Question

The rate constant for a first order reaction

is 2.80-10-35-1 at 60.0°C. The rate
constant at 45.0°C is 4.80.10-4 5-1. What
is the activation energy for the reaction?
[ ? ]kJ/mol
Recall that R = 8.314 J/mol K

Answer 1

E_a = 103.626 × 10³ KJ/mol

Step-by-step explanation:

Formula to solve this is given by;

Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))

Where;

k2 is rate constant at second temperature

k1 is rate constant at first temperature

R is universal gas constant

T1 is first temperature

T2 is second temperature

We are given;

k1 = 2.8 × 10^(-3) /s

k2 = 4.8 × 10^(-4) /s

R = 8.314 J/mol.k

T1 = 60°C = 333.15 K

T2 = 45°C = 318.15 K

Thus;

Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))

We now have;

-0.76592 = -0.00000739121E_a

E_a = -0.76592/-0.00000739121

E_a = 103.626 × 10³ KJ/mol