Question

The rate of a reaction quadruples when the temperature changes from $293 \ K$ to $313\ K$. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Answer 1

$From\ Arrhenius\ equation,\ we\ obtain$

$log \ \frac {k_2}{k_1} = \frac {E_a}{2.303\ R} (\frac {T_2-T_1}{T_1T_2})$

$It \ is\ given\ that,$ $k_2 = 4k_1$

$T_1 = 293 \ K$

$T_2 = 313\ K$

$Therefore,$

$log \ \frac {4k_1}{k_1} = \frac {E_a}{2.303 \times 8.314} (\frac {313-293}{293 \times 313})$

⇒ $0.6021 = \frac {20\times E_a}{2.303\times8.314\times293\times313}$

⇒ $E_a = \frac {0.6021\times 2.303\times 8.314\times293x313}{20}$

⇒ $E_a = 52863.33 \ J mol^{-1}$

⇒ $E_a = 52.86\ kJ mol^{-1}$

$Hence,\ the\ required\ energy\ of\ activation \ is$ $52.86\ kJ mol^{-1}$.