Consider the decomposition of ammonium hydrogen sulfide: $$\mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g}\) of \(\mathrm{NH}_{4} \mathrm{HS}\), ammonia with a partial pressure of \(0.692 \mathrm{~atm}\), and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by 12.4\%. Calculate \(K\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

The equilibrium constant expression for the given reaction is: $$K = \frac{[\mathrm{NH}_3][\mathrm{H}_2\mathrm{S}]}{[\mathrm{NH}_{4}\mathrm{HS}]}$$ However, since we are dealing with partial pressures, we need to write the expression in terms of partial pressures: $$K_p = \frac{P_{\mathrm{NH}_3}P_{\mathrm{H}_2\mathrm{S}}}{P_{\mathrm{NH}_{4}\mathrm{HS}}}$$ Note that the partial pressure of solid \(\mathrm{NH}_{4}\mathrm{HS}\) does not contribute to the equilibrium, so the expression becomes: $$K_p = P_{\mathrm{NH}_3}P_{\mathrm{H}_2\mathrm{S}}$$

First, we need to find the moles of \(\mathrm{NH}_{4}\mathrm{HS}\): $$n_{\mathrm{NH}_{4}\mathrm{HS}} = \frac{10.0\ \mathrm{g}}{M_{\mathrm{NH}_{4}\mathrm{HS}}}$$ The molar mass of \(\mathrm{NH}_{4}\mathrm{HS}\) is: $$M_{\mathrm{NH}_{4}\mathrm{HS}} = 14.01 + 4.03 + 1.01 + 32.07 + 1.01 = 52.13\ \mathrm{g/mol}$$ So the moles of \(\mathrm{NH}_{4}\mathrm{HS}\) are: $$n_{\mathrm{NH}_{4}\mathrm{HS}} = \frac{10.0\ \mathrm{g}}{52.13\ \mathrm{g/mol}} = 0.192\ \mathrm{mol}$$ The partial pressure of ammonia has increased by \(12.4 \%\), so the number of moles consumed of \(\mathrm{NH}_{4}\mathrm{HS}\), and produced of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2}\mathrm{S}\) will be the same. Let's denote that change in moles as \(\Delta n\). The relationship between moles and partial pressure is given by the ideal gas law: $$PV = nRT$$ Since temperature, volume, and gas constant (\(R\)) are constant, the change in partial pressure is proportional to the change in moles. So, \(\Delta P_\mathrm{NH_{3}} = 0.124P_\mathrm{NH_{3, initial}}\), and \(\Delta n_\mathrm{NH_{3}} = \Delta n_\mathrm{H_{2}\mathrm{S}} = \Delta n_\mathrm{NH_{4}\mathrm{HS}}\). We can find \(\Delta n\) using the initial partial pressure of ammonia (\(0.692\ \mathrm{atm}\)): $$\Delta n_\mathrm{NH_{3}} =\frac{\Delta P_{\mathrm{NH}_3}}{RT} = \frac{0.124 \times 0.692\ \mathrm{atm}}{0.0821 \times (273.15 + 25) \mathrm{atm\ L\ K^{-1}\ mol^{-1}}} \approx 0.00377\ \mathrm{mol}$$

Now we will find the number of moles of each species at equilibrium and calculate the equilibrium partial pressures: \(n_\mathrm{NH_{4}\mathrm{HS(equil)}} = n_\mathrm{NH_{4}\mathrm{HS(initial)}} - \Delta n = 0.192 - 0.00377 = 0.188\ \mathrm{mol}\) \(P_\mathrm{NH_{4}\mathrm{HS(equil)}}\) does not contribute to the equilibrium as it is a solid. \(n_\mathrm{NH_{3}(equil)} = n_\mathrm{NH_{3}(initial)} + \Delta n = 0.692\ \mathrm{atm} + 0.124 \cdot 0.692\ \mathrm{atm} \approx 0.776\ \mathrm{atm}\) \(n_\mathrm{H_{2}\mathrm{S(equil)} = n_\mathrm{H_{2}\mathrm{S(initial)} + \Delta n = 0.0532\ \mathrm{atm} + \frac{0.00377\ \mathrm{mol}}{0.192\ \mathrm{mol}} \times 0.0532\ \mathrm{atm} \approx 0.0568\ \mathrm{atm}\)

Finally, we will substitute the equilibrium partial pressures into the equilibrium constant expression and solve for \(K_p\): $$K_p = P_{\mathrm{NH}_3}P_{\mathrm{H}_2\mathrm{S}} = (0.776\ \mathrm{atm})(0.0568\ \mathrm{atm}) \approx 0.0441$$ Thus, the equilibrium constant for the decomposition of \(\mathrm{NH}_{4}\mathrm{HS}\) at \(25^{\circ} \mathrm{C}\) is approximately \(0.0441\).