The concentrations of `N_(2)O_(5)` decomposing in first order kinetics after 800 s is `1.45 mol L^(-1)` and after `1600` s is `0.88 mol L^(-1)`. Calculate the rate constant.

Answer

To calculate the rate constant \( k \) for the decomposition of \( N_2O_5 \) under first-order kinetics, we can use the first-order rate equation:

\[ k = \frac{2.303}{t_2 - t_1} \log_{10} \left( \frac{[A_1]}{[A_2]} \right) \]

Where:
- \( t_1 \) and \( t_2 \) are the times at which the concentrations \( [A_1] \) and \( [A_2] \) are measured.
- \( [A_1] \) is the concentration at time \( t_1 \).
- \( [A_2] \) is the concentration at time \( t_2 \).

### Step 1: Identify the given values
- \( [A_1] = 1.45 \, \text{mol L}^{-1} \) at \( t_1 = 800 \, \text{s} \)
- \( [A_2] = 0.88 \, \text{mol L}^{-1} \) at \( t_2 = 1600 \, \text{s} \)

### Step 2: Calculate the time difference
\[ t_2 - t_1 = 1600 \, \text{s} - 800 \, \text{s} = 800 \, \text{s} \]

### Step 3: Calculate the ratio of concentrations
\[ \frac{[A_1]}{[A_2]} = \frac{1.45}{0.88} \]

### Step 4: Calculate the logarithm of the concentration ratio
Using a calculator:
\[ \log_{10} \left( \frac{1.45}{0.88} \right) \approx \log_{10}(1.6455) \approx 0.215 \]

### Step 5: Substitute the values into the rate constant equation
\[ k = \frac{2.303}{800} \times 0.215 \]

### Step 6: Calculate the rate constant \( k \)
\[ k \approx \frac{2.303 \times 0.215}{800} \approx \frac{0.49545}{800} \approx 0.0006193 \, \text{s}^{-1} \]

### Step 7: Convert to scientific notation
\[ k \approx 6.19 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer
The rate constant \( k \) for the decomposition of \( N_2O_5 \) is approximately:
\[ k \approx 6.19 \times 10^{-4} \, \text{s}^{-1} \]

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