Question

The rate constant for the decomposition of hydrocarbons is 2.418 × 10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Answer 1

Hey Dear,

◆ Answer -

Pre-exponential factor = 3.9×10^12 s⁻¹

● Explaination -

# Given -

k = 2.418×10^-5 s⁻¹

Ea = 179.9 kJ/mol = 1.8×10^5 J/mol

# Solution -

We know the Arrhenius equation,

k = A.e^(-Ea/RT)

A/k = e^(Ea/RT)

Taking log on both sides -

log(A/k) = Ea / 2.303RT

log(A/k) = 1.8×10^5 / (2.303 × 8.314 × 546)

log(A/k) = 17.218

Taking antilog,

A/k = antilog(17.218)

A/k = 1.652×10^17

A = 1.652×10^17 × 2.418×10^-5

A = 3.9×10^12 s⁻¹

Thus, value of pre-exponential factor is 3.9×10^12 s⁻¹.

Hope this helped you. Keep asking..

Answer 2

Explanation:

`=>In k = In A - E_a/"RT"`

`=>log k = log A - E_a/(2.303 RT)`

`=> logA = log k + E_a/(2.303 RT)`

`=log(2.418 xx10^(-5) s^(-1))+ (179.9xx10^3 "J mol"^-1)/(2.303xx8.314 " JK"^(-1) mol^(-1)xx546K)`

= (0.3835 − 5) + 17.2082

= 12.5917