devarshi-dt-logo

Question:

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

12

6

4

8

Solution:

The correct option is C 4
We know
K = Ae^(-Ea/RT)
K1 = Ae^(-Ea1/RT)
K2 = Ae^(-Ea2/RT)
K2/K1 = e^(Ea1 - Ea2)/RT
ln(K2/K1) = (Ea1 - Ea2)/RT = 10 × 1000 / (8.314 × 300) = 4.00
The correct option is C.