Q24 / 75 · JEE Main 2020 (04 Sep Shift 2) · Chemistry

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from $27^{o} C$ to $42^{o} C$ . Its energy of activation in $J / mol$ is _________ (Take $\ln 5 = 1.6094; R = 8.314 J {mol}^{- 1}$ )

Correct answer: 84297

Solution

$k = {Ae}^{-} \frac{Ea}{RT}$

$\ln (\frac{K_{2}}{K_{1}}) = \frac{Ea}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$\ln (5) = \frac{Ea}{8.314} [\frac{1}{300} - \frac{1}{315}]$

$1.6094 = \frac{Ea}{8.314} [\frac{15}{300 \times 315}]$

$Ea = 84297 J$

Hence, the correct answer is 84297.