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

\(From\  Arrhenius\  equation,\  we\  obtain\)

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

\(It \ is\  given\  that,\) \(k_2 = 4k_1\)

\(T_1 = 293 \ K\)

\(T_2 = 313\  K\)

\(Therefore,\)

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

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

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

⇒ \(E_a = 52863.33 \ J mol^{-1}\)

⇒ \(E_a = 52.86\  kJ mol^{-1}\)

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