Chapter 18, Problem 18.1P

Interpretation Introduction

Interpretation:

The activation energy for the reaction has to be calculated.

Concept Introduction:

Arrhenius relation:

This relation talks about the dependence of absolute temperature on the rate constant of chemical reaction.

The minimum amount of energy required for the reactants to start a chemical reaction is known as activation energy. The activation energy is represented as $\text{Ea}$.

Arrhenius equation: ${\text{k = A e}}^{-\frac{\text{Ea}}{\text{RT}}}$

$\text{k}$ is the rate constant for the reaction.

$\text{A}$ is the Arrhenius constant.

${\text{E}}_{\text{a}}$ is the activation energy.

$\text{R}$ is universal gas constant.

$\text{T}$ is the temperature.

Activation energy for two rate constants with two different temperatures can be calculated by Arrhenius equation:

  $\text{ln}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{\text{Ea}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)$

  $\begin{array}{l}{\text{k}}_{\text{1}}\text{,}{\text{k}}_{\text{2}}\text{=}\text{Rate}\text{constants}\\ \\ \text{Ea}\text{=}\text{Activation}\text{energy}\\ \\ \text{R}\text{=}\text{gas}\text{constant}\text{(8}\text{.314}\text{J/mol}\text{.K)}\\ \\ {\text{T}}_{\text{1}}{\text{,T}}_{\text{2}}\text{=}\text{temperatures}\end{array}$

Answer to Problem 18.1P

The activation energy of the given reaction is $100.3kJ/mol.$

Explanation of Solution

Given:

  $\begin{array}{l}{\text{E}}_{\text{a}}\text{ =}\text{? }\text{;}{\text{k}}_{\text{1}}\text{=}\left(2.35\times {10}^{-4}\right){\text{s}}^{-1}\text{;}{\text{T}}_{\text{1}}\text{=}293.0\text{K}\text{;}\\ \\ {\text{k}}_{\text{2}}\text{=}\left(9.15\times {10}^{-4}\right){\text{s}}^{-1}\text{;}{\text{T}}_{\text{2}}\text{= }303.0\text{K}\text{.}\end{array}$

Calculating the Activation energy as follows,

$\begin{array}{l}\text{ln}\frac{{\text{k}}_{\text{2}}}{{\text{k}}_{\text{1}}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{R}}\left(\frac{\text{1}}{{\text{T}}_{\text{1}}}\text{-}\frac{\text{1}}{{\text{T}}_{\text{2}}}\right)\\ \\ \text{ln}\frac{9.15\times {10}^{-4}{\text{s}}^{-1}}{2.35\times {10}^{-4}{\text{s}}^{-1}}\text{=}\frac{{\text{E}}_{\text{a}}}{\text{8}\text{.314}\text{J/mol}\text{.K}}\left(\frac{\text{1}}{293.0\text{K}}-\frac{\text{1}}{303.0\text{K}}\right)\\ \\ 1.359=\frac{{\text{E}}_{\text{a}}}{\text{8}\text{.314}\text{J/mol}\text{.K}}\left(1.126\times {10}^{-4}\right)\\ \\ {\text{E}}_{\text{a}}=\frac{\left(1.359\right)\left(\text{8}\text{.314}\text{J/mol}\text{.K}\right)}{1.126\times {10}^{-4}{K}^{-1}}\\ \\ \text{=}100.3kJ/mol.\end{array}$

Therefore, the activation energy of the given reaction is $100.3kJ/mol.$