The rate constant fro a second order reaction is `8 xx 10^(-5) M^(-1) "min"^(-1)`. How long will in take a `1 M` solution to be reduced to `0.5 M`

Options

A. `8.0 xx 10^(-5) min`
B. `8.665 xx 10^(5) min`
C. `4 xx 10^(-5) min`
D. `1.25 xx 10^(4) min`

Answer

To solve the problem, we need to use the integrated rate law for a second-order reaction. The integrated rate equation for a second-order reaction is given by:

\[ \frac{1}{A_t} - \frac{1}{A_0} = kt \]

Where:
- \( A_t \) is the concentration at time \( t \)
- \( A_0 \) is the initial concentration
- \( k \) is the rate constant
- \( t \) is the time

### Step-by-Step Solution:

1. **Identify the given values:**
- Rate constant \( k = 8 \times 10^{-5} \, M^{-1} \, min^{-1} \)
- Initial concentration \( A_0 = 1 \, M \)
- Final concentration \( A_t = 0.5 \, M \)

2. **Substitute the values into the integrated rate equation:**
\[ \frac{1}{A_t} - \frac{1}{A_0} = kt \]
Substituting the known values:
\[ \frac{1}{0.5} - \frac{1}{1} = (8 \times 10^{-5}) t \]

3. **Calculate the left side of the equation:**
\[ \frac{1}{0.5} = 2 \quad \text{and} \quad \frac{1}{1} = 1 \]
Therefore:
\[ 2 - 1 = (8 \times 10^{-5}) t \]
This simplifies to:
\[ 1 = (8 \times 10^{-5}) t \]

4. **Solve for \( t \):**
\[ t = \frac{1}{8 \times 10^{-5}} \]
Calculating this gives:
\[ t = 1.25 \times 10^{4} \, min \]

### Final Answer:
The time taken for the concentration to be reduced from \( 1 \, M \) to \( 0.5 \, M \) is \( 1.25 \times 10^{4} \, minutes \).

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