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 need to determine the equilibrium constant \( K_p \) for the decomposition of \( NH_4SH \) into \( NH_3 \) and \( H_2S \) at a temperature of 327°C. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Molar Mass of \( NH_4SH \)
The molar mass of \( NH_4SH \) can be calculated as follows:
- Molar mass of \( N \) = 14 g/mol
- Molar mass of \( H \) = 1 g/mol (4 H atoms contribute 4 g)
- Molar mass of \( S \) = 32 g/mol

\[ \text{Molar mass of } NH_4SH = 14 + 4 + 32 = 50 \text{ g/mol} \]

### Step 2: Calculate the Moles of \( NH_4SH \)
Using the mass of \( NH_4SH \) given (5.1 g), we can calculate the number of moles:

\[ \text{Moles of } NH_4SH = \frac{\text{mass}}{\text{molar mass}} = \frac{5.1 \text{ g}}{50 \text{ g/mol}} = 0.102 \text{ mol} \]

### Step 3: Determine the Amount Decomposed
We know that 30% of the solid \( NH_4SH \) decomposes:

\[ \text{Moles decomposed} = 0.102 \text{ mol} \times 0.30 = 0.0306 \text{ mol} \]

### Step 4: Calculate Moles at Equilibrium
At equilibrium:
- Moles of \( NH_4SH \) remaining = \( 0.102 - 0.0306 = 0.0714 \text{ mol} \)
- Moles of \( NH_3 \) produced = \( 0.0306 \text{ mol} \)
- Moles of \( H_2S \) produced = \( 0.0306 \text{ mol} \)

### Step 5: Total Moles at Equilibrium
Total moles of gases at equilibrium:

\[ \text{Total moles} = \text{Moles of } NH_3 + \text{Moles of } H_2S = 0.0306 + 0.0306 = 0.0612 \text{ mol} \]

### Step 6: Calculate the Total Pressure Using Ideal Gas Law
Using the ideal gas law \( PV = nRT \):

- \( V = 3.0 \text{ L} \)
- \( R = 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1} \)
- \( T = 327 + 273 = 600 \text{ K} \)

\[ P = \frac{nRT}{V} = \frac{0.0612 \text{ mol} \times 0.082 \text{ L atm K}^{-1} \text{ mol}^{-1} \times 600 \text{ K}}{3.0 \text{ L}} \]

Calculating this gives:

\[ P = \frac{3.008064}{3.0} \approx 1.0027 \text{ atm} \]

### Step 7: Calculate Partial Pressures
Since the moles of \( NH_3 \) and \( H_2S \) are equal, their partial pressures are:

\[ P_{NH_3} = P_{H_2S} = \frac{1.0027 \text{ atm}}{2} \approx 0.5014 \text{ atm} \]

### Step 8: Calculate \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:

\[ K_p = P_{NH_3} \times P_{H_2S} = (0.5014 \text{ atm})^2 \approx 0.2514 \text{ atm}^2 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) at 327°C is approximately:

\[ K_p \approx 0.2514 \text{ atm}^2 \]