Chapter 18.2, Problem 12LC

Interpretation Introduction

Interpretation: The order of the reaction with respect to A and the overall order of the reaction when the initial concentration of A is increased from $\text{0}\text{.35}\text{mol/L}\text{to}\text{0}\text{.70}\text{mol/L}$ , the initial rate increases from $\text{1}\text{.6}\text{mol/}\left(\text{L}\text{.s}\right)\text{to}\text{3}\text{.2}\text{mol/}\left(\text{L}\text{.s}\right)$ is to be stated.

Concept Introduction: The rate of a chemical reaction is generally expressed as the reactant’s concentration consumed and the product’s concentration formed per unit time.

Answer to Problem 12LC

The reaction is first order in A. Since A is the only reactant that means the overall order of the reaction is first order.

Explanation of Solution

Given;

  $\begin{array}{c}{\left[\text{A}\right]}_{\text{1}}=0.35\text{mol/L}\\ {\left[\text{A}\right]}_2=0.70\text{mol/L}\\ {\text{Rate}}_1=1.6\text{mol/}\left(\text{L}\text{.s}\right)\\ {\text{Rate}}_2=3.2\text{mol/}\left(\text{L}\text{.s}\right)\end{array}$

Assuming,

  $\begin{array}{c}{\text{Rate}}_1=k{\left[{\text{A}}_{\text{1}}\right]}^x\\ {\text{Rate}}_2=k{\left[{\text{A}}_2\right]}^x\end{array}$

Now, divide the above rates as;

  $\begin{array}{c}\frac{{\text{Rate}}_{\text{2}}}{{\text{Rate}}_1}=\frac{k{\left[{\text{A}}_2\right]}^x}{k{\left[{\text{A}}_{\text{1}}\right]}^x}\\ \frac{3.2\text{mol/}\left(\text{L}\text{.s}\right)}{1.6\text{mol/}\left(\text{L}\text{.s}\right)}=\frac{{\left[0.70\text{mol/L}\right]}^x}{{\left[0.35\text{mol/L}\right]}^x}\\ 2={\left(2\right)}^x\\ x=1\end{array}$

Thus, the reaction is first order in A. Since A is the only reactant that means the overall order of the reaction is first order.