A subtance `'A'` decomposes by a first-order reaction starting initially with `[A] = 2.00 m` and after `200 min` `[A] = 0.15 m`. For this reaction what is the value of `k`

Options

A. `1.29 xx 10^(-2) mi n^(-1)`
B. `2.29 xx 10^(-2) mi n^(-1)`
C. `3.29 xx 10^(-2) mi n^(-1)`
D. `4.40 xx 10^(-2) mi n^(-1)`

Answer

To find the rate constant \( k \) for the first-order decomposition of substance \( A \), we can use the first-order rate equation:

\[ k = \frac{1}{t} \ln \left( \frac{[A]_0}{[A]_t} \right) \]

### Step-by-Step Solution:

1. **Identify the Initial and Final Concentrations**:
- Initial concentration, \([A]_0 = 2.00 \, \text{M}\)
- Final concentration after 200 minutes, \([A]_t = 0.15 \, \text{M}\)

2. **Identify the Time Interval**:
- Time, \( t = 200 \, \text{minutes} \)

3. **Substitute Values into the First-Order Rate Equation**:
\[ k = \frac{1}{200} \ln \left( \frac{2.00}{0.15} \right) \]

4. **Calculate the Ratio**:
\[ \frac{2.00}{0.15} = 13.33 \]

5. **Calculate the Natural Logarithm**:
\[ \ln(13.33) \approx 2.59 \]

6. **Substitute the Natural Logarithm Back into the Equation**:
\[ k = \frac{1}{200} \times 2.59 \]

7. **Calculate \( k \)**:
\[ k = \frac{2.59}{200} = 0.01295 \, \text{min}^{-1} \]

8. **Express \( k \) in Scientific Notation**:
\[ k \approx 1.29 \times 10^{-2} \, \text{min}^{-1} \]

### Final Answer:
The value of the rate constant \( k \) is approximately \( 1.29 \times 10^{-2} \, \text{min}^{-1} \).

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