Chapter 11, Problem 58E

Consider the hypothetical reaction

$\text{A}\text{+}\text{B}\text{+}\text{2C }\to \text{2D}\text{+}\text{3E}$

where the rate law is

$\text{Rate}=-\frac{\Delta \left[\text{A}\right]}{\Delta t}=k\left[\text{A}\right]{\left[\text{B}\right]}^2$

An experiment is carried out where [A]₀ = 1.0 × 10⁻² M, [B]₀ = 3.0 M, and [C]₀ = 2.0 M. The reaction is started, and after 8.0 seconds, the concentration of A is 3.8 × 10⁻³ M.

a. Calculate the value of k for this reaction.

b. Calculate the half-life for this experiment.

c. Calculate the concentration of A after 13.0 seconds.

d. Calculate the concentration of C after 13.0 seconds.

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{8}\text{.0}\text{s}$ are given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The value of the rate constant $\left(k\right)$ for the given reaction.

Answer to Problem 58E

Answer

1. (a) The value of rate constant is $\underset{\_}{1.34\times {10}^{-2}{M}^{-2}\cdot s^{-1}}$.

Explanation of Solution

Explanation

Given

The initial concentration of ${\left[\text{A}\right]}_{\text{0}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\text{M}$.

The initial concentration of ${\left[\text{B}\right]}_{\text{0}}$ is $\text{3}\text{.0}\text{M}$.

The initial concentration of ${\left[\text{C}\right]}_{\text{0}}$ is $\text{2}\text{.0}\text{M}$.

The initial concentration of $\text{A}$ after $\text{8}\text{.0}\text{s}$ is $\text{3}\text{.8}\times {\text{10}}^{-\text{3}}\text{M}$.

The stated reaction is,

$\text{A}+\text{B}+\text{2C}\to \text{2D}+\text{3E}$

The rate law is represented as,

$\text{Rate}=-\frac{\Delta \left[\text{A}\right]}{\Delta t}=k\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

The above reaction is assumed to be pseudo first order. Since, the initial concentration of $\text{B}$ is very less as compared to $\text{A}$, therefore, change in its concentration is negligible, that is,

${\left[\text{B}\right]}_{\text{0}}\approx \left[\text{B}\right]$

The simplified rate law is,

$\frac{\text{d}\left[\text{A}\right]}{\text{dt}}=-k^{\text{'}}\left[\text{A}\right]$

Where, $k^{\text{'}}$ is rate constant, and it is equal to,

$k^{\text{'}}=k{\left[\text{B}\right]}_{\text{0}}^{\text{2}}$ (1)

The integral rate law equation of the first order reaction is,

$\begin{array}{c}\ln \left[\text{A}\right]=-k^{\text{'}}t+\ln {\left[\text{A}\right]}_{\text{0}}\\ k^{\text{'}}=\frac{\ln {\left[\text{A}\right]}_{\text{0}}-\ln \left[\text{A}\right]}{t}\end{array}$

Where,

• ${\left[\text{A}\right]}_{\text{0}}$ is the initial concentration of reactant $\text{A}$.
• $\left[\text{A}\right]$ is the concentration of reactant $\text{A}$ after $\text{8}\text{.0}\text{s}$.
• $t$ is the time.

Substitute the values of ${\left[\text{A}\right]}_{\text{0}},\left[\text{A}\right]$ and $t$ in the above equation.

$\begin{array}{c}{k}^{\text{'}}=\frac{\ln {\left[\text{A}\right]}_{\text{0}}-\ln \left[\text{A}\right]}{t}\\ =\frac{\ln \left(\frac{\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\text{M}}{\text{3}\text{.8}\times {\text{10}}^{-\text{3}}\text{M}}\right)}{\text{8}\text{s}}\\ =\text{0}\text{.121}{\text{s}}^{-\text{1}}\end{array}$

Substitute the values of $k^{\text{'}}$ and ${\left[\text{B}\right]}_{\text{0}}$ in equation (1).

$\begin{array}{c}{k}^{\text{'}}=k{\left[\text{B}\right]}_{\text{0}}^{\text{2}}\\ \text{0}\text{.121}{\text{s}}^{-\text{1}}=k{\left(\text{3}\text{.0}\text{M}\right)}^{\text{2}}\\ k=\underset{\_}{1.34\times {10}^{-2}{M}^{-2}\cdot s^{-1}}\end{array}$

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{8}\text{.0}\text{s}$ are given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The half life of the given reaction.

Answer to Problem 58E

Answer

The half life of the given reaction is $\underset{\_}{5.73s}$.

Explanation of Solution

Explanation

The value of rate constant is $\text{0}\text{.121}{\text{s}}^{-\text{1}}$.

Formula

The half-life is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k^{\text{'}}}$

Where,

• $t_{\frac{\text{1}}{\text{2}}}$ is half life.
• $k^{\text{'}}$ is rate constant.

Substitute the values of $k^{\text{'}}$ in the above equation.

$\begin{array}{c}{t}_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k^{\text{'}}}\\ =\frac{\text{0}\text{.693}}{\text{0}\text{.121}{\text{s}}^{-\text{1}}}\\ =\underset{\_}{5.73s}\end{array}$

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{8}\text{.0}\text{s}$ are given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The concentration of $\text{A}$ after $\text{13}\text{.0}\text{s}$.

Answer to Problem 58E

Answer

The concentration of $\text{A}$ after $\text{13}\text{.0}\text{s}$ is $\underset{\_}{2.07\times 10^{-3}M}$.

Explanation of Solution

Explanation

Given

The concentration of ${\left[\text{A}\right]}_{\text{0}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\text{M}$.

The concentration of ${\left[\text{B}\right]}_{\text{0}}$ is $\text{3}\text{.0}\text{M}$.

The concentration of ${\left[\text{C}\right]}_{\text{0}}$ is $\text{2}\text{.0}\text{M}$.

The integral rate law equation of first order reaction is,

$\begin{array}{c}\ln \left[\text{A}\right]=-k^{\text{'}}t+\ln {\left[\text{A}\right]}_{\text{0}}\\ \left[\text{A}\right]={\text{e}}^{\left\{\ln {\left[\text{A}\right]}_{\text{0}}\right\}}{\text{e}}^{\left\{-k^{\text{'}}t\right\}}\\ ={\left[\text{A}\right]}_{\text{0}}{\text{e}}^{\left\{-k^{\text{'}}t\right\}}\end{array}$

Where,

• $k^{\text{'}}$ is the rate constant.
• ${\left[\text{A}\right]}_{\text{0}}$ is the initial concentration of reactant.
• $\left[\text{A}\right]$ is the concentration of reactant.
• $t$ is the time.

Substitute the values of ${\left[\text{A}\right]}_{\text{0}},k^{\text{'}}$ and $t$ in the above equation.

$\begin{array}{c}\left[\text{A}\right]={\left[\text{A}\right]}_{\text{0}}{\text{e}}^{\left\{-k^{\text{'}}t\right\}}\\ =\text{1}\text{.0}\times {\text{10}}^{-\text{2}}{\text{e}}^{\left\{-\text{0}\text{.121}{\text{s}}^{-\text{1}}\left(\text{13}\text{s}\right)\right\}}\\ =\underset{\_}{2.07\times 10^{-3}M}\end{array}$

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{8}\text{.0}\text{s}$ are given. The answers are to be given for each option.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

The half-life of the first order reaction is calculated using the formula,

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{0}\text{.693}}{k}$

To determine: The concentration of $\text{C}$ after $\text{13}\text{.0}\text{s}$.

Answer to Problem 58E

Answer

The concentration of $\text{C}$ after $\text{13}\text{.0}\text{s}$ is $\underset{\_}{1.984M}$.

Explanation of Solution

Explanation

Given

The concentration of ${\left[\text{A}\right]}_{\text{0}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\text{M}$.

The concentration of ${\left[\text{B}\right]}_{\text{0}}$ is $\text{3}\text{.0}\text{M}$.

The concentration of ${\left[\text{C}\right]}_{\text{0}}$ is $\text{2}\text{.0}\text{M}$.

The concentration of $\text{A}$ after $\text{13}\text{.0}\text{s}$ is $\text{2}\text{.07}\times {\text{10}}^{-\text{3}}\text{M}$.

The stated reaction is,

$\text{A}+\text{B}+\text{2C}\to \text{2D}+\text{3E}$

The rate law is represented as,

$\begin{array}{c}\Delta \left[\text{A}\right]=\frac{\Delta \left[\text{C}\right]}{\text{2}}\\ \left[\text{A}\right]-{\left[\text{A}\right]}_{\text{0}}=\frac{\left[\text{C}\right]-{\left[\text{C}\right]}_{\text{0}}}{\text{2}}\end{array}$

Substitute the values of ${\left[\text{A}\right]}_{\text{0}},\left[\text{A}\right]$ and ${\left[\text{C}\right]}_{\text{0}}$ in the above equation.

$\begin{array}{c}\left[\text{A}\right]-{\left[\text{A}\right]}_{\text{0}}=\frac{\left[\text{C}\right]-{\left[\text{C}\right]}_{\text{0}}}{\text{2}}\\ \text{2}\left(\text{2}\text{.07}\times {\text{10}}^{-\text{3}}\text{M}-\text{1}\text{.0}\times {\text{10}}^{-\text{2}}\text{M}\right)=\left[\text{C}\right]-\text{2}\text{.0}\text{M}\\ \left[\text{C}\right]=\underset{\_}{1.984M}\end{array}$

Conclusion

Conclusion

The required value of rate constant is $\underset{\_}{1.34\times {10}^{-2}{M}^{-2}\cdot s^{-1}}$. The half life of the given reaction is $\underset{\_}{5.73s}$. The required concentration of $\text{A}$ after $\text{13}\text{.0}\text{s}$ is $\underset{\_}{2.07\times 10^{-3}M}$. The required concentration of $\text{C}$ after $\text{13}\text{.0}\text{s}$ is $\underset{\_}{1.984M}$.