5.1 g `NH_(4)SH` is introduced in 3.0 L evacuated flask at `327^(@)C`, 30% of the solid `NH_(4)SH` decomposed to `NH_(3)` and `H_(2)S` as gases. The `K_(p)` of the reaction at `327^(@)C` is :
`(R = 0.082 L atm mol^(-1)K^(-1), "Molar mass of S =" 32 g mol^(-1) , "molar mass of N" = 14 g mol^(-1))`

Options

A. `0.242 atm^(2)`
B. `4.9 xx 10^(-3) atm^(2)`
C. `0.242 xx 10^(-4) atm^(2)`
D. `1 xx 10^(-4) atm^(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of NH₄SH
We start with the mass of NH₄SH given in the problem, which is 5.1 g. We need to convert this mass into moles using the molar mass of NH₄SH.

**Molar mass of NH₄SH:**
- N: 14 g/mol
- H: 4 × 1 g/mol = 4 g/mol
- S: 32 g/mol
- H: 1 g/mol

Total molar mass = 14 + 4 + 32 + 1 = 51 g/mol

**Calculating moles:**
\[ \text{Moles of NH₄SH} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.1 \text{ g}}{51 \text{ g/mol}} \approx 0.1 \text{ mol} \]

### Step 2: Determine the amount decomposed
We know that 30% of NH₄SH decomposes.

**Calculating the moles decomposed:**
\[ \text{Moles decomposed} = 0.3 \times 0.1 \text{ mol} = 0.03 \text{ mol} \]

### Step 3: Calculate the moles at equilibrium
Initially, we had 0.1 moles of NH₄SH. After decomposition, the moles at equilibrium will be:

- Moles of NH₄SH remaining = \(0.1 - 0.03 = 0.07 \text{ mol}\)
- Moles of NH₃ produced = 0.03 mol
- Moles of H₂S produced = 0.03 mol

### Step 4: Calculate the partial pressures of NH₃ and H₂S
Using the ideal gas law \(PV = nRT\), we can find the partial pressures of the gases.

**Temperature conversion:**
\[ T = 327 + 273 = 600 \text{ K} \]

**Calculating the partial pressure of NH₃:**
\[ P_{NH₃} = \frac{nRT}{V} = \frac{0.03 \text{ mol} \times 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 600 \text{ K}}{3.0 \text{ L}} \]
Calculating this gives:
\[ P_{NH₃} \approx 0.492 \text{ atm} \]

**Calculating the partial pressure of H₂S:**
Since the moles of H₂S produced are the same as NH₃, we have:
\[ P_{H₂S} = P_{NH₃} = 0.492 \text{ atm} \]

### Step 5: Calculate Kp
The equilibrium constant \(K_p\) for the reaction can be expressed as:
\[ K_p = P_{NH₃} \times P_{H₂S} \]
Substituting the values:
\[ K_p = 0.492 \text{ atm} \times 0.492 \text{ atm} = 0.242 \text{ atm}^2 \]

### Final Answer:
The value of \(K_p\) at 327°C is approximately **0.242 atm²**.

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