Question 4.23: 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.

Question 4.23: 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
Given that
Activation energy , E_a = 179.9 kJ mol⁻¹ = 179.9 × 10³ J mol⁻¹
Use the formula of Arrhenius equation,

Plug the values in this formula we get
Log A = (0.3835 − 5) + 17.2082 = 12.5917
Take antilog both sides we get
A = 3.9 × 10¹² s⁻¹ (approx)