Question

The activation energy for a reaction is 25 $kJmo{l}^{-1}$. If the rate of the reaction should be increased 10 times, what should be the temperature to which it should be raised, if the initial temperature is 300 K?

• A. 389.6

Answer 1

The correct option is A 389.6
Let the rate constant at temperature T be $k_T$.

Given,

$\frac{k_T}{k_{300}}=10$

$log\frac{k_T}{k_{300}}$ = $\frac{Ea[T_2-T_1]}{\mathrm{2.303.}R.\left[T_1{T}_2\right]}$

Solving for $T_2$, we get,

$T_2=389.6K$