Question

The rate constant of the first-order reaction, i.e., decomposition of ethylene oxide into $C{H}_4$ and $CO$ may be described by the following equation: $\log k{s}^{-1}=14.34$ $-\frac{1.25\times 10^4}{T}K$
Find the energy of activation (in $kJ/mole$ ). Report answer till the nearest integer.

Answer 1

Answer: 239

$\log k=14.34-\frac{1.25\times 10^4}{T}K$
Comparing with the standard form of Arrhenius equation:
$\log k=\log A-\frac{Ba}{2.303RT}$
$\frac{Ea}{2.303R}=1.25\times 10^4$
$Ea=\frac{2.303\times 8.314\times 1.25\times 10^4}{1000}kJ/mole$
$Ea=239.34kJ/mole$