Question

A chemical reaction has an energy of activation Ea = 1∙104 J mol-1 at T = 300 K. The first-order rate constant for this reaction was found to be 15.0 s-1. In the presence of a catalyst, the activation energy is reduced to 1∙103 J mol-1. Calculate the pre-exponential factor in the Arrhenius equation

Answer 1

Answer : The pre-exponential factor in the Arrhenius equation is, 14.99 s⁻¹

Explanation :

Using Arrhenius equation,

`K=A* e^{(-Ea)/(RT)}`

Taking ln on both the sides, we get:

`\ln K=(-Ea)/(RT)+\ln A`

where,

K = rate constant =
`15.0s^(-1)`

Ea = activation energy = 1.104 J/mol

T = temperature = 300 K

R = gas constant = 8.314 J/mole.K

A = Arrhenius constant

Now put all the given values in the above formula, we get the value of \ln a.

`\ln (15.0s^(-1))=(1.104J/mol)/((8.314J/mole.K)* (300K))+\ln A`

`\ln A=2.708s^(-1)`

`A=14.99s^(-1)`

Therefore, the pre-exponential factor in the Arrhenius equation is, 14.99 s⁻¹