2A →[k] B is a zero-order reaction with k=1.0 mol L⁻¹min⁻¹. If the initial concentration of A is 2 M, the time taken to complete 75% of the reaction is:

Solution & Explanation

Concept: zero-order kinetics. For a zero-order reaction the rate is independent of concentration, so concentration falls linearly with time. For the reaction 2A → B written with rate = k: rate = -(1/2) d[A]/dt = k, so [A]₀ - [A]_t = 2kt. (Equivalently, A is consumed at 2k per unit time.) Initial [A]₀ = 2 M. After 75% completion, the amount of A reacted = 0.75 × 2 = 1.5 M, leaving [A]_t = 0.5 M. So [A]₀ - [A]_t = 2 - 0.5 = 1.5 = 2kt. With k = 1.0 mol L⁻¹ min⁻¹: t = 1.5 / (2 × 1.0) = 0.75 min. Hence the time for 75% reaction is 0.75 min — answer (B). Trap: the stoichiometric 2 in 2A → B doubles the consumption rate of A; missing it gives t = 1.5 min (a distractor).