What is the activation energy and rate constant?

Question

A reaction has a rate constant of 1.23×10⁻² /s/s at 400 K and 0.668 / s at 450 K.

Determine the activation barrier for the reaction.

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

Answer 1

k1= 1.23×10^-2 K2= 0.668 T2=450k

T1=400k

As ln(K2/K1) = - Ea/R (1/T2- 1/T1)

1.19× 10^5

In 425k

as ln(k2/k) = -Ea/R (1/T2- 1/T1)

ln (k2)0.0123)= - 1.19×10^5 /8.314 (1/425- 1/400)

K2= 0.100