The decomposition of `N_(2)O_(5)` is a first order reaction represented by
`N_(2)O_(5) rarr N_(2)O_(4)+1/2 O_(2)`
After 15 minutes the volume of `O_(2)` produced is 9 cc and at the end of the reaction 35 cc The rate constant is equal to

Options

A. `1/15 ln 35/44`
B. `1/15 ln 44/55`
C. `1/15 ln 44/26`
D. `1/15 ln 35/26`

Explanation

To find the rate constant \( k \) for the first-order decomposition of \( N_2O_5 \), we can follow these steps:

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

### Step 2: Determine the Initial and Final Volumes of \( O_2 \)
From the problem statement:
- After 15 minutes, the volume of \( O_2 \) produced is 9 cc.
- At the end of the reaction, the total volume of \( O_2 \) produced is 35 cc.

### Step 3: Relate the Volume of \( O_2 \) to the Decomposition of \( N_2O_5 \)
Since the reaction produces \( \frac{1}{2} \) mole of \( O_2 \) for every mole of \( N_2O_5 \) that decomposes, we can find the amount of \( N_2O_5 \) decomposed:
- At the end of the reaction (when 35 cc of \( O_2 \) is produced), the amount of \( N_2O_5 \) decomposed is:
\[ \text{Volume of } N_2O_5 \text{ decomposed} = 2 \times 35 \text{ cc} = 70 \text{ cc} \]

### Step 4: Calculate the Volume of \( N_2O_5 \) Decomposed after 15 Minutes
- After 15 minutes, 9 cc of \( O_2 \) is produced, which corresponds to:
\[ \text{Volume of } N_2O_5 \text{ decomposed} = 2 \times 9 \text{ cc} = 18 \text{ cc} \]

### Step 5: Use the First-Order Rate Constant Formula
The formula for the rate constant \( k \) for a first-order reaction is:
\[ k = \frac{1}{t} \ln \left( \frac{A_0}{A_t} \right) \]
Where:
- \( A_0 \) = initial concentration (or volume) of \( N_2O_5 \)
- \( A_t \) = concentration (or volume) of \( N_2O_5 \) at time \( t \)
- \( t \) = time in minutes

### Step 6: Substitute the Values into the Formula
From our calculations:
- Initial volume of \( N_2O_5 \) \( A_0 = 70 \) cc
- Volume of \( N_2O_5 \) remaining after 15 minutes \( A_t = 70 - 18 = 52 \) cc
- Time \( t = 15 \) minutes

Now substituting these values into the formula:
\[ k = \frac{1}{15} \ln \left( \frac{70}{52} \right) \]

### Step 7: Calculate the Natural Logarithm
Calculating the natural logarithm:
\[ k = \frac{1}{15} \ln \left( \frac{35}{26} \right) \]

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