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The rate constant of a reaction at 200K is 10 times less than the rate constant at 400K what is the activation energy $(E a)$ of the reaction (R = gas constant)

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921.2 R

460.6 R

230.3 R

1842.4 R

answer is B.

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Detailed Solution

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

$T_{1} = 200 T_{2} = 400$

$K_{1} = \frac{1}{10} K_{2} K_{2}$

$log \frac{K_{2}}{\frac{1}{10} K_{2}} = \frac{Ea}{2.303R} [\frac{1}{200} - \frac{1}{400}]$

$log 10 = \frac{Ea}{2.303R} \frac{200}{200 × 400}$

$E a = 2.303 × 400 R$

= 921.2 R

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