Question

The slope of the line for the graph of $logk$versus$\frac{1}{T}$ 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: $\left(kJ{k}^{-1}mo{l}^{-1}\right)$

• A. $95.7$
• B. $9.57$
• C. $957$
• D. none

Answer 1

The correct option is A $95.7$

As we know,
$k=A{e}^{-Ea/RT}$
$lnk=lnA-\frac{Ea}{RT}$
$2.303logk=-\frac{Ea}{RT}+2.303logA$
$logk=-\frac{Ea}{2.303RT}+logA$
$slope\frac{Ea}{2.303R}=5000$
$Ea=5000\times 8.31\times 2.303=95.7kJ{k}^{-1}mo{l}^{-1}$