The rate of a reaction decreased by 3.555 times when the temperature was changed from $40^{\circ} \mathrm{C}$ to $30^{\circ} \mathrm{C}$. The activation energy (in $\mathrm{kJ} \mathrm{mol}^{-1}$) of the reaction is ____.

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Chemical Kinetics

The rate of a reaction decreased by 3.555 times when the temperature was changed from

4 0^{\circ} C

3 0^{\circ} C

. The activation energy (in

kJ mol^{- 1}

) of the reaction is ____.

Answer: 100

Use the two-temperature Arrhenius relation

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]

. The rate falls by a factor

3.555

on cooling from

4 0^{\circ} C

(

313 K

) to

3 0^{\circ} C

(

303 K

), so

\frac{k _{2}}{k _{1}} = 3.555

with

lo g 3.555 = 0.5508

. Then

0.5508 = \frac{E _{a}}{2.303 \times 8.314} \cdot \frac{313 - 303}{303 \times 313}

, which solves to

E_{a} \approx 1.0 \times 1 0^{5} J mol^{- 1} = 100 kJ mol^{- 1}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Numerical answer