The following reaction descomposes `N_(2)O_(5)rarr 2NO_(2)+(1)/(2)O_(2)`. At a 25 degree centigrade the rate constant of the reaction is `5xx10^(-3)sec^(-1)`. The intial pressure of `N_(2)O_(5)` is 0.2 atm. If total pressure of gaseous mixture becomes 0.35 atm, then calculate the time of decomposition of `N_(2)O_(5)`.

Options

A. 238.6 seconds
B. 138.6 seconds
C. 69.6 seconds
D. 89.3 seconds

Answer

To solve the problem step by step, we will follow the given reaction and the provided data to find the time of decomposition of \( N_2O_5 \).

### Step 1: Write the balanced reaction and identify the changes in pressure
The reaction is:
\[ N_2O_5 \rightarrow 2NO_2 + \frac{1}{2} O_2 \]

### Step 2: Define initial and final pressures
- Initial pressure of \( N_2O_5 \) = \( P_0 = 0.2 \) atm
- Let the change in pressure of \( N_2O_5 \) be \( x \).
- At time \( t \):
- Pressure of \( N_2O_5 \) = \( P_0 - x = 0.2 - x \)
- Pressure of \( NO_2 \) = \( 2x \) (since 2 moles of \( NO_2 \) are produced for every mole of \( N_2O_5 \))
- Pressure of \( O_2 \) = \( \frac{x}{2} \) (since \( \frac{1}{2} \) mole of \( O_2 \) is produced for every mole of \( N_2O_5 \))

### Step 3: Write the equation for total pressure
The total pressure at time \( t \) is given as:
\[ (0.2 - x) + 2x + \frac{x}{2} = 0.35 \]

### Step 4: Simplify the equation
Combining the terms:
\[ 0.2 - x + 2x + \frac{x}{2} = 0.35 \]
This simplifies to:
\[ 0.2 + x + \frac{x}{2} = 0.35 \]
Combining \( x \) terms:
\[ 0.2 + \frac{3x}{2} = 0.35 \]

### Step 5: Solve for \( x \)
Subtract \( 0.2 \) from both sides:
\[ \frac{3x}{2} = 0.35 - 0.2 \]
\[ \frac{3x}{2} = 0.15 \]
Multiply both sides by \( \frac{2}{3} \):
\[ x = 0.15 \times \frac{2}{3} = 0.1 \]

### Step 6: Determine the pressure of \( N_2O_5 \) at time \( t \)
Now, substitute \( x \) back to find the pressure of \( N_2O_5 \):
\[ P = 0.2 - x = 0.2 - 0.1 = 0.1 \text{ atm} \]

### Step 7: Use the first-order reaction formula to find time
For a first-order reaction, the time \( t \) can be calculated using the formula:
\[ t = \frac{2.303}{k} \log\left(\frac{P_0}{P}\right) \]
Where:
- \( k = 5 \times 10^{-3} \text{ sec}^{-1} \)
- \( P_0 = 0.2 \text{ atm} \)
- \( P = 0.1 \text{ atm} \)

### Step 8: Substitute the values into the formula
\[ t = \frac{2.303}{5 \times 10^{-3}} \log\left(\frac{0.2}{0.1}\right) \]
\[ t = \frac{2.303}{5 \times 10^{-3}} \log(2) \]
The logarithm of 2 is approximately \( 0.3010 \):
\[ t = \frac{2.303}{5 \times 10^{-3}} \times 0.3010 \]

### Step 9: Calculate \( t \)
\[ t = \frac{2.303 \times 0.3010}{5 \times 10^{-3}} = \frac{0.69303}{5 \times 10^{-3}} = 138.606 \text{ seconds} \]

### Final Answer
Thus, the time of decomposition of \( N_2O_5 \) is approximately:
\[ t \approx 138.64 \text{ seconds} \]