1. Mole Fractions & Partial Pressures in Gas Equilibria
For reactions taking place entirely in the gas phase, it is far more practical to measure pressures rather than molar concentrations. To formulate equilibrium expressions for gases, we use partial pressures.
x_A = n_A / n_total
The sum of all mole fractions in a gaseous mixture is always equal to 1: sum(x_i) = 1.
p_A = x_A * P_total
According to Dalton's Law of Partial Pressures, the total pressure of a mixture of gases equals the sum of the individual partial pressures:
P_total = p_A + p_B + p_C + ...
2. The Equilibrium Constant Kp & Unit Deduction
For a reversible gas-phase equilibrium:
a A(g) + b B(g) <=> c C(g) + d D(g)
The equilibrium constant in terms of partial pressures (Kp) is defined as:
Kp = (p(C)^c * p(D)^d) / (p(A)^a * p(B)^b)
Always write Kp expressions using round brackets with lower-case 'p' preceding the chemical formula: e.g. p(NH3)^2 or (p_NH3)^2. Never use square brackets [NH3] in a Kp expression, as square brackets strictly denote molar concentration (which belongs to Kc, not Kp).
Deducing the Units of Kp
Substitute the pressure unit given in the question (usually kPa, Pa, or MPa) into the Kp expression and simplify algebraic powers:
- Haber Process: N2(g) + 3H2(g) <=> 2NH3(g)
Kp = p(NH3)^2 / (p(N2) * p(H2)^3)
Units = (kPa)^2 / (kPa * (kPa)^3) = (kPa)^2 / (kPa)^4 = kPa-2 - Contact Process: 2SO2(g) + O2(g) <=> 2SO3(g)
Kp = p(SO3)^2 / (p(SO2)^2 * p(O2))
Units = (kPa)^2 / ((kPa)^2 * kPa) = kPa-1 - Ester/HI Decomposition: 2HI(g) <=> H2(g) + I2(g)
Kp = (p(H2) * p(I2)) / p(HI)^2
Units = (kPa * kPa) / (kPa)^2 = no units (dimensionless)
3. Gas Phase Equilibrium Diagram
The diagram below models an industrial Haber synthesis vessel at equilibrium, illustrating the relationship between mole fractions and partial pressures:
4. ICE Table Calculations for Kp
Quantitative problems are solved systematically using an ICE table (Initial, Change, Equilibrium):
- List initial moles of each gaseous reactant and product.
- Use stoichiometric ratios to determine the change in moles in terms of x.
- Calculate equilibrium moles for each component.
- Sum the equilibrium moles to find n_total.
- Divide each mole quantity by n_total to find mole fractions (x_i).
- Multiply mole fractions by P_total to calculate partial pressures (p_i).
- Substitute partial pressures into the Kp expression and evaluate.
5. Factors Affecting Kp: Temperature, Pressure & Catalysts
| Variable Changed | Effect on Equilibrium Position | Effect on Value of Kp | Chemical Explanation |
|---|---|---|---|
| Temperature Increase (Exothermic Reaction, delta H < 0) | Shifts left (towards reactants) | Decreases | System absorbs heat. Backward endothermic direction favored; partial pressures of products decrease while reactants increase, lowering Kp. |
| Temperature Increase (Endothermic Reaction, delta H > 0) | Shifts right (towards products) | Increases | System absorbs heat. Forward endothermic direction favored; product partial pressures rise, increasing Kp. |
| Pressure Increase | Shifts toward side with fewer gas moles | NO CHANGE | Kp is constant at constant temperature. An increase in total pressure changes the ratio of partial pressures momentarily; equilibrium shifts so the ratio returns to Kp. |
| Addition of Catalyst | No change | NO CHANGE | Catalyst increases the rates of both forward and reverse reactions by the same factor. Equilibrium is reached faster, but yields and Kp remain unaltered. |
6. Worked Calculations
N2O4(g) <=> 2NO2(g)
At equilibrium, 0.40 mol of NO2 has formed. The total pressure is 150 kPa.
Calculate Kp and state its units.
Step 1: Set up the ICE table:
- Initial moles: N2O4 = 1.00 mol; NO2 = 0.00 mol
- Change in moles: NO2 increases by +0.40 mol; N2O4 decreases by -(0.40 / 2) = -0.20 mol
- Equilibrium moles: N2O4 = 1.00 - 0.20 = 0.80 mol; NO2 = 0.40 mol
- Total equilibrium moles: n_total = 0.80 + 0.40 = 1.20 mol
Step 2: Calculate mole fractions:
- x(N2O4) = 0.80 / 1.20 = 0.6667
- x(NO2) = 0.40 / 1.20 = 0.3333
Step 3: Calculate partial pressures (p = x * P_total):
- p(N2O4) = 0.6667 * 150 kPa = 100.0 kPa
- p(NO2) = 0.3333 * 150 kPa = 50.0 kPa
Step 4: Substitute into Kp and find units:
Kp = p(NO2)^2 / p(N2O4) = (50.0)^2 / 100.0 = 2500 / 100.0 = 25.0 kPa
If the equilibrium partial pressure of PCl5 is 25 kPa and the partial pressures of PCl3 and Cl2 are equal, calculate the total pressure of the mixture.
Step 1: Write Kp expression:
Kp = (p(PCl3) * p(Cl2)) / p(PCl5)
Let p(PCl3) = p(Cl2) = y:
4.0 = y^2 / 25
y^2 = 4.0 * 25 = 100 => y = 10.0 kPa
Step 2: Calculate total pressure:
P_total = p(PCl5) + p(PCl3) + p(Cl2) = 25.0 + 10.0 + 10.0 = 45.0 kPa
Exam-Style Practice Questions
Test your understanding of these core syllabus concepts with targeted questions and detailed explanations.
Question 1: Which change will increase the numerical value of Kp for an exothermic gas reaction?
Show Answer & Explanation
Correct Answer: B
Temperature is the only factor that alters Kp. For an exothermic reaction (delta H < 0), lowering the temperature shifts the equilibrium in the exothermic forward direction to oppose the cooling, increasing the product yield and raising Kp.
Question 2: A gas mixture contains 0.20 mol of gas A, 0.30 mol of gas B, and 0.50 mol of gas C at a total pressure of 500 kPa. What is the partial pressure of gas B?
Show Answer & Explanation
Correct Answer: B
Total moles = 0.20 + 0.30 + 0.50 = 1.00 mol. Mole fraction of B = 0.30 / 1.00 = 0.30. Partial pressure p(B) = 0.30 * 500 kPa = 150 kPa.
Question 3: What happens to the numerical value of Kp when total pressure on the equilibrium 2SO2(g) + O2(g) <=> 2SO3(g) is doubled at constant temperature?
Show Answer & Explanation
Correct Answer: C
Kp is independent of pressure. Although the position of equilibrium shifts to the right (towards fewer moles of gas), the value of Kp itself remains strictly constant.
Question 4: For the gas equilibrium H2(g) + CO2(g) <=> H2O(g) + CO(g), what are the units of Kp?
Show Answer & Explanation
Correct Answer: D
Kp = (p(H2O) * p(CO)) / (p(H2) * p(CO2)) = (kPa * kPa) / (kPa * kPa) = 1. The pressure units cancel out completely, so Kp is dimensionless.