The decomposition of dinitrogen pentoxide `(N_(2)O_(5))` follows first order rate law. Calculate the rate constant from the given data:
`t=800sec[N_(2)O_(5)]=1.45 "mol"L^(-1)=[A_()]`
`t=600 sec [N_(2)O_(5)]=0.88"mol"L^(-1)=[A_(2)]`

Options

A. `3.12 xx 10^(-4) s^(-1)`
B. `6.24 xx 10^(-4) s^(-1)`
C. `2.84 xx 10^(-4) s^(-1)`
D. `8.14 xx 10^(-4) s^(-1)`

Explanation

To calculate the rate constant for the decomposition of dinitrogen pentoxide (N₂O₅) which follows a first-order rate law, we can use the following formula:

\[ k = \frac{2.303}{t_2 - t_1} \log \left( \frac{[A_1]}{[A_2]} \right) \]

Where:
- \( k \) is the rate constant
- \( t_1 \) and \( t_2 \) are the times at which the concentrations are measured
- \( [A_1] \) is the initial concentration
- \( [A_2] \) is the final concentration

### Step-by-step Solution:

1. **Identify the given data**:
- At \( t_1 = 800 \) sec, \( [A_1] = 1.45 \) mol/L
- At \( t_2 = 600 \) sec, \( [A_2] = 0.88 \) mol/L

2. **Calculate the time difference**:
\[ t_2 - t_1 = 600 \, \text{sec} - 800 \, \text{sec} = -200 \, \text{sec} \]
(Note: Here we should take \( t_1 \) as the earlier time, so we should use \( t_1 = 600 \) sec and \( t_2 = 800 \) sec for the calculation.)

3. **Re-calculate the time difference correctly**:
\[ t_2 - t_1 = 800 \, \text{sec} - 600 \, \text{sec} = 200 \, \text{sec} \]

4. **Plug the values into the rate constant formula**:
\[ k = \frac{2.303}{200} \log \left( \frac{1.45}{0.88} \right) \]

5. **Calculate the logarithm**:
\[ \log \left( \frac{1.45}{0.88} \right) = \log (1.6455) \approx 0.2169 \]

6. **Substitute the logarithm value into the equation**:
\[ k = \frac{2.303}{200} \times 0.2169 \]

7. **Calculate \( k \)**:
\[ k = \frac{2.303 \times 0.2169}{200} \approx \frac{0.4994}{200} \approx 0.002497 \, \text{sec}^{-1} \]

8. **Convert to scientific notation**:
\[ k \approx 2.497 \times 10^{-3} \, \text{sec}^{-1} \]

### Final Answer:
The rate constant \( k \) for the decomposition of dinitrogen pentoxide is approximately \( 2.497 \times 10^{-3} \, \text{sec}^{-1} \).