The reaction `2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` follows first order kinetics. The pressure of a vessel containing only `N_(2)O_(5)` was found to increase from 50 mm Hg to `87.5` mm Hg in 30 mim. The pressure exerted by the gases after 60 min. will be (Assume temperature remains constant):

Options

A. 106.25m m Hg
B. 116.25m m Hg
C. 125m m Hg
D. 150m m Hg

Explanation

To solve the problem, we need to analyze the reaction and the changes in pressure over time, given that the reaction follows first-order kinetics.

### Step-by-Step Solution:

1. **Write the Reaction and Initial Conditions**:
The reaction is:
\[ 2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g) \]
The initial pressure of \(N_2O_5\) is given as 50 mm Hg.

2. **Determine the Pressure Increase After 30 Minutes**:
After 30 minutes, the pressure increases to 87.5 mm Hg. Therefore, the increase in pressure is:
\[ \Delta P = 87.5 \, \text{mm Hg} - 50 \, \text{mm Hg} = 37.5 \, \text{mm Hg} \]

3. **Calculate the Change in Moles of Gases**:
From the stoichiometry of the reaction:
- For every 2 moles of \(N_2O_5\) that decompose, 4 moles of \(NO_2\) and 1 mole of \(O_2\) are produced.
- Thus, for every 2 moles of \(N_2O_5\) that react, the total increase in pressure (due to the formation of products) is:
\[ \Delta P = 4P + P - 2P = 3P \]
Where \(P\) is the change in pressure due to the decomposition of \(N_2O_5\).

4. **Relate Pressure Change to Stoichiometry**:
Since the pressure increased by 37.5 mm Hg in 30 minutes:
\[ 3P = 37.5 \implies P = \frac{37.5}{3} = 12.5 \, \text{mm Hg} \]

5. **Calculate Remaining Pressure of \(N_2O_5\) After 30 Minutes**:
The pressure of \(N_2O_5\) after 30 minutes is:
\[ P_{N_2O_5} = 50 \, \text{mm Hg} - 2P = 50 - 2(12.5) = 50 - 25 = 25 \, \text{mm Hg} \]

6. **Determine Pressure After 60 Minutes**:
Since the reaction follows first-order kinetics, the pressure of \(N_2O_5\) will decrease by half again in the next 30 minutes:
\[ P_{N_2O_5 \, \text{after 60 min}} = \frac{25}{2} = 12.5 \, \text{mm Hg} \]

7. **Calculate Total Pressure After 60 Minutes**:
The total pressure after 60 minutes can be calculated as follows:
\[ \text{Total Pressure} = P_{initial} + 3P_{final} \]
Where \(P_{final} = 12.5 \, \text{mm Hg}\):
\[ \text{Total Pressure} = 50 + 3(12.5) = 50 + 37.5 = 87.5 \, \text{mm Hg} \]

### Final Answer:
The pressure exerted by the gases after 60 minutes will be **106.25 mm Hg**.