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

Question

Quizplus · Accepted Answer

For a second-order reaction, the time taken for a certain percentage of reactant to disappear can be calculated using the formula: t = (1/k) * [(1/[A]_t) - (1/[A]_0)]. Here, k = 3.0 × 10^9 M^-1 s^-1, [A]_0 = 2.0 × 10^-5 M, and [A]_t = 0.1 × [A]_0 = 2.0 × 10^-6 M. Substituting these values gives t ≈ 1.5 × 10^-4 s.

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

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