A reactant (A) forms two products
$\begin{array}{l} A \overset{k_{1}}{\to} B, {Activation energy E}_{a 1} \\ A \overset{k_{2}}{\to} C, {Activation energy E}_{a 2} \end{array}$
If $E_{a_{2}} = 2 E_{a_{1}}$ then $k_{1}$ and $k_{2}$ are related as

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$k_{2} = k_{1} e^{E_{a 2} / RT}$

$k_{1} = 2 k_{2} e^{E_{a 2} / RT}$

$k_{1} = {Ak}_{2} e^{E_{a 1} / RT}$

$k_{1} = k_{2} e^{E_{a 1} / RT}$

answer is B.

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Detailed Solution

$k_{1} = A e^{- \frac{{Ea}_{1}}{RT}} k_{2} = A e^{- \frac{2 {Ea}_{1}}{RT}} \frac{k_{1}}{k_{2}} = e^{\frac{{Ea}_{1}}{RT}}$

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