# Question Video: Identifying the Order of the Reaction from the Change in Concentration of the Reactants

In a reaction between compounds A and B, doubling the concentration of A increases the rate of reaction by a factor of 4. However, tripling the concentration of reactant B increases the rate by a factor of 3. a) What is the order of reaction with respect to A? b) What is the order of reaction with respect to B?

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### Video Transcript

In a reaction between compounds A and B, doubling the concentration of A increases the rate of reaction by a factor of four. However, tripling the concentration of reactant B increases the rate by a factor of three. What is the order of reaction with respect to A?

In this question, we’re looking at a reaction between chemicals A and B. For such a reaction, we expect the rate equation to be of this form. Where the rate of reaction is equal to the rate constant multiplied by the concentration of A raised to a power, which is its order of reaction, multiplied by the same for B. At this point, it’s impossible to predict what form the rate equation will take. It could be as simple as the rate directly equal to the rate constant. Or it could be the rate equals the rate constant multiplied by some exotic powers of A and B.

It is impossible to deduce the rate equation simply from the stoichiometry of the equation. Thankfully, there are some experimental details in the question. The question tells us that doubling the concentration of A increases the overall rate of the reaction by a factor of four. The factor four is equal to the rate when we have twice the concentration of A divided by the rate when we have the initial concentration of A. What we need to do is interpret this equation and rearrange it so that we can find the value for 𝑥, which is the order of reaction with respect to A.

We know the general form for the rate equation. The rate when the concentration of A is at its initial value is equal to 𝑘 times the concentration of A to the power of 𝑥 multiplied by the concentration of B to the power 𝑦. So what do we plug in when we want to calculate the rate when we double the concentration of A? All we need to do is plug in two times the initial concentration of A. We don’t need to come up with imagined values. We can just use the algebraic representation. Just remember that we’re raising two to the power of 𝑥 as well as the initial concentration of A.

This is what we get when we evaluate the power. Two to the 𝑥 is multiplied by a fraction where the top and the bottom are identical. If we cancel common terms, we are left with simply two to the power of 𝑥. The factor by which the rate increased because of doubling the concentration was four. So two to the power of 𝑥 is equal to four. Two squared is also equal to four. Therefore, 𝑥 is equal to two. If you’re confident with your logarithms, you may have noticed that we could’ve taken log to the base two of both sides. This gives 𝑥 equal log to the base two of four, which gives us the same answer. Either way, we’ve demonstrated that since doubling the concentration of A increases the rate of reaction by a factor of four, the order of reaction with respect to A must be two.

What is the order of reaction with respect to B?

This question is very similar to part a). In this case, we’re trying to find 𝑦, the power to which the concentration of B is raised in the rate equation. For B, tripling the concentration increases the rate of the reaction by a factor of three. As we did with part a), we can see that the factor can be determined by taking the rate when the concentration of B is triple its initial value and dividing it by the rate when the concentration of B was at its initial value. So at the bottom of the fraction we’ll have the rate when the concentration of B is at its initial value. And it’s raised to the power of 𝑦, its order of reaction.

In this example, I’ve inserted the value two for the order of reaction with respect to A. This isn’t necessary, you could use 𝑥 instead. It doesn’t really matter as we’ll come to see because the value will cancel either way. The rate when we have triple the initial concentration of B is equal to 𝑘 times the concentration of A squared multiplied by three times the initial concentration of B all to the power of 𝑦. Again, it’s vitally important that we include the three inside the brackets because it’s part of a concentration term.

If we evaluate this expression, we can see that our factor is equal to three to the power of 𝑦 multiplied by a fraction where the numerator and denominator are exactly the same. So we can cancel common terms, leaving us with the fact that three to the power of 𝑦 is equal to three. Any number to the power of one is equal to itself. So three to the power of one is also equal to three. Therefore, our value for 𝑦 is one.

If you wanted to, you could also use log rules, taking log to the base three of both sides and evaluating log to the base three of three as one. Whichever way you do it, you should get that the order of reaction with respect to B is one. This leaves us with our final rate equation of the rate equal to the rate constant multiplied by the concentration of A squared multiplied by the concentration of B.