Kinetic data for the reaction 2A + B - Products: Experiment I: [A] = 0.10 mol L⁻¹, [B] = 0.20 mol L⁻¹, initial rate = 6.93 x 10⁻³ mol L⁻¹ min⁻¹ Experiment II: [A] = 0.10 mol L⁻¹, [B] = 0.25 mol L⁻¹, initial rate = 6.93 x 10⁻³ mol L⁻¹ min⁻¹ Experiment III: [A] = 0.20 mol L⁻¹, [B] = 0.30 mo

Question

Kinetic data for the reaction 2A + B -> Products: Experiment I: [A] = 0.10 mol L⁻¹, [B] = 0.20 mol L⁻¹, initial rate = 6.93 x 10⁻³ mol L⁻¹ min⁻¹ Experiment II: [A] = 0.10 mol L⁻¹, [B] = 0.25 mol L⁻¹, initial rate = 6.93 x 10⁻³ mol L⁻¹ min⁻¹ Experiment III: [A] = 0.20 mol L⁻¹, [B] = 0.30 mol L⁻¹, initial rate = 1.386 x 10⁻² mol L⁻¹ min⁻¹ The time (in minutes) required to consume half of A is:

Accepted Answer

10. Comparing I and II shows rate is independent of [B] (order 0); comparing I and III shows rate doubles when [A] doubles (order 1 in A), so it is first order with k = rate/[A] = 6.93x10⁻³/0.10 = 0.0693 min⁻¹, giving t(1/2) = 0.693/0.0693 = 10 min.

Solution

Related JEE Advanced Chemistry questions