Question

The rate constant for a first order reaction is given by the following equation:
$\ln k =33.24-\frac{2.0 \times 10^{4} K }{ T }$
The Activation energy for the reaction is given by _______$kJ\, mol ^{-1}$. (In Nearest integer)
(Given: $R =8.3 \,J \,K ^{-1}\, mol ^{-1}$ )

Answer 1

$\ln k =\ln A -\frac{ E _{ A }}{ RT }$
Given: $\ln k =33.24-\frac{2.0 \times 10^{4}}{ T }$
$\therefore $ on comparing $\frac{ E _{ A }}{ R }=2.0 \times 10^{4}$
$\therefore E _{ A }=2.0 \times 10^{4} \times R $
$\Rightarrow E _{ A }=2.0 \times 10^{4} \times 8.3\, J $
$\Rightarrow E _{ A }=16.6 \times 10^{4} J =166\, kJ$