In a first order reaction the concentration of reactant decreases from 800 mol/`dm^3` to 50 mol/dm3 in `2xx 10^4 s`. The rate constant of reaction in `s^(–1)` is

Options

A. `2 xx 10^4`
B. `3.45 xx 10^(-5)`
C. `1.386 xx 10^(-4)`
D. `2xx 10^(-4)`

Explanation

To find the rate constant \( k \) for the first-order reaction where the concentration of the reactant decreases from 800 mol/dm³ to 50 mol/dm³ in \( 2 \times 10^4 \) seconds, we can use the first-order rate equation:

\[ k = \frac{2.303}{T} \log \left( \frac{[A_0]}{[A_t]} \right) \]

Where:
- \( k \) is the rate constant,
- \( T \) is the time in seconds,
- \( [A_0] \) is the initial concentration,
- \( [A_t] \) is the final concentration after time \( T \).

### Step-by-step Solution:

1. **Identify the initial and final concentrations**:
- Initial concentration, \( [A_0] = 800 \, \text{mol/dm}^3 \)
- Final concentration, \( [A_t] = 50 \, \text{mol/dm}^3 \)

2. **Identify the time duration**:
- Time, \( T = 2 \times 10^4 \, \text{s} \)

3. **Substitute the values into the formula**:
\[ k = \frac{2.303}{2 \times 10^4} \log \left( \frac{800}{50} \right) \]

4. **Calculate the ratio of concentrations**:
\[ \frac{800}{50} = 16 \]

5. **Calculate the logarithm**:
\[ \log(16) = 1.2041 \quad (\text{using a calculator}) \]

6. **Substitute the logarithm back into the equation**:
\[ k = \frac{2.303}{2 \times 10^4} \times 1.2041 \]

7. **Calculate \( k \)**:
\[ k = \frac{2.303 \times 1.2041}{2 \times 10^4} \]
\[ k = \frac{2.776}{2 \times 10^4} = 1.388 \times 10^{-4} \, \text{s}^{-1} \]

8. **Final result**:
\[ k \approx 1.386 \times 10^{-4} \, \text{s}^{-1} \]

### Conclusion:
The rate constant \( k \) for the reaction is approximately \( 1.386 \times 10^{-4} \, \text{s}^{-1} \).