Final answer:
The decomposition of solid NH₄HS into NH₃ and H₂S cannot provide exact partial pressures for each gas without additional information or the assumption of complete decomposition. Kp helps relate partial pressures of reactants and products at equilibrium but is not sufficient alone to determine partial pressures.
Step-by-step explanation:
The partial pressure of a component in a reaction can be determined using the equilibrium constant, Kp, which relates the partial pressures of the reactants and products at equilibrium for a given temperature. In the decomposition reaction of ammonium hydrosulfide, NH₄HS, into ammonia, NH₃, and hydrogen sulfide, H₂S, at a pressure of 2 atm, we use the provided Kp to calculate the partial pressures of the products assuming that NH₄HS(s) decomposes fully into equal parts NH₃(g) and H₂S(g).
For the reaction NH₄HS(s) ⇌ NH₃(g) + H₂S(g), at equilibrium the partial pressures of NH₃ and H₂S will be equal, because they are produced in a 1:1 ratio. If Kp is 3, it implies that the product of the partial pressures of NH₃ and H₂S is 3 when they are at equilibrium. We must keep in mind that total pressure at equilibrium is the sum of the partial pressures of NH₃ and H₂S.
However, the original question does not provide enough information to calculate the partial pressures of NH₃ and H₂S without additional data, such as the degree to which NH₄HS decomposes or assuming it decomposes completely. Hence, without assumptions or additional information, it isn't possible to provide a numerical value for the partial pressure of each product gas.