# How do you find rate law for a reaction?

Aug 6, 2014

You do a series of experiments to determine how the rate depends on the concentration of each component in the reaction.

For a reaction aA +bB → products, the rate law is

rate = $k {\left[\text{A"]^m["B}\right]}^{n}$

You must determine the values of $k$, $m$, and $n$ experimentally for a given reaction at a given temperature. $m$ and $n$ are usually integers.

One way is to use initial concentrations. It is called the method of initial rates.

Example

Find the rate law for the reaction aA +bB → products at 27 °C, given the data below.

Trial; [A]₀; [B]₀; rate
1; 1.0 mol·L⁻¹; 1.0 mol·L⁻¹; 2.0 mol·L⁻¹s⁻¹
2: 1.0 mol·L⁻¹; 2.0 mol·L⁻¹; 8.1 mol·L⁻¹s⁻¹
3; 2.0 mol·L⁻¹; 2.0 mol·L⁻¹; 15.9 mol·L⁻¹s⁻¹

Solution:

Look for two trials in which all but one concentration stays constant.

Step 1

In Trials 1 and 2, [A] remains constant, but [B] changes.

$\left(\text{rate"_2)/("rate"_1) = (k["A"]_2^m["B"]_2^n)/(k["A"]_1^m["B"]_1^n) = ((1.0"mol·L⁻¹")/(1.0"mol·L⁻¹"))^m × ((2.0"mol·L⁻¹")/(1.0"mol·L⁻¹"))^n = (8.1"mol·L⁻¹s⁻¹")/(2.0"mol·L⁻¹s⁻¹}\right)$

${\left(2.0\right)}^{n}$ = 4.05

$n \log 2.0$ = log 4.05

$n = \frac{\log 4.05}{\log 2.0} = \frac{0.607}{0.30}$ = 2.0 ≈ 2

Step 2

In Trials 2 and 4, [B] remains constant, but [A] changes.

$\left(\text{rate"_3)/("rate"_2) = (k["A"]_3^m["B"]_3^n)/(k["A"]_3^m["B"]_3^n) = ((2.0"mol·L⁻¹")/(1.0"mol·L⁻¹"))^m × ((2.0"mol·L⁻¹")/(2.0"mol·L⁻¹"))^n = (15.9"mol·L⁻¹s⁻¹")/(8.1"mol·L⁻¹s⁻¹}\right)$

${\left(2.0\right)}^{m}$ = 1.96

$m \log 2.0$ = log 1.96

$m = \frac{\log 1.96}{\log 2.0} = \frac{0.292}{0.30}$ = 0.97 ≈ 1

So the rate law is

rate = $k {\left[\text{A"]["B}\right]}^{2}$