another rate law question please?

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Jan 27, 2007
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For the reaction A + 2B = Products, the following rate data is obtained:

[A] Rate (M/min)
.10 .10 .010
.20 .20 .040
.30 .20 .060

What is the rate law for this reaction? I do'nt understand, can anyone explain please? Thank you a lot

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The rate law for any equation is always rate = k * [reactant 1]^n * [reactant 2]^m * and so on for every reactant (usually no more than three since it's difficult for four molecules to collide in the right fashion) The exponents m and n must be found through experimentation only and not simply taking the stoichiometry of the balanced equation.

To find the exponents = reaction order, we want to see the extent of change when the other reactants are held constant. If reactant A doubles in rate when you double its concentration, then the reaction order is 1. If reactant A quadruples in rate when you double its concentration, then the reaction order is 2 (logically this makes sense because squaring something you doubled results in a four-fold increase of whatever your product is)

Normally, most PCAT questions set up the data so that you can hold one reactant constant while varying the other. So for your data, you can use a proportion to see how much the rate changes when you change the concentration of a reactant when hold the other reactant constant:

1. Hold constant at 0.2 M and find the proportional change after increasing [A]

[A-Experiment 2]^n\[A-Experiment 3]^n = Rate Exp 2/ Rate Exp 3

=([A-Experiment 2]\[A-Experiment 3]) ^ n = Rate Exp 2/ Rate Exp 3

(3/2)^n = 3/2

n =1

2. To find m requires more work because you aren't given two sets of experiments where [A] is constant. However, you now know the reaction order of A, so you can use it to find the reaction order of B

In reality, I didn't really tell you how I used my shortcut to find the reaction order of A. What actually happens when you use proportions to solve for rate orders is:

rate 2/ rate 1 = rate law 2 product/rate law 1 product

rate 2/rate 1 = (k[A-exp 2]^n[B-exp 2]^m)/(k[A-exp 1]^n[B-exp 1]^m)

**the rate constant "k" cancels out, leaving you:

rate 2/rate 1 = ([A-exp 2]^n[B-exp 2]^m)/([A-exp 1]^n[B-exp 1]^m)

You now know "n", so you pick two experiments, and plug in the neccessary data to find m. Use the first two experiments because the quotient of the [A] terms will be a whole number that is easier to manipulate:

0.040/0.010 = 0.2 * [0.20]^m / 0.1 * [0.10]^m

4 = 2 * ([0.20]^m / [0.10]^m)

2 = ([0.20]/ [0.10])^m

2 = 2^m


3. Knowing both reaction orders, now the last step is to find the rate constant of the rate law. Simply write out the rate law as much as you know:

rate = k * [A]^1 * ^1

**plug in numbers from one experiment only, and solve for rate constant k:

0.01 = k * 0.1^1 * 0.1^1

k = 1

4. Write the final rate law

Rate = 1 * [A]^1 * ^1

5. Optional: Checking your work if you have time, you find that this law holds for every experiment given to you.

6. The quickest way: You may have been able to solve this question in 5 seconds if you noticed that [A] * = rate for every experiment. That mandates a reaction order of 1 for both reactants, and a rate constant of 1.