Question

An irreversible chemical reaction occurs on a porous catalyst. All the pores are of same size. In strong pore diffusion regime, the observed activation energy is 120 kJ mol^-1. The activation energy of diffusion is 10 kJ mol^-1. Assuming Arrhenius temperature dependency for both reaction and diffusion, which one of the following is the true activation energy (in kJ mol^-1) of the reaction?

• A. 65
• B. 110
• C. 130
• D. 230

Hint:

In the strong pore diffusion regime, E(observed) is the average of E(true) and E(diffusion); solve 120 = (E_true + 10)/2.

Answer 1

The Correct Option is D

Step 1: Recall the effectiveness factor behaviour in the strong pore diffusion limit.
For a first-order irreversible reaction in a porous catalyst pellet, the Thiele modulus is phi = L sqrt(k/D_e). When pore diffusion resistance is strong, the effectiveness factor simplifies to eta ~ 1/phi = (1/L) sqrt(D_e/k).
Step 2: Express the observed rate constant.
k_obs = eta k, which substitutes to k_obs = (1/L) sqrt(D_e k), so k_obs is proportional to the square root of the product of diffusivity and true rate constant.
Step 3: Apply Arrhenius dependence.
Let k = A exp(-E_true/RT) and D_e = A_D exp(-E_D/RT). Then k_obs is proportional to exp(-(E_D+E_true)/(2RT)), giving E_obs = (E_D + E_true)/2. This is the well known result that under strong pore diffusion control, the apparent activation energy is the average of the true kinetic activation energy and the diffusion activation energy.
Step 4: Solve for E_true.
Given E_obs = 120 kJ/mol and E_D = 10 kJ/mol: 120 = (10 + E_true)/2, so E_true = 230 kJ/mol.
Step 5: Sanity check.
Strong pore diffusion resistance suppresses the observed rate's temperature sensitivity, so the true activation energy must be larger than observed, confirming 230 as the only consistent value. \[ \boxed{E_{true} = 230\ \text{kJ mol}^{-1}} \]