Consider two different first order reactions R1 and R2 . The pre-exponential factor of R1 is twice that of R2 Activation energy of R2 is lesser that of R1 by 700 Cal/mol. The temperature at which rate constant of R1 and R2are equal, will be: [Given ln 2=0.7]

For Reaction R1:A1:Ea1:K1 For Reaction R2:A2:Ea2:K2Given: :A1=2A2:Ea2=Ea1−700K1=K2A1e−Ea1/RT=A2e−Ea2/RT2A2e−Ea1/RT=A2e−Ea1−700/RTe−Ea1/RTe−Ea1−700/RT=12e−700/RT=12−700RT=−ln⁡2RT=7000.7=1000T=500K

Consider two different first order reactions R1 and R2 . The pre-exponential factor of R1 is twice that of R2 Activation energy of R2 is lesser that of R1 by 700 Cal/mol. The temperature at which rate constant of R1 and R2are equal, will be: [Given ln 2=0.7]

For Reaction R1:A1:Ea1:K1 For Reaction R2:A2:Ea2:K2Given: :A1=2A2:Ea2=Ea1−700K1=K2A1e−Ea1/RT=A2e−Ea2/RT2A2e−Ea1/RT=A2e−Ea1−700/RTe−Ea1/RTe−Ea1−700/RT=12e−700/RT=12−700RT=−ln⁡2RT=7000.7=1000T=500K

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