Chapter 14, Problem 14.95SP

Interpretation Introduction

(a)

Interpretation:

The activation energy in $\text{kJ/mol}$ when the reaction rate doubles and temperature is increased to $35°\text{C}$ needs to be determined.

Concept introduction:

The following Arrhenius equation is used to calculate activation energy.

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

Where $K_2$ 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.

Interpretation Introduction

(b)

Interpretation:

The activation energy in $\text{kJ/mol}$ needs to be determined when the reaction rate triples and temperature is increased to $35°\text{C}$.

Concept introduction:

The following Arrhenius equation is used 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.