WEBVTT chemistry/ap-chemistry/hovasapian
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Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.
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In our last lesson, we concluded our discussion of kinetics, but I thought it would be a good idea to actually spend some time doing some practice problems, directly from the AP exam, and just to give you an idea of what it is that you are going to be doing on the exam, because ultimately, again, this is an AP chemistry course, and the final thing is the AP Chemistry test.
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There is nothing altogether different as far as the example problems that we did, but I thought it would be nice to have some real, specific examples of what it is that you are going to see.
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So, we're going to do a couple of multiple-choice and a couple of free-response questions, and things like that.
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Let's just jump right in and see what we can do.
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OK, so our first series of questions is going to be three multiple-choice questions, and it is going to be based on the following hypothetical reaction and the following choices.
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OK, so let's see: we'll say: Consider the following hypothetical reaction.
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Now, mind you, these multiple choice questions--you can't use your calculator in these, so the numbers are going to be very, very simple.
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Often, you will be able to just look at something and see what it is that is going on, so if you find yourself having to do multiple strange computations, or if you think that you can't do it because of a computation, chances are that there is something wrong.
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These multiple-choice questions--they test basic understanding: can you follow a logical train of thought, where the numbers are completely secondary?
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It is the free-response questions where you can use your calculator to actually find specific answers.
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Hypothetical reaction: we have: our reactant R, plus reactant S, goes to product T.
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Now, the following are our choices.
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A) Rate = K times concentration of R to the first power.
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B) we have: The rate = K times the concentration of S to the second power.
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Our C) choice is: The rate is equal to K, times the concentration of R to the first, the concentration of S to the first.
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D) Our choice is: The rate is equal to K, times the concentration of R to the second power, the concentration of S to the first.
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And E), our final choice, because you have ABCDE to bubble in: The rate is equal to K times R squared, S squared.
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So, we have a hypothetical reaction, and we have these choices to choose from, based on the questions that we are going to ask.
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The first question: #1: OK, it says: When R and S (in other words, when the concentrations of R and S) are doubled, the rate increases by a factor of 8 (so they are telling you this; they are telling you that they double the concentration of R; they double the concentration of S; the rate increases be a factor of 8).
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The question is: What is the rate law?
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A through E are our choices.
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OK, so how do we deal with this?
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Again, we should be able to do this reasonably quickly.
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Well, they are telling us that they are doubling R and they are doubling S, and that the rate is increasing by a factor of 8.
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Well, there are a couple of ways that we can do this.
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I could say to myself, "Well, if I double the rate..." I can say by doubling something, I am just making it twice that; well, if the rate increases by a factor of 8, I can say, "Two to what power equals 8?"
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But, I think that may be a little bit confusing, so let's go ahead and do this a more natural way.
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Let's just go ahead and stick the doubled rate into these to see what happens.
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Well, notice what happens: if I put in twice R into the first one, I get K2R; the 2 comes out--I get 2 KR.
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Let me just do that; let's try choice A.
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Try choice A: I'm a little bit longer on this first one, but it shouldn't take this long--I just want you to see the process.
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Try choice A: Well, I get that the rate is equal to K, times twice the rate of R.
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2 just comes out; so twice the KR.
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Well, KR is the rate; so by doubling the rate, all I have done is multiply the rate by a factor of 2; that is not the one.
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I do the same thing with S; so, if I try choice B, they are telling me that the rate is equal to K2S--just double the rate and put it in there; and this time, it's 2S², right?--because our choice is S to the second power.
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Well, this is equal to 4 times KS².
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So now, by doubling the concentration of S, I put it in the choice, and I run the mathematics, and I get that my final answer is 4 times K to the S squared.
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Well, K to the S squared was the original rate; 4 times that--that is not it, either.
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So, as it turns out, the only one of these, when I put it in, where I double R and I double S and I run the mathematics--the only thing that gives me an actual increase by a factor of 8--is choice D.
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The answer is D, and the best way to see that--just in terms of by looking at it without having to run through this--is just: take the number 2--they double the rate--and put it in here.
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2 squared is 4; times 2 is 8.
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You get 8 on this side.
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That is how you solved this problem: just put it in and see which one actually gives you what it is that they are asking for.
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I hope that makes sense.
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OK, under the same circumstances now, problem #2: When R and S are doubled (again, the concentrations are doubled), the rate increases by a factor of 2.
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Well, they say that both R and S are doubled; so, if you go back to your choices, and if you look through your choices--if you put 2 in for R, 2 in for S, well, the only one that actually gives you double the rate is choice A.
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So, A is the answer.
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And again, if you don't see that: the rate they told me was K, times R to the first power.
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Well, if I put in 2, they tell me R and S are doubled; in this case, this choice A is not dependent on S at all--it only depends on R; so when I put 2 in there, I get K times 2R equals 2 times K times R.
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Well, 2 times R--that equals 2 times the rate; that means the rate is doubled; that is the only choice.
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If I put it into the other choices, I get that the rate is increased by a factor of 4, by a factor of 8, by a factor of 16...so the only one that is doubled is choice A, so our answer is A.
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That is how we do it; these are very, very quick--you just need to be able to know what to put where and what to multiply.
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OK, well, let's take a look at #3.
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This time, they say, R is doubled; so this time, they are only doubling R.
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R is doubled; S is unchanged.
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Well, let's write this a little differently.
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This time, what we have is: When R is doubled (the concentration of R), and S is unchanged, the rate is unchanged.
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OK, well, they tell me that R is doubled, and they tell me that S is unchanged, and they tell me that the rate is unchanged.
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Well, if I go back to my choices and if I take a look--well, if R is doubled--notice that choice A, choice C, choice D, and choice E all have R in them.
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So, if I stick a 2 in there, I am going to get an increase; there isn't going to be no change; but they tell me that S is unchanged.
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Well, if S is unchanged, the only choice (which is choice B)--the rate on this one is equal to K, times the concentration of S squared.
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Well, they are telling me that S is unchanged; well, if I don't change S, I don't change the rate.
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S unchanged; rate unchanged; my only choice is B, because everything else has an R in it, and if the rate is doubled, that means there is going to be some sort of a change to the rate.
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In this case, my only choice is choice B, under these circumstances.
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I hope that makes sense.
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OK, let's try a free-response question here.
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No, actually, I'm sorry; let's try one more multiple-choice question.
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OK, so now: For the hypothetical reaction A + B → C, based on the following data, what is the rate law?
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OK, so we are going to give you some data here; and they give you some data: experiment, and then we have the concentration of A; we have the concentration of B; and then we have the rate.
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Experiment 1, 2, 3; 0.20, 0.20, 0.40, 0.10, 0.20, 0.10; and we have 2.5x10^-2; we have 5.0x10^-2; and we have 5.0x10^-2.
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Our choices are the following: K times the concentration of A; K times the concentration of A squared; C is K times the concentration of B; D is K times the concentration of B squared; and E: K times the concentration of A, times the concentration of B.
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OK, so now, let's take a look at this data.
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Whenever we compare two experiments, when we are given concentration data and rate information, whenever we compare two experiments, we need to find something where only one of the concentrations changes--because again, we are only dealing with one independent variable for anything that we compare.
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So, in this case, when we change B from point 1 to point 2, when we double it, we ended up doubling the rate.
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OK, so that means that it is going to be first-order in B.
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The rate is K, so so far we know that it is first-order in B, because again, when you double something and you double the rate, that is first-order in that reactant.
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Well, now let's take a look at Experiment 1 and Experiment 3.
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Let me do this in red; now, we look at 1, and we look at 3.
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Here, the B concentration stays the same, but we double up A and A.
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Let me see here: .2, .4, .2, point...oh, wait; I'm sorry; our data is wrong.
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It is not .1; this is .2; that is .20; OK, let's try this again, shall we?
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Let me erase this, and let me put in blue again: this is .10, .20; this is .20; OK.
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Yes, I think that is going to change some things.
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OK, so now let's take a look at Experiment 1 and 2 again; that is fine; so .2, .2; that is the same; we double up the concentration of B, and then we end up doubling the rate; so yes, not a problem.
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Our rate is...that part is the same; it is going to be first-order in B.
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And now, let's see; now, let's look at Experiment 2 and Experiment 3.
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.2, .2; B is the same; when I end up going from changing the concentration of A from .2 to .4, when I double that, the rate goes from 5.0x10^-2 to 5.0x10^-2; I double the concentration, but the rate didn't change.
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Therefore, the rate is not dependent on A at all; so we are done.
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Rate is KB; our choice is C; that is it.
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Any time you have data like this, you take a look at some doubling, some changing, concentration, and you see how the rate changes.
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That will tell you the order of that particular concentration term in the differential rate law; and then, you just look through your choices.
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Good; OK, let's see; let's try another one here.
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#5: Reactant P underwent decomposition, and the concentration was measured at different times.
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Now, based on the data, what is the rate law?
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OK, so we have time data, and we have concentration of P data.
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0 time (this is in hours, and the concentration is in moles per liter, as always): 0, 1, 2, 3; so we started off at 0.4, and an hour later, we measured the concentration; we have 0.2; an hour later, we have 0.1; an hour later, we have 0.05.
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Now, our choices are the following.
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First-order in P (I'm getting a little sloppy here); second-order in P; twice; D--we have half KP; and E--we have K.
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OK, so they are saying that some reactant, P, underwent decomposition, and the concentration was measured at different times.
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Based on the data, what is the rate law?
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OK, take a look at the time: 1 hour, 1 hour, 1 hour.
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The time increment is uniform; it's 1 hour.
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Now, notice the concentration: after 1 hour, there is half of the initial concentration.
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After another hour, there is half of this concentration.
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After another hour, there is half of this concentration.
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So, every hour, it is diminishing by half.
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That should immediately tell you something; we are talking about half-life here; and what is interesting is that the time--the half-life for it to diminish by half, diminish by half, diminish by half--is uniform.
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The t < font size="-6" > 1/2 < /font > is clearly 1 hour, and it is constant.
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Any time you have a constant half-life, you are talking about a first-order reaction.
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Remember that the t < font size="-6" > 1/2 < /font > of a first-order reaction is equal to the logarithm of 2, over K, or the logarithm of 1/2, over -K.
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It is a constant; it doesn't depend on the concentration.
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Because it doesn't depend on the concentration, it stays constant.
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That is what this data is telling me.
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One hour, one hour, one hour; half-life is constant in order for it to diminish by half, diminish by half, diminish by half.
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So, that automatically tells me that I am talking about a first-order rate law; that is my answer.
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My choice is A; I hope that makes sense.
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OK, let's see what we can do here.
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Let's see: OK, let's try a free response question.
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This is #6: We have the following reaction: we have 2 nitrogen monoxide, plus chlorine gas, forming 2 moles of NOCl.
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OK, we have some kinetic data; we run a series of experiments--we run 3 experiments--1, 2, 3.
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We measure the concentration of nitrogen monoxide (initial concentration); we measure the initial concentration of Cl₂; and we also measure...this time, we are measuring the rate of appearance of NOBr.
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That is fine; we can measure the rate of appearance--a rate is a rate; rate of appearance, rate of disappearance--it's the same; it's just that one is negative and one is positive.
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We have: the following data were obtained: .02; 0.04; 0.02; 0.02; 0.02; 0.04.
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We have 9.6x10^-2; we have 3.8x10^-1; and we have 1.9x10^-1.
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Our first question to you is: what is the rate law?
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What is the rate law?--OK, it should be easy enough.
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We have some concentration data; we have rate data; we hold one of them fixed; we check the other one to see what is going on.
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When we check Experiment 1 and 2, we notice that the chlorine concentration is the same.
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We doubled the nitrogen monoxide concentration; when we doubled the concentration, we quadrupled the rate.
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OK, I hope you see that: double, from .02 to .04, the concentration of nitrogen monoxide; this 3.8x10^-1 is four times 9.6x10^-2.
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OK, I hope that you see that; watch these exponents very, very carefully.
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So now, double NO concentration means quadruple the rate.
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This implies that it is second-order in NO.
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2 squared is 4; that is where this 2 comes from here.
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OK, now, let's take a look at Experiment 2 and 3.
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When we look at 2 and 3, we notice that we have...actually, not 2 and 3; we will have to look at 1 and 3; I'm sorry.
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1 and 3; and the reason is because .02...now, we're going to leave the NO concentration the same; we are going to double the Cl₂ concentration.
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So, in this particular case, when we double the chlorine concentration, we end up doubling the rate: 1.9x10^-1 is twice the 9.6x10^-2.
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We double the rate.
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Well, that means--when you double something and the rate doubles, that means it is first-order.
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Therefore, our rate law is equal to some K, times NO to the second power, times Cl₂ to the first power; that is our answer.
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That is it--nice and simple.
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OK, let's do our next phase.
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Part B: what is the value of the rate constant? Include units.
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OK, what is the value of the rate constant? Include units: I do my units and my numbers separately; that is just something that I like to do.
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You are welcome to do it any way that you like.
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OK, so let's see: we have "What is the value of the rate constant? Include units."
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OK, well, we just said that the rate is equal to K times NO squared, times the Cl₂ concentration.
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Well, just pick any one of the experiments; you have the rate; you have the NO concentration; you have the Cl₂ concentration; just solve for K.
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So, K is equal to the rate, divided by the NO concentration squared, times the Cl₂ concentration.
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Let's just take (I think I'm going to use Experiment 3) 1.9x10^-1, divided by 0.02 squared, times 0.04; and when we do that, we end up with 1.2x10⁴.
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That is the numerical value; now, let's do the units.
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The rate: OK, the rate is in moles per liter per second, because that is how much was showing up, or that is how much was disappearing.
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That is this up here; it is in moles per liter per second.
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These are concentrations; you get moles per liter squared, times moles per liter; you get moles per liter cubed.
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You get moles cubed over liters cubed.
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Well, let's see what we have: this becomes 2; this becomes 2; that cancels that and cancels that; you flip that over, and you get liters squared over moles squared-seconds.
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That is your unit; so your answer is 1.2x10⁴ liters squared/moles squared-second.
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That is your answer.
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So again, my recommendation, as far as kinetics is concerned, is: in one of the lessons (not the last lesson, but the one right before that), we did a summary of first-order reaction, second-order reaction, zero-order reaction.
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On that summary, we discussed the differential rate law; we discussed the integrated rate law; we discussed the half-life; and we discovered the quality of the graph that gives us a straight line (whether it's the logarithm versus time; whether it's 1 over the concentration versus time; or whether it's just concentration versus time).
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That is what you want to know, as far as kinetics is concerned.
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If you understand that summary, if you understand that table and where each thing is coming from, all of these problems will fall out naturally.
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OK, thank you for joining us here at Educator.com, and thank you for joining us at AP Chemistry.
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See you next time; goodbye.