Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln (k2/k1) is equal to : (R = 8.314 J mol–1 K

Engineering

Chemistry

Arrhenius Equation

Question

Two reactions R₁ and R₂ have identical pre-exponential factors. Activation energy of R₁ exceeds that of R₂ by 10 kJ mol^–1. If k₁ and k₂ are rate constants for reactions R₁ and R₂ respectively at 300 K, then ln (k₂/k₁) is equal to : (R = 8.314 J mol^–1 K^–1)

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$Given [E_{a_{1}} - E_{a_{2}} = 10] K_{1} = A . e^{- E_{a_{1}} / RT}$

$\frac{K_{2}}{K_{1}} = e^{- (E_{a_{2}} - E_{a_{1}}) / RT} K_{2} = A . e^{- E_{a_{2}} / RT}$

$\ln (\frac{K_{2}}{K_{1}}) = - E_{a_{2}} + E_{a_{1}} / RT$

$= (\frac{10 \times 1000}{8 .314 \times 300}) = 4$