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

According to Arrhenius equation

$log\frac{k_2}{k_1}=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$; given,

$[\therefore T_1=293K;T_2=313K;T_2-T_1=20K]$

$log\frac{4}{1}=\frac{E_a}{2.303\times \left(8.314Jmo{l}^{-1}{K}^{-1}\right)}\times \frac{20}{293\times 313}$

$0.6021=\frac{E_a}{2.303\times \left(8.314Jmo{l}^{-1}\right)}\times \frac{20}{293\times 313}$

$E_a=\frac{0.6021\times 2.303\times 8.314\times 293\times 313}{20}Jmo{l}^{-1}$

$=52863Jmo{l}^{-1}$

$E_a=52.863kJmo{l}^{-1}$