Erika Tan walks through a past AP Chemistry free response problem regarding chemical kinetics.
This question actually has a couple parts to it. So first of all, it’s asking us to find the order with respect to reactant A, find the order with respect to reactant B, find the rate law, calculate the rate constant, and calculate the initial rate of experiment #4 which is the question mark. I just copied this chart from the question that I took it from, so let’s start on the first part: find the order of the reaction with respect to A. So here I’m going to look down this column, which gives me the initial concentration of A, and the units are molarity (which is moles/liter), and then, I’m going to see, since I’m trying to find the order with respect to A, I’ll have to find two experiments that have about the same concentration of B. So I’m looking at experiments 2 and 3 right now because .137 and .136 are basically the same thing. Since I’m keeping B constant, I’ll be able to see what happened to A to change the rate. But by how much?
So I’m just going to do .136 divided by .034, that’s what I’m getting here, and I see that this, experiment #3, is just the concentration of A times 4. So let’s not forget that. This is times 4. Since we know that this is being constant, and we know that this is being multiplied by 4, let’s see what happens to the rate. So I see 1.08 times 10^-2 and 1.07 times 10^-2. Well, those are basically the same thing, I just see the 8 and the 7 which only has a difference of 1. So, we know the concentration of A has no effect on the rate, since rate stays constant. So that means the order of reactant A is 0, since changing the concentration of A does not change the rate. Now let’s look at B. We have to find 2 experiments where A is being held constant. I see experiments 1 and 2 because this change in concentration is almost nothing. Let’s see what happens when we do .137 divided by .034. That gives us 4 again, so we’re multiplying .034 by 4 to get .137. And let’s see what happens to the rate. Again, these two don’t look the same so I’m going to divide and see what happened to it. .108 times 10^-2 divided by 6.67 times 10^-4. And that gives us about 16. So this times 16 gives you that. Obviously, there’s been a change in rate, so what is the order of B? Well, this concentration of B has been multiplied by 4. The rate has been multiplied by 16. To find the order, we’ll have to do 4 to what exponent gives you 16? X has to be 2, because 4 squared gives you 16. That means 2 is the order of B. So now we have A is 0 order, and B is second order. Let’s write the rate law. It’s just rate=k (which is a constant) and we’ll have concentration of A to the 0, which anything to the 0th power is just 1, so you don’t even have to write that, and then we have concentration of B squared. That’s because the order of B is 2, and we have to write it as an exponent. Now it’s asking us to calculate the rate constant next. All we have to do is take the rate law that we just wrote, plug in some numbers that you can just pick from the table, and calculate k, which is the rate constant. So, I’m just going to choose experiment #1 because it’s the first one on the chart. So we have our rate law, rate=k[B]^2, and if you plug in what you have for experiment 1, we see that the rate is this, 6.67 times 10^-4 = k times [B] which is .034, and don’t forget we have to square it from the rate law.
We have our answer for k. And all we have to do is calculate the initial rate for experiment number 4 which it doesn’t give you on the table so let’s use the rate law and rate constant that we just got. So rate = .577 because that’s what we got for k, times concentration of B squared. So we see that for experiment #4, the concentration of B is .233, so all we have to do is rate = .577 times .233 squared. Easy. Now let’s plug it into our calculator and that gives us about 3.13 times 10^-2.