Question

A reaction has a rate constant of 1.20×10⁻² s−1 at 400.0 K and 0.684 s−1 at 450.0 K.

A) What is activation energy?

B)What is the value of the rate constant at 425 K?

Answer 1

A)
Given:
T1 = 400 K
T2 = 450 K
K1 = 1.2*10^-2 s-1
K2 = 0.684 s-1

use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(0.684/1.2*10^-2) = ( Ea/8.314)*(1/400 - 1/450)
4.0431 = (Ea/8.314)*(2.778*10^-4)
Ea = 121010 J/mol
Ea = 121 KJ/mol
Answer: 121 KJ/mol

B)
Given:
T1 = 400 K
T2 = 425 K
K1 = 1.2*10^-2 s-1
Ea = 121.01 KJ/mol
= 121010 J/mol

use:
ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)
ln(K2/1.2*10^-2) = (121010.0/8.314)*(1/400 - 1/425.0)
ln(K2/1.2*10^-2) = 14555*(1.471*10^-4)
ln(K2/1.2*10^-2) = 2.14
(K2/1.2*10^-2) = e^(2.14)
(K2/1.2*10^-2) = 8.503
K2 = 0.102 s-1
Answer: 0.102 s-1