Chapter 14, Problem 42QP

The rate constant of a first-order reaction is $\text{4}{\text{.60 × 10}}^{\text{-4}}{\text{s}}^{\text{-1}}\text{at 350°C}$. If the activation energy is $\text{104 kJ/mol}$, calculate the temperature at which its rate constant is $\text{8}{\text{.80 × 10}}^{\text{-4}}{\text{s}}^{\text{-1}}$.

Interpretation Introduction

Interpretation:

The final temperature of the given reaction is to be calculated

Concept introduction:

The rate of reaction is the measurement of the concentration change of reactants within a period of time during the course of a reaction. The concentration-versus-time graph of the reaction is used to determine the rate constant of the reaction.

The rate of reaction increases with increase in the temperature of the reaction.

Answer to Problem 42QP

Solution: $371°\text{C}$

Explanation of Solution

Given information:

Rate constant of the first-order reaction is,

$\begin{array}{l}{\text{k}}_1=4.60\times {10}^{-4}{\text{s}}^{-1}\\ {\text{k}}_2=8.80\times {10}^{-4}{\text{s}}^{-1}\end{array}$

The activation energy of the reaction is $104\text{kJ}/\text{mol}$.

Temperature, ${\text{T}}_1=350°\text{C}$

The conversion of activation energy into joules is given as follows:

$\begin{array}{c}{E}_a=104\text{kJ}/\text{mol}\\ =104000\text{J}/\text{mol}\end{array}$

The conversion of temperature into kelvin is given as follows:

$\begin{array}{c}{\text{T}}_1=\left(350+273.15\right)\text{K}\\ =\text{623}\text{.15}\text{K}\end{array}$

The temperature of the given reaction is calculated by the following expression:

$\begin{array}{l}\ln \frac{{\text{k}}_2}{{\text{k}}_1}=\frac{E_a}{\text{R}}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]\\ {\text{T}}_2=\frac{E_a{T}_1}{E_a-T_1\text{R}\ln \frac{{\text{k}}_2}{{\text{k}}_1}}\end{array}$

Here, ${\text{k}}_1$ is the initial rate constant, ${\text{k}}_2$ is the final rate constant, $E_a$ is the activation energy, $\text{R}$ is the gas constant, $T_1$ is the initial temperature, and $T_2$ is the final temperature of the reaction.

Substitute the values of ${\text{k}}_1,{\text{k}}_2,E_a,\text{R}$, and $T_1$ in the above equation,

$\begin{array}{c}{\text{T}}_2=\frac{104000\text{J}/\text{mol}\times \text{623}\text{.15}\text{K}}{104000\text{J}/\text{mol}-\left(\text{623}\text{.15}\text{K}\right)\left(8.314\text{J}/\text{K}\text{.mol}\right)\ln \frac{8.80\times {10}^{-4}{\text{s}}^{-1}}{4.60\times {10}^{-4}{\text{s}}^{-1}}}\\ =\frac{104000\text{J}/\text{mol}\times \text{623}\text{.15}\text{K}}{104000\text{J}/\text{mol}-\left(\text{623}\text{.15}\text{K}\right)\left(8.314\text{J}/\text{K}\text{.mol}\right)\left(\ln 1.9130\right)}\\ =\frac{104000\cancel{\text{J}/\text{mol}}\times 623.15\text{K}}{\left(104000-3360.688\right)\cancel{\text{J}/\text{mol}}}\\ =\frac{64807600\text{K}}{\left(104000-3360.688\right)}\end{array}$

Simplify the above equation to obtain the value of temperature.

$\begin{array}{c}{\text{T}}_2=\frac{64807600\text{K}}{100639.31}\\ =643.959\text{K}\\ \cong 644\text{K}\end{array}$

The conversion of temperature into $°\text{C}$ is given as follows:

$\begin{array}{c}{T}_2=\left(644-273.15\right)°\text{C}\\ =370.85°\text{C}\\ \cong 371°\text{C}\end{array}$

The value of temperature at rate constant $8.80\times {10}^{-4}{\text{s}}^{-1}$ is $371°\text{C}$.

Conclusion

The final temperature of the given reaction is $371°\text{C}$.