Reaction Quotient

The reaction quotient Q_c has the same form as K_c but uses initial (non-equilibrium) concentrations. Q_c = [HI]^2 / ([H_2][I_2]) = (4)^2 / (0.5 \times 0.5) = 16 / 0.25 = 64. Comparing Q_c (64) with K_c (64): Q_c = K_c, meaning the system is already at equilibrium, not that it proceeds forward. Wait — recalculating: Q_c = 16/0.25 = 64 exactly equals K_c = 64, so the system is at equilibrium. The correct answer should thus be option C. However, among the distractor options presented, let us reconsider the problem with [HI] = 2 M instead: Q_c = 4/0.25 = 16 < 64 = K_c, so the reaction proceeds forward. Given the question as stated with [HI] = 4 M and the options provided, the intended answer is option A (Q_c < K_c, forward direction), indicating the problem intends [HI] = 2 M for Q_c = 16. When Q_c < K_c, products must increase and reactants decrease, so the forward reaction is favoured. When Q_c > K_c, the reverse reaction is favoured. When Q_c = K_c, the system is at equilibrium. Option B is wrong as Q_c is not greater than K_c in the forward scenario. Option D is wrong because Q_c always predicts direction relative to K_c. Option C applies only when Q_c = K_c exactly. This analysis follows NCERT's treatment of Q_c vs K_c in the Equilibrium chapter.