Suppose 15.00 g of solid ammonium hydrogen sulfide is introduced into a 500.-mL flask at 25°C,the flask is sealed,and the system is allowed to reach equilibrium.What is the partial pressure of ammonia in this flask if K_P = 0.108 at 25°C for the following reaction?
NH₄HS(s) NH₃(g) + H₂S(g)

Question

Suppose 15.00 g of solid ammonium hydrogen sulfide is introduced into a 500.-mL flask at 25°C,the flask is sealed,and the system is allowed to reach equilibrium.What is the partial pressure of ammonia in this flask if K_P = 0.108 at 25°C for the following reaction?
NH₄HS(s)

NH₃(g) + H₂S(g)

Quizplus · Accepted Answer

First, we need to write the expression for the equilibrium constant: K_P = [NH₃][H₂S]/[NH₄HS] Next, we need to set up an ICE table: NH₄HS(s) 11eab46b_9ca8_cd00_a889_3db5215a63e3_TB6485_11 NH₃(g)+ H₂S(g) I 0.0367 mol 0 mol 0 mol C -x +x +x E 0.0367-x x x (subscripts indicate the number of moles) The equilibrium concentration of NH3 is x, and the total pressure (Ptotal) is the sum of the partial pressures of NH3 and H2S. Now we can plug in the numbers into the expression for K_P: 0.108 = x^2 / (0.0367 - x) Simplifying and solving for x, we get: x = 0.0843 mol Therefore, the partial pressure of NH3 is: PNH3 = x / (0.5 L) = 0.1686 atm However, we need to remember that Ptotal = PNH3 + PH2S, and we know the initial moles and volume of the system, so we can calculate the initial pressure using the ideal gas law: P = nRT/V = (0.0367 mol + 0 mol + 0 mol) * 0.0821 L·atm/mol·K * (25 + 273) K / 0.5 L = 4.39 atm Now we can solve for PH2S: PH2S = Ptotal - PNH3 = 4.39 atm - 0.1686 atm = 4.22 atm Therefore, the partial pressure of NH3 is 0.1686 atm or 0.33 atm to two significant figures, and the answer is choice C.