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

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Chemistry

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Question

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)

53.6 kJ mol^–1

58.5 kJ mol^–1

60.5 kJ mol^–1

48.6 kJ mol^–1

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$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2 .303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$0 .301 = \frac{E_{a}}{2 .303 \times 8 .314} [\frac{1}{300} - \frac{1}{310}]$

= 53.6 kJ/mol

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