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)

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) (A) 6 (B) 4 (C) 8 (D) 12

Tardigrade Admin · Accepted Answer

k1=Ae-Ea1/RT k2=Ae-Ea2/RT (k2/k1)=e(1/RT)(Ea1-Ea2) In(k2/k1)= (Ea1-Ea2/RT) =(10 × 103/8.314 × 300) ≈ 4

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

6

4

8

12

k1​=Ae−Ea1​​/RT k2​=Ae−Ea2​​/RT k1​k2​​=eRT1​(Ea1​−Ea2​​​) Ink1​k2​​=RTEa1​​−Ea2​​​ =8.314×30010×103​≈4

$k_{1} = A e^{- E_{a_{1}} / RT}$

$k_{2} = A e^{- E_{a_{2}} / RT}$

$\frac{k _{2}}{k _{1}} = e^{\frac{1}{RT} (E_{a_{1} - E_{a_{2}}})}$

In $\frac{k _{2}}{k _{1}} = \frac{E _{a_{1}} - E _{a_{2}}}{RT}$

$= \frac{10 \times 1 0 ^{3}}{8.314 \times 300} \approx 4$