The second-order reaction 2 Mn(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 rate law for the reaction is:

Rate = k[Mn(CO)S1U1B15S1U1B0]S1U1P2[S1U1B0]S1U1P1

where k = 3.0 × 10S1U1P19S1S1P0 L molS1U1P1-1S1S1P0 sS1U1P1-1S1S1P0.

The half-life for a second-order reaction is:

tS1U1P1/2S1S1P0 = 1 / k[Mn(CO)S1U1B15S1U1B0]S1U1P1

We can use this equation to find the time it takes for 90% of the reactant to disappear:

ln ([Mn(CO)S1U1B15S1U1B0]S1U1P1 / [Mn(CO)S1U1B15S1U1B0]S1U1B0]) = -kt

0.9[Mn(CO)S1U1B15S1U1B0]S1U1B0 = [Mn(CO)S1U1B15S1U1B0]S1U1P1(1 - 0.1)

[Mn(CO)S1U1B15S1U1B0]S1U1B0 / [Mn(CO)S1U1B15S1U1B0]S1U1P1 = 9

ln 9 = - k t

t = ln 9 / k

t = ln 9 / (3.0 × 10S1U1P19S1S1P0 L molS1U1P1-1S1S1P0 sS1U1P1-1S1S1P0 × 2.0 × 10S1U1P1-5S1S1P0 mol LS1U1P1-1S1S1P0) = 1.5 × 10S1U1P1-S1S1P0S1U1P14S1S1P0 s

Therefore, the best choice is C.

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