Question

A reactant $\left(A\right)$ forms two products
$A \xrightarrow{k_1} B$ Activation energy $E_{a1}$
$A \xrightarrow{k_2} B$ Activation energy $E_{a2}$
If $E_{a2} = 2E_{a1}$ then $k_1$ and $k_2$ are related as

• A. $k_{1}=2k_{2}e^{E_{a2/RT}}$
• B. $k_{1}=k_{2}e^{E_{a1/RT}}$
• C. $k_{2}=k_{1}e^{E_{a2/RT}}$
• D. $k_{1}=A k_{2}e^{E_{a1/RT}}$

Answer 1

Answer: (B)

$A \xrightarrow{k_1} B, A \xrightarrow{k_2} B$
By Arrhenius equation, $k_{1}=Ae^{E_{a1/RT}}$
and $k_{2}=Ae^{E_{a2/RT}}$
[A' is Arrhenius constant]
$\because\frac{k_{1}}{k_{2}}=\frac{Ae^{-E_{a1}/RT}}{Ae^{-2E_{a_1/RT}}}=e^{E_{a1/RT}}$
$\therefore k_{1}=k_{2}e^{E_{a1/RT}}$