Mole Fractions, Partial Pressures & Equilibrium Constant Kp Calculations

1

Mole Fraction x(A) & Partial Pressure p(A)

In a mixture of gases, the mole fraction of gas A is: x(A) = (Equilibrium moles of gas A) ÷ (Total equilibrium moles of all gases in the mixture). Note: Mole fractions have no units, and the sum of mole fractions of all gases in a mixture equals 1 (Σx = 1). The partial pressure of gas A is the pressure it would exert if it occupied the container alone: p(A) = x(A) × P_total. The sum of all partial pressures equals the total pressure (Σp = P_total).

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2

Writing Kp Expressions & Heterogeneous Equilibria

For a homogeneous gaseous equilibrium aA(g) + bB(g) ⇌ cC(g) + dD(g), Kp = [p(C)^c × p(D)^d] ÷ [p(A)^a × p(B)^b]. IMPORTANT notation rule: Always use round brackets p(A) or p_A for partial pressures in Kp - never use square brackets [A], which strictly denote concentration in mol dm⁻³ and will lose marks. For heterogeneous equilibria (e.g. CaCO₃(s) ⇌ CaO(s) + CO₂(g)), solids and pure liquids are omitted completely, so Kp = p(CO₂).

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3

Ice Tables for Gas Equilibria & Units of Kp

1. Set up an I-C-E table (Initial moles, Change in moles using stoichiometric ratios, Equilibrium moles). 2. Sum the equilibrium moles of all gases to find n_total. 3. Calculate the mole fraction x = n_i / n_total and partial pressure p_i = x_i × P_total for each gas. 4. Substitute into Kp and determine units by cancelling pressure units (kPa, Pa, or atm) using the power Δn = (moles of gas on product side) - (moles of gas on reactant side).

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4

Effect of Temperature on Kp (The only Factor That Changes Kp)

Kp (and Kc) is a constant at a given temperature and is only changed by changing the temperature: 1. If the forward reaction is exothermic (ΔH < 0): increasing temperature shifts equilibrium to the left (endothermic direction), decreasing product partial pressures and increasing reactant partial pressures, so Kp decreases. 2. If the forward reaction is endothermic (ΔH > 0): increasing temperature shifts equilibrium to the right, so Kp increases.

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5

Why changing pressure does not change kp (quotient qp argument)

If total pressure P_total is increased on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (4 moles of gas → 2 moles of gas): 1. Kp stays constant because temperature has not changed. 2. Immediately upon compression, all partial pressures increase, and since there are more gas moles on the denominator (power 4) than the numerator (power 2), the reaction quotient Qp falls below Kp. 3. To restore Qp back up to Kp, the system shifts to the right (toward fewer gas moles), increasing the mole fraction of NH₃ and decreasing the mole fractions of N₂ and H₂.

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6

Effect of Catalysts on Kp

A catalyst (such as solid iron in the Haber process or V₂O₅ in the Contact process) provides an alternative pathway of lower activation energy (Eₐ), increasing the rate of both the forward and reverse reactions by the exact same factor. Therefore, a catalyst has zero effect on the position of equilibrium, the equilibrium mole fractions, or the value of Kp.

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1. Q1

   In an experiment, 1.00 mol of nitrogen (N₂) and 3.00 mol of hydrogen (H₂) were heated in a sealed vessel at 700 K and a total equilibrium pressure of 20.0 MPa: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). At equilibrium, 0.80 mol of ammonia (NH₃) had been formed. Calculate the mole fraction and partial pressure of each gas at equilibrium, and calculate the value of Kp (stating its units). (6 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Use the 1:3:2 stoichiometry to find equilibrium moles: since +0.80 mol of NH₃ formed, 0.40 mol of N₂ reacted and 1.20 mol of H₂ reacted; therefore at equilibrium: n(N₂) = 1.00 - 0.40 = 0.60 mol, n(H₂) = 3.00 - 1.20 = 1.80 mol, and n(NH₃) = 0.80 mol [1 mark]. 2. Total equilibrium moles of gas = 0.60 + 1.80 + 0.80 = 3.20 mol [1 mark]. 3. Mole fractions: x(N₂) = 0.60 ÷ 3.20 = 0.1875; x(H₂) = 1.80 ÷ 3.20 = 0.5625; x(NH₃) = 0.80 ÷ 3.20 = 0.2500 [1 mark]. 4. Partial pressures (x × 20.0 MPa): p(N₂) = 3.75 MPa; p(H₂) = 11.25 MPa; p(NH₃) = 5.00 MPa [1 mark]. 5. Substitute into Kp = (p(NH₃))² ÷ [p(N₂) × (p(H₂))³] = (5.00)² ÷ [3.75 × (11.25)³] = 25.0 ÷ 5339.36 = 4.68 × 10⁻³ [1 mark]. 6. Units of Kp = (MPa)² ÷ (MPa)⁴ = MPa⁻² [1 mark].

2. Q2

   Dinitrogen tetroxide dissociates into nitrogen dioxide in an endothermic equilibrium: N₂O₄(g) ⇌ 2NO₂(g) (ΔH = +58 kJ mol⁻¹). At temperature T₁ and a total pressure of 100 kPa, the equilibrium mixture has Kp = 48.0 kPa. Explain in terms of Kp why increasing the total pressure to 200 kPa at constant temperature T₁ causes the equilibrium yield of NO₂ to decrease. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Kp = (p(NO₂))² ÷ p(N₂O₄), and because temperature T₁ is kept constant, the numerical value of Kp remains unchanged at 48.0 kPa [1 mark]. 2. Since p(gas) = x(gas) × P_total, we can write Kp = [(x(NO₂))² ÷ x(N₂O₄)] × P_total [1 mark]. 3. When P_total doubles from 100 kPa to 200 kPa, the ratio (p(NO₂))² / p(N₂O₄) temporarily becomes larger than Kp (because partial pressure is squared in the numerator) [1 mark]. 4. To keep Kp constant at 48.0 kPa when P_total increases, the mole fraction ratio (x(NO₂))² / x(N₂O₄) must decrease, so the equilibrium shifts to the left (converting NO₂ back into N₂O₄) [1 mark].

3. Q3

   At 1000 K, the equilibrium constant Kp for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is 3.40 × 10⁻² kPa⁻¹. In an equilibrium mixture at 1000 K, the partial pressure of SO₂ is 42.0 kPa and the partial pressure of SO₃ is 28.0 kPa. Calculate the equilibrium partial pressure of O₂ and the total pressure of the mixture. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Write the Kp expression: Kp = (p(SO₃))² ÷ [(p(SO₂))² × p(O₂)] [1 mark]. 2. Rearrange for p(O₂): p(O₂) = (p(SO₃))² ÷ [(p(SO₂))² × Kp] [1 mark]. 3. Substitute values: p(O₂) = (28.0)² ÷ [(42.0)² × 3.40 × 10⁻²] = 784 ÷ (1764 × 0.0340) = 784 ÷ 59.976 = 13.1 kPa [1 mark]. 4. Calculate total pressure P_total = p(SO₂) + p(SO₃) + p(O₂) = 42.0 + 28.0 + 13.1 = 83.1 kPa [1 mark].

4. Q4

   Methanol is manufactured industrially from synthesis gas: CO(g) + 2H₂(g) ⇌ CH₃OH(g) (ΔH = -91 kJ mol⁻¹). Operating conditions of 520 K, 5-10 MPa pressure, and a Cu/ZnO/Al₂O₃ catalyst are used. Explain the economic and kinetic compromises involved in choosing these conditions. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Temperature compromise (520 K): Because the forward reaction is exothermic (ΔH = -91 kJ mol⁻¹), a lower temperature increases Kp and gives a higher equilibrium yield of methanol, but too low a temperature makes the rate of reaction too slow; 520 K achieves a fast rate with an acceptable yield [1 mark]. 2. Pressure compromise (5-10 MPa): There are 3 moles of gas on the left and 1 mole of gas on the right, so high pressure shifts equilibrium right (increasing yield) and increases collision frequency (faster rate) [1 mark]. 3. However, extremely high pressures require expensive thick-walled steel pipes, high energy costs for gas compressors, and pose safety risks, so a moderate pressure of 5-10 MPa is used [1 mark]. 4. The heterogeneous catalyst lowers the activation energy (Eₐ), allowing a fast rate of reaction at a lower temperature (where Kp and equilibrium yield are higher) without affecting Kp itself [1 mark].

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