The half life of decompoistion of `N_(2)O_(5)` is a first order reaction represented by `N_(2)O_(5)rarrN_(2)O_(4)+1//2O_(2)`
After `15`min the volume of `O_(2)` profuced is `9ml` and at the end of the reaction `35ml`. The rate constant is equal to

Options

A. `(1)/(15)log_(e).(35)/(26)`
B. `(1)/(15)log_(e).(44)/(26)`
C. `(1)/(15) log_(e).(35)/(36)`
D. None of these

Explanation

To find the rate constant \( k \) for the first-order decomposition of \( N_2O_5 \), we can use the given data about the volume of \( O_2 \) produced at different times. Here's a step-by-step solution:

### Step 1: Understand the Reaction
The decomposition reaction is:
\[ N_2O_5 \rightarrow N_2O_4 + \frac{1}{2} O_2 \]
From this reaction, we can see that for every mole of \( N_2O_5 \) that decomposes, half a mole of \( O_2 \) is produced.

### Step 2: Identify Given Data
- Volume of \( O_2 \) produced after 15 minutes (\( V_t \)): 9 ml
- Volume of \( O_2 \) produced at the end of the reaction (\( V_{\infty} \)): 35 ml
- Time (\( T \)): 15 minutes

### Step 3: Use the First-Order Rate Constant Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{T} \log \left( \frac{V_{\infty}}{V_{\infty} - V_t} \right) \]
However, since the question asks for the natural logarithm, we will use:
\[ k = \frac{1}{T} \ln \left( \frac{V_{\infty}}{V_{\infty} - V_t} \right) \]

### Step 4: Substitute the Values
Substituting the values into the equation:
- \( T = 15 \) minutes
- \( V_{\infty} = 35 \) ml
- \( V_t = 9 \) ml

We can calculate:
\[ V_{\infty} - V_t = 35 - 9 = 26 \text{ ml} \]

Now substituting these values into the formula:
\[ k = \frac{1}{15} \ln \left( \frac{35}{26} \right) \]

### Step 5: Calculate the Rate Constant
Now, we can compute \( k \):
\[ k = \frac{1}{15} \ln \left( \frac{35}{26} \right) \]

### Final Answer
Thus, the rate constant \( k \) is:
\[ k = \frac{1}{15} \ln \left( \frac{35}{26} \right) \text{ min}^{-1} \]