it one more time--well, two more times, actually): P < font size="-6" > NH₃ < /font > , squared, over P < font size="-6" > O₂ < /font > , times the P < font size="-6" > H₂ < /font > , cubed, times 1/(1/RT, squared) (because this cancels two of those, leaving that) and 1/(1/RT squared) is RT squared.
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So, it equals the partial pressure of NH₃, squared, over the partial pressure of N₂ and the partial pressure of H₂, cubed, times RT, squared.
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Well, this is the K < font size="-6" > eq < /font > ; that is this expression.
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This expression right here is the K < font size="-6" > P < /font > (RT)².
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So as it turns out, K < font size="-6" > eq < /font > equals K < font size="-6" > P < /font > , in this particular case, times RT squared.
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There is a relationship between expressing it with concentrations and expressing it with pressures.
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The relationship is that the K < font size="-6" > eq < /font > equals K < font size="-6" > P < /font > (for this particular reaction--we'll do the general one in a minute), times RT squared.
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I can go back and forth between the two; so, if I'm working with K < font size="-6" > P < /font > and I want K < font size="-6" > eq < /font > , in concentrations, I can do that.
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If I have K < font size="-6" > eq < /font > and I want K < font size="-6" > P < /font > , I can do that by multiplying by the square of RT.
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OK, so now, let's do the general version; I just wanted to show you where the math came from.
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For the general expression, aA + bB in equilibrium with cC + dD, our relationship is the following.
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From now on, I'm not going to put the K < font size="-6" > eq < /font > ; I'm just going to put K; whenever you see K with no subscript on it, it just means moles per liter, concentration; K equals K < font size="-6" > eq < /font > .
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So, K is equivalent to K < font size="-6" > eq < /font > ; when we speak of P, we will write K < font size="-6" > P < /font > ; that means we are dealing with pressures.
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When we just see a K, without this eq, they are the same--simply to avoid writing the eq over and over again.
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As it turns out, the K is equal to K < font size="-6" > P < /font > , times RT to the negative Δn, where Δn equals the sum of the coefficients of the products, minus the sum of the stoichiometric coefficients of the reactants.
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We can write it that way; or, another way that it is written is in terms of K < font size="-6" > P < /font > .
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K < font size="-6" > P < /font > is equal to K times (RT)^Δn; either one of these is fine.
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If you have K, you can find K < font size="-6" > P < /font > ; if you have K < font size="-6" > P < /font > , you can find K by just using this expression; that is it--nothing more than that.
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Δn is this coefficient plus that coefficient, minus that plus that; that is all.
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Let's just do an example, and everything will make sense.
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Example 1: OK, at 427 degrees Celsius, a 2.0-liter flask contains 40.0 moles of H₂, 36.0 moles of CO₂, 24.0 mol of H₂O, and 11.8 mol of CO, carbon monoxide, at equilibrium.
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So, at equilibrium, we measure 427 degrees; the system has come to equilibrium; we measure in a 2-liter flask; we find that we have this many moles of each of these species.
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Now, the reaction is as follows: CO₂ gas + H₂ gas goes to carbon monoxide gas + H₂O gas.
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All of these are gaseous species: that means all of them are involved in the equilibrium expression.
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Now, our task here is to find K and K < font size="-6" > P < /font > ; so, find the equilibrium constant expression, in terms of moles per liter, and find the K < font size="-6" > P < /font > constant, in terms of partial pressure.
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Well, we have the relationship, so we can find K, and then we can find K < font size="-6" > P < /font > , based on this--no problem.
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OK, so now, this is our first example of a real equilibrium problem, in the sense that we really have to watch everything that is going on.
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No two problems are going to read the same way; we can't follow an algorithmic procedure for solving every problem.
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Physical systems--now, it just depends; different things can happen--you can have different data.
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It can be worded in a certain way; you have to be able to extract the information that is necessary.
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Yes, there are certain things that are universal, that you can always count on, but you have to watch every single little thing.
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Notice here: they are giving you the volume of a flask, and they are giving you the amounts in moles; so you actually have to calculate the moles per liter, before you put it into the equilibrium expression, because the equilibrium expression requires that it be in concentrations and not in moles.
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You have to sort of watch for that.
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Let's write the equilibrium expression, K; we will do K first.
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It is going to equal the products over the reactants, raised to their stoichiometric coefficients.
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This is a balanced reaction, so everything is 1:1:1:1.
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We have the concentration of CO, times the concentration of H₂O (everything is a gas), over the concentration of CO₂, times the concentration of H₂ gas.
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Well, the concentration is the number of moles per volume; our volume is 2 liters (I'm going to go ahead and do this in red), and our moles are 40, 36, 24, and 11.8.
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So, CO--the concentration of CO, moles per liter of carbon monoxide--is 11.8; so this is going to be 11.8 moles per 2 liters.
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I take the number of moles and divide by the liters; it has to be concentration.
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Times the concentration of H₂O: H₂O is 24 moles, so it's 24.0 divided by 2.
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The concentration of CO₂: well, CO₂ is 36 moles, 36.0 moles; it is sitting in a 2-liter flask, so its concentration is 36/2, or 18 moles per liter.
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And now, the H₂ is 40 moles in a 2-liter flask.
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Now, you might think to yourself, "Well, wait a minute; 2, 2, 2, 2: don't the 2's cancel?"
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Yes, in *this* case they cancel; that is because all of these coefficients are 1, 1, 1, 1.
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But, you can't guarantee that all of the coefficients are 1, 1, 1, 1, so you can't just use the mole values in here.
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The equilibrium expression explicitly requires that you use concentrations, moles per liter, so let's just use moles per liter.
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Yes, they will cancel, but at least we know what is going on; we won't lose our way.
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That is what is important: write down everything--don't do anything in your head; don't cut corners; don't take shortcuts.
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I promise, it will go badly for you.
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OK, when we calculate this: 0.197; that is what we wanted--we wanted the K.
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Now, we want the K < font size="-6" > P < /font > , because that is the other thing that we have to find.
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Well, we know that K < font size="-6" > P < /font > is equal to K times RT^Δn.
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Well, what is Δn?--Δn is adding the product coefficients and subtracting the other coefficients.
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Well, Δn is (let me go ahead and write it here) equal to 1+1 (is 2), minus 1+1 (2), is equal to 0.
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So, in this case, Δn is 0.
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K < font size="-6" > P < /font > is equal to K, times RT, to the Δn, equals K, times RT to the 0 (anything raised to the 0 power is 1); is equal to K.
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So, in this case, K < font size="-6" > P < /font > is equal to 0.197...in *this* case; and the only reason it is this way is because everything is in a 1:1:1:1.
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2 reactants, 2 products: 1+1 is 2; minus 1+1 (is 2) is 0; that is it.
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Watch what you are doing very, very carefully; do not cut corners on equilibrium--do not cut corners ever, in any problem that you do.
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Write everything out; it's very, very important.
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OK, let's see: now, let's do a slightly more complicated problem, and again, this is going to be an example of taking information that is written in a particular problem and reasoning it out.
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It isn't just the math--the math is actually pretty simple once you know what is going on.
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That is the biggest problem with chemistry, or physics, or anything else: it's not how to turn it into math; it is, "What is going on, so that I know which math to use?"
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That is the real issue, and no two problems are the same; no two problems will be worded the same; you cannot count on that, especially at this level.
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OK, so Example 2 (and again, we are going to do a lot of problems, like I said, from here on in--from equilibrium all the way through at least electrochemistry, because this is the heart and soul of chemistry, not to mention the free-response questions that you are going to face on the AP exam)...
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OK, so the question is a bit long: A sample of gaseous PCl₅ (phosphorus pentachloride) was placed in an evacuated flask so the pressure of pure PCl₅ would be 0.5 atmospheres.
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So, a sample of gaseous phosphorus pentachloride was placed in an evacuated flask so the pressure of the pure PCl₅ would be .5 atmospheres.
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We stuck it in there so that it would be .5 atmospheres.
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But, PCl₅ decomposes according to: PCl₅ goes to PCl₃ + Cl₂.
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The final total pressure in the flask was 0.84 atmospheres at 250 degrees Celsius.
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Calculate K at this temperature.
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OK, so let's make sure we understand exactly what this problem is asking.
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Problems are going to be very, very specific; do not read into it anything that is there--read exactly what is there--very, very important.
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Don't cut corners.
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A sample of gaseous PCl₅ was placed in an evacuated flask so the pressure of pure PCl₅ would be .5 atmospheres.
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In other words, they filled it up with gas, and the pressure of the PCl₅ gas was .5 atmospheres.
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The problem is: "But, PCl₅ decomposes"--in other words, the PCl₅ they put there at .5 atmospheres--all of a sudden, it starts to come apart.
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It decomposes into PCl₃ and Cl₂.
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Well, now there is an equilibrium that exists; now, you not only have PCl₅ at .5 atmospheres; now, you have also produced some phosphorus trichloride gas and some chlorine gas, and there is also some PCl₅ gas left over.
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This is an equilibrium expression; so now, it is becoming a little bit more complicated.
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Now, you have three things in the flask, when you only introduced one.
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We measure the final pressure of the flask, and it comes out to .84 atmospheres at 250 degrees Celsius.
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Calculate the K at this temperature (K, we said, is K equilibrium, moles per liter).
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They have given this to us in pressures; so, the first thing we have to do is: we have to find K < font size="-6" > P < /font > and then calculate K.
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So, we want to find (well, if it's OK, I'm not going to write out everything; we know what we are doing)...so notice: they didn't write and say "Find K < font size="-6" > P < /font > ."
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They said, "Find K"; but the problem as written, since we are dealing with atmospheres and gases (PCl₅ gas, PCl₃ gas, Cl₂ gas)...we are going to deal with pressures, and then from pressures, we are going to use the RT and Δn expression that we just worked with to find the K.
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Make sure that you understand what it is that they are asking for.
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Don't just stop by getting the K < font size="-6" > P < /font > .
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OK, let's see how we are going to do this.
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We are going to introduce something called an ICE chart; and this ICE chart is something that we are going to use for absolutely the rest of the time that we discuss this.
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All equilibrium problems--acid-base problems, further aspects of acid-base equilibria, solubility-product equilibria, electrochemistry--we are going to deal with these things called ICE charts.
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This is sort of an introduction to them, and it is a way of dealing with what is going on in the problem.
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If you understand these, everything should fall out naturally.
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OK, so let's write the equation.
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PCl₅ is in equilibrium with PCl₃ + Cl₂.
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ICE stands for Initial concentration, before anything happens; C stands for Change--what changes take place; and E stands for Equilibrium concentrations.
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Well, it is these equilibrium concentrations that go into the equilibrium constant expression, right?
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That is what we said: the K < font size="-6" > eq < /font > : those values that we put in there are concentrations at equilibrium.
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I means Initial; C means change.
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Even if you are not sure what to do in a problem, just start, and just write down what you know; write down what is happening; eventually, the solution will fall out.
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The single biggest mistake that kids make is: they think that they are supposed to just look at a problem and automatically know what is going on.
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Even I don't just look at a problem and know what is going on!
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After all of these years of experience, I have to sit there and stare at it, and sometimes just see where I am going.
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ICE chart is a great place to start with equilibrium problems.
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It will give you a sense of what is happening; then, you can put the math together.
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Don't ever feel that you have to just know what is happening; you are extracting information.
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And...sorry about that--E stands for equilibrium.
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OK, so our initial concentration of PCl₅ was .5 atmospheres.
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Remember what we said: pressure and concentration--they are the same; they are just different sides of the same coin, so we can deal in pressures the same way we deal with concentrations.
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We start with 0.5 atmospheres, where, before anything happens--before PCl₅ decomposes--there is no PCl₃, and there is no Cl₂; so this is our initial condition.
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It's nice; well, a certain amount of PCl₅ decomposes.
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Well, look at our equation; it's 1:1:1.
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For every 1 mole that decomposes, 1 mole is produced, and 1 mole is produced of the Cl₂.
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We can say that, if -x is the amount that disappears (or, again...concentration and pressure are the same thing)...so, if the pressure drops--PCl₅--by a certain amount, that means it has to increase here by that same amount for each.
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For every 1 atmosphere that drops, that means a certain amount has been used; well, that means that a certain amount has been produced of the PCl₃ and the Cl₂.
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That is what the stoichiometry tells me.
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That is why I have -x, +x, +x; I hope that makes sense--because, again, when this decomposes, this is forming; that is what is going on.
00:26:50.000 --> 00:26:54.000
So, if 1 mole of this decomposes, 1 mole of this is formed; 1 mole of this is formed.
00:26:54.000 --> 00:27:01.000
If 5.2 moles of this decomposes, 5.2 moles of PCl₃ is formed; 5.2 moles of Cl₂ is formed.
00:27:01.000 --> 00:27:07.000
We don't know how much is formed yet, so that is why we use x: -x here, +x, +x.
00:27:07.000 --> 00:27:13.000
Now, we add: the initial plus the change gives us the equilibrium condition.
00:27:13.000 --> 00:27:17.000
At equilibrium, I have 0.50-x.
00:27:17.000 --> 00:27:25.000
Here, I have 0+x is x; 0+x is x; so at equilibrium, this is how much I have.
00:27:25.000 --> 00:27:31.000
Now, they want us to find K; well, what other information do they give us?
00:27:31.000 --> 00:27:37.000
They give us the fact that our total pressure is equal to .84 atmospheres.
00:27:37.000 --> 00:27:45.000
Well, our total pressure is the pressure at equilibrium: this plus this plus this.
00:27:45.000 --> 00:28:02.000
The total pressure is the sum of the individual pressures; so, we write: 0.50-x+x+x (write it out; don't do it in your head first) =0.84 atm.
00:28:02.000 --> 00:28:13.000
Well, this -x cancels with that x, and I'm left with 0.50+x=0.84; this is simple arithmetic--there is nothing hard about this.
00:28:13.000 --> 00:28:21.000
x=0.84-0.5; it equals 0.34 atmospheres; look at that.
00:28:21.000 --> 00:28:30.000
I have just solved for x, x, x; I did it; that is nice.
00:28:30.000 --> 00:28:44.000
I found the value of x: .34 atmospheres of PCl₃ show up; .34 atmospheres of Cl₂ show up; and .34 atmospheres of PCl₅ is lost.
00:28:44.000 --> 00:28:47.000
So now, we can actually find our K.
00:28:47.000 --> 00:28:52.000
That is the best part, which is ultimately what we want; so we are solving for K < font size="-6" > P < /font > here.
00:28:52.000 --> 00:29:27.000
So now, I have partial pressures: I have the partial pressure of chlorine gas, is equal to x, which is 0.34 atm; I have the partial pressure of the PCl₃ gas, which, again, is x, which is 0.34 atm; and I have the partial pressure of the PCl₅ gas, which is 0.50-x (that is the equilibrium) minus 0.34, which is 0.16 atm.
00:29:27.000 --> 00:29:32.000
Now, I can put these values into my equilibrium expression.
00:29:32.000 --> 00:29:52.000
I know what my equilibrium expression is; my equilibrium expression is K < font size="-6" > P < /font > equals the partial pressure of Cl₂, times the partial pressure of PCl₃, divided by the partial pressure of PCl₅, each raised to the first power, because the stoichiometric coefficients are 1.
00:29:52.000 --> 00:30:15.000
Well, that equals (let's go down here) 0.34, times 0.34, divided by 0.16; I do the multiplication, and I get 0.72; this is my K < font size="-6" > P < /font > ; K < font size="-6" > P < /font > equals 0.72.
00:30:15.000 --> 00:30:22.000
They didn't ask for K < font size="-6" > P < /font > ; they asked for K, which means they asked for K < font size="-6" > eq < /font > .
00:30:22.000 --> 00:30:26.000
Well, the relationship between K and K < font size="-6" > P < /font > is the following.
00:30:26.000 --> 00:30:47.000
K is equal to K < font size="-6" > P < /font > times (RT)^Δn, if that is correct; let me double-check; K < font size="-6" > P < /font > equals...-Δn.
00:30:47.000 --> 00:31:02.000
OK, so we have...let's see...yes, OK; so now, what is Δn, first of all?
00:31:02.000 --> 00:31:11.000
Δn is equal to...well, we take the...again, let's write out the equation, so we have it on this page.
00:31:11.000 --> 00:31:27.000
We have PCl₅ in equilibrium with PCl₃ plus Cl₂; Δn is equal to 1+1-1; 2-1=1.
00:31:27.000 --> 00:32:00.000
So, K is equal to K < font size="-6" > P < /font > (which is 0.72), times RT...OK, so here is where we have to be careful; R is not...well, the R that we are going to be using here is 0.08206; that is liter-atmosphere/mole-Kelvin; times temperature in Kelvin; it has to be in Kelvin, because the unit of this is liter-atmosphere...
00:32:00.000 --> 00:32:05.000
You know, let me write out the units here, so you can see it.
00:32:05.000 --> 00:32:15.000
This is liter-atmosphere per mole-Kelvin; so the temperature has to be in Kelvin.
00:32:15.000 --> 00:32:28.000
At this particular temperature, add 273; you get 523 Kelvin to the -1 power (negative Δn; that was the relationship).
00:32:28.000 --> 00:32:37.000
You put in K < font size="-6" > P < /font > ; R is .08206; temperature in Kelvin is 523; negative Δn...Δn was 1; negative 1.
00:32:37.000 --> 00:32:46.000
We do the math, and we end up with 0.017.
00:32:46.000 --> 00:32:56.000
There you go; in this particular problem, there was a lot going on; we handled it by just sort of stopping, taking a look, and making sure we wrote everything out.
00:32:56.000 --> 00:33:08.000
We introduced this thing called an ICE chart; we write out the equation on top; underneath, we write the initial concentrations, we discuss the changes that take place, and we add to get our equilibrium.
00:33:08.000 --> 00:33:11.000
From there, we take a look at what the problem is asking.
00:33:11.000 --> 00:33:16.000
Sometimes, they might give us a K, and they might ask for a particular concentration.
00:33:16.000 --> 00:33:21.000
I take those equilibrium concentrations, and I put them into my expression, and I solve that way.
00:33:21.000 --> 00:33:25.000
In this case, they gave me a total pressure so I could find x.
00:33:25.000 --> 00:33:29.000
I used that to find the K; it just depends on what they are asking.
00:33:29.000 --> 00:33:38.000
This is why we are going to do a lot of different types of problems for these equilibrium and so on...at least through electrochemistry.
00:33:38.000 --> 00:33:47.000
OK, so let's do another example.
00:33:47.000 --> 00:34:46.000
Example 3: We have: Solid ammonium chloride, solid NH₄Cl, was placed in an evacuated chamber, then heated; it decomposed according to: NH₄Cl, solid, decomposes into ammonia, NH₃, gas, plus hydrogen chloride gas.
00:34:46.000 --> 00:34:50.000
It's not hydrochloric acid; it's hydrogen chloride gas.
00:34:50.000 --> 00:35:19.000
Now, after heating, the total pressure was found to be 4.4 atmospheres.
00:35:19.000 --> 00:35:25.000
Our task is to calculate the K < font size="-6" > P < /font > , the equilibrium constant with respect to pressures.
00:35:25.000 --> 00:35:32.000
OK, well, let's write out the equilibrium expression first; it's always a great thing to do--write everything out.
00:35:32.000 --> 00:35:37.000
Write out the equilibrium expression; write out the ICE chart; and then, just see where you go from there.
00:35:37.000 --> 00:35:43.000
So, the equilibrium expression here, based on this equation--well, we have a gas; we have a gas; and we have a solid.
00:35:43.000 --> 00:35:48.000
Solids don't show up in the equilibrium expression, so in this case, it is just these two.
00:35:48.000 --> 00:35:57.000
The coefficients are 1, so we have: the K < font size="-6" > P < /font > is equal to the partial pressure of NH₃ gas, times the partial pressure of HCl gas.
00:35:57.000 --> 00:36:03.000
That is it; if we find the partial pressures, we are done--we plug them in, we multiply them, and we are done.
00:36:03.000 --> 00:36:05.000
OK, now let's do our ICE chart.
00:36:05.000 --> 00:36:09.000
Do our ICE chart: always do it this way.
00:36:09.000 --> 00:36:23.000
NH₄Cl (write out the equation, and do it underneath; don't do it separately--you want to be able to keep everything straight) goes to NH₃ + HCl.
00:36:23.000 --> 00:36:28.000
Our initial concentration; our change; and our equilibrium concentration...
00:36:28.000 --> 00:36:38.000
OK, solid NH₄Cl--we don't care; it doesn't even matter, so we just put lines there; it doesn't show up in the equilibrium expression--it doesn't matter.
00:36:38.000 --> 00:36:48.000
Our initial NH₃ and HCl concentration--well, we started off with solid NH₄Cl; so, initially, there is none of these.
00:36:48.000 --> 00:36:54.000
The change--well, a certain amount shows up; that certain amount is what we want.
00:36:54.000 --> 00:37:00.000
So, +x+x, right?--so again, you have to be able to see what the question is saying.
00:37:00.000 --> 00:37:09.000
It is telling you that you start off with NH₄Cl; it decomposes--when something decomposes, that means it is going away; the products are showing up.
00:37:09.000 --> 00:37:15.000
That is why you have a +x and a +x here.
00:37:15.000 --> 00:37:19.000
This + goes here; it's not that +.
00:37:19.000 --> 00:37:31.000
0+x is +x; 0+x is +x; well, they are telling me that the total pressure in this flask is 4.4 atmospheres.
00:37:31.000 --> 00:37:36.000
Well, which gases are in the flask?
00:37:36.000 --> 00:37:41.000
Well, the solid is a solid; that doesn't matter--that doesn't do anything for the gas.
00:37:41.000 --> 00:37:45.000
The gases in here are NH₃ gas and HCl gas.
00:37:45.000 --> 00:38:03.000
So, I basically have: x+x=4.4; they are telling me that the total pressure in there is 4.4 atmospheres; that has to be made up of the amount of NH₃ gas and the amount of HCl gas.
00:38:03.000 --> 00:38:07.000
Well, that is even, because they are forming a 1:1 ratio.
00:38:07.000 --> 00:38:27.000
So, 2x is equal to 4.4; x is equal to 2.2; well, 2.2--there you go; that is the partial pressure of HCl and the partial pressure of NH₃.
00:38:27.000 --> 00:38:39.000
I have 2.2 and 2.2 (2.2, not 2.4...oh, numbers and arithmetic!); these numbers are the one that I put back in here.
00:38:39.000 --> 00:38:52.000
So, K < font size="-6" > P < /font > is equal to 2.2, times...OK, now watch this: even though 2.2 and 2.2 is the same, please don't write 2.2².
00:38:52.000 --> 00:39:12.000
They are different species; I promise you, if you write 2.2², and if somewhere along the way you get lost and you have to come back, you will spend five minutes trying to figure out what happened, because again, stoichiometric coefficients--they show up in the equation, so write them separately; write 2.2 times 2.2.
00:39:12.000 --> 00:39:15.000
Even though they are the same, they represent different species.
00:39:15.000 --> 00:39:17.000
Don't mix them up.
00:39:17.000 --> 00:39:30.000
We multiply that; we get 4.84; 4.84--that is the K < font size="-6" > P < /font > , and we double-check: "Calculate K < font size="-6" > P < /font > "--that is what we wanted; we're done; that is it.
00:39:30.000 --> 00:39:40.000
So, they gave us a certain amount of information; they gave us an equation to work with; we wrote down the K < font size="-6" > P < /font > expression; we wrote down the equation.
00:39:40.000 --> 00:39:48.000
We wrote an ICE chart (Initial, Change, Equilibrium--ultimately, it is the equilibrium that we are concerned with, because the system has come to equilibrium).
00:39:48.000 --> 00:39:52.000
We followed it; we get x and x.
00:39:52.000 --> 00:39:57.000
They tell us that the total pressure is 4.4; well, the total pressure is the sum of the individual pressures.
00:39:57.000 --> 00:40:01.000
There are only 2 gases in here (the NH₃ and the HCl); each one is x.
00:40:01.000 --> 00:40:13.000
2x=4.4; x=2.2; at equilibrium, 2.2 atmospheres is hydrogen chloride gas; 2.2 atmospheres is ammonia gas.
00:40:13.000 --> 00:40:19.000
You plug that into the equilibrium expression; we multiply, and we get 4.84: a standard equilibrium problem.
00:40:19.000 --> 00:40:25.000
OK, this sort of gets us going with the types of problems we are going to be dealing with with equilibrium.
00:40:25.000 --> 00:40:43.000
We are going to systematize this, and this whole idea of the ICE chart--our problems are going to start to become a little bit more complex, but this whole idea of using an ICE chart, writing the equilibrium expression, and then seeing what they want, based on what they give us--the ICE chart itself is going to be different.
00:40:43.000 --> 00:40:45.000
That is what is going to change.
00:40:45.000 --> 00:40:48.000
The approach does not change.
00:40:48.000 --> 00:40:52.000
OK, so thank you for joining us here at Educator.com for AP Chemistry and equilibrium.
00:40:52.000 --> 00:40:53.000
We will see you next time; goodbye.