For a gaseous reaction: A(g) → Products (order = n), the rate may be expressed as: (i) -dPA/dt = K1 (PA)^n (ii) -1/V dnA/dt = K2 (CA)^n. The rate constants, K1 and K2 are related as (PA and CA are the

Question

For a gaseous reaction: A(g) → Products (order = n), the rate may be expressed as: (i) -dPA/dt = K1 (PA)^n (ii) -1/V dnA/dt = K2 (CA)^n. The rate constants, K1 and K2 are related as (PA and CA are the partial pressure and molar concentration of A at time 't', respectively)

Prashant Kotian · Accepted Answer

Correct answer: D. Ideal gas law gives PA = CA * RT. Differentiating gives dPA/dt = RT * dCA/dt. Substituting into the first rate equation: RT * (-dCA/dt) = K1 (CA * RT)^n. Since -dCA/dt = K2 (CA)^n, we get RT * K2 (CA)^n = K1 (RT)^n (CA)^n. Thus, K2 = K1 (RT)^(n-1). Therefore, correct answer is D.

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