Consider two different first order reactions $R_{1} and R_{2}$ . The pre-exponential factor of $R_{1}$ is twice that of R₂Activation energy of $R_{2}$ is lesser that of $R_{1}$ by 700 Cal/mol. The temperature at which rate constant of $R_{1} and R_{2}$ are equal, will be: [Given ln 2=0.7]

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Detailed Solution

For Reaction $R_{1} : A_{1} : E_{a_{1}} : K_{1}$
For Reaction $R_{2} : A_{2} : E_{a_{2}} : K_{2}$
Given: $: A_{1} = 2 A_{2} : E_{a_{2}} = E_{a_{1}} - 700$
$\begin{aligned} K_{1} = K_{2} \\ A_{1} e^{- E_{a_{1} / RT}} = A_{2} e^{- E_{a_{2} / RT}} \\ 2 A_{2} e^{- E_{a_{1} / RT}} = A_{2} e^{- (E_{a_{1}} - 700) / RT} \\ \frac{e^{- E_{a_{1}} / RT}}{e^{- (E_{a_{1}} - 700) / RT} = \frac{1}{2}} \end{aligned}$
$\begin{aligned} e^{- 700 / RT} = \frac{1}{2} \\ - \frac{700}{RT} = - \ln 2 \\ RT = \frac{700}{0.7} = 1000 \\ T = 500 K \end{aligned}$