Chapter 12, Problem 101AE

Consider the reaction

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

where the rate law is defined as

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

An experiment is carried out where [B]₀ = [C]₀ = 1.00 M and [A]₀ = 1.00 × 10⁻⁴M.

a. If after 3.00 min, [A] = 3.26 × 10⁻⁵M, calculate the value of k.

b. Calculate the half-life for this experiment.

c. Calculate the concentration of B and the concentration of A after 10.0 min.

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{3}\text{.00}\text{min}$ 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,

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

Answer to Problem 101AE

The value of rate constant is $\underset{\_}{114.86s^{-1}}$.

Explanation of Solution

Explanation

Given

The initial concentration of ${\left[\text{A}\right]}_{\text{0}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\text{M}$.

The initial concentration of ${\left[\text{B}\right]}_{\text{0}}$ is $\text{1}\text{.0}\text{M}$.

The initial concentration of ${\left[\text{C}\right]}_{\text{0}}$ is $\text{1}\text{.0}\text{M}$.

The initial concentration of $\text{A}$ after $\text{3}\text{.00}\text{min}$ is $\text{3}\text{.26}\times {\text{10}}^{-\text{5}}\text{M}$.

The stated reaction is,

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

The rate law is represented as,

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

The graph of $\frac{\text{1}}{\text{A}}$ versus time (min) is plotted for the given concentration of $\text{A}$

Since, the obtained graph is a straight line, hence, the order of the reaction is second. The slope of graph is $\text{6891}\text{.6}$.

The integral rate law equation of second order reaction is,

$\frac{\text{1}}{\left[\text{A}\right]}=kt+\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$ (1)

Where,

• $k$ is the rate constant.
• ${\left[\text{A}\right]}_{\text{0}}$ is the initial concentration of reactant $\text{A}$.
• $\left[\text{A}\right]$ is the initial concentration of reactant $\text{A}$ after $\text{3}\text{.00}\text{min}$.
• $t$ is the time.

The equation (1) is similar to the equation of a straight line, that is,

$y=mx+c$ (2)

Where,

• $y$ is the y-intercept.
• $x$ is the x-intercept.
• $m$ is the slope.
• $c$ is a constant.

Compare equation (1) and (2).

$m=k$

Substitute the value of slope in the above equation.

$\begin{array}{c}m=k\\ k=6891.6\cdot {\text{min}}^{-\text{1}}\left(\frac{1\min }{60\text{s}}\right)\\ =\underset{\_}{114.86s^{-1}}\end{array}$

Conclusion

Conclusion

The required value of rate constant is $\underset{\_}{114.86s^{-1}}$.

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{3}\text{.00}\text{min}$ 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 101AE

The half life of the given reaction is $\underset{\_}{87.06s}$.

Explanation of Solution

Explanation

The value of rate constant is $\text{114}\text{.86}{\text{s}}^{-\text{1}}$.

Formula

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

$t_{\frac{\text{1}}{\text{2}}}=\frac{\text{1}}{k{\left[\text{A}\right]}_{\text{0}}}$

Where,

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

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

$\begin{array}{c}{t}_{\frac{\text{1}}{\text{2}}}=\frac{\text{1}}{k{\left[\text{A}\right]}_{\text{0}}}\\ =\left[\frac{\text{1}}{\text{114}\text{.86}\left(\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\right)}\right]\text{s}\\ =\underset{\_}{87.06s}\end{array}$

Conclusion

Conclusion

The half life of the given reaction is $\underset{\_}{87.06s}$.

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{3}\text{.00}\text{min}$ 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 $\text{A}$ $\text{10}\text{min}$.

Answer to Problem 101AE

The concentration of $\text{A}$ after $\text{10}\text{min}$ is $\underset{\_}{1.27\times {10}^{-5}M}$.

Explanation of Solution

Explanation

Given

The initial concentration of ${\left[\text{A}\right]}_{\text{0}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\text{M}$.

The initial concentration of ${\left[\text{B}\right]}_{\text{0}}$ is $\text{1}\text{.0}\text{M}$.

The initial concentration of ${\left[\text{C}\right]}_{\text{0}}$ is $\text{1}\text{.0}\text{M}$.

The initial concentration of $\text{A}$ after $\text{3}\text{.00}\text{min}$ is $\text{3}\text{.26}\times {\text{10}}^{-\text{5}}\text{M}$.

The integral rate law equation of second order reaction is,

$\frac{\text{1}}{\left[\text{A}\right]}=kt+\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}$

Where,

• $k$ is the rate constant.
• ${\left[\text{A}\right]}_{\text{0}}$ is the initial concentration of reactant $\text{A}$.
• $\left[\text{A}\right]$ is the initial concentration of reactant $\text{A}$ after $\text{10}\text{min}$.
• $t$ is the time.

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

$\begin{array}{c}\frac{\text{1}}{\left[\text{A}\right]}=kt+\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}\\ =\text{114}\text{.86}{\text{s}}^{-\text{1}}\times \text{600}\text{s}+\frac{\text{1}}{\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\text{M}}\\ \left[\text{A}\right]=\underset{\_}{1.27\times {10}^{-5}M}\end{array}$

Conclusion

Conclusion

The required concentration of $\text{A}$ after $\text{10}\text{min}$ is $\underset{\_}{1.27\times {10}^{-5}M}$.

Interpretation Introduction

Interpretation: A hypothetical reaction and its rate law, initial concentration of reactants and the concentration $\text{A}$ after $\text{3}\text{.00}\text{min}$ 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 $\text{B}$ after $\text{10}\text{min}$.

Answer to Problem 101AE

The concentration of $\text{B}$ after $\text{10}\text{min}$ is $\underset{\_}{0.99M}$.

Explanation of Solution

The initial concentration of ${\left[\text{A}\right]}_{\text{0}}$ is $\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\text{M}$.

The initial concentration of ${\left[\text{B}\right]}_{\text{0}}$ is $\text{1}\text{.0}\text{M}$.

The initial concentration of ${\left[\text{C}\right]}_{\text{0}}$ is $\text{1}\text{.0}\text{M}$.

The initial concentration of $\text{A}$ after $\text{3}\text{.00}\text{min}$ is $\text{3}\text{.26}\times {\text{10}}^{-\text{5}}\text{M}$.

The stated reaction is,

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

The rate law is represented as,

$\begin{array}{c}\frac{{\left[\text{A}\right]}_{\text{0}}-\left[\text{A}\right]}{\text{3}}={\left[\text{B}\right]}_{\text{0}}-\left[\text{B}\right]\\ \left[\text{B}\right]={\left[\text{B}\right]}_{\text{0}}-\frac{{\left[\text{A}\right]}_{\text{0}}-\left[\text{A}\right]}{\text{3}}\end{array}$

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

$\begin{array}{c}\left[\text{B}\right]={\left[\text{B}\right]}_{\text{0}}-\frac{{\left[\text{A}\right]}_{\text{0}}-\left[\text{A}\right]}{\text{3}}\\ =\text{1}\text{.0}\text{M}-\frac{\left[\text{1}\text{.0}\times {\text{10}}^{-\text{4}}\text{M}-\text{1}\text{.27}\times {\text{10}}^{-\text{5}}\text{M}\right]}{\text{3}}\\ \left[\text{B}\right]=\underset{\_}{0.99M}\end{array}$

Conclusion

Conclusion

The required concentration of $\text{B}$ after $\text{13}\text{.0}\text{s}$ is $\underset{\_}{0.99M}$.