`5.1g` of solid `NH_(4)HS` is introduced in a `16.4` lit. vessel & heated upto `500 K` `K_(B)` for equilibrium `NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g)` is `0.16`. The maximum pressure developed in the vessel will be `:`

Options

A. `0.8 atm`
B. `0.40 atm`
C. `0.5 atm`
D. None of these

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the number of moles of NH₄HS
We are given the mass of NH₄HS and need to find the number of moles using the formula:
\[ \text{Number of moles (n)} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of NH₄HS is calculated as follows:
- N: 14 g/mol
- H: 1 g/mol (4 H atoms = 4 g/mol)
- S: 32 g/mol

Thus, the molar mass of NH₄HS = 14 + 4 + 32 = 50 g/mol.

Now, substituting the values:
\[ n = \frac{5.1 \text{ g}}{51 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: Set up the equilibrium expression
The reaction is:
\[ \text{NH}_4\text{HS}(s) \rightleftharpoons \text{NH}_3(g) + \text{H}_2\text{S}(g) \]
At equilibrium, if we start with 0.1 moles of NH₄HS, we can assume that it completely dissociates into 0.1 moles of NH₃ and 0.1 moles of H₂S.

### Step 3: Calculate the total number of moles at equilibrium
At equilibrium:
- Moles of NH₄HS = 0 (since it completely dissociates)
- Moles of NH₃ = 0.1
- Moles of H₂S = 0.1

Total moles at equilibrium:
\[ \text{Total moles} = 0.1 + 0.1 = 0.2 \text{ moles} \]

### Step 4: Use the ideal gas law to find the pressure
The ideal gas law is given by:
\[ PV = nRT \]
We need to rearrange this to find pressure \(P\):
\[ P = \frac{nRT}{V} \]
Where:
- \(n = 0.2\) moles (total moles from equilibrium)
- \(R = 0.0821 \text{ L atm/(K mol)}\)
- \(T = 500 \text{ K}\)
- \(V = 16.4 \text{ L}\)

Substituting the values:
\[ P = \frac{0.2 \times 0.0821 \times 500}{16.4} \]

### Step 5: Perform the calculation
Calculating the numerator:
\[ 0.2 \times 0.0821 \times 500 = 8.21 \]
Now divide by the volume:
\[ P = \frac{8.21}{16.4} \approx 0.5 \text{ atm} \]

### Conclusion
The maximum pressure developed in the vessel is approximately **0.5 atm**.

---