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

Options

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

Explanation

To find the rate constant (k) for the first-order reaction where the concentration of the reactant decreases from \(800 \, \text{mol dm}^{-3}\) to \(50 \, \text{mol dm}^{-3}\) in \(2 \times 10^4 \, \text{s}\), we can use the first-order rate equation:

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

Where:
- \(A_0\) = initial concentration = \(800 \, \text{mol dm}^{-3}\)
- \(A_t\) = concentration at time \(t\) = \(50 \, \text{mol dm}^{-3}\)
- \(t\) = time = \(2 \times 10^4 \, \text{s}\)

### Step-by-Step Solution:

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

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

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

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

5. **Calculate the logarithm:**
\[ \log(16) \approx 1.2041 \]

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

7. **Calculate \(k\):**
- First, calculate \(2 \times 10^4 = 20000\).
- Now calculate:
\[ k = \frac{2.303 \times 1.2041}{20000} \]
\[ k \approx \frac{2.775}{20000} \approx 1.3875 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer:
The rate constant \(k\) is approximately \(1.3875 \times 10^{-4} \, \text{s}^{-1}\).