Question

The rate of a reaction becomes four times when the temperature changes from 293K to 313 K. Calculate the energy of activation $\left(E_a\right)$ of the reaction assuming that it does not change with temperature.
$[R=8.314J{K}^{-1}mo{l}^{-1},log4=0.6021]$

Answer 1

Let $k_1=k,T_1=293K$
$\therefore k_2=4k,T_2=313K$
$log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$log\frac{4k}{k}=\frac{E_a}{2.303\times 8.314}\left[\frac{313-293}{293\times 313}\right]$
$log4=\frac{E_a}{2.303\times 8.314}\left[\frac{20}{91709}\right]$
$0.6021=\frac{E_a\times 20}{2.303\times 8.314\times 91709}$
$\therefore E_a=\frac{6021\times 2.303\times 8.314\times 91709}{20}$
$=52863Jmo{l}^{-1}$
$=52.863kJmo{l}^{-1}$