Gas Equilibrium with Changing Mole Number

Introduction

For gas-phase equilibria, the natural variable is partial pressure rather than concentration. If the number of gas molecules changes during reaction, the total amount of gas depends on how far the reaction has gone, and so do the mole fractions. Handling this correctly is the main source of difficulty in gas equilibrium numericals. This page builds the method from the ideal-gas law and Dalton's law of partial pressures, both met earlier in the unit.

Core explanation

Partial pressures from mole fractions. In an ideal-gas mixture, each gas contributes pᵢ = xᵢP, where xᵢ = nᵢ / n total and P is the total pressure. The equilibrium constant in pressure terms is, for aA ⇌ bB, Kp = p B^b / p A^a.

Total amount changes with extent. Consider N₂O₄(g) ⇌ 2NO₂(g), starting from 1 mol of N₂O₄ with a degree of dissociation α.

N₂O₄ NO₂ Total --- --- --- --- Initial / mol 1 0 1 Change / mol −α +2α +α Equilibrium / mol 1 − α 2α 1 + α

The mole fractions are (1 − α)/(1 + α) and 2α/(1 + α). Multiplying by P and substituting:

Kp = [2αP/(1 + α)]² / [(1 − α)P/(1 + α)] = 4α²P / (1 − α²)

Pressure dependence at constant total pressure. Take Kp = 0.115 (pressures in bar) at 298 K. At P = 1.00 bar: 4α²/(1 − α²) = 0.115, so 4.115α² = 0.115, α² = 0.0279 and α = 0.167. At P = 10.0 bar: 4α²/(1 − α²) = 0.0115, giving α = 0.054. Raising the pressure suppresses the dissociation that increases the number of molecules, exactly as Le Chatelier's principle predicts — but now with numbers. Kp itself has not changed; only the composition has.

Constant volume is different. In a rigid vessel at fixed temperature, partial pressure is proportional to amount (pᵢ = nᵢRT/V). You can then run the ICE table directly in partial pressures. Starting with N₂O₄ at an initial pressure p₀, the equilibrium partial pressures are p₀ − y and 2y, and the total pressure is p₀ + y. A measured rise in total pressure therefore gives y directly. At constant pressure, by contrast, the volume changes as the reaction proceeds and you must go through mole fractions.

Choosing a basis. It is usually convenient to take 1 mol of feed, or the actual moles given, and carry the extent as a symbol. Only at the end convert to mole fractions and partial pressures. Always check that the partial pressures add up to the total pressure.

Formulae

pᵢ = xᵢP; xᵢ = nᵢ / n total; n total = n₀ + Δν·ξ, where Δν is the change in gas moles per unit extent. For A ⇌ 2B from pure A: Kp = 4α²P / (1 − α²). For A ⇌ B + C from pure A: Kp = α²P / (1 − α²). In a rigid vessel: P total = Σpᵢ and each pᵢ changes in the coefficient ratio.

Step-by-step reasoning

1. Write the ICE table in moles, including a total column. 2. Express each mole fraction in terms of the extent or α. 3. Multiply by the total pressure to get partial pressures. 4. Substitute into Kp and solve for the extent. 5. Check that partial pressures sum to P and that all amounts are positive.

Visual explanation

Imagine a piston holding the gas at constant pressure. As each N₂O₄ molecule splits into two NO₂ molecules, the piston rises because there are more particles. In a rigid box the walls cannot move, so the same splitting shows up as a rising pressure gauge instead. The two pictures correspond to the two calculation routes.

Real-world analogy

A lift that holds a fixed number of people per square metre (constant pressure) gets bigger as groups split into individuals; a lift with fixed walls (constant volume) simply becomes more crowded. The fraction of the crowd who are "individuals" depends on which kind of lift you are in.

Real-world example

Ammonia synthesis, N₂ + 3H₂ ⇌ 2NH₃, reduces the number of gas molecules from four to two. Plant designers exploit this by operating at high pressures, commonly around 100–250 bar, because the equilibrium mole fraction of ammonia rises with pressure even though Kp does not change.

Why?

Why does pressure change the composition but not Kp? Kp depends only on temperature. When Δn ≠ 0, the total pressure appears in the expression for Kp in terms of mole fractions (as P^Δn), so the mole fractions must adjust to keep Kp constant when P changes.

Common misconception

"The partial pressure of each gas is its initial pressure minus the change." That shortcut only works at constant volume and temperature. At constant total pressure the volume changes, so you must use mole fractions of the new total amount.

Worked example

Question: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At a certain temperature and a total pressure of 2.00 bar, PCl₅ is 50.0% dissociated. Calculate Kp.

Reasoning: Take 1 mol PCl₅. Equilibrium: PCl₅ 0.500, PCl₃ 0.500, Cl₂ 0.500 mol; total 1.500 mol. Each mole fraction is 1/3, so each partial pressure is 2.00/3 = 0.667 bar. Kp = (0.667 × 0.667) / 0.667 = 0.667 bar. Check with the formula: α²P/(1 − α²) = 0.25 × 2.00 / 0.75 = 0.667.

Answer: Kp = 0.667 bar.

Quick check

1. A rigid flask initially contains N₂O₄ at 0.500 bar. At equilibrium the total pressure is 0.600 bar. What is the partial pressure of NO₂? Answer: Total = 0.500 + y = 0.600, so y = 0.100 bar and p(NO₂) = 2y = 0.200 bar.

Exam focus

Show the total-moles column explicitly; many candidates lose marks by dividing by the initial amount instead of the equilibrium total. State pressure units and specify whether the vessel is at constant pressure or constant volume, because this determines the route.

Advanced insight

The general result Kp = Kx·P^Δn, where Kx is the constant written in mole fractions, shows why pressure matters only when Δn ≠ 0. For real gases at hundreds of bar, fugacity coefficients replace partial pressures; for ammonia synthesis these corrections are significant, so industrial design uses measured non-ideal data.

Summary

When a gas reaction changes the number of molecules, the total amount depends on the extent, so mole fractions must be computed over the equilibrium total. Partial pressures are mole fraction times total pressure, and they give Kp. At constant pressure, composition shifts with P while Kp stays fixed; in a rigid vessel, partial pressures change directly in the coefficient ratio and the total pressure reveals the extent.

Practice questions

1. Using the Quick check data, calculate Kp for N₂O₄ ⇌ 2NO₂ at that temperature. Answer: p(N₂O₄) = 0.400 bar and p(NO₂) = 0.200 bar, so Kp = 0.200² / 0.400 = 0.100 bar. 2. A mixture of 1.00 mol N₂ and 3.00 mol H₂ reacts until the extent is 0.40 mol. Find the mole fractions. Answer: N₂ 0.60, H₂ 1.80, NH₃ 0.80 mol; total 3.20 mol; mole fractions 0.188, 0.563 and 0.250. 3. For PCl₅ ⇌ PCl₃ + Cl₂ with Kp = 0.667 bar, find α at a total pressure of 1.00 bar. Answer: α²/(1 − α²) = 0.667, so α² = 0.667/1.667 = 0.400 and α = 0.632. 4. Explain why the degree of dissociation of N₂O₄ falls when the total pressure is raised from 1 bar to 10 bar. Answer: Kp = 4α²P/(1 − α²) is fixed, so a larger P requires a smaller α; the equilibrium shifts towards fewer gas molecules.