### Video Transcript

In a reaction between compounds A and B, quadrupling the concentration of A doubles the rate of reaction. However, doubling the concentration of B increases the rate by a factor of eight. What is the order of reaction with respect to A?

So we’re dealing with a reaction between chemicals A and B, nothing else. And we don’t know the products. Based on this equation, we expect the rate equation to have the general form of 𝑘 multiplied by the concentration of A to the 𝑥, multiplied by the concentration of B to the 𝑦. 𝑘 is the rate constant. 𝑥 is the order of reaction with respect to A. And 𝑦 is the order of reaction with respect to B. We’re being asked what the order of reaction is with respect to A. So by the end of this, we need to find a value for 𝑥.

We’ve been told that quadrupling the concentration of A doubles the rate of the reaction. Quadrupling means increasing by a factor of four, and doubling means increasing by a factor of two. We can write this as the final rate, rate f, equal to twice the initial rate, rate i. So the final rate is when we have four times the concentration as when we have the initial rate. We can now substitute these expressions into our rate equation. So our final rate is equal to 𝑘 multiplied by four times the initial concentration of A to the power of 𝑥 multiplied by the concentration of B to the power of 𝑦.

Remember it’s vitally important to put the four inside the parentheses because we’re raising that to the power of 𝑥 as well. This is equal two times our initial rate, which is two times 𝑘 times the initial concentration of A to the power of 𝑥 multiplied by the concentration of B to the power of 𝑦. And this is what our expressions evaluate too.

If you look at the equation, you can see that we have the same expression on both sides. If we divide through by this term, we’re left with four to the power of 𝑥 equals two. We can work out 𝑥 by realizing that four to the power of a half is equal to two. Four to the power of a half is the same as the square root of four, which, of course, equals two. Therefore, 𝑥 equals a half. If you’re confident with using logarithms, you could take log to the base four of both sides and get 𝑥 equals log to the base four of two, which equals 0.5 or a half. Whichever way doesn’t really matter. We’ve demonstrated that the order of reaction with respect to A in this case is a half.

What is the order of reaction with respect to B?

We’re going to go through a very similar process, as we did with part a, and find the value for 𝑦. The question tells us that doubling the concentration of B increases the rate by a factor of eight. We can write this this way where our final rate, where we have twice the initial concentration, is equal to eight times the initial rate, where we just have the initial concentration. And this is what you get when you substitute in the values into the general rate equation. Here, I’ve used the power of a half with the concentration of A. But it really doesn’t matter because in the end it’ll cancel out. What matters is that we include the two in the parentheses when we raise the concentration of B to the power of 𝑦.

When we evaluate the expression, this is what we get: an equation where both sides have lots of common terms. If we divide through by this, we get two to the power of 𝑦 is equal to eight. Two to the power of three — that’s two times two times two — is also equal to eight. Therefore, 𝑦, our order reaction with respect to B, is equal to three. You could also have taken log to the base two of both sides so that 𝑦 is equal to log to the base two of eight, which is equal to three. Either way, we’ve demonstrated that the order of reaction with respect to B is three.