Chapter 14, Problem 14.97SP

Interpretation Introduction

Interpretation:

The activation energy in $\text{kJ/mol}$ needs to be determined, when the rate of the reaction increases by a factor of 6.37 on raising the temperature from $15°\text{C}$ to $45°\text{C}$.

Concept introduction:

The Arrhenius equation is applied to calculate activation energy.

$\ln \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$

Here, $K_2$ and $K_1$ are rate constant, $E_a$ is activation energy, R is gas constant, $T_1\text{and}{T}_2$ are initial and finale temperatures respectively.