A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 rain respectively. Star — Chemical Kinetics Chemistry Question
Question
A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 rain respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ____min (Round off to the Nearest Integer).
💡 Solution & Explanation
**Step 1: Set up first-order kinetics equations** For first-order decomposition: [A] = [A]₀e^(-kt) or [A] = [A]₀(1/2)^(t/t₁/₂) Given: - t₁/₂(A) = 54.0 min - t₁/₂(B) = 18.0 min - Initial: [A]₀ = [B]₀ = C (equimolar) - Find: t when [A] = 16[B] **Step 2: Express concentrations at time t** [A] = C(1/2)^(t/54) [B] = C(1/2)^(t/18) **Step 3: Apply the condition [A] = 16[B]** C(1/2)^(t/54) = 16 × C(1/2)^(t/18) (1/2)^(t/54) = 16 × (1/2)^(t/18) **Step 4: Simplify using exponent rules** (1/2)^(t/54) = 2⁴ × (1/2)^(t/18) (1/2)^(t/54) = (1/2)^(-4) × (1/2)^(t/18) (1/2)^(t/54) = (1/2)^(t/18 - 4) **Step 5: Equate exponents** t/54 = t/18 - 4 t/54 - t/18 = -4 t(1/54 - 1/18) = -4 t(1/54 - 3/54) = -4 t(-2/54) = -4 t = 4 × 27 = 108 min Therefore, the answer is 108.00.