When 5.1 g of solid NH$$_4$$HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K$$_p$$ for the reaction at 27°C is $$x \times 10^{-2}$$. The value of x is _________. (Integer answer)
[Given 1R = 0.082 L atm K$$^{-1}$$ mol$$^{-1}$$]

We consider the equilibrium

$$\text{NH}_4\text{HS(s)} \;\rightleftharpoons\; \text{NH}_3(\text{g}) + \text{H}_2\text{S}(\text{g})$$

The solid does not appear in the expression for $$K_p$$, only the gaseous products do.

First we calculate how many moles of solid are initially taken. The molar mass of NH$$_4$$HS is

$$M = 14 \;(\text{N}) + 5 \;(\text{H}) + 32 \;(\text{S}) = 51 \,\text{g mol}^{-1}$$

Hence, from 5.1 g of the solid the number of moles present initially is

$$n_0 = \frac{5.1\ \text{g}}{51\ \text{g mol}^{-1}} = 0.10\ \text{mol}$$

Only 20 % of this solid decomposes. Therefore the amount that actually undergoes decomposition is

$$n_{\text{decomp}} = 0.20 \times 0.10 = 0.02\ \text{mol}$$

Because the stoichiometric coefficients of NH$$_3$$ and H$$_2$$S are each 1, the decomposition of 0.02 mol of solid produces

$$n_{\text{NH}_3} = 0.02\ \text{mol},\qquad n_{\text{H}_2\text{S}} = 0.02\ \text{mol}$$

The gases are contained in a 2 L flask at 27 °C, i.e. 300 K. From the ideal-gas equation $$P = \dfrac{nRT}{V}$$ their partial pressures are

$$P_{\text{NH}_3} = \frac{(0.02)\,(0.082)\,(300)}{2}$$

First multiply the numerator:

$$0.02 \times 0.082 = 0.00164$$

$$0.00164 \times 300 = 0.492$$

Now divide by 2 L:

$$P_{\text{NH}_3} = \frac{0.492}{2} = 0.246\ \text{atm}$$

The amount of H$$_2$$S is the same, so

$$P_{\text{H}_2\text{S}} = 0.246\ \text{atm}$$

The equilibrium constant in terms of pressure is defined by

$$K_p = P_{\text{NH}_3}\;P_{\text{H}_2\text{S}}$$

Substituting the values just obtained,

$$K_p = (0.246)\,(0.246) = 0.060516 \,\text{atm}^2$$

This can be written in scientific notation as

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

The question states $$K_p = x \times 10^{-2}$$, so clearly

$$x \approx 6.05 \; \Longrightarrow \; x = 6\ \text{(nearest integer)}$$

So, the answer is $$6$$.