The rate of constant of a first order reaction is $6.4 \times 10^{- 5} S^{- 1}$ at $47^{\circ} C$ and $1.6 \times 10^{- 5} S^{- 1}$ at $27^{\circ} C .$ . The activation energy for the reaction is around $(\log 4 = 0.6021, R = 8.314 {jK}^{- 1} {mol}^{- 1})$

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$110.0 kJ {mol}^{- 1}$

$27.0 kJ {mol}^{- 1}$

$55.0 kJ {mol}^{- 1}$

$220.0 kJ {mol}^{- 1}$

answer is C.

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Detailed Solution

$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$\log \frac{6.4 \times 10^{- 5}}{1.6 \times 10^{- 5}} = \frac{E_{a}}{2.303 \times 8.314} [\frac{1}{300} - \frac{1}{320}]$

$\log 4 = \frac{E_{a}}{2.303 \times 8.314} [\frac{20}{300 \times 320}]$

$E_{a} = 55336.8 J / mole = 55.34 kJ / mole .$

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