Q16.25 P - Chemistry: Molecular Nature Of Matter And Change

Question

By what factor does the rate change in each of the following cases (assuming constant temperature)? 

(a) A reaction is first order in reactant A, and [A] is doubled. 

(b) A reaction is second order in reactant B, and [B] is halved.

(c) A reaction is second order in reactant C, and [C] is tripled.

Step-by-Step Solution

Answer

1. When the reaction is doubled, the rates increase by a factor of 2.
2. When the reaction is halved, the rate reduces by a factor of 4.
3. When the reaction is tripled, the rate rises by a factor of 9.

Step 1: (a) When the reaction is doubled

For reactant A, write the rate law for the reaction in a first-order:

$\text{rate}=\text{k}{\left[\text{A}\right]}^1$

When the concentration of the reactant doubles, write a new rate rule for the reaction:

$\begin{array}{rcl}\text{rate}& =& \text{k}{\left[2\times \text{A}\right]}^1\\ & =& 2\times \text{k}{\left[\text{A}\right]}^1\end{array}$

As a result, the rates increase by a factor of 2.

Step 2: (b) When the reaction is halved

For reactant B, write the rate law for the reaction in a second order:

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

When the concentration of the reactant is half, write the following revised rate law for the reaction:

$\begin{array}{rcl}\text{rate}& =& \text{k}{\left[\frac{1}{2}\times \text{B}\right]}^2\\ & =& \frac{1}{4}\times \text{k}{\left[\text{B}\right]}^2\end{array}$

 As a result, the rate reduces by a factor of 4.

Step 3: (c) When the reaction is triple

For reactant C, write the rate rule for the reaction in a second order:

$\text{rate}=\text{k}{\left[\text{C}\right]}^2$

Write a new rate rule for the reaction when the reactant concentration is tripled:

$\begin{array}{rcl}\text{rate}& =& \text{k}{\left[3\times \text{C}\right]}^2\\ & =& 9\times \text{k}{\left[\text{C}\right]}^2\end{array}$

As a result, the rate rises by a factor of 9.