Mr Cole Chemistry

Kp Calculation Questions

Kp Calculations

Worksheet

Kp Calculations

PCl₅ is placed in a container and left to reach equilibrium with PCl₃ and Cl₂. Once equilibrium was reached, there were 12.2 moles of PCl₅ and 3.6 moles of PCl₃ and Cl₂. The pressure in the container was 120 kPa.
a) Calculate the Kp for the equilibrium and deduce its units.
PCl5(g) ⇌ PCl3(g) + Cl2(g)

Total Moles = 12.2 + 3.6 + 3.6 = 19.4 moles.

Partial Pressures (pp = mole fraction × Total P):
pp(PCl5) = (12.2 / 19.4) × 120 = 75.46 kPa
pp(PCl3) = (3.6 / 19.4) × 120 = 22.27 kPa
pp(Cl2) = (3.6 / 19.4) × 120 = 22.27 kPa

$$K_p = \frac{p(PCl_3)p(Cl_2)}{p(PCl_5)} = \frac{22.27 \times 22.27}{75.46} = 6.57$$

Answer: 6.6 kPa (2 s.f.)

b) State what would happen if the pressure were to be increased, and explain your answer. In your answer, make reference to both Le Chatelier’s principle and the expression for Kp.

The value of Kp would not change as Kp is a constant for the given temperature.

According to Le Chatelier’s principle, the equilibrium will shift towards the side with the smallest number of gas molecules (the left) to counteract the pressure increase. The yield of PCl3 and Cl2 would decrease.

Mathematically, in the Kp expression, the numerator (products) has two pressure terms while the denominator has one. If pressure increases, the numerator increases more than the denominator. To restore the constant Kp, the mole fractions of products must decrease and the reactant must increase.

Hydrogen and bromine were mixed in a container. Once equilibrium was obtained, there were 6 moles of hydrogen, 6 moles of bromine, and 3.6 moles of hydrogen bromide.
a) Calculate the Kp for the equilibrium and deduce its units.
H2(g) + Br2(g) ⇌ 2HBr(g)

$$K_p = \frac{p(HBr)^2}{p(H_2)p(Br_2)}$$

Since there are equal moles of gas on both sides, the pressure (P) and total moles terms cancel out. We can calculate Kp using moles directly.

$$K_p = \frac{3.6^2}{6 \times 6} = \frac{12.96}{36} = 0.36$$

Answer: 0.36 (No Units)

b) Explain why it was not necessary to determine the pressure to calculate the Kp.

Pressure appears twice in the numerator ($P^2$) and twice in the denominator ($P \times P$). They cancel out completely, so the partial pressure calculation simplifies to the mole ratio.

An equilibrium was set up for the equation below and it was found that the mole fractions of SO₂ and O₂ were 0.20 and 0.40 respectively. Given that the Kp was 0.10 kPa⁻¹ and the pressure was 100kPa, calculate the mole fraction of SO₃ in the equilibrium.
2SO₂ + O₂ ⇌ 2SO₃
Answer

$$K_p = \frac{p(SO_3)^2}{p(SO_2)^2 p(O_2)} = \frac{(x_{SO3} P)^2}{(x_{SO2} P)^2 (x_{O2} P)} = \frac{x_{SO3}^2}{x_{SO2}^2 x_{O2} P}$$

Substitute known values:

$$0.10 = \frac{x_{SO3}^2}{(0.20)^2 \times 0.40 \times 100}$$

$$0.10 = \frac{x_{SO3}^2}{0.04 \times 40} = \frac{x_{SO3}^2}{1.6}$$

$$x_{SO3}^2 = 0.10 \times 1.6 = 0.16$$

$$x_{SO3} = \sqrt{0.16} = 0.40$$

Answer: 0.40

Check: Sum of mole fractions = 0.20 + 0.40 + 0.40 = 1.0.

2.00 moles of PCl₃ and 2.00 of Cl₂ are placed in a closed container and allowed to reach equilibrium. The equilibrium mixture has a pressure of 100kPa, and contains 0.810 moles of PCl₅. Calculate the value of Kp for this equilibrium.
Answer
PCl3 + Cl2 ⇌ PCl5
Species PCl3 Cl2 PCl5
Initial 2.00 2.00 0.00
Change -0.810 -0.810 +0.810
Equilibrium 1.19 1.19 0.810

Total Moles = 1.19 + 1.19 + 0.810 = 3.19 mol.

Partial Pressures (P = 100):
pp(PCl5) = (0.810/3.19) * 100 = 25.39 kPa
pp(PCl3) = (1.19/3.19) * 100 = 37.30 kPa

$$K_p = \frac{25.39}{37.30 \times 37.30} = \frac{25.39}{1391.29} = 0.0182…$$

Answer: 0.0182 kPa-1

100 moles of methane and 100 moles of steam are reacted together in a sealed container. The equilibrium mixture formed contains an unknown amount of carbon monoxide and 103 moles of hydrogen. The pressure is 1000kPa.
a) Calculate the Kp value for this equation.
CH4 + H2O ⇌ CO + 3H2

Change in H2 = +103 mol. Ratio H2:CO is 3:1.

Species CH4 H2O CO 3H2
Initial 100 100 0 0
Change -34.33 -34.33 +34.33 +103
Eq. Mol 65.67 65.67 34.33 103

Total Moles = 268.67 mol. P = 1000 kPa.

pp(H2) = (103/268.67)*1000 = 383.4 kPa
pp(CO) = (34.33/268.67)*1000 = 127.8 kPa
pp(CH4) = (65.67/268.67)*1000 = 244.4 kPa

$$K_p = \frac{127.8 \times (383.4)^3}{244.4 \times 244.4} = \frac{7.20 \times 10^9}{59731} = 1.20 \times 10^5$$

Answer: 1.20 × 105 kPa2

b) Given that the forward reaction has an enthalpy change of +210kJmol⁻¹, state what would happen to the value of Kp if you were to increase the temperature. Explain your answer.

The forward reaction is endothermic (+210 kJmol⁻¹). Increasing the temperature favors the endothermic direction. The position of equilibrium shifts to the right, increasing the yield of products. Therefore, the value of Kp increases.

30 moles of nitrogen and 90 moles of hydrogen are reacted in a sealed container until they reach equilibrium. The equilibrium mixture contained 45 moles of ammonia when the pressure was at 500kPa. Calculate the value of Kp at this temperature.
Answer
N2 + 3H2 ⇌ 2NH3
Species N2 3H2 2NH3
Initial 30 90 0
Change -22.5 -67.5 +45
Eq. Mol 7.5 22.5 45

Total Moles = 75. P = 500 kPa.

pp(NH3) = (45/75)*500 = 300 kPa
pp(N2) = (7.5/75)*500 = 50 kPa
pp(H2) = (22.5/75)*500 = 150 kPa

$$K_p = \frac{300^2}{50 \times 150^3} = \frac{90000}{1.6875 \times 10^8} = 5.33 \times 10^{-4}$$

Answer: 5.33 × 10-4 kPa-2

25 moles of ammonia and 25 moles of oxygen were left to reach equilibrium. Once equilibrium was reached there were 10 moles of water. The value of Kp was 0.0316. What pressure was the reaction? The equation is below.
4NH₃ + 5O₂ ⇌ 4NO + 6H₂O
Answer
4NH3 + 5O2 ⇌ 4NO + 6H2O
Species 4NH3 5O2 4NO 6H2O
Initial 25 25 0 0
Change -6.67 -8.33 +6.67 +10
Eq. Mol 18.33 16.67 6.67 10
Mole Frac (x) 0.355 0.323 0.129 0.194

Total Moles = 51.67.

$$K_p = \frac{x_{NO}^4 x_{H2O}^6}{x_{NH3}^4 x_{O2}^5} \times P$$

Calculating the mole fraction term (Kx):

$$K_x = \frac{0.129^4 \times 0.194^6}{0.355^4 \times 0.323^5} = 2.58 \times 10^{-4}$$

$$P = \frac{K_p}{K_x} = \frac{0.0316}{0.000258} = 122.5$$

Answer: 123 kPa

A 0.56 mol sample of sulfur dioxide is mixed with oxygen in a 2:1 mol ratio and allowed to reach equilibrium in a sealed flask at a constant temperature. At equilibrium the mixture contains 0.30 mol of sulfur dioxide. The total pressure of this mixture is 1.04 × 10⁴ kPa. Calculate the partial pressure, in kPa, of sulfur trioxide in the equilibrium mixture.
Answer
2SO2 + O2 ⇌ 2SO3

Initial O2 = 0.56 / 2 = 0.28 mol.

Species 2SO2 O2 2SO3
Initial 0.56 0.28 0.00
Change -0.26 -0.13 +0.26
Equilibrium 0.30 0.15 0.26

Total Moles = 0.30 + 0.15 + 0.26 = 0.71 mol.

Mole fraction SO3 = 0.26 / 0.71 = 0.366

Partial Pressure = 0.366 × 1.04 × 104 = 3808 kPa.

Answer: 3.81 × 103 kPa

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