The second-order reaction, 2 Mn(CO)5 → Mn2(CO)10 has a rate constant equal to 3.0 × 109 M-1s-1 at 25°C. If the initial concentration of Mn(CO)5 is 1.0 × 10-5 M, how long will it take for 90.% of the reactant to disappear?
A) 3.3 × 10-16 s
B) 3.7 × 10-15 s
C) 3.0 × 10-4 s
D) 3.0 × 103 s
The correct answer and explanation is :
To solve this problem, we need to apply the integrated rate law for a second-order reaction. The general form of the integrated rate law for a second-order reaction is:
$$
\frac{1}{[A]} – \frac{1}{[A_0]} = kt
$$
Where:
- $[A]$ is the concentration of the reactant at time $t$,
- $[A_0]$ is the initial concentration of the reactant,
- $k$ is the rate constant,
- $t$ is the time it takes for the reaction to proceed.
Step 1: Identify the given values
- Initial concentration, $[A_0] = 1.0 \times 10^{-5} \, M$
- Rate constant, $k = 3.0 \times 10^9 \, M^{-1}s^{-1}$
- We are asked to find the time $t$ for 90% of the reactant to disappear, meaning that 10% of the original reactant remains. Therefore, the final concentration will be:
$$
[A] = 0.1 \times [A_0] = 0.1 \times 1.0 \times 10^{-5} \, M = 1.0 \times 10^{-6} \, M
$$
Step 2: Substitute into the integrated rate law
We can now substitute the known values into the integrated rate law:
$$
\frac{1}{[A]} – \frac{1}{[A_0]} = kt
$$
$$
\frac{1}{1.0 \times 10^{-6}} – \frac{1}{1.0 \times 10^{-5}} = (3.0 \times 10^9) \times t
$$
Simplifying the left side:
$$
\frac{1}{1.0 \times 10^{-6}} = 1.0 \times 10^6
$$
$$
\frac{1}{1.0 \times 10^{-5}} = 1.0 \times 10^5
$$
So the equation becomes:
$$
(1.0 \times 10^6) – (1.0 \times 10^5) = (3.0 \times 10^9) \times t
$$
$$
9.0 \times 10^5 = (3.0 \times 10^9) \times t
$$
Step 3: Solve for $t$
$$
t = \frac{9.0 \times 10^5}{3.0 \times 10^9} = 3.0 \times 10^{-4} \, s
$$
Step 4: Conclusion
The time it takes for 90% of Mn(CO)5 to disappear is $3.0 \times 10^{-4} \, s$, so the correct answer is:
C) 3.0 × 10^-4 s
This answer is derived from the application of the second-order rate law, considering the given concentration, rate constant, and the percentage of the reactant that has disappeared.