Answer:
E_a = 103.626 × 10³ KJ/mol
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
