1. Write the equilibrium reaction

The decomposition is:

$NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g)$

For this reaction,

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

since the solid does not appear in the equilibrium expression.

2. Calculate initial moles of $NH_4HS$

Molar mass of $NH_4HS$:

$14 + 5 + 32 = 51\, \text{g mol}^{-1}$

Given mass = $5.1$ g,

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

20% decomposes, so moles decomposed:

$0.1 \times 0.20 = 0.02\, \text{mol}$

From stoichiometry,

$n_{NH_3} = n_{H_2S} = 0.02\, \text{mol}$

3. Find partial pressures

Using

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

Given:

• $R = 0.082\, \text{L atm K}^{-1}\text{mol}^{-1}$
• $T = 27^\circ C = 300\, K$
• $V = 2\, L$

So for each gas,

$P_{NH_3} = P_{H_2S} = \frac{0.02 \times 0.082 \times 300}{2}$

$= 0.02 \times 12.3$

$= 0.246\, \text{atm}$

4. Calculate $K_p$

$K_p = P_{NH_3} \cdot P_{H_2S} = (0.246)^2$

$K_p = 0.060516 \approx 6.05 \times 10^{-2}$

Thus,

$K_p = x \times 10^{-2}$

so

$x \approx 6$

5. Final answer

The integer value is:

$\boxed{6}$

6. Comparison with stored answer

Stored correct answer = $6$

Our derived answer also = $6$, so they agree.