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Choose the correct rate law for the elementary reversible reaction 3A ⇌ B + 2C.(a) –rA = kCA^3 – CBCC^2/KC(b) –rA = k[CA^3 – CBCC^2KC](c) –rA = k[CA^3 – CBCC^2]/KC(d) –rA = k[CA^3 – CBCC^2/KC]The question was posed to me in an online interview.I would like to ask this question from Reversible Reactions Rate Laws topic in portion Rate Laws and Stoichiometry of Chemical Reaction Engineering

Answer»

Correct choice is (d) –rA = K[CA^3 – CBCC^2/KC]

EASIEST explanation: Let’s say the rate constant for the forward reaction is k1 and for the backward reaction it is k2.

Rate of disappearance of A = k1CA^3

Rate of formation of A = k2CBCC^2

-rA = k1CA^3 – k2CBCC^2. We KNOW that KC = k1/k2. So, -rA = k[CA^3 – CBCC^2/KC].



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