Question

The slope of the line for the graph of $logkversus\frac{1}{T}$ for the reaction,

$N_2{O}_5\to 2N{O}_2+\frac{1}{2}{O}_2$ is $-5000.$

Calculate the energy of activation of the reaction $\left(kJ{K}^{-1}mo{l}^{-1}\right)$.

Write the value of nearest integer.

Answer 1

$k=A{e}^{-Ea/RT}$

$lnk=lnA-\frac{Ea}{RT}$

$2.303logk=-\frac{Ea}{RT}+2.303logA$

$logk=-\frac{Ea}{2.303RT}+logA$

$slope=-\frac{Ea}{2.303R}=-5000$

$Ea=5000\times 8.31\times 2.303=95.73kJ{K}^{-1}mo{l}^{-1}$