Chapter 14, Problem 48QP

Variation of the rate constant with temperature for the first-order reaction:

$2{\text{N}}_2{\text{O}}_5(g)\underset{}{\to }{\text{ 2N}}_2{\text{O}}_4(g){\text{ + O}}_2(g)$

is given in the following table. Determine graphically the activation energy for the reaction.

T(K)	k(s-1)
298	$1.74\times {\text{ 10}}^{-5}$
308	$6.61\times {\text{ 10}}^{-5}$
318	$2.51\times {\text{ 10}}^{-4}$
328	$7.59\times {\text{ 10}}^{-4}$
338	$2.40\times {\text{ 10}}^{-3}$

Interpretation Introduction

Interpretation:

The activation energy of the given reaction is to be determined graphically.

Concept introduction:

Activation energy is the minimum energy required to initiate a reaction. It is calculated by Arrhenius equation. The higher the activation energy of the reaction, the more is the energy required for the initiation of the reaction, and slower is the rate of reaction.

Activation energy can be calculated by Arrhenius equation. $ln\frac{k_1}{k_2}=\frac{E_a}{R}\left[\frac{T_1-T_2}{T_1{T}_2}\right]$

Here, $\frac{k_1}{k_2}$ is the ratio of the rate constant, $E_a$ is the activation energy, $R$ is the gas constant, and $T_1\text{ and }{T}_2$ are the absolute temperatures.

Slope of the graph is calculated by the expression given as: $\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

Here, $x_1$ and $x_2$ are the coordinates on the x-axis and $y_1$ and $y_2$ are the coordinates on the y-axis.

According to Arrhenius equation, the slope is equal to the expression given as: $\begin{array}{l}\text{slope}=\frac{-E_a}{\text{R}}\\ E_a=-\left(\text{slope}\right)\text{R}\end{array}$

Here, $E_a$ is the activation energy and $\text{R}$ is the gas constant.

Answer to Problem 48QP

Solution: $103\text{kJ}/\text{mol}$

Explanation of Solution

Given information:

The data corresponding to the reaction is given as follows:

$\begin{array}{cc}T\left(\text{K}\right)& k\left({\text{s}}^{-1}\right)\\ 298& 1.74\times {10}^{-5}\\ 308& 6.61\times {10}^{-5}\\ 318& 2.51\times {10}^{-4}\\ 328& 7.59\times {10}^{-4}\\ 338& 2.40\times {10}^{-3}\end{array}$

Taking the natural log for each value of $k$ and inverse of each value of $T$,

$\begin{array}{cc}\ln k& \frac{1}{T}\left({\text{K}}^{-1}\right)\\ -10.959& 3.3\times {10}^{-3}\\ -9.6243& 3.2\times {10}^{-3}\\ -8.2900& 3.1\times {10}^{-3}\\ -7.183& 3.0\times {10}^{-3}\\ -6.0322& 2.9\times {10}^{-3}\end{array}$

The plot between $\ln k$ on y-axis and $\frac{1}{T}$ on x-axis is given below.

Slope of the graph is calculated by the expression given as follows:

$\text{slope}=\frac{y_2-y_1}{x_2-x_1}$

Here, $x_1$ and $x_2$ are the coordinates on the x-axis and $y_1$ and $y_2$ are the coordinates on the y-axis.

Substitute the values of $x_1,x_2,y_2$, and $y_1$ in the above equation,

$\begin{array}{c}\text{slope}=\frac{-9.6243+7.183}{3.2\times {10}^{-3}{\text{K}}^{-1}-3.0\times {10}^{-3}{\text{K}}^{-1}}\\ =\frac{2.4413}{-0.0002}\\ =-1.24\times {10}^4\text{K}\end{array}$

According to Arrhenius equation, the slope is equal to the expression given as follows:

$\begin{array}{l}\text{slope}=\frac{-E_a}{\text{R}}\\ E_a=-\left(\text{slope}\right)\text{R}\end{array}$

Here, $E_a$ is the activation energy and $\text{R}$ is the gas constant.

Rearrange the above equation for activation energy as,

$\begin{array}{c}{E}_a=-\left(-1.24\times {10}^4\cancel{\text{K}}\right)\left(8.314\text{J}/\cancel{\text{K}}.\text{mol}\right)\\ =1.03\times {10}^5\text{J}/\text{mol}\\ =103\text{kJ}/\text{mol}\end{array}$

Conclusion

The value of activation energy is $103\text{kJ}/\text{mol}$.