Question

For first order gaseous reaction, $\log k$ when plotted against $\frac{1}{T}$ gives a straight line with a slope of $-8000K$. Calculate the activation energy of the reaction.

Answer 1

since we know arrhenius equation is given by $k=A{e}^{-E_a/RT}$
By taking log on both sides $\log k=\frac{-E_a}{RT}+\log A$
So by comparing with $y=mx+c$

$⟹m=\frac{-E_a}{R}$
$⟹-8000=\frac{-E_a}{8.314}$
$⟹E_a=66.5J$