Question

Rate constant for the decomposition of ethylene oxide into $C{H}_4$ and $CO$ may be described by the equation, $\log \left[k\left(s^{-1}\right)\right]=14.34-\frac{1.25\times {10}^4}{T}$.
(a) What is the energy of activation of this reaction?
(b) What is the value of $k$ at $670$ $K$?

Answer 1

(a) We know that,
${\log }_{10}k={\log }_{10}A-\frac{E}{2.303RT}$ ...(i)
${\log }_{10}k\left(s^{-1}\right)=14.34-\frac{1.25\times {10}^4}{T}$ ...(ii)
Comparing equation (i) and equation (ii), we get
$\frac{E}{2.303R}=1.25\times {10}^4$
$E=1.25\times {10}^4\times 2.303\times 8.314\times {10}^{-3}$
$E=239.339kJ/mol$
(b) Substituting the value of $T$, i.e., $670$ $K$, in equation ?(ii), we get,
$\log k\left(s^{-1}\right)=14.34-\frac{1.25\times {10}^4}{670}=-4.3167$
$k=4.82\times {10}^{-5}{s}^{-1}$