The pre-equilibrium approximation | Kinetics | AP Chemistry | Khan Academy

The pre-equilibrium approximation is used to find the rate law for a mechanism with a fast initial step. As an example, let's look at the reaction between nitric oxide and bromine.

In the first step of the mechanism, nitric oxide combines with bromine to form NOBr2, and in the second step of the mechanism, NOBr2 reacts with NO to form our product, 2NOBr. NOBr2 is generated from the first elementary step of the mechanism, and then NOBr2 is used up in the second step. Since NOBr2 wasn't there in the beginning and it's not there in the end, we call NOBr2 an intermediate.

The first step of the mechanism is fast, and the second step of the mechanism is slow. Since the second step of the mechanism is slow, this is the rate-determining step. We can write the rate law for the overall reaction by writing the rate law for this elementary reaction that makes up step two of our mechanism.

So we can write the rate of reaction is equal to—for step 2, our rate constant is k2, and we multiply k2, the rate constant, by the concentration of our two reactants, which would be the concentration of NOBr2 and the concentration of NO. Since the coefficients in our balanced equation are ones for NOBr2 and one for NO, we can take the coefficients and turn them into exponents in our rate law.

So we can do this because this is an elementary reaction. However, we can't leave the rate law for the overall reaction in terms of the concentration of our intermediate NOBr2; it's preferable to have rate laws written in terms of the concentration of our reactants, which were NO and Br2. So we need some way of substituting in for the concentration of NOBr2.

We can do that by assuming that the first elementary step in our mechanism comes to a fast equilibrium. If we assume the first step comes to a fast equilibrium or a pre-equilibrium, we know at equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction.

In the forward reaction for step one, NO combines with Br2 to form NOBr2, and the reverse reaction, NOBr2 breaks apart to form NO and Br2. So if the rate of the forward reaction is equal to the rate of the reverse reaction at equilibrium, let's go ahead and write the rate laws for the forward and the reverse reaction.

The rate constant for the forward reaction is k1, so we can go ahead and write the rate of the forward reaction is equal to k1, and our two reactants are NO and Br2. So we have k1 times the concentration of NO times the concentration of Br2. Since the coefficients in our balanced equation are both ones for these two reactants, we can raise the power of these two concentrations to the first power since this is an elementary reaction.

We can do this, and we set this rate of the forward reaction equal to the rate of the reverse reaction. The reverse reaction has a rate constant of k−1, and we have only NOBr2 with a coefficient of 1 in it, so we multiply k−1 times the concentration of NOBr2 to the first power.

Next, our goal is to substitute in for the concentration of our intermediate. We can divide both sides of the equation by k−1. So if we decide to divide both sides of the equation by k−1, on the right side, k−1 cancels out, and we get that the concentration of our intermediate NOBr2 is equal to k1 times the concentration of NO to the first power times the concentration of Br2 to the first power divided by k−1.

Next, we can substitute all of this in for the concentration of our intermediate. That gives us the rate of reaction is equal to—we still have this k2 in here, so we have to make sure to include it, and we're going to substitute everything in for the concentration of our intermediate.

So that would be times k1 times the concentration of NO to the first power times the concentration of Br2 to the first power divided by k−1, and then we still have this concentration of NO to the first power. To make sure to include that in our rate law, let's think about what we would get if we multiply two constants together and then divide by a third constant.

So multiplying k2 times k1 and then dividing by k−1, that would just give us another constant, which we could just call k. So k is now the rate constant for the overall reaction. We have the rate law for the overall reaction equal to k times the concentration of NO to the first power times NO to the first power, which is just the concentration of NO to the second power, and we still have to include the concentration of bromine to the first power.

So now we have a rate law for our overall reaction in terms of the concentration of our two reactants. The rate of reaction is equal to the rate constant k times the concentration of NO squared times the concentration of bromine to the first power.

The experimentally determined rate law matches the rate law that we found using the pre-equilibrium approximation. If you look at the coefficients for the overall equation, there's a 2 in front of NO and a 1 in front of Br2. It might be tempting just to say, "Can't we just take those coefficients and turn them into exponents?" Because in this case, they happen to match the exponents in our rate law.

That's just a coincidence for this reaction. We can't just take the coefficients for an overall equation and turn them into exponents in the rate law; we can only do that for elementary reactions like the elementary reactions in the two steps of our mechanism. It's important to point out that if the rate of the forward reaction is equal to the rate of the reverse reaction, the concentration of our intermediate NOBr2 remains constant, and therefore we can use this pre-equilibrium approximation to find the rate law for a reaction with a fast initial step.