The compoistion of `N_(2)O_(5)` is a first order reaction represented by: `N_(2)O_(5) rarr N_(2)O_(4) + 1//2O_(2)`.
After `20 min` the volume of `O_(2)` profuced is `10 mL` and at the end of the reaction `40 mL`. The rate constant is equal to

Options

A. `(1)/(20)ln.(30)/(50)`
B. `(1)/(20)ln.(50)/(30)`
C. `(1)/(20)ln.(50)/(40)`
D. `(1)/(20)ln.(40)/(30)`

Explanation

To solve the problem, we need to find the rate constant \( k \) for the first-order reaction of \( N_2O_5 \) decomposing into \( N_2O_4 \) and \( O_2 \). The reaction is represented as:

\[ N_2O_5 \rightarrow N_2O_4 + \frac{1}{2} O_2 \]

### Step-by-Step Solution:

1. **Identify the given data:**
- Time \( t = 20 \) minutes
- Volume of \( O_2 \) produced at \( t = 20 \) min, \( V_t = 10 \) mL
- Final volume of \( O_2 \) produced at the end of the reaction, \( V_{\infty} = 40 \) mL

2. **Use the first-order rate constant formula:**
The formula for the rate constant \( k \) for a first-order reaction can be expressed as:
\[ k = \frac{2.303}{t} \log \left( \frac{V_{\infty}}{V_{\infty} - V_t} \right) \]
However, since we want to express this in natural logarithm (ln), we can use:
\[ k = \frac{1}{t} \ln \left( \frac{V_{\infty}}{V_{\infty} - V_t} \right) \]

3. **Substitute the values into the equation:**
- Substitute \( t = 20 \) minutes, \( V_{\infty} = 40 \) mL, and \( V_t = 10 \) mL into the equation:
\[ k = \frac{1}{20} \ln \left( \frac{40}{40 - 10} \right) \]

4. **Calculate the expression inside the logarithm:**
- Calculate \( 40 - 10 = 30 \):
\[ k = \frac{1}{20} \ln \left( \frac{40}{30} \right) \]

5. **Simplify the fraction:**
- The fraction simplifies to:
\[ \frac{40}{30} = \frac{4}{3} \]
Thus, we have:
\[ k = \frac{1}{20} \ln \left( \frac{4}{3} \right) \]

6. **Final expression for the rate constant:**
- Therefore, the rate constant \( k \) is:
\[ k = \frac{1}{20} \ln \left( \frac{4}{3} \right) \]