Attempt 3, Question 8. Given the data below, find k.

Question

Experiment	Initial conc of A, mol/L	Initial conc of B, mol/L	Initial conc of C, mol/L	Initial rate, mol/L.s
1	0.1	0.2	0.4	2 x 10-3
2	0.2	0.2	0.4	4 x 10-3
3	0.3	0.4	0.4	6 x 10-3
4	0.4	0.6	0.2	2 x 10-3

I have found first order in A, but I'm getting stuck on the order for B. If i use experiments 1 and 3 where A triples (3^1) and where B doubles (2^x) and the rate increases by 3. I get the questions 6^x = 3 which doesn't make sense. If I use experiments 2 and 3, I end with the equation 3^x = 1.5 which also doesn't make sense. How do I find the rate for B.

Answer 1

The answer given by @Mark L is correct and makes sense to me, but it may not be as clear to you, since you are beginning with these types of problems. So, my take is as follows, without using any math or exponents, etc:

Here's how to find the order in B, in as simple a way as I can explain. We can use experiments 1 and 3 as you did. Yes, [A] triples, [B] doubles and rate triples. Since we already know that the rate is first order in A, that means that the rate will triple just by tripling [A], so that must mean that [B] has NO EFFECT on the rate. If it did, the rate would more than triple (or less than triple). So, it has to be zero order in B.

Answer 2

Consider first three lines. [C] does not change, so the equation is just v = k'[A]^n[B]^m, where k'=is k[C].

As you can see, v is growing linearly relative to [A] (0.1 -> 2, 0.2 -> 4, 0.3 -> 6) , and it is not sensitive to the change of [B], which means m=0. Or, the reaction has a zeroth order by B (which is the same). The rest is simple: extrapolation of [A] to 0.4 (assuming [C] does not change) gives 8. But we have just 2, which means the rate drops four times when [C] decreases by 2. The conclusion is the reaction has a first order by A zeroth order by B and second order by C.