AP Chemistry AP Practice for Kinetics

Hello, and welcome back to Educator.com, and welcome back to AP Chemistry.0000

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.0005

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.0025

So, we're going to do a couple of multiple-choice and a couple of free-response questions, and things like that.0035

Let's just jump right in and see what we can do.0040

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.0044

OK, so let's see: we'll say: Consider the following hypothetical reaction.0055

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.0072

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.0078

These multiple-choice questions--they test basic understanding: can you follow a logical train of thought, where the numbers are completely secondary?0091

It is the free-response questions where you can use your calculator to actually find specific answers.0102

Hypothetical reaction: we have: our reactant R, plus reactant S, goes to product T.0109

Now, the following are our choices.0117

A) Rate = K times concentration of R to the first power.0124

B) we have: The rate = K times the concentration of S to the second power.0134

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.0143

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.0152

And E), our final choice, because you have ABCDE to bubble in: The rate is equal to K times R squared, S squared.0163

So, we have a hypothetical reaction, and we have these choices to choose from, based on the questions that we are going to ask.0176

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).0183

The question is: What is the rate law?0219

A through E are our choices.0228

OK, so how do we deal with this?0232

Again, we should be able to do this reasonably quickly.0235

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 8.0239

Well, there are a couple of ways that we can do this.0248

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?"0249

But, I think that may be a little bit confusing, so let's go ahead and do this a more natural way.0268

Let's just go ahead and stick the doubled rate into these to see what happens.0274

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.0279

Let me just do that; let's try choice A.0293

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.0297

Try choice A: Well, I get that the rate is equal to K, times twice the rate of R.0304

2 just comes out; so twice the KR.0313

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.0319

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.0328

Well, this is equal to 4 times KS².0345

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.0352

Well, K to the S squared was the original rate; 4 times that--that is not it, either.0363

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.0369

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.0386

2 squared is 4; times 2 is 8.0398

You get 8 on this side.0402

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.0404

I hope that makes sense.0411

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 2.0414

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.0448

So, A is the answer.0469

And again, if you don't see that: the rate they told me was K, times R to the first power.0477

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.0492

Well, 2 times R--that equals 2 times the rate; that means the rate is doubled; that is the only choice.0513

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.0520

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.0534

OK, well, let's take a look at #3.0539

This time, they say, R is doubled; so this time, they are only doubling R.0543

R is doubled; S is unchanged.0552

Well, let's write this a little differently.0562

This time, what we have is: When R is doubled (the concentration of R), and S is unchanged, the rate is unchanged.0569

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.0608

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.0618

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.0631

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.0641

Well, they are telling me that S is unchanged; well, if I don't change S, I don't change the rate.0658

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.0663

In this case, my only choice is choice B, under these circumstances.0675

I hope that makes sense.0682

OK, let's try a free-response question here.0686

No, actually, I'm sorry; let's try one more multiple-choice question.0697

OK, so now: For the hypothetical reaction A + B → C, based on the following data, what is the rate law?0700

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.0737

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^-2.0752

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.0778

OK, so now, let's take a look at this data.0804

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.0811

So, in this case, when we change B from point 1 to point 2, when we double it, we ended up doubling the rate.0828

OK, so that means that it is going to be first-order in B.0841

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.0848

Well, now let's take a look at Experiment 1 and Experiment 3.0862

Let me do this in red; now, we look at 1, and we look at 3.0866

Here, the B concentration stays the same, but we double up A and A.0870

Let me see here: .2, .4, .2, point...oh, wait; I'm sorry; our data is wrong.0880

It is not .1; this is .2; that is .20; OK, let's try this again, shall we?0891

Let me erase this, and let me put in blue again: this is .10, .20; this is .20; OK.0900

Yes, I think that is going to change some things.0911

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.0913

Our rate is...that part is the same; it is going to be first-order in B.0927

And now, let's see; now, let's look at Experiment 2 and Experiment 3.0934

.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.0939

Therefore, the rate is not dependent on A at all; so we are done.0956

Rate is KB; our choice is C; that is it.0961

Any time you have data like this, you take a look at some doubling, some changing, concentration, and you see how the rate changes.0969

That will tell you the order of that particular concentration term in the differential rate law; and then, you just look through your choices.0977

Good; OK, let's see; let's try another one here.0987

#5: Reactant P underwent decomposition, and the concentration was measured at different times.1001

Now, based on the data, what is the rate law?1038

OK, so we have time data, and we have concentration of P data.1055

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 0.05.1064

Now, our choices are the following.1088

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.1094

OK, so they are saying that some reactant, P, underwent decomposition, and the concentration was measured at different times.1132

Based on the data, what is the rate law?1140

OK, take a look at the time: 1 hour, 1 hour, 1 hour.1142

The time increment is uniform; it's 1 hour.1148

Now, notice the concentration: after 1 hour, there is half of the initial concentration.1151

After another hour, there is half of this concentration.1157

After another hour, there is half of this concentration.1160

So, every hour, it is diminishing by half.1163

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.1168

The t_1/2 is clearly 1 hour, and it is constant.1184

Any time you have a constant half-life, you are talking about a first-order reaction.1189

Remember that the t_1/2 of a first-order reaction is equal to the logarithm of 2, over K, or the logarithm of 1/2, over -K.1199

It is a constant; it doesn't depend on the concentration.1217

Because it doesn't depend on the concentration, it stays constant.1220

That is what this data is telling me.1225

One hour, one hour, one hour; half-life is constant in order for it to diminish by half, diminish by half, diminish by half.1228

So, that automatically tells me that I am talking about a first-order rate law; that is my answer.1237

My choice is A; I hope that makes sense.1247

OK, let's see what we can do here.1254

Let's see: OK, let's try a free response question.1260

This is #6: We have the following reaction: we have 2 nitrogen monoxide, plus chlorine gas, forming 2 moles of NOCl.1264

OK, we have some kinetic data; we run a series of experiments--we run 3 experiments--1, 2, 3.1278

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.1286

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.1309

We have: the following data were obtained: .02; 0.04; 0.02; 0.02; 0.02; 0.04.1317

We have 9.6x10^-2; we have 3.8x10^-1; and we have 1.9x10^-1.1333

Our first question to you is: what is the rate law?1346

What is the rate law?--OK, it should be easy enough.1352

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.1357

When we check Experiment 1 and 2, we notice that the chlorine concentration is the same.1366

We doubled the nitrogen monoxide concentration; when we doubled the concentration, we quadrupled the rate.1372

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^-2.1383

OK, I hope that you see that; watch these exponents very, very carefully.1395

So now, double NO concentration means quadruple the rate.1401

This implies that it is second-order in NO.1413

2 squared is 4; that is where this 2 comes from here.1423

OK, now, let's take a look at Experiment 2 and 3.1430

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.1435

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.1448

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^-2.1459

We double the rate.1476

Well, that means--when you double something and the rate doubles, that means it is first-order.1479

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.1487

That is it--nice and simple.1504

OK, let's do our next phase.1507

Part B: what is the value of the rate constant? Include units.1513

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.1517

You are welcome to do it any way that you like.1545

OK, so let's see: we have "What is the value of the rate constant? Include units."1548

OK, well, we just said that the rate is equal to K times NO squared, times the Cl₂ concentration.1553

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.1565

So, K is equal to the rate, divided by the NO concentration squared, times the Cl₂ concentration.1575

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 1.2x10⁴.1587

That is the numerical value; now, let's do the units.1615

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.1619

That is this up here; it is in moles per liter per second.1635

These are concentrations; you get moles per liter squared, times moles per liter; you get moles per liter cubed.1638

You get moles cubed over liters cubed.1648

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.1652

That is your unit; so your answer is 1.2x10⁴ liters squared/moles squared-second.1672

That is your answer.1684

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.1687

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).1704

That is what you want to know, as far as kinetics is concerned.1728

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.1731

OK, thank you for joining us here at Educator.com, and thank you for joining us at AP Chemistry.1741

See you next time; goodbye.1747