Question

Given that $K(se{c}^{-1}=5\times {10}^{14}{e}^{-124080/RT})$, where activation energy is expressed in joule. The temperature at which reaction has $t_{1/2}$ equal to 25 minute. Assume Ist order reaction in ${}^{o}C$ is

Answer 1

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{25\times 60}=4.62\times {10}^{-4}se{c}^{-1}$......(1)
Also $K=5\times {10}^{14}{e}^{-124080/RT}$......(2)
From (1) and (2)
$4.62\times {10}^{-4}=5\times {10}^{14}{e}^{-124080/RT}$
$e^{-124080/RT}=9.24\times {10}^{-19}$
$-124080/8.314T=ln9.24\times {10}^{-19}=-41.53$
$T=359K=86{}^{o}C$