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

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.

Philoid · Accepted Answer

Given,

k = 2.418 × 10^-5 s^-1

T = 546 K

E_a = 179.9 kJ mol^-1 = 179.9 × 10³J mol^-1

The Arrhenius equation is given by k = Ae^-Ea/RT

Taking natural log on both sides,

Ln k = ln A-(E_a/RT)

Substituting the values,

ln(2.418 × 10^-5 ) = ln A-179.9/(8.314 × 546)

ln A = 12.5917

A = 3.9 × 10¹² s^-1(approximately)

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

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