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

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Chemical Kinetics

The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from

2 7^{\circ} C

4 2^{\circ} C

. Its energy of activation in

J / mol

is ............. (Take in

5 = 1.6094; R = 8.314 J mol^{- 1} K^{- 1}

)

Answer: 84297

The fraction of molecules past the threshold follows the Arrhenius form, so

ln \frac{k _{2}}{k _{1}} = \frac{E _{a}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})

with

\frac{k _{2}}{k _{1}} = 5

T_{1} = 300 K

and

T_{2} = 315 K

. Then

ln 5 = 1.6094 = \frac{E _{a}}{8.314} \cdot \frac{15}{300 \times 315}

, which gives

E_{a} = 1.6094 \times 8.314 \times \frac{300 \times 315}{15} \approx 84297 J mo l^{- 1}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer