Question

In a decompostion reaction, the reaction decays 21.2 times more rapidly at 22°C then at 4 °C. What is the overall activation energy (kJ/mol) of the process ?

Answer 1

Explanation:

The given data is as follows.

      $T_{1} = 4^{o}C$ = (4 + 273) K = 277 K

      $T_{2} = 22^{o}C$ = (22 + 273) K = 295 K

      $\frac{k_{2}}{k_{1}}$ = 21.2

Formula to calculate activation energy will be as follows.

   $ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

   $ln (21.2) = \frac{E_{a}}{8.314 J/mol K} [\frac{1}{277} - \frac{1}{295}]$

      3.05 = $\frac{E_{a}}{8.314 J/mol K} \times \frac{18}{81715}$

          $E_{a}$ = 115116.914 J/mol

or                     = 115.116 kJ/mol

Thus, we can conclude that overall activation energy (kJ/mol) of the process is 115.116 kJ/mol.