Q. 24 The activation energy (Ea) values for two reactions carried out at 25°C differ by 5.0 kJ mol−1. If the pre-exponential factors (A1 and A2) for these two reactions are of the same magnitude, the ratio of rate constants (k1/k2) is ____ (up to two decimal places). (Given: Gas constant R = 8.314 J K−1 mol−1)
Arrhenius Equation Basics
The Arrhenius equation k = A e^(-Ea/RT) links rate constant k to pre-exponential factor A, activation energy Ea, gas constant R = 8.314 J K⁻¹ mol⁻¹, and temperature T in Kelvin.
Lower Ea yields higher k due to the exponential term, explaining why reactions with smaller energy barriers proceed faster.
At constant T and equal A, k ratio depends solely on Ea difference.
Step-by-Step Calculation
Assume reaction 1 has lower Ea (so k1 > k2): k1/k2 = e^( (Ea2 - Ea1)/RT ). Here, ΔEa = Ea2 - Ea1 = 5000 J/mol, T = 298 K.
Exponent calculation:
5000 / (8.314 × 298) = 2.016
Thus, k1/k2 = e^2.016 ≈ 7.50. The problem states “differ by” without specifying which is larger; convention takes k1/k2 > 1 for the faster reaction.
Common Options Explained
- 7.50: Correct if
Ea1 < Ea2by 5 kJ/mol; reaction 1 faster. - 0.13: Inverse (
k2/k1), if labels swapped. - 1.56: Possible miscalculation with
R = 2 cal/K molor temperature error; incorrect here. - e^1: Wrong exponent handling.
Exam Tips for Kinetics
Use log form for precision: log(k1/k2) = ΔEa/(2.303RT) ≈ 0.876, so ratio 10^0.876 ≈ 7.50.
