Question

It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be _________kJ mol-1. [nearest integer] (Given ln10 = 2.3, R = 8.3 J K-1 mol-1, log 2 = 0.30)

Answer 1

The correct answer is 59
T1 = 300 K (Rate constant)
K2 = 2K1, on increase temperature by 9K
T2 = 309 K
Ea = ?
$log\frac{K_2}{K_1}=\frac{Ea}{2.3R}[\frac{T_2−T_1}{T_2.T_1}]$
$log 2=\frac{Ea}{2.3×8.3}[\frac{9}{309×300}]$
$Ea=\frac{0.3×309×300×2.3×8.3}{9}$
= 58988.1 J/mole ≃ 59 kJ/mole