Question

The rate constant of a first order reaction is 6 times when the temperature is increased from 350 K to 410 K. The energy of activation for the reaction is ($inkJmo{l}^{-1}$). Write it in 10x form.( write your answer to nearest integer)

Answer 1

$k_2=6k_{1}or\frac{k_2}{k_1}=6$
log $\frac{k_2}{k_1}\frac{E_a}{2.3R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
log 6 $=\frac{E_a}{2.3\times 8.314J{K}^{-1}mo{l}^{-1}}\left[\frac{410-350}{410\times 350}\right]K$.
log 3 = log 2 $=\frac{E_a}{2.3\times 8.314J{K}^{-1}mo{l}^{-1}}\times \left(\frac{60}{410\times 350}\right)K$
$0.48+0.3=\frac{E_a}{45733.92}$
$\therefore E_a=35672.4Jmo{l}^{-1}=35.67kJmo{l}^{-1}$