Question

The rate constant of a reaction at 33 ∘C was measured to be 6.0×10⁻² /s .

If the frequency factor is 1.2×1013/s, what is the activation barrier?

Answer 1

k = A x e^-Ea/RT

k = rate constant given 6.0×10−2 /s

T = temerature 33 ∘C + 273 = 306 K

Ea = activation energy

R = gas constant 8.314 x J/mol-K

A = frequency factor = 1.2×10¹³ s-1

put all these values in above equation

6.0×10⁻² = 1.2×10¹³ x e^{-(Ea/8.314 x 306 K)}

e^{-(Ea/2544.084)} = 6.0×10⁻² / 1.2×10¹³

e^{-(Ea/2544.084)} = 5 x 10¹¹

^{-(Ea/2544.084)} = ln (5 x 10¹¹)

-Ea /2544.084  = 26.9378

-Ea = 26.9378 x 2544.084

-Ea = 68532.0259

Ea = - 68532.0259 Joules

Ea = -68.532 kJ