Question

Based on the data in the following table, what will be the value of k at 655 K?

T (K) k (s−1)

625 1.1 × 10^−4

635 1.5 × 10^−4

645 2.0 × 10^−4

Answer 1

use,
ln(k2/k1) = (Ea/R)*{1/T1 - 1/T2}
apply this equation for T1 = 625 K and T2 = 635 K
ln{(1.5*10^-4)/(1.1*10^-4)} = (Ea/R)*{1/625 - 1/635}
ln1.36 = (Ea/R)*2.5*10^-5
0.31 = (Ea/R)*2.5*10^-5
Ea/R = 1.24*10^-6

ln(k2/k1) = (Ea/R)*{1/T1 - 1/T2}
apply this equation for T1 = 645 K and T2 = 655 K
ln{K2/(2.0*10^-4)} = (Ea/R)*{1/645 - 1/655}
put value of Ea/R
ln{K2/(2.0*10^-4)} = (1.24*10^-6)*2.4*10^-5
ln{K2/(2.0*10^-4)} = 0.298
K2/(2.0*10^-4) = exp(0.298)
K2/(2.0*10^-4) = 1.35
K2 = 1.35*2.0*10^-4
K2 = 2.7*10^-4

Answer : 2.7*10^-4