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The rate constant of a reaction at 200K is 10 times less than rate constant at 400K. Value of $E_{a}$ ? [R = gas constant]

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1842.4 R

921.2 R

460.6 R

230.3 R

answer is B.

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Detailed Solution

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

$\log \frac{10}{1} = \frac{E_{a}}{2.303 R} [\frac{200}{200 \times 400}]$

$1 = \frac{E_{a}}{2.303 R \times 400}$

$E_{a} = 921.2 R$

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