`1`mole `NH_(3)` gas is introduced in a `4 litre` vessel containing `5gram` of He and heated at `400K` temperature. `NH_(3)` is `40%` dissociated and following equilubrium is established `:-`
`NH_(3)hArr(1)/(2)N_(2)+(3)/(2)H_(2)`
Find total pressure at equilibrium `:-`

Options

A. `2.87atm`
B. `21.76atm`
C. `2.17atm`
D. `28.7atm`

Explanation

To find the total pressure at equilibrium for the given reaction, we will follow these steps:

### Step 1: Determine Initial Moles of Gases
- We start with 1 mole of \( NH_3 \) gas.
- The mass of helium \( He \) is given as 5 grams. To find the number of moles of helium, we use the formula:
\[ \text{Moles of } He = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \text{ g}}{4 \text{ g/mol}} = 1.25 \text{ moles} \]

### Step 2: Calculate the Change in Moles Due to Dissociation
- The dissociation of \( NH_3 \) is given as 40%. This means that:
\[ \text{Dissociated moles of } NH_3 = 0.4 \times 1 \text{ mole} = 0.4 \text{ moles} \]
- Therefore, at equilibrium:
- Moles of \( NH_3 \) remaining = \( 1 - 0.4 = 0.6 \) moles
- Moles of \( N_2 \) produced = \( \frac{1}{2} \times 0.4 = 0.2 \) moles
- Moles of \( H_2 \) produced = \( \frac{3}{2} \times 0.4 = 0.6 \) moles

### Step 3: Calculate Total Moles at Equilibrium
- Now we can calculate the total moles of gas at equilibrium:
\[ \text{Total moles} = \text{Moles of } He + \text{Moles of } NH_3 + \text{Moles of } N_2 + \text{Moles of } H_2 \]
\[ \text{Total moles} = 1.25 + 0.6 + 0.2 + 0.6 = 2.65 \text{ moles} \]

### Step 4: Use Ideal Gas Law to Find Pressure
- We will use the ideal gas equation \( PV = nRT \) to find the pressure at equilibrium.
- Rearranging the equation gives us:
\[ P = \frac{nRT}{V} \]
- Where:
- \( n = 2.65 \) moles (total moles calculated)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 400 \, K \)
- \( V = 4 \, L \)

### Step 5: Substitute Values and Calculate Pressure
- Now substituting the values into the equation:
\[ P = \frac{2.65 \times 0.0821 \times 400}{4} \]
\[ P = \frac{87.068}{4} = 21.767 \, \text{atm} \]
- Rounding this gives us approximately \( 21.75 \, \text{atm} \).

### Final Answer
The total pressure at equilibrium is approximately \( 21.75 \, \text{atm} \).

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