For the general reaction,

A + 2B → C + D,

the rate law will have the form

Rate = [A]^m[B]ⁿ

To determine the exponents, m and n, we run a series of experiments in which one reactant concentration changes while the other is kept constant. We then measure the effect on the initial rate in each case.

For example, the reaction below has the following rate law.

2 HgCl₂ (aq) + C₂O₄^2- (aq) → 2 Cl^– (aq) + CO₂ (g) + Hg₂Cl₂ (s)

Rate = k[HgCl₂]^m[C₂O₄^2-]ⁿ

The initial rate data for this reaction is in the table below:

From this initial rate data, we can determine:

We see there are three experiments with varying concentrations of reactants. If we look at experiments 2 and 3, we see the concentration of oxalate ion is held constant while the concentration of mercuric chloride is 0.210 M in experiment 2 and 0.104 M in experiment 3. We can compare these two experiments to determine the exponent, m.

\(\displaystyle \frac{rate_2}{rate_3}\;=\;\frac{k[0.210\;M]^m[0.60\;M]^n}{k[0.104\;M]^m[0.60\;M]^n}\;=\;\frac{7.1\times 10^{-5}M/s}{3.5\times 10^{-5}M/s}\)
 
The k values and the oxalate ion concentrations cancel.

\(\displaystyle \frac{rate_2}{rate_3}\;=\require{cancel}\;\frac{\cancel{k}[0.210\;M]^m\cancel{[0.60\;M]^n}}{\cancel{k}[0.104\;M]^m\cancel{[0.60\;M]^n}}\;=\;\frac{7.1\times 10^{-5}M/s}{3.5\times 10^{-5}M/s}\)
 
We are left with,

\(\displaystyle \frac{rate_2}{rate_3}\;=\;\frac{[0.210\;M]^m}{[0.104\;M]^m}\;=\;\frac{7.1\times 10^{-5}M/s}{3.5\times 10^{-5}M/s}\)
 
Divide the concentrations and the initial rates.

2^m = 2

The exponent m is equal to 1. The reaction is first order in HgCl₂.

Next we can compare experiments 1 and 2 to determine the exponent n. I always put the larger rate in the numerator.

\(\displaystyle \frac{rate_2}{rate_1}\;=\;\frac{k[0.210\;M]^m[0.60\;M]^n}{k[0.210\;M]^m[0.30\;M]^n}\;=\;\frac{7.1\times 10^{-5}M/s}{1.8\times 10^{-5}M/s}\)
 

The concentrations of HgCl₂ and the k values cancel.

\(\displaystyle \frac{rate_2}{rate_1}\;=\require{cancel}\;\frac{\cancel{k}\cancel{[0.210\;M]^m}[0.60\;M]^n}{\cancel{k}\cancel{[0.210\;M]^m}[0.30\;M]^n}\;=\;\frac{7.1\times 10^{-5}M/s}{1.8\times 10^{-5}M/s}\)
 
\(\displaystyle \frac{rate_2}{rate_1}\;=\;\frac{[0.60\;M]^n}{[0.30\;M]^n}\;=\;\frac{7.1\times 10^{-5}M/s}{1.8\times 10^{-5}M/s}\)
 

2ⁿ = 3.9 ≅ 4

The value of n is equal to 2 and the reaction is second order in oxalate ion. The rate law is:

Rate = k[HgCl₂][C₂O₄^2-]²

Parts a and b have been answered. The reaction is third order overall. For part c, it states to find the value of the rate constant, k. For this part we can use any of the experiments. Here, I will use the values from experiment 1, and solve the rate law for k.

\(\displaystyle k\;=\;\frac{Rate}{[HgCl_2][C_2O_4^{2-}]^2}\;=\;\frac{1.8\times 10^{-5}\;M/s}{[0.210\;M][0.30\;M]^2}\;=\;9.5\times 10^{-4}\;M^{-2}s^{-1}\)
 
Sometimes, the value of the exponent can be more difficult to determine. You can use the properties of logs to determine the exponent. For example, 2^m = 4. You can use, log x^m = m log(x).

log 2^m = 4

m log(2) = log (4)

m = \(\displaystyle\frac{log (4)}{log (2)} = 2\)

Keep in mind the differences between the rate constant, k, and the rate of reaction. The reaction rate is

 
The rate constant, k