1. Write the equilibrium reaction

$NH_4SH(s) \rightleftharpoons NH_3(g) + H_2S(g)$

For this equilibrium,

$K_p = P_{NH_3}\, P_{H_2S}$

because the solid does not appear in the equilibrium expression.

2. Find molar mass of $NH_4SH$

Given:

• $N = 14$
• $H = 1$
• $S = 32$

So,

$M(NH_4SH) = 14 + 5(1) + 32 = 51\text{ g mol}^{-1}$

3. Calculate initial moles of $NH_4SH$

Mass given = $5.1\,$g

$n_0 = \frac{5.1}{51} = 0.1\text{ mol}$

30% decomposes, so moles decomposed:

$n_{\text{decomp}} = 0.30 \times 0.1 = 0.03\text{ mol}$

From stoichiometry:

$NH_4SH(s) \to NH_3(g) + H_2S(g)$

Therefore,

$n_{NH_3} = 0.03\text{ mol}, \qquad n_{H_2S} = 0.03\text{ mol}$

4. Calculate partial pressures

Given:

• $V = 3.0\,$L
• $T = 327^\circ C = 600\,$K
• $R = 0.082\,$L atm mol$^{-1}$ K$^{-1}$

Using

$P = \frac{nRT}{V}$

For $NH_3$:

$P_{NH_3} = \frac{0.03 \times 0.082 \times 600}{3}$

$P_{NH_3} = 0.492\text{ atm}$

Similarly,

$P_{H_2S} = 0.492\text{ atm}$

5. Calculate $K_p$

$K_p = P_{NH_3} \cdot P_{H_2S} = (0.492)(0.492)$

$K_p = 0.242\text{ atm}^2$

6. Match with the options

The value is:

$K_p = 0.242\text{ atm}^2$

So the correct option is:

D: $0.242\text{ atm}^2$

7. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.