Question

A certain reaction has an activation energy of 66.0 kJ/mol and a frequency factor of A1 = 8.30×10¹² M−1s−1 . What is the rate constant, k, of this reaction at 27.0 ∘C ?

An unknown reaction was observed, and the following data were collected:

T (K)	k (M−1⋅s−1)
352	109
426	185

Determine the activation energy for this reaction.

Answer 1

1)
Given:
T = 27.0 oC
=(27.0+273)K
= 300.0 K
A = 8.3*10^12 M-1.s-1
Ea = 66.0 KJ/mol
= 66000.0 J/mol
R = 8.314 J/mol.K

use:
K = A*e^(-Ea/RT)
= 8.3*10^12*e^(-66000.0/(8.314*300.0))
= 8.3*10^12*e^(-26.4614)
= 8.3*10^12*3.221*10^-12
= 26.73 M-1.s-1
Answer: 26.7 M-1.s-1

2)
Given:
T1 = 352 K
T2 = 426 K
K1 = 1.09*10^2 M-1.s-1
K2 = 1.85*10^2 M-1.s-1

use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(1.85*10^2/1.09*10^2) = ( Ea/8.314)*(1/352 - 1/426)
0.529 = (Ea/8.314)*(4.935*10^-4)
Ea = 8912 J/mol
Ea = 8.9124 KJ/mol
Answer: 8.91 KJ/mol