Chapter 13, Problem 13.50QE

(a)

Interpretation Introduction

Interpretation:

Concentration of the reactant has to be calculated that will be present after $60\sec$ in the reaction.

Answer to Problem 13.50QE

The concentration of the reactant that will be present after $60\sec$ is $0.16M$.

Explanation of Solution

Given initial concentration of the reactant is $0.64M$.  The half-life of the reactant is given as $30\sec$.  Rate constant can be calculated using the relationship between the rate constant and half-life as shown below;

    $\begin{array}{c}\text{Rate}\text{constant}(k)=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{30\text{s}}\\ =0.0231{\text{s}}^{-1}\end{array}$

Therefore, the rate constant of the reaction is $0.0231{\text{s}}^{-1}$.

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.

Substituting the values in above equation, the concentration of the reactant that will be present after $60\sec$ can be calculated as follows;

    $\begin{array}{c}\ln {[}_{\text{R}}=\ln {[}_{\text{R}}-kt\\ =\ln (0.64)-(0.0231{\text{s}}^{-1})(60\sec )\\ =-0.446-1.386\\ \ln {[}_{\text{R}}=-1.832\\ {[}_{\text{R}}=e^{-1.832}\\ =0.16M\end{array}$

Therefore, the concentration of reactant that remains after $60\sec$ is $0.16M$.

(b)

Interpretation Introduction

Interpretation:

The time that will be taken for the concentration of the reactant to decrease to one-eighth of the initial value has to be calculated.

Answer to Problem 13.50QE

Time taken for the reactant concentration to get reduced to one-eighth of initial value is $90\sec$.

Explanation of Solution

Given initial concentration of the reactant is $0.64M$.  The half-life of the reactant is given as $30\sec$.  Rate constant can be calculated using the relationship between the rate constant and half-life as shown below;

    $\begin{array}{c}\text{Rate}\text{constant}(k)=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{30\text{s}}\\ =0.0231{\text{s}}^{-1}\end{array}$

Therefore, the rate constant of the reaction is $0.0231{\text{s}}^{-1}$.

Concentration of the reactant reduced to one-eighth is calculated as shown below;

    $\begin{array}{c}{[}_R=\frac{1}{8}\times 0.64M\\ =0.080M\end{array}$

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.

Substituting the values in above equation, the time taken for the concentration of the reactant to get reduced to one-eighth of the initial value can be calculated as follows;

    $\begin{array}{c}\ln {[}_{\text{R}}=\ln {[}_{\text{R}}-kt\\ kt=\ln {[}_{\text{R}}-\ln {[}_{\text{R}}\\ =\ln \frac{{[\text{R}]}_0}{{[\text{R}]}_t}\\ t=\frac{1}{0.0231{\text{s}}^{-1}}\ln \frac{0.64}{0.080}\\ =\frac{2.08}{0.0231{\text{s}}^{-1}}\\ =90\sec \end{array}$

Therefore, the time taken for the concentration of the reactant to get reduced to one-eighth of the initial value is $90\sec$.

(c)

Interpretation Introduction

Interpretation:

The time that will be taken for the concentration of the reactant to decrease to $0.040\text{mol}\cdot {\text{L}}^{-1}$ has to be calculated.

Answer to Problem 13.50QE

Time taken for the reactant concentration to get reduced to $0.040\text{mol}\cdot {\text{L}}^{-1}$ is $120\sec$.

Explanation of Solution

Given initial concentration of the reactant is $0.64M$.  The half-life of the reactant is given as $30\sec$.  Rate constant can be calculated using the relationship between the rate constant and half-life as shown below;

    $\begin{array}{c}\text{Rate}\text{constant}(k)=\frac{0.693}{t_{1/2}}\\ =\frac{0.693}{30\text{s}}\\ =0.0231{\text{s}}^{-1}\end{array}$

Therefore, the rate constant of the reaction is $0.0231{\text{s}}^{-1}$.

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.

Substituting the values in above equation, the time taken for the concentration of the reactant to get reduced to $0.040\text{mol}\cdot {\text{L}}^{-1}$ can be calculated as follows;

    $\begin{array}{c}\ln {[}_{\text{R}}=\ln {[}_{\text{R}}-kt\\ kt=\ln {[}_{\text{R}}-\ln {[}_{\text{R}}\\ =\ln \frac{{[\text{R}]}_0}{{[\text{R}]}_t}\\ t=\frac{1}{0.0231{\text{s}}^{-1}}\ln \frac{0.64}{0.040}\\ =\frac{2.77}{0.0231{\text{s}}^{-1}}\\ =120\sec \end{array}$

Therefore, the time taken for the concentration of the reactant to get reduced to $0.040\text{mol}\cdot {\text{L}}^{-1}$ is $120\sec$.