The activation energy for a reaction is $9.0 k c a l / m o l$ . The increase in the rate constant when its temperature is increased from $298 K$ to $308 K$ is:

63

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50

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100

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10

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Solution

The correct option is A $63$ %
$2.303 log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$log \frac{K_{2}}{K_{1}} = \frac{9.0 \times 10^{3}}{2.303 \times 2} [\frac{308 - 298}{308 \times 298}]$

$\frac{K_{2}}{K_{1}} = 1.63; K_{2} = 1.63 K_{1}; \frac{1.63 K_{1} - K_{1}}{K_{1}} \times 100 = 63.0 %$

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