Chapter 14, Problem 14.140AP

(a)

Interpretation Introduction

To determine: The rate law and the value of the rate constant for the given reaction at $\text{298}\text{K}$.

Interpretation: The rate law and the value of the rate constant for the given reaction at $\text{298}\text{K}$ are to be calculated.

Concept introduction: Rate law expression for the given reaction is,

$\text{Rate}=k{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}$

Answer to Problem 14.140AP

The rate law for the given reaction is,

$\text{Rate}=k\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{O}}_{\text{3}}\right]$

The value of the rate constant for the given reaction is $\underset{\_}{10^3{M}^{-1}{s}^{-1}}$.

Explanation of Solution

The given reaction is,

${\text{H}}_{\text{2}}\text{C}={\text{CH}}_{\text{2}}+2{\text{O}}_3\to 2{\text{H}}_{\text{2}}\text{C}=\text{O}+2{\text{O}}_2$

The given Table is below.

$\begin{array}{cccc}\text{Experiment}& \left[{\text{O}}_3\right]\text{M}& \left[{\text{C}}_{\text{2}}{\text{H}}_4\right]\text{M}& \text{Rate}\left(\text{M/s}\right)\\ 1& 0.86\times {10}^{-2}& 1.00\times {10}^{-2}& 0.0877\\ 2& 0.43\times {10}^{-2}& 1.00\times {10}^{-2}& 0.0439\\ 3& 0.22\times {10}^{-2}& 0.50\times {10}^{-2}& 0.0110\end{array}$

Table 1

The rate law is,

$\text{Rate}=k{\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]}^{\text{m}}{\left[{\text{O}}_{\text{3}}\right]}^{\text{n}}$ (1)

Substitute the values of rate and concentrations of ${\text{O}}_3$ and ${\text{C}}_{\text{2}}{\text{H}}_4$ of the experiment 1 from the above Table in the above expression.

$0.0877=k{\left[1.00\times {10}^{-2}\right]}^{\text{m}}{\left[0.86\times {10}^{-2}\right]}^{\text{n}}$ (2)

Substitute the values of rate and concentrations of ${\text{O}}_3$ and ${\text{C}}_{\text{2}}{\text{H}}_4$ of the experiment 2 from the above Table in the equation (1).

$0.0439=k{\left[1.00\times {10}^{-2}\right]}^{\text{m}}{\left[0.43\times {10}^{-2}\right]}^{\text{n}}$ (3)

Substitute the values of rate and concentrations of ${\text{O}}_3$ and ${\text{C}}_{\text{2}}{\text{H}}_4$ of the experiment 3 from the above Table in the equation (1).

$0.0110=k{\left[0.50\times {10}^{-2}\right]}^{\text{m}}{\left[0.22\times {10}^{-2}\right]}^{\text{n}}$ (4)

The equation (3) is divided by the equation (2).

$\begin{array}{c}\frac{0.0439}{0.0877}={\left[\frac{1.00\times {10}^{-2}}{1.00\times {10}^{-2}}\right]}^{\text{m}}{\left[\frac{0.43\times {10}^{-2}}{0.86\times {10}^{-2}}\right]}^{\text{n}}\\ 0.5={\left(1\right)}^{\text{m}}{\left(0.5\right)}^{\text{n}}\\ \text{n}=1\end{array}$

The equation (4) is divided by the equation (3).

$\begin{array}{c}\frac{0.0110}{0.0439}={\left[\frac{0.50\times {10}^{-2}}{1.00\times {10}^{-2}}\right]}^{\text{m}}{\left[\frac{0.22\times {10}^{-2}}{0.43\times {10}^{-2}}\right]}^{\text{n}}\\ 0.25={\left(0.50\right)}^{\text{m}}{\left(0.50\right)}^{\text{n}}\\ \text{m}=1\end{array}$

Therefore, the rate law for the given reaction is,

$\text{Rate}=k\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{O}}_{\text{3}}\right]$

The value of rate constant is calculated by the equation (1).

Substitute any value of rate and concentrations of ${\text{O}}_3$ and ${\text{C}}_{\text{2}}{\text{H}}_4$ from the above Table in the above expression.

$\begin{array}{c}0.0110=k\left[0.50\times {10}^{-2}\right]\left[0.22\times {10}^{-2}\right]\\ k=\frac{0.0110}{\left(0.50\times {10}^{-2}\right)\left(0.22\times {10}^{-2}\right)}\\ =\underset{\_}{10^3{M}^{-1}{s}^{-1}}\end{array}$

Therefore, the value of the rate constant for the given reaction is $\underset{\_}{10^3{M}^{-1}{s}^{-1}}$.

(b)

Interpretation Introduction

To determine: The activation energy of the given reaction.

Interpretation: The activation energy of the given reaction is to be calculated.

Concept introduction: The rate constant at different temperatures is calculated with the formula,

$\ln \frac{k_1}{k_2}=\frac{E_{\text{a}}}{2.303\times R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Answer to Problem 14.140AP

The activation energy of the given reaction is $\underset{\_}{20.621kJ/mol}$.

Explanation of Solution

The value of ideal gas constant is $8.314\text{J}/\text{mol}\cdot \text{K}$.

The given Table is below.

$\begin{array}{cc}T\left(\text{K}\right)& k\left({\text{M}}^{-1}{\text{s}}^{-1}\right)\\ 263& 3.28\times {10}^2\\ 273& 4.73\times {10}^2\\ 283& 6.65\times {10}^2\\ 293& 9.13\times {10}^2\end{array}$

Table 2

The activation energy of the given reaction is calculated by the formula,

$\ln \frac{k_1}{k_2}=\frac{E_{\text{a}}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$ (1)

Where,

• $k_1$ is rate constant at first temperature.
• $k_2$ is rate constant at second temperature.
• ${\text{E}}_{\text{a}}$ is activation energy.
• $\text{R}$ is ideal gas constant.
• $T_1$ is first temperature.
• $T_2$ is second temperature.

Substitute the value of different rate constants and temperatures from the above Table in the equation (1).

$\begin{array}{c}\ln \left(\frac{4.73\times {10}^2{\text{M}}^{-1}{\text{s}}^{-1}}{3.28\times {10}^2{\text{M}}^{-1}{\text{s}}^{-1}}\right)=\left(\frac{E_{\text{a}}}{2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}}\right)\left(\frac{1}{263\text{K}}-\frac{1}{273\text{K}}\right)\\ 0.15=\left(\frac{E_{\text{a}}}{2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}}\right)\left(\frac{273-263}{273\times 263}\right)\\ 0.15=\left(\frac{E_{\text{a}}}{2.303\times 8.314\text{J}/\text{mol}\cdot \text{K}}\right)\left(\frac{\left(10\right)}{71799}\right)\\ E_{\text{a}}=\underset{\_}{20.621kJ/mol}\end{array}$

Therefore, the activation energy of the given reaction is $\underset{\_}{20.621kJ/mol}$.

Conclusion

1. a. The rate law for the given reaction is,

$\text{Rate}=k\left[{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\right]\left[{\text{O}}_{\text{3}}\right]$

The value of the rate constant for the given reaction is $\underset{\_}{10^3{M}^{-1}{s}^{-1}}$.

1. b. The activation energy of the given reaction is $\underset{\_}{20.621kJ/mol}$.