For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$,\(K=400 .\) at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and NH,HS are placed in a 5.00 -L vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

The reaction is given as: \(\mathrm{NH}_{3}(g)+\mathrm{H}_{2}\mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4}\mathrm{HS}(s)\) Since NH4HS is a solid, its concentration does not affect K. So, K expression is written in terms of the concentrations of NH3 and H2S only: \(K = \dfrac{[\mathrm{NH}_{3}][\mathrm{H}_{2}\mathrm{S}]}{[\mathrm{NH}_{4}\mathrm{HS}]}\)

Given 2.00 moles of NH3, H2S, and NH4HS in a 5.00 L vessel, we find their initial concentrations as: \([NH_3]_0 = \frac{2.00\,\text{mol}}{5.00\,\text{L}} = 0.400\,\text{M}\) \([H_2S]_0 = \frac{2.00\,\text{mol}}{5.00\,\text{L}} = 0.400\,\text{M}\) \([NH_4HS]_0 = \frac{2.00\,\text{mol}}{5.00\,\text{L}} = 0.400\,\text{M}\)

The ICE table shows the initial concentrations, changes during the reaction, and equilibrium concentrations. Let "x" be the change in moles going towards the product: \(\begin{array}{c|ccc} & [NH_3] & [H_2S] & [NH_4HS]\\ \hline \text{Initial} & 0.400\,\text{M} & 0.400\,\text{M} & 0.400\,\text{M}\\ \text{Change} & -x & -x & +x\\ \text{Equilibrium} & 0.400-x & 0.400-x & 0.400+x \end{array}\)

Substitute the equilibrium concentrations into the K expression, given K = 400: \(400 = \dfrac{(0.400-x)(0.400-x)}{0.400+x}\) Solving for x, we get \(x = 0.267\,\text{M}\). Now, we can find the equilibrium concentrations: \([NH_3]_{eq} = 0.400 - 0.267 = 0.133\,\text{M}\) \([H_2S]_{eq} = 0.400 - 0.267 = 0.133\,\text{M}\) \([NH_4HS]_{eq} = 0.400 + 0.267 = 0.667\,\text{M}\)

To find the mass of NH4HS at equilibrium, we can use its equilibrium concentration and the volume: Mass of NH4HS \(= (0.667\,\text{M})(5.00\,\text{L}) = 3.33\,\text{mol}\) Now we can convert moles into grams using the molar mass of NH4HS (53.11 g/mol): Mass of NH4HS \(= (3.33\,\text{mol})(53.11\,\text{g/mol}) = 176.84\,\text{g}\) To find the pressure of H2S at equilibrium, we can use the ideal gas law (PV=nRT) since H2S is a gas. We will use the equilibrium concentration of H2S and convert it back to moles: Moles of H2S \(= (0.133\,\text{M})(5.00\,\text{L}) = 0.667\,\text{mol}\) Now, we can calculate the pressure of H2S using the ideal gas law at 35.0°C (308.15 K): \(P = \frac{nRT}{V} = \frac{(0.667\,\text{mol})(0.0821\,\text{L\,atm/mol\,K})(308.15\,\text{K})}{5.00\,\text{L}} = 3.38\,\text{atm}\)

The mass of NH4HS present at equilibrium is 176.84 g, and the pressure of H2S at equilibrium is 3.38 atm.