Question

The rate constant of a reaction is 6 × 10^-3 s^-1  at 50° and 9 × 10^-3  s^-1  at 100° C. Calculate the energy of activation of the reaction.

(a) 6.123 kJ mol^-1

(b) 8.124 kJ mol^-1

(c) 12.357 kJ mol^-1

(d) 18.256 kJ mol^-1

This question was posed to me during an internship interview.

I'd like to ask this question from Chemical Kinetics topic in division Chemical Kinetics of Chemistry – Class 12

Answer 1

The correct choice is (b) 8.124 kJ mol^-1

Easiest explanation: Given,

k1=6 × 10^-3s^-1                          T1=50 + 273=323 K

k2=9 × 10^-3s^-1                         T2=100 + 273=373 K

Substituting these values in the equation:

log (k2 / k1 ) = (Ea / 2.303 R) × ((T2 – T1) / T1 T2), we get

log (9 × 10^-3  s^-1 / 6 × 10^-3 s^-1  ) = ((Ea / (2.303 × 8.314)) × ((373 – 323) / (373 × 323))

log 9 / 6 = ((Ea / (2.303 × 8.314)) × (50 / (373 × 323))

Ea = 8.124 kJ mol^-1.