`N_(2)O_(5)` decomposes to `NO_(2)` and`O_(2)` and follows first order kinetics. After `50` minutes the pressure inside the vessel increases from 50 mm Hg to `87.5 mmHg`. The pressure of the gaseous mixture afte 100 minute at constant temperature will be :

Options

A. `116.25 mm Hg`
B. `106.25 mmHg`
C. `136.25 mmHg`
D. `175.0 mmHg`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The decomposition of \( N_2O_5 \) can be represented as:
\[ 2 N_2O_5 \rightarrow 4 NO_2 + O_2 \]
This means that for every 2 moles of \( N_2O_5 \) that decompose, 4 moles of \( NO_2 \) and 1 mole of \( O_2 \) are produced.

### Step 2: Initial and Final Pressures
Given:
- Initial pressure \( P_0 = 50 \, \text{mm Hg} \)
- Pressure after 50 minutes \( P = 87.5 \, \text{mm Hg} \)

The increase in pressure after 50 minutes is:
\[ \Delta P = P - P_0 = 87.5 \, \text{mm Hg} - 50 \, \text{mm Hg} = 37.5 \, \text{mm Hg} \]

### Step 3: Relate Pressure Change to Decomposition
From the stoichiometry of the reaction, we can see that:
- For every 2 moles of \( N_2O_5 \) that decompose, 4 moles of \( NO_2 \) and 1 mole of \( O_2 \) are produced.
- Therefore, the total change in pressure due to the decomposition of \( N_2O_5 \) can be expressed as:
\[ \Delta P = 4x + y \]
Where \( x \) is the amount of \( N_2O_5 \) decomposed and \( y \) is the amount of \( O_2 \) produced.

### Step 4: Calculate the Change in Pressure
From the reaction stoichiometry:
- If \( x \) moles of \( N_2O_5 \) decompose, then \( 2x \) moles of \( NO_2 \) are produced and \( x/2 \) moles of \( O_2 \) are produced.
- Thus, the total pressure increase can be calculated as:
\[ \Delta P = 4 \left(\frac{x}{2}\right) + \left(\frac{x}{2}\right) = 2x + \frac{x}{2} = \frac{5x}{2} \]

### Step 5: Find the Value of x
From the increase in pressure after 50 minutes:
\[ \frac{5x}{2} = 37.5 \implies x = \frac{37.5 \times 2}{5} = 15 \, \text{mm Hg} \]

### Step 6: Calculate the Pressure After 100 Minutes
Since the reaction follows first-order kinetics, we can use the first-order rate equation:
\[ k = \frac{2.303}{t} \log \left(\frac{P_0}{P_t}\right) \]
Using the data from 50 minutes, we can find \( k \) and then use it to find the pressure after 100 minutes.

### Step 7: Calculate the Total Pressure at 100 Minutes
After 100 minutes, the pressure will be:
\[ P_{100} = P_0 + 3 \left(\frac{P_{50} - P_0}{2}\right) = 50 + 3 \left(\frac{87.5 - 50}{2}\right) \]
Calculating this gives:
\[ P_{100} = 50 + 3 \left(\frac{37.5}{2}\right) = 50 + 3 \times 18.75 = 50 + 56.25 = 106.25 \, \text{mm Hg} \]

### Final Answer
The pressure of the gaseous mixture after 100 minutes at constant temperature will be:
\[ \boxed{106.25 \, \text{mm Hg}} \]