Question

The rate constant of a reaction at 32∘C is 0.055 s−1. If the frequency or pre-exponential factor is 1.2 × 1013 s−1, what is the activation energy?

Answer 1

Answer : The activation energy is, 83.71 kJ

Solution :

Using Arrhenius equation,

$K=a\times e^{\frac{-Ea}{RT}}$

Taking 'ln' on both the sides, we get

$\ln K=\frac{-Ea}{RT}+\ln a$

where,

K = rate constant = $0.055s^{-1}$

Ea = activation energy = ?

T = temperature = $32^oC=273+32=305K$

R = gas constant = 8.314 J/mole/K

a = Arrhenius constant  = $1.2\times 10^{13}s^{-1}$

Now put all the given values in the above formula, we get the value of activation energy.

$\ln (0.055s^{-1})=\frac{-Ea}{(8.314J/moleK)\times (305K)}+\ln (1.2\times 10^{13}s^{-1})$

$Ea=83.71kJ$

Therefore, the activation energy is, 83.71 kJ