The second-order reaction 2Mn(CO)₅ → Mn₂(CO)₁₀ has a rate constant equal to 3.0 × 10⁹ L mol^-1 s^-1 at 25 °C. If the initial concentration of Mn(CO)₅ is 2.0 × 10^-5 mol L^-1, how long will it take for 90.% of the reactant to disappear?

Question

Quizplus · Accepted Answer

The integrated rate law for a second-order reaction is:1/[A]t - 1/[A]0 = ktwhere [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is time.We want to find the time it takes for 90% of the reactant to disappear, so [A]t = 0.1[A]0. Substituting this into the integrated rate law and solving for t, we get:t = (1/[A]0 - 1/0.1[A]0) / k= (1 - 10) / (3.0 × 10^9 L mol^-1 s^-1 * 2.0 × 10^-5 mol L^-1)= 1.5 × 10^-4 s

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The Second-Order Reaction 2Mn(CO)₅ → Mn₂(CO)₁₀ Has a Rate Constant