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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Today, we are going to continue our detailed discussion of equilibrium.
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We are just going to be doing examples.
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We have talked about what equilibrium is, the expression; we have actually done a fair number of problems so far, and problems that are reasonably complicated.
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But, I wanted to do some more that go a little bit more in-depth, as far as--they have some slight variations; they are a little more detailed and require a little bit more care--for the purposes of showing you where things can possibly go wrong, but mostly just to get you really, really accustomed to these equilibrium problems.
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It has been my experience, from all of my students who have had a solid grounding in equilibrium--the rest of chemistry is a complete breeze for them.
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They know exactly what to do; they know where to go; I virtually didn't have to teach the class after this.
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We hammered the equilibrium so much that they knew exactly what was going on; that is why this is important.
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If you can get your head around this, and be able to handle at least a good 70 or 80 percent of these problems, the AP test should be absolutely a breeze for you, I promise.
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OK, so let's jump right on in.
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Our first example is going to be a synthesis of hydrogen fluoride from hydrogen and fluorine gas.
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Let's go: Example 1: So, in a synthesis of hydrogen fluoride gas...
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Now notice, I didn't say hydrofluoric acid, because this is hydrogen fluoride gas--it's in gaseous form; remember, an acid is something that is when you take the thing and you drop it in the water, and it dissociates into a hydrogen ion and fluoride ion, or hydrogen ion and any other anion.
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That is when it is an acid--the acid is the H^+; when HF is together, it is not an acid--it won't do any damage.
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Well, I won't say it won't do any damage, but it won't do any damage as an acid--let's put it that way.
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In the synthesis of HF from H₂ and F₂ at a given temperature, 3.000 moles of H₂ and 6.000 moles of F₂ are mixed (it might be nice if I actually knew how to spell--yes, that would be good) in a 3.0-liter flask.
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So again, we have moles, and we have a 3-liter flask; this time it is not 1-liter, so we are going to have to actually calculate the concentration.
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OK, the eq constant at this temperature is 1.15x10².
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What are the equilibrium concentrations of all species?
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Or I should say, "What is the equilibrium concentration of each species?"
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Being able to handle these ICE charts is the real key to the majority of the second half of chemistry.
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The ICE chart itself is always going to be a ubiquitous feature of the problems that we solve.
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The difference is what the ICE chart looks like: acid-base problems, equilibrium problems, thermodynamic problems, whatever it is...the ICE chart itself is going to change, depending on what the problem is asking.
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Being able to handle that--that is the real deal; that is when you know you actually know what is going on--when you know how to arrange the ICE chart as necessary; the rest is just math, basic algebra.
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OK, so in the synthesis of HF from H₂ and F₂ at a given temperature, 3 moles of hydrogen and 6 moles of fluorine are mixed in a 3-liter flask.
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The equilibrium constant is 1.15x10², or 115.
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What is the equilibrium concentration of each species?
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OK, so let's write our equation; like we said, that is our process: H₂ + F₂ goes to 2 HF.
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We want to make sure that the equation is balanced; we want to write our equilibrium expression--it is going to be the concentration of HF squared (stoichiometric coefficient), over the concentration of H₂, times the concentration of F₂.
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OK, now let's go ahead and calculate the initial concentration of H₂.
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Well, it says we have 3.00 mol of H₂ in a 3.000-liter flask; so we have 1.000 Molar H₂.
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That is what this little 0 here, down at the bottom is: H₀; it just means the initial concentration.
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If you want to put an i, that is fine, too.
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F₂, the initial concentration--it says we have 6 moles of fluorine gas, over (again, it's in the same flask, so it's) 3 liters, so our concentration is 2.00 Molar.
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Well, Q--the next thing we want to do is, we want to calculate the Q in this case, the reaction quotient, to tell us what direction it is in.
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There are a couple of ways to do this in this particular problem: we can just plug it in; the concentration of HF to begin with...well, there is no HF, so it's just 0...over 1, times 2, which is 0.
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That is fine: you can do it that way, or you can just say, "Well, since there is no HF to begin with, I know the reaction is going to move in that direction; there is none of this yet--there is only this and this--so it has to move forward; there is nowhere else for it to go."
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But, I think it is best to stick with the process, and just go ahead and put the numbers in, and do it that way.
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OK, so this is 0, which implies that the reaction will move to the right.
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Moving to the right means it will form product and it will deplete reactant.
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H₂ and F₂ concentration will go down; HF concentration will rise.
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Now, we can do our ICE chart: H₂ + F₂ (I'll give myself plenty of room here) goes to 2 HF; Initial, Change, Equilibrium.
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We start off with 1 Molar of hydrogen; 2 Molar of fluorine; 0 Molar of HF.
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This is going to deplete by x; this is going to deplete by a certain amount, x; this is going to increase by an amount, 2x, because of that 2.
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That is the whole idea: 1, 1, 2.
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We get equilibrium concentrations of 1.000-x, 2.000-x, and we get 2x.
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Now, we plug in these equilibrium concentrations into this expression, and since we already know what that is, we solve the algebraic equation.
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So, let's write: K is equal to...well, it's equal to the concentration of HF squared, so it's 2x squared, over 1.000-x, times 2.000-x.
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That is our equation, and we know it is equal to 1.15x10².
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They gave us the K; now, let's solve for x.
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Well, this is just a quadratic equation; so you are just going to have to sort of do it.
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You can...there are several ways you can do this: you can do it by hand, do the quadratic formula; you can go ahead and use your graphic utility, for those of you that have the TI-83s and 84s and 87 calculators.
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You can handle these really, really easily; that is what I prefer; that is what I use; I have a TI-84--that is how I solve these.
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Let's just go ahead and at least work out the algebra, and then we'll just write out the answer, presuming that we actually used a graphical utility or something like that to do it.
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We end up with 4x²=1.15x10² (2.000-3.000x+x²), and we get 4x²=230-345x+115x².
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And then, we end up with 111x² - 345 x + 230 = 0.
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And again, when I put this into my graphical utility, and I solve, I end up with (again, this is a quadratic equation, so I have 2 values): the first x-value is 0.968 Molar, and the second root is going to be 2.14 Molar.
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Now, we can't have both of these be true; there is only one equilibrium condition; one of these has to be true.
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Here is how you decide: well, take a look at the original concentration--1 Molar of hydrogen gas, I think it was.
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Well, 1 molarity - 2.14 molarity--that is going to give you a negative 1.14 molarity; you can't have a negative concentration, so that one drops out.
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Let's do that in red; that one drops out--this is the x-value.
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So, we have that .968 Molar is the x-value; we plug those back into the equilibrium concentrations in our ICE chart to get the following.
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Our H₂ concentration is 1.000-0.968=3.2x10^-2 Molar at equilibrium.
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Our F₂ concentration is 2.000-0.968, is equal to (let's get a better-looking equals sign than that) 1.032 molarity.
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And our final HF concentration is 2x, so it's 2 times 0.968, is equal to 1.936 molarity (you know what, these numbers are getting strange again...1.936 molarity).
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And my friends, we have our final solution.
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Notice, we started off with 1 Molar of H; we end up with 3.2x10^-2; that means most of the H is used up.
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That makes sense, because again, you are looking at a very, very large K < font size="-6" > eq < /font > ; the K < font size="-6" > eq < /font > is 115--that means the reaction is very far to the right.
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It favors the product formation, not reactants.
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When it has come to equilibrium, virtually no reactants are left over.
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The only reason that you have 1.32 moles per liter left over of the other reactant: because you started off with 2 moles per liter of that, and the stoichiometry of the H₂ to the F₂ is 1:1.
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You have used up half of it; that is the only reason that it looks like this number is so big, compared to this number.
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The K < font size="-6" > eq < /font > tells you that it is mostly to the right; you started off with no hydrogen fluoride; you ended up virtually 2 moles per liter of hydrogen fluoride.
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That is confirmed; so, the numbers match up; everything is good.
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OK, let's see what is next.
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Example 2: OK, this is going to be an example of an equilibrium problem where the K < font size="-6" > eq < /font > is actually very, very small.
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And again, depending on the equation, you might run into some rather complicated things, like cubic, quartic, quintal equations, which you can certainly handle with your graphical utility--it's not a problem--that is the great thing about having graphical utilities.
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But, the method we are going to show is going to be a slightly simplified version, if you just want to do the math quickly.
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And then, we will give you a way of checking to see whether any simplifications you made were actually viable--whether you could actually get away with it.
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There might be situations where you might simplify something, and you can't get away with it.
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We will give you a rule of thumb for doing that.
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OK, so Example 2: Gaseous NOCl decomposes to form gaseous NO and Cl₂.
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At 40 degrees Celsius, the eq constant is 1.4x10^-5; it is very small--that means there is virtually no product at equilibrium.
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OK, initially, 1.0 mol of NOCl is placed in a 2.0-liter flask.
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Here we go again: what is the equilibrium concentration of each species?
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OK, let's write our equation: 2 NOCl decomposes into 2 NO + Cl₂.
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Let's write our equilibrium expression--that is what we always do: it's going to be the concentration of NO squared, times the concentration of Cl₂, divided by the concentration of NOCl squared (is that correct?--yes, that is correct).
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OK, so now we do that, and let's see what else we have.
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Initial concentrations--we have to do initial concentrations.
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The NOCl equals 1.0 mol over...it looks like a 2.0-liter flask: is that correct?--yes.
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Let me circle my numbers: 2 (oops, let me use blue) liters, 1 mole; that is my K < font size="-6" > eq < /font > ; OK.
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Let's go back to red; that is going to equal 0.50 moles per liter, and then I have my initial NO concentration, which is 0, and my initial Cl₂ concentration, which is also 0.
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Therefore, Q is equal to 0 squared times 0, over 0.5, equals 0, which is definitely smaller--the K < font size="-6" > eq < /font > is small, but it is still bigger than (the 1.4x10^-5 is still bigger than) 0, which implies that the reaction moves to the right.
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Moving to the right means we are forming product.
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We are depleting reactant.
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OK, so let's go ahead and set up our ICE chart.
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We write our equation: 2 NOCl goes to 2 NO + Cl₂; Initial, Change, Equilibrium.
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We start off with 0.50 Molar and none of those; we are going to deplete this by 2x; we are going to form 2x, and we are going to form x.
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Therefore, the equilibrium concentrations...just add them straight down: what you started with, what you lost, and what you end up with.
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0.50 minus 2x; 2x; and x; now, let's go ahead and put it into our expression.
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Our K is equal to 2x squared (that is the squared part--2x squared) times x to the 1 power, divided by that squared, (0.50-2x)².
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That equals 1.4x10^-5; OK.
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Well, all right, see now: it is getting a little complicated.
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You are going to end up with 4x² times x; it is going to be 4x³; I don't know...you are certainly welcome to go ahead and solve this, just because you have a graphical utility; it's not a problem--you can do that.
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But, let me give you an alternate procedure, which actually makes things at least a little bit more tractable mathematically.
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Because we notice that the K < font size="-6" > eq < /font > is very small, which means that the reaction is over here, mostly; that means not a lot of product; this is small, because that is small, meaning that there is not a lot of product--in fact, there is virtually no product at all...it is mostly NOCl still.
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Well, because of that--because it is mostly NOCl still--that means that NOCl hasn't lost very much.
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Very little of it has actually decomposed.
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Well, very little of it has decomposed...well, we started with .5; that means x is probably really, really, really small compared to .5.
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Because it is so small compared to .5, it is possible to take this term and just leave it out--ignore it.
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Solve the problem, and then check to see whether it is actually valid or not that we did what we did.
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So, this is how we do it: so again, to our first approximation, we are going to presume because this is small, it means that most of it is here; that means very little of this is going to decompose; in other words, very little of this is going to form.
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Because this is small, we are saying it is so small compared to .5 that it is probable that the .5 minus the 2x is going to go unnoticed, so let me just ignore it and solve the easier problem.
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OK, well, let's see what we can do.
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We have: It's going to be 2x² times x, over 0.50 squared, equals 1.4x10^-5.
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Well, what we end up with here is (what we are going to end up with, once we do all the multiplication): we are going to end up with: 4x³ is equal to 3.5x10^-6; x³ is equal to 8.75x10^-7; x is going to equal 9.6x10^-3.
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We found a value of x, 9.6x10^-3; now, we want to check to see whether we were actually justified in ignoring it.
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Well, if I take (yes, that is fine) 9.6x10^-3, and if I divide by the 0.50, and I multiply by 100, I am trying to see what percentage of the .5 this 9.6x10^-3 really is.
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That is what I'm doing; I am trying to see how much of the .5 this is.
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Well, as it turns out, it actually equals about 1.91%.
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Now, we did, actually, 2x: so 2x would put us roughly at twice that; but, as you can see, the 1.91% is actually really, really small.
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As it turns out, if you make an approximation of this nature, and the value that you get ends up being anywhere from about...anywhere less than 5 percent of your total value, it is actually (as a good rule of thumb) a valid approximation.
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That means, virtually, the mathematics is not going to notice that you actually eliminated that.
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So, because of the 1.91 percent, the x compared to the .5 (or, in this case, it's going to be 2x, but again, you are still going to be below 5%), you are actually pretty good.
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And this allows us to sort of keep this value of x, as opposed to having to solve this entire equation, this cubic equation.
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We just did it in a nice, simple way to get an answer, instead of having to use a graphical utility, and we got this value for x.
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Given that value of x, we can go ahead and use it now to find our equilibrium concentrations.
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Our NOCl concentration, final, is equal to 0.50, minus the 9.6...oops, this is going to be minus *two* times the 9.6x10^-3.
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Our final NO concentration (I will go ahead and let you finish off the arithmetic here) is going to equal 2 times 9.6x10^-3.
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Our final Cl₂ concentration is going to equal, well, just x: which is 9.6x10^-3.
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The rest is just arithmetic, which I will leave to you--nice and straightforward.
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Hopefully you are getting a sense of the general procedure and of the things that you have to sort of watch out for.
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OK, so let's see: let's go ahead and close this off with an interesting type of problem; it should go pretty quickly, actually.
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Let's do...yes, that is fine; we can start it on this page.
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OK, Example 3: Calculate the value of the equilibrium constant for the reaction O₂ gas + oxygen gas goes to ozone gas, given reaction 1 (this is going to be sort of a Hess's Law kind of problem): NO₂ in equilibrium with NO + O, and K₁ is equal to 6.8x10^-49 (wow, that is really, really small) and that one, which is O₃ + NO goes to NO₂ + O₂; K₂ equals 5.8x10^-34.
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Let's see what they are asking: they are asking you to calculate the value of the equilibrium constant for this reaction (let's do this in blue), given these two reactions and their corresponding equilibrium constants.
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Well, we know from Hess's Law that if we want to find a final reaction that involves some of the reactants that we have, we have to rearrange them by either flipping them or multiplying them by constants.
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In doing so, if we add all of the equations together, and then get our final equation--well, when we did ΔHs, when we did enthalpies, we just added the enthalpy; with Ks, with equilibrium constants, it is actually different.
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When we add equations to get a final equation, what we do to equilibrium constants is: we actually multiply them.
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So, let's go ahead and do this one.
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In order to actually come up with this, I am going to flip Equation 1.
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I am going to flip Equation 1, and that will give me: NO + O goes to NO₂.
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Now, when I flip an equation, I take the reciprocal of the equilibrium constant, right?--because you are just flipping products and reactants.
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That equilibrium constant, now, is 1.47x10^48--huge!
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I am also going to flip the second reaction; I am going to make the reactants the products and the products the reactants; when I do that, I end up with NO₂ + O₂ in equilibrium with NO + O₃.
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Well, this equilibrium constant--again, I flipped it, so I take the reciprocal of that, and I get 1.72x10^33.
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Now I add these two equations; NO₂ cancels NO₂; NO cancels NO; I am left with O + O₂ goes to O₃; this was the equation that we wanted.
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Now, in order to get the final equilibrium constant, I have to...I don't add these; I multiply them; it becomes K₁' times K₂'.
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It equals 1.47x10^48, times 1.72x10^33, and I end up with getting some huge number, if I am not mistaken.
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2.53x10^88: that is huge!--that means that any time oxygen gas and free oxygen atom come together, you will not find them separately!
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This huge number tells me that the reaction is way to the right; there is no equilibrium here--not really.
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Any time you have a bunch of equations, if you add them together to come up with a final equation, you multiply the equilibrium constants for all of the individual equations.
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So, when we add equations to get a final net equation, our final K is equal to K₁ times K₂ times...all the way to K < font size="-6" > n < /font > , if we had n equations; and that is it.
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OK, so we have gone ahead and dealt with a fair number of problems in equilibrium.
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Next time, we actually are going to continue our discussion of equilibrium; we are going to talk about Le Chatelier's Principle, and then we are going to do some more problems involving Le Chatelier's Principle.
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And Le Chatelier's Principle, just to give you a little bit of a preamble, is basically just: If I have a system at equilibrium, if I stress that system out somehow--if I put pressure on it, meaning if I add this or heat it up or cool it down, what happens to that equilibrium.
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So, I can shift the equilibrium; well, we know (well, I can tell you) that a system will always seek out equilibrium.
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When I apply stress to a particular system, the system is going to respond by doing whatever is necessary to relieve that stress.
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You know this from just being a human being: any time, any system that you apply stress to, the response of that system will be doing the things that relieve that stress.
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Well, a chemical system behaves in exactly the same way, and it is a very deep, fundamental thing called Le Chatelier's Principle.
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It is actually quite beautiful; so we look forward to seeing you next time.
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Thank you for joining us at Educator.com and AP Chemistry.
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We'll see you next time; goodbye.