# Question #37a30

Aug 19, 2016

See below.

#### Explanation:

Let us assume that the reaction is under the following general form:

$A + B \to C$

The rate law is determined experimentally to be:

$R = k \left[A\right] {\left[B\right]}^{2}$

The rate law with the first concentrations could be written as:

$R = k \left[{A}_{1}\right] {\left[{B}_{1}\right]}^{2}$

If the concentration of $A$ is doubled, this means that the new concentration is: $\left[{A}_{2}\right] = 2 \times \left[{A}_{1}\right]$

and the concentration of $B$ is tripled: $\left[{B}_{2}\right] = 3 \times \left[{B}_{1}\right]$

Replacing the new concentration in the expression of the rate law: $R ' = k \left[{A}_{2}\right] {\left[{B}_{2}\right]}^{2}$

Therefore, we have:
$R ' = k \left[{A}_{2}\right] {\left[{B}_{2}\right]}^{2} = k \left(2 \times \left[{A}_{1}\right]\right) {\left(3 \times \left[{B}_{1}\right]\right)}^{2}$

$\implies R ' = \left(2\right) {\left(3\right)}^{2} \times {\underbrace{k \left[{A}_{1}\right] {\left[{B}_{1}\right]}^{2}}}_{R}$

$\implies R ' = \left(2\right) {\left(3\right)}^{2} \times R$

$\implies R ' = 2 \times 9 \times R$

$\implies R ' = 18 \times R$

Which indicates that the rate of the reaction will increase by a factor of $18$.

Here is a complete lesson on reaction rate:
Chemical Kinetics | Reaction Rates & Rate Law.