Chapter 13, Problem 13.56QE

Interpretation Introduction

Interpretation:

Half-life of the first-order reaction has to be calculated if the reactant concentration is $0.0451M$ at $30.5\sec$ decreases to $0.0321M$ at $45.0\sec$ and also the time that is taken for the reactant concentration to decrease to $0.0100M$ has to be calculated.

Answer to Problem 13.56QE

Half-life of the reactant is $\text{29}\text{.6}\text{s}$ and the time taken for the reactant concentration to decrease to $0.0100M$ is $\text{95}\text{s}$.

Explanation of Solution

Integrated rate law for the first order reaction is given as follows;

    $\ln {[\text{R}]}_t=\ln {[\text{R}]}_0-kt$

Where,

    ${[\text{R}]}_0$ is the initial concentration of the reactant.

    $k$ is the rate constant.

For the concentration of the reactant of $0.0451M$ at $30.5\sec$:

The integrated rate law can be given as shown below;

    $\ln {[\text{R}]}_0=\ln {[\text{R}]}_t+kt$

Substituting the values in the equation as shown below;

    $\ln {[\text{R}]}_0=\ln (0.0451)+k(30.5\sec )$        (1)

For the concentration of the reactant of $0.0321M$ at $45.0\sec$:

The integrated rate law can be given as shown below;

    $\ln {[\text{R}]}_0=\ln {[\text{R}]}_t+kt$

Substituting the values in the equation as shown below;

    $\ln {[\text{R}]}_0=\ln (0.0321)+k(45.0\sec )$        (2)

Considering the equations (1), and (2), it is found that the left side is equal.  Therefore, equating the right side of both equations, the rate constant can be calculated as shown below;

    $\begin{array}{c}\ln (0.0451)+k(30.5\sec )=\ln (0.0321)+k(45.0\sec )\\ -3.099+k(30.5\sec )=-3.439+k(45.0\sec )\\ k(45.0\sec )-k(30.5\sec )=3.439-3.099\\ k(14.5\sec )=0.340\\ k=\frac{0.340}{14.5\sec }\\ =0.0234{\text{s}}^{-1}\end{array}$

Therefore, the rate constant of the reaction is $0.0234{\text{s}}^{-1}$.

The relationship between half-life and the rate constant of first order reaction is given as follows;

    $\begin{array}{c}\text{Half-life}(t_{1/2})=\frac{0.693}{k}\\ =\frac{0.693}{0.0234{\text{s}}^{-1}}\\ =29.6\text{s}\end{array}$

Thus, half-life of the reactant is calculated as $\text{29}\text{.6}\text{s}$.

Initial concentration can be calculated by substituting the rate constant value in equation (1) as follows;

    $\begin{array}{c}\ln {[}_{\text{R}}=\ln (0.0451)+k(30.5\sec )\\ =\ln (0.0451)+(0.0234{\text{s}}^{-1})(30.5\sec )\\ =-3.099+0.7137\\ \ln {[}_{\text{R}}=-2.3853\\ {[}_{\text{R}}=e^{-2.3853}\\ =0.092M\end{array}$

Therefore, the initial concentration of the reactant is $0.092M$.

Time taken for the reactant concentration to decrease from $0.092M$ to $0.0100M$ can be calculated using the integrated rate law as shown below;

    $\begin{array}{c}\ln {[}_{\text{R}}=\ln {[}_{\text{R}}-kt\\ kt=\ln {[}_{\text{R}}-\ln {[}_{\text{R}}\\ =\ln \frac{{[\text{R}]}_0}{{[\text{R}]}_t}\\ t=\frac{1}{0.0234{\text{s}}^{-1}}\ln \frac{0.092}{0.0100}\\ =\frac{2.219}{0.0234{\text{s}}^{-1}}\\ =95\text{s}\end{array}$

Therefore, the time taken for the reactant concentration to decrease to $0.0100M$ is $95\text{s}$.