The rate expression for a heterogeneous catalytic reaction is given by, – rA = K.KA PA(1 + KA.PA + KR.PR), where K is surface reaction rate constant and KA and KR are absorption equilibrium constants of A and R respectively. If KR PR (1 + KA PA), the apparent activation energy EA is equal to (given E is the activation energy for the reaction and ΔHR and ΔHA are the activation energies of adsorption of R and A) ?

Question

The rate expression for a heterogeneous catalytic reaction is given by, – rA = K.KA PA(1 + KA.PA + KR.PR), where K is surface reaction rate constant and KA and KR are absorption equilibrium constants of A and R respectively. If KR PR >> (1 + KA PA), the apparent activation energy EA is equal to (given E is the activation energy for the reaction and ΔHR and ΔHA are the activation energies of adsorption of R and A) ?

Accepted Answer

E. Under the given condition (KR*PR >> 1 + KA*PA), the rate expression simplifies such that the apparent activation energy EA equals E.