Question

The rate of a reaction quadruples when the temperature changes from 293K to 313K. Calculate the energy of activation (in kJ/mol) of the reaction assuming that is does not change with temperature (nearest integer).

Answer 1

53

Answer 2

The integrated Arrhenius equation relates the rate constants ($k_1, k_2$) at two different temperatures ($T_1, T_2$) to the activation energy ($E_a$) and the ideal gas constant ($R$):

$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Given $T_1 = 293$ K, $T_2 = 313$ K, and $\frac{k_2}{k_1} = 4$. Using $R = 8.314$ J/mol·K, we substitute these values:

$$\ln(4) = \frac{E_a}{8.314}\left(\frac{1}{293} - \frac{1}{313}\right)$$

Rearranging for $E_a$:

$$E_a = \frac{8.314 \times \ln(4)}{\left(\frac{1}{293} - \frac{1}{313}\right)} \approx 52857.6 \text{ J/mol}$$

Converting to kJ/mol and rounding to the nearest integer gives $53$ kJ/mol.