The decomposition of `N_2O_5` is a first order reaction represented by : `N_2O_5 to N_2O_4 + ½O_2`. After 15 minutes the volume of `O_2` produced is 9 ml and at the end of the reaction is 35 ml. The rate constant given by the expression:

Options

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

Answer

To solve the problem of finding 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 of \( N_2O_5 \) is given by the reaction:
\[ N_2O_5 \rightarrow N_2O_4 + \frac{1}{2} O_2 \]

### Step 2: Identify the Given Data
- After 15 minutes, the volume of \( O_2 \) produced is 9 ml.
- At the end of the reaction, the volume of \( O_2 \) produced is 35 ml.

### Step 3: Relate the Volume of \( O_2 \) to the Decomposition of \( N_2O_5 \)
From the stoichiometry of the reaction:
- For every 1 mole of \( N_2O_5 \) that decomposes, \( \frac{1}{2} \) mole of \( O_2 \) is produced.
- Therefore, if \( X \) ml of \( N_2O_5 \) decomposes, then \( \frac{X}{2} \) ml of \( O_2 \) is produced.

### Step 4: Calculate the Amount of \( N_2O_5 \) Decomposed
At \( t = 15 \) minutes, the volume of \( O_2 \) produced is 9 ml:
\[ \frac{X}{2} = 9 \]
Thus,
\[ X = 18 \text{ ml} \]
This means that 18 ml of \( N_2O_5 \) has decomposed after 15 minutes.

### Step 5: Determine the Initial Volume of \( N_2O_5 \)
At the end of the reaction, the total volume of \( O_2 \) produced is 35 ml:
\[ \frac{X_{total}}{2} = 35 \]
Thus,
\[ X_{total} = 70 \text{ ml} \]
This means the initial volume of \( N_2O_5 \) (\( A_0 \)) was 70 ml.

### Step 6: Calculate the Remaining Volume of \( N_2O_5 \) After 15 Minutes
The remaining volume of \( N_2O_5 \) after 15 minutes is:
\[ A_0 - X = 70 - 18 = 52 \text{ ml} \]

### Step 7: Use the First-Order Rate Constant Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{1}{t} \ln \left( \frac{A_0}{A_t} \right) \]
Where:
- \( A_0 \) is the initial concentration (70 ml),
- \( A_t \) is the concentration at time \( t \) (52 ml),
- \( t \) is the time (15 minutes).

Substituting the values:
\[ k = \frac{1}{15} \ln \left( \frac{70}{52} \right) \]

### Step 8: Calculate the Value of \( k \)
Now we compute:
\[ k = \frac{1}{15} \ln \left( \frac{70}{52} \right) \]
Calculating \( \frac{70}{52} \):
\[ \frac{70}{52} \approx 1.346 \]
Now, take the natural logarithm:
\[ \ln(1.346) \approx 0.297 \]
Finally, calculate \( k \):
\[ k \approx \frac{0.297}{15} \approx 0.0198 \text{ min}^{-1} \]

### Conclusion
The rate constant \( k \) for the decomposition of \( N_2O_5 \) is approximately \( 0.0198 \text{ min}^{-1} \).

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