# 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) ? with Answer - Chemical Engineering Mcqs | TestPrep

> Solved MCQ: 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) ? Correct Answer: C. E + ΔHA – ΔHR). Verified answer explanation for FPSC, PPSC, MDCAT, NTS, and CSS competitive exam preparation.

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# 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) ?

Direct Answer & Quick Reference

The correct option is **C. E + ΔHA – ΔHR)**.

A. E

B. E + ΔHA

C. E + ΔHA – ΔHR) **(Correct Answer)**

D. ΔHA + ΔHR

✨ Explain with AI

Verified Key: C. E + ΔHA – ΔHR)
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