Formation of `NO_(2)F` from `NO_(2)` and `F_(2)` as per the reaction `2NO_(2)(g)+F_(2)(g)rarr2NO_(2)F(g)` is a second order reaction, first order with respect to `NO_(2)` and first order with respect to `F_(2)`. If `NO_(2)` and `F_(2)` are present in a closed vessel in ratio `2:1` maintained at a constant temperature with an initial total pressure of 3 atm, what will be the total pressure in the vessel after the reaction is complete?

Options

A. 1 atm
B. 2 atm
C. `2.5` atm
D. 3 atm

Answer

To solve the problem, we need to analyze the reaction and the initial conditions given. The reaction is:

\[ 2NO_2(g) + F_2(g) \rightarrow 2NO_2F(g) \]

### Step 1: Determine Initial Moles and Pressures
Given that the initial total pressure is 3 atm and the ratio of \( NO_2 \) to \( F_2 \) is 2:1, we can denote the moles of \( NO_2 \) and \( F_2 \) as follows:

Let the pressure of \( NO_2 \) be \( P_{NO_2} \) and the pressure of \( F_2 \) be \( P_{F_2} \).

From the ratio:
- \( P_{NO_2} = 2x \)
- \( P_{F_2} = x \)

The total pressure is:
\[ P_{NO_2} + P_{F_2} = 3 \]
\[ 2x + x = 3 \]
\[ 3x = 3 \]
\[ x = 1 \]

Thus:
- \( P_{NO_2} = 2 \, \text{atm} \)
- \( P_{F_2} = 1 \, \text{atm} \)

### Step 2: Identify the Limiting Reagent
In the reaction, 2 moles of \( NO_2 \) react with 1 mole of \( F_2 \). Since we have:
- 2 atm of \( NO_2 \) (which corresponds to 2 moles)
- 1 atm of \( F_2 \) (which corresponds to 1 mole)

The limiting reagent is \( F_2 \) because it will be completely consumed first.

### Step 3: Calculate the Change in Pressure
From the stoichiometry of the reaction:
- 2 moles of \( NO_2 \) produce 2 moles of \( NO_2F \).
- 1 mole of \( F_2 \) will react with 2 moles of \( NO_2 \).

Since \( F_2 \) is the limiting reagent, it will completely react:
- \( F_2 \) will decrease by 1 atm.
- \( NO_2 \) will decrease by 2 atm (since 2 moles of \( NO_2 \) are needed for every mole of \( F_2 \)).

### Step 4: Calculate Final Pressures
Initial pressures:
- \( P_{NO_2} = 2 \, \text{atm} \)
- \( P_{F_2} = 1 \, \text{atm} \)

After the reaction:
- \( P_{NO_2} \) will decrease by 2 atm: \( 2 - 2 = 0 \, \text{atm} \)
- \( P_{F_2} \) will decrease by 1 atm: \( 1 - 1 = 0 \, \text{atm} \)
- \( P_{NO_2F} \) will increase by 1 atm (produced from the reaction of \( F_2 \)): \( 0 + 2 = 2 \, \text{atm} \)

### Step 5: Calculate Total Pressure After Reaction
The total pressure after the reaction is complete is:
\[ P_{total} = P_{NO_2F} + P_{NO_2} + P_{F_2} \]
\[ P_{total} = 2 + 0 + 0 = 2 \, \text{atm} \]

### Final Answer
The total pressure in the vessel after the reaction is complete will be **2 atm**.