Mole Fractions after a Gas Reaction

Introduction

When gases react, their final composition is rarely the same as their starting composition. A limiting-reagent calculation gives final amounts of each species; only then can mole fractions and partial pressures be calculated. A reaction may also change the total number of gas moles. Using initial total moles in a final-state pressure or fraction is a subtle but serious error.

Core explanation

For a single reaction, write nᵢ,final = nᵢ,initial + νᵢξ, where νᵢ is negative for a reactant and positive for a product. Include only gas-phase species when summing n gas,final. Final gas mole fraction is xᵢ,final = nᵢ,final/n gas,final. If a gas mixture is ideal, its final partial pressure is Pᵢ = xᵢ,final P total, and P total = n gas,final RT/V at a stated T and V. These equations describe a final state after the extent ξ has been determined; they do not by themselves establish how far a reversible reaction proceeds.

For 2CO(g) + O₂(g) → 2CO₂(g), each mole of reaction as written consumes three moles of gaseous reactants and forms two moles of gaseous product. If all named substances remain gaseous, the net gas-mole change per mole extent is Δν gas = 2 − 2 − 1 = −1. Therefore n gas,final = n gas,initial − ξ. If a solid or liquid appears in a different reaction, its coefficient affects chemical amount and mass balance but does not contribute to n gas for a gas-phase fraction or ideal-gas pressure.

Take initial 2.00 mol CO and 1.50 mol O₂ with no CO₂. Maximum extents are 2.00/2 = 1.00 mol and 1.50/1 = 1.50 mol, so CO limits at ξ = 1.00 mol for complete reaction. Final amounts are CO = 0, O₂ = 1.50 − 1.00 = 0.50 mol, and CO₂ = 2.00 mol. The final gas total is 2.50 mol, compared with 3.50 mol initially. Final mole fractions are xO2 = 0.50/2.50 = 0.20 and xCO2 = 2.00/2.50 = 0.80. Dividing by the original 3.50 mol would make the fractions add to only 0.714, signalling an incorrect denominator.

If the vessel is rigid and returns to its initial temperature, total pressure changes in proportion to total gas moles in the ideal model: P final/P initial = 2.50/3.50 = 5/7. If temperature also changes during an exothermic reaction, include T final/T initial; reaction heating can initially raise pressure even though the number of gas moles falls. State which state is being compared. For a vented vessel, pressure may remain externally controlled while volume or gas discharge changes instead.

Reaction extent may be supplied directly, inferred from a product measurement, or limited by a reactant. For an equilibrium problem, a final extent must satisfy the equilibrium relation rather than simply reaching the limiting-reagent maximum. Whatever determines ξ, the inventory-and-fraction calculation is the same after ξ is known.

Step-by-step reasoning

1. Balance the reaction and mark the phases of every species. 2. Convert starting gas amounts to moles and determine ξ from the stated conversion or reaction condition. 3. Calculate each final mole amount with its signed stoichiometric change. 4. Sum only final gas-phase amounts, including inert gas if present. 5. Divide by the new total for mole fractions, then calculate partial pressures from the correct final total pressure.

Visual explanation

Sketch a before-and-after table for 2CO + O₂ → 2CO₂. Before: two CO counters and one-and-a-half O₂ counters. After complete CO consumption: zero CO, half an O₂ counter and two CO₂ counters. Put a bracket around 3.50 total gas moles before and 2.50 after, then write xCO2 = 2.00/2.50 below the after table.

Real-world analogy

Suppose a club has 35 members, then several leave and others join. The share belonging to one group after those changes must use the new membership total. Using the old membership count gives fractions that no longer sum to one. Gas composition has the same denominator rule after a reaction changes the population of molecules.

Real-world example

An exhaust-gas calculation begins with fuel and oxygen amounts, uses combustion stoichiometry to find products and unused oxygen, and then reports dry exhaust mole fractions. If water forms, a dry exhaust analysis excludes condensed water while a wet analysis includes water vapour. The chosen phase basis changes every fraction denominator.

Why?

Why can total pressure drop at fixed V and T after this reaction? Three gas molecules on the reactant side are replaced by two on the product side for each reaction packet. In the ideal model, fewer final gas particles per volume produce lower total pressure at the same absolute temperature.

Common misconception

“Use the initial total gas moles when finding final fractions because mass is conserved.” Mass is conserved, but molecule count need not be. The balanced coefficients explicitly show a change in total gas moles. Recalculate the total after reaction and phase changes.

Worked example

In a 25.0 L rigid vessel at 300 K, 2.00 mol CO and 1.50 mol O₂ react completely by 2CO + O₂ → 2CO₂; assume the vessel returns to 300 K. CO limits at ξ = 1.00 mol. Final gas amounts are 0 mol CO, 0.50 mol O₂ and 2.00 mol CO₂, total 2.50 mol. Therefore xO2 = 0.200 and xCO2 = 0.800. With R = 0.08206 L atm mol⁻¹ K⁻¹, P final = 2.50(0.08206)(300)/25.0 = 2.46 atm. Thus P O2 = 0.492 atm and P CO2 = 1.97 atm. Their sum equals total pressure within rounding.

Quick check

1. For 2CO + O₂ → 2CO₂, what is the net change in gas moles per mole of reaction as written? Answer: Three gaseous reactant moles become two gaseous product moles, so Δν gas = −1 mol per mole extent.

Exam focus

Write a final inventory before calculating a mole fraction. Include inert gas in total if present, and omit condensed liquid from a gas-phase denominator. Distinguish the pressure change due to total gas amount from any temperature change.

Advanced insight

At constant V and T, a measured pressure change can reveal reaction extent when Δν gas is known: ΔP = Δν gas ξRT/V for a closed ideal-gas system. This inference fails if gas leaks, condenses, dissolves or changes temperature without correction. It is a useful bridge between reaction stoichiometry and observable pressure.

Summary

After finding ξ, update every gas amount, recalculate the final gas total, and form final mole fractions from that new denominator. Partial pressures follow from the final total pressure. Mass conservation does not imply conservation of gas-molecule count.

Practice questions

1. With 2.00 mol CO and 1.50 mol O₂ initially, how much O₂ remains after complete reaction? Answer: CO permits ξ = 1.00 mol, consuming 1.00 mol O₂, so 0.50 mol remains. 2. What fraction of the final gas is CO₂ in that mixture? Answer: CO₂ is 2.00 mol out of 2.50 mol total, so xCO2 = 0.800. 3. If an inert 1.00 mol Ar had been present throughout, what would xCO2 become? Answer: Final total would be 2.50 + 1.00 = 3.50 mol; xCO2 = 2.00/3.50 ≈ 0.571.