Answer:
Answer has been given below
Explanation:
Every molecules contain electrons. So, London dispersion force exists in all molecules.For polar aprotic molecules, additional dipole-dipole force exists between molecules.For ionic compounds, additional ionic interaction force exist between ions.For non polar molecules, only London dispersion force exists between molecules.Benzene is a non-polar molecule. So London dispersion force exists between benzene molecules.[tex]CH_{3}Cl[/tex] is a polar molecule. So, dipole-dipole force and London dispersion force exists between [tex]CH_{3}Cl[/tex] molecules.[tex]PF_{3}[/tex] is a polar molecule. So, dipole-dipole force and London dispersion force exists between [tex]PF_{3}[/tex] molecules.NaCl is an ionic compound. So London dispersion force and ionic interaction force exists between NaCl molecules.[tex]CS_{2}[/tex] is a nonpolar molecule. So, London dispersion force exists between [tex]CS_{2}[/tex] molecules.A container of hydrogen gas has a volume of 1.46 liters, a pressure of 2.18 atm, and a temperature of 185 Kelvin. How many moles of gas are in the container?
Assume ideal gas behavior.
The answer is ______ mol.
A 12.0 L gas cylinder is filled with 8.00 moles of gas. The tank is stored at 35°C. What is the pressure in the tank?
A.)The equilibrium constant, Kp, for the following reaction is 0.110 at 298 K:
NH4HS(s) NH3(g) + H2S(g)
Calculate the equilibrium partial pressure of H2S when 0.371 moles of NH4HS(s) is introduced into a 1.00 L vessel at 298 K.
PH2S = ?? atm
B.) The equilibrium constant, Kp, for the following reaction is 0.497 at 500 K:
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium partial pressures of all species when PCl5(g) is introduced into an evacuated flask at a pressure of 1.47 atm at 500 K .
PPCl5 = ?? atm
PPCl3 = ?? atm
PCl2 = ?? atm
Final answer:
To calculate the equilibrium partial pressure of H2S, use the equilibrium constant expression and rearrange to solve for PH2S.
Explanation:
To calculate the equilibrium partial pressure of H2S, we need to use the equilibrium constant, Kp. The equilibrium constant expression for the given reaction is:
Kp = [H2S] / [NH4HS]
Given that Kp = 0.110, we can establish the relationship:
Kp = (PH2S) / (PNH4HS)
Since the initial moles of NH4HS is 0.371 moles and the volume of the vessel is 1.00 L, we can calculate the initial partial pressure of NH4HS:
PNH4HS = (0.371 mol) / (1.00 L) = 0.371 atm
Now, we can rearrange the equilibrium constant expression and solve for PH2S:
PH2S = Kp * PNH4HS = (0.110)(0.371 atm) = 0.0408 atm
The equilibrium partial pressure of H₂S is 0.332 atm. The equilibrium partial pressures for PCl₅, PCl₃, and Cl₂ are 0.83 atm, 0.64 atm, and 0.64 atm, respectively.
A.) To determine the equilibrium partial pressure of H₂S for the reaction NH₄HS(s) ⇌ NH₃(g) + H₂S(g) with Kp = 0.110 at 298 K, follow these steps:
Recognize that solid NH₄HS does not affect the equilibrium expression since its activity is 1.Let the equilibrium partial pressures of NH₃ and H₂S be PNH₃ = PH₂S = P since they are produced in a 1:1 ratio.Use the equation for Kp: Kp = PNH₃ × PH₂S = P².Substitute the given Kp value: 0.110 = P².Solve for P: P = √0.110 = 0.332 atm.Therefore, the equilibrium partial pressure of H₂S is 0.332 atm.
B.) For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) with Kp = 0.497 at 500 K:
Set up an ICE table. Let the initial pressure of PCl₅ be 1.47 atm, and the changes in pressure be -x for PCl₅ and +x for both PCl₃ and Cl₂.At equilibrium: PCl₅ = 1.47 - x, PCl₃ = x, and Cl₂ = x.The equilibrium expression is: Kp = (PCl₃ × PCl₂) / PPCl₅.Substitute the values: 0.497 = (x × x) / (1.47 - x).Solve the quadratic equation: 0.497 = x² / (1.47 - x).Rearrange to get 0.497(1.47 - x) = x² which gives 0.73 - 0.497x = x².Rearrange into standard quadratic form: x² + 0.497x - 0.73 = 0.Use the quadratic formula x = [-b ± √(b² - 4ac)] / 2a where a = 1, b = 0.497, and c = -0.73.Solve for x to get two possible solutions, but only the positive value is physically meaningful: x ≈ 0.64 atm.Calculate the equilibrium pressures: PPCl₃ = 0.64 atm, PCl₂ = 0.64 atm, and PPCl₅ = 1.47 - 0.64 = 0.83 atm.Therefore, the equilibrium partial pressures are PPCl₅ = 0.83 atm, PPCl₃ = 0.64 atm, and PCl₂ = 0.64 atm.