Question

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the higher temperature to f at the lower temperature

Answer 1

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Step-by-step explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

`Log((f_2)/(f_1) ) = (E_a)/(2.303 * R) [(1)/(T_1) -(1)/(T_2) ]`

Where;

`(f_2)/(f_1)` is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

`Log((f_2)/(f_1) ) = (E_a)/(2.303 * R) [(1)/(T_1) -(1)/(T_2) ]\\\\Log((f_2)/(f_1) ) = (185,000)/(2.303 * 8.314) [(1)/(505) -(1)/(525) ]\\\\Log((f_2)/(f_1) ) = 0.7289\\\\(f_2)/(f_1) = 10^(0.7289)\\\\(f_2)/(f_1) = 5.356`

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356