For the first order reaction. `N_2O_5(g) rightarrow N_2O_4(g) + 1/2 O_2`(g), the volume of `O_2` produced is 15 mL and 40 mL after 8 minutes and at the end of the reaction respectively. The rate constant is equal to

Options

A. `1/8` ln `80/50`
B. `1/8` ln `40/25`
C. `1/8` ln`40/10`
D. `1/8` ln`80/65`

Answer

To find the rate constant \( K \) for the first-order reaction \( N_2O_5(g) \rightarrow N_2O_4(g) + \frac{1}{2} O_2(g) \), we can use the formula for first-order reactions:

\[ K = \frac{2.303}{T} \log \left( \frac{V_{\infty}}{V_{\infty} - V_T} \right) \]

Where:
- \( K \) is the rate constant,
- \( T \) is the time in minutes,
- \( V_{\infty} \) is the volume of \( O_2 \) produced at the end of the reaction,
- \( V_T \) is the volume of \( O_2 \) produced at time \( T \).

### Step-by-step Solution:

1. **Identify the given values**:
- Volume of \( O_2 \) produced after 8 minutes (\( V_T \)) = 15 mL
- Volume of \( O_2 \) produced at the end of the reaction (\( V_{\infty} \)) = 40 mL
- Time (\( T \)) = 8 minutes

2. **Substitute the values into the formula**:
\[ K = \frac{2.303}{8} \log \left( \frac{40}{40 - 15} \right) \]

3. **Calculate \( V_{\infty} - V_T \)**:
\[ V_{\infty} - V_T = 40 - 15 = 25 \text{ mL} \]

4. **Rewrite the equation**:
\[ K = \frac{2.303}{8} \log \left( \frac{40}{25} \right) \]

5. **Calculate the logarithm**:
\[ \frac{40}{25} = 1.6 \]
Thus,
\[ K = \frac{2.303}{8} \log(1.6) \]

6. **Convert logarithm to natural logarithm**:
Since \( \log(x) \) can be converted to natural logarithm \( \ln(x) \) using the relation \( \log(x) = \frac{\ln(x)}{2.303} \), we can write:
\[ K = \frac{1}{8} \ln \left( \frac{40}{25} \right) \]

7. **Final expression for \( K \)**:
\[ K = \frac{1}{8} \ln \left( \frac{40}{25} \right) \]

### Conclusion:
The rate constant \( K \) is equal to \( \frac{1}{8} \ln \left( \frac{40}{25} \right) \), which corresponds to option B.