Q16 / 90 · JEE Main 2013 (07 Apr) · Chemistry

The rate of a reaction doubles when its temperature changes from $300 K to 310 K$ . Activation energy of such a reaction will be:

$(R = {8.314 JK}^{-1} {mol}^{-1} and log 2 = 0.301)$

A
$58.5 kJ {mol}^{- 1}$
B
$60.5 kJ {mol}^{- 1}$
Correct answer
C
$53.6 kJ mol^{- 1}$
D
$48.6 kJ {mol}^{- 1}$

Solution

\log \frac{k_{2}}{k_{1}} = - \frac{E_{a}}{2 .303 R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})

log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})

log 2 = \frac{E_{a}}{2.303 \times 8.314} \times \frac{10}{300 \times 310}

E_{a} = \frac{0.301 \times 2.303 \times 8.314 \times 300 \times 310}{10} = 53603.9 J / mol

E_{a} = {53.6 kJ mol}^{- 1}

.

Hence, the correct answer is option: 'C'.

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