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

Answers

Answer 1

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

Answer 2

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.


Related Questions

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?

Answers

Hi Again!
this question is another one for the PV=nRT formula
P is pressure, V is volume, n is moles, R is 8.314 and T is temperature
first this is turn your temperature in Kelvin, by adding 273 and you should get 308
Now you can plug into the formula
if you're solving for P, that means you are doing nRT divided by V
so you would sub in 8 for "n", 8.314 for "R", 308 for "T" and 12.0 for "V"
 you would now end up getting P= 8.00×8.314×308 and then all of that divided by 12 L
 and you should get 1707.141333 but with significant digits is turns into 1.70 ×10^1
and since you're finding pressure, your units are KpA
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