For the equilibrium: A(g) ⇌ nB(g), the equilibrium constant, K_p, is related with the degree of diss — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Let initial moles of A be 1. At equilibrium, moles of A = 1-α, moles of B = nα. Total moles = 1-α+nα = 1+(n-1)α. Partial pressures: P_A = [(1-α) / (1+(n-1)α)] P, P_B = [(nα) / (1+(n-1)α)] P. K_p = (P_B)ⁿ / P_A = [(nα P) / (1+(n-1)α)]ⁿ / [(1-α)P / (1+(n-1)α)]. Simplifying gives K_p = (nα)ⁿ Pⁿ⁻¹ / [(1-α)(1+(n-1)α)ⁿ⁻¹]. Therefore, correct answer is A.