Given initial rates data, what is the rate law? What is the rate constant?

Question

Trial 1: [A] = 0.3 M, [B] = 0.3 M, [C] = 0.3 M, rate = 9.0 x 10^-5 M/s

Trial 2: [A] = 0.3 M, [B] = 0.3 M, [C] = 0.9 M, rate = 2.7 x 10^-4 M/s

Trial 3: [A] = 0.6 M, [B] = 0.3 M, [C] = 0.3 M, rate = 3.6 x 10^-4 M/s

Trial 4: [A] = 0.6 M, [B] = 0.6 M, [C] = 0.3 M, rate = 3.6 x 10^-4 M/s

1. What is the rate law for this reaction? A + B + C --> D + E
2. What is the rate constant? (and what are it's units?)

Answer 1

In order to determine the rate constant, we must first determine the rate equation and to do this we must determine the order of the reaction. There are 3 reactants, A, B and C, and we must find the order with respect to each one.

For A: compare trials 1 and 3. [A] doubles from 0.3 to 0.6 while [B] and [C] remain constant at 0.3 M. The rate increases 4 times from 9x10^-5 to 3.6x10^-4 M/s. Thus it is 2nd order in A.

For B: compare trials 3 and 4. [B] doubles from 0.3 to 0.6 while [A] remains constant at 0.6 and [C] remains constant at 0.3 M. The rate stays the same at 3.6x10^-4 M/s. Thus it is zero order in B.

For C: compare trials 1 and 2. [C] triples from 0.3 to 0.9 while [A] and [B] remain constant at 0.3 M. The rate triples from 9.0x10^-5 to 2.7x10^-4 M/s. Thus it is 1st order in C.

1). Now that we have determined the orders with respect to each reactant, we can write the rate law.

rate = k[A]²[C] ... note that B is not included since rate is independent of the concentration of B.

2). The rate constant, k, can be found by using the data from any of the trials, and plugging the values into the rate equation. We will use trial 1 as an example:

rate = k[A]²[C]

9.0x10^-5 M/s = k [0.3 M]²[0.3 M]

9.0x10^-5 M/s = 0.027 M³ k

k = 3.33x10^-3 M^-2s^-1

Answer 2

I've written out an explanation here: https://imgur.com/a/i9Hr3Py