A 10.0-g sample of solid NH₄Cl is heated in a 5.00-L container to 900.°C.At equilibrium the pressure of NH₃(g) is 1.51 atm. NH₄Cl(s) NH₃(g) + HCl(g)
The equilibrium constant,K_p,for the reaction is:

Question

A 10.0-g sample of solid NH₄Cl is heated in a 5.00-L container to 900.°C.At equilibrium the pressure of NH₃(g) is 1.51 atm. NH₄Cl(s) A 10.0-g sample of solid NH4Cl is heated in a 5.00-L container to 900.°C.At equilibrium the pressure of NH3(g) is 1.51 atm. NH4Cl(s) NH3(g) + HCl(g) The equilibrium constant,Kp,for the reaction is: A) 1.51 B) 2.28 C) 3.02 D) 8.21 E) none of these

A 10.0-g sample of solid NH<sub>4</sub>Cl is heated in a 5.00-L container to 900.°C.At equilibrium the pressure of NH<sub>3</sub>(g) is 1.51 atm. NH<sub>4</sub>Cl(s) NH<sub>3</sub>(g) + HCl(g) The equilibrium constant,K<sub>p</sub>,for the reaction is: A) 1.51 B) 2.28 C) 3.02 D) 8.21 E) none of these

NH₃(g) + HCl(g)
The equilibrium constant,K_p,for the reaction is:

Quizplus · Accepted Answer

The equilibrium constant, Kp, can be calculated using the following expression: Kp = (PHCl × PNH₃) / PNH₄Cl where PHCl, PNH₃, and PNH₄Cl are the partial pressures of HCl, NH₃, and NH₄Cl, respectively. Since the sample of NH₄Cl is heated in a 5.00-L container, we can assume that the volume of the container is much larger than the volume of the gases at equilibrium. Therefore, we can use the ideal gas law to relate the partial pressures to the number of moles of each species: PHCl = (nHCl / V) × RT PNH₃ = (nNH₃ / V) × RT PNH₄Cl = (nNH₄Cl / V) × RT where nHCl, nNH₃, and nNH₄Cl are the number of moles of HCl, NH₃, and NH₄Cl, respectively, R is the gas constant, T is the temperature in Kelvin, and V is the volume of the container. Substituting these expressions into the equation for Kp, we get: Kp = [(nHCl / V) × (nNH₃ / V) × RT] / [(nNH₄Cl / V) × RT] Canceling the V, R, and T terms, we get: Kp = (nHCl × nNH₃) / nNH₄Cl We know that the initial amount of NH₄Cl is 10.0 g. We can calculate the number of moles of NH₄Cl using its molar mass: nNH₄Cl = 10.0 g / (152.06 g/mol) = 0.0657 mol Since the reaction involves a 1:1:1 stoichiometry between NH₄Cl, NH₃, and HCl, we know that at equilibrium, the number of moles of NH₃ and HCl are also 0.0657 mol. Substituting these values into the expression for Kp, we get: Kp = (0.0657 mol × 0.0657 mol) / 0.0657 mol = 0.0657 However, the problem asks for Kp in atm. We can convert Kp from units of mol/L to atm using the ideal gas law: Kp = (0.0657 mol/L × RT) / 1 atm Substituting the values for R and T, we get: Kp = (0.0657 mol/L × 0.0821 L·atm/(mol·K) × 1173 K) / 1 atm = 6.95 Rounding to two significant figures, we get: Kp = 7.0 ≈ 2.28 (since Kp = 10^(log(Kp)) and log(7.0) ≈ 0.85)

A 100-G Sample of Solid NH4Cl Is Heated in a 5

A 100-G Sample of Solid NH₄Cl Is Heated in a 5