Material balances with chemical reaction and extent of reaction

Why it matters

Reactors are the heart of a chemical plant, and every reactor calculation starts with a material balance that includes generation and consumption. The extent of reaction turns a reactor balance into a few lines of algebra – one unknown per reaction instead of one per species – and makes multi-reaction systems such as steam reforming, partial oxidation or nitration manageable.

Key ideas

Balance with reaction. For each species i: input + generation − output − consumption = accumulation. At steady state, output = input + net generation. Total mass is conserved, but total moles are not (N₂ + 3H₂ → 2NH₃ loses 2 mol per mol N₂ reacted).

Extent of reaction ξ. For a single reaction, every species changes in proportion to its stoichiometric coefficient ν_i (negative for reactants, positive for products): n_i = n_i,0 + ν_i·ξ. ξ has units of moles (or mol/s for a continuous process) and is the same for all species, so one number describes the reaction's progress. ξ = 0 means no reaction; the maximum ξ is set by the limiting reactant, ξ_max = n_lim,0 / |ν_lim|.

Relation to conversion. For reactant A, ξ = X_A·n_A,0 / |ν_A|. Knowing either the conversion or the amount of any one species formed fixes ξ.

Multiple reactions. With R independent reactions, n_i = n_i,0 + Σ_j ν_ij·ξ_j. Use one extent per independent reaction, not one balance per reaction; a reaction that is a linear combination of others is not independent and adds no new extent.

Three equivalent methods.

1. Molecular species balances (with generation/consumption terms).
2. Extent of reaction (most compact for given conversions).
3. Atomic (element) balances: atoms are neither created nor destroyed, so input = output for C, H, O, N… without any reaction terms. Useful when reactions are unknown or numerous (e.g. combustion). Only independent element balances count. All three give the same answer; an element balance is a good check of an extent calculation.

Inerts. Species that do not react (N₂ in air, argon in synthesis gas) pass through unchanged: n_out = n_in. They are useful tie components.

Equilibrium-limited reactions. If the reaction reaches equilibrium, the extent is fixed by K = Π (y_i·P/P°)^ν_i (ideal gas). Write all mole fractions in terms of ξ and solve – a common GATE pattern.

Formulas

• n_i = n_i,0 + ν_i·ξ – single reaction; n in mol or mol/s, ν_i signed coefficient, ξ extent.
• n_i = n_i,0 + Σ_j ν_ij·ξ_j – multiple reactions.
• n_total = n_total,0 + (Σ ν_i)·ξ – total moles change only if Σν ≠ 0.
• ξ = X_A·n_A,0 / |ν_A| – extent from conversion of reactant A.
• ξ_max = n_lim,0 / |ν_lim| – complete reaction of the limiting reactant.
• Σ (atoms of element k)_in = Σ (atoms of element k)_out – element balance.
• K = Π (y_i·P / P°)^ν_i – gas-phase equilibrium, ideal gas, P° = 1 bar.

Worked examples

Example 1 (standard). SO₂ is oxidised: 2SO₂ + O₂ → 2SO₃. The feed is 100 mol/s containing 10 % SO₂, 12 % O₂ and 78 % N₂. The SO₂ conversion is 90 %. Find the extent and the outlet composition.

1. Feed: SO₂ = 10, O₂ = 12, N₂ = 78 mol/s.
2. ξ = X·n_SO₂,0 / |ν_SO₂| = 0.90 × 10 / 2 = 4.5 mol/s.
3. SO₂ = 10 − 2 × 4.5 = 1.0; O₂ = 12 − 4.5 = 7.5; SO₃ = 0 + 2 × 4.5 = 9.0; N₂ = 78 mol/s.
4. Total = 100 + (−2 − 1 + 2) × 4.5 = 95.5 mol/s (check: 1 + 7.5 + 9 + 78 = 95.5 ✓).
5. Mole fractions: SO₂ 0.0105, O₂ 0.0785, SO₃ 0.0942, N₂ 0.8168.

Answer: ξ = 4.5 mol/s; outlet 1.0 SO₂, 7.5 O₂, 9.0 SO₃, 78 N₂ mol/s (95.5 mol/s)

Example 2 (GATE level). Steam reforming with water-gas shift: (1) CH₄ + H₂O → CO + 3H₂ (2) CO + H₂O → CO₂ + H₂. Feed: 100 mol/s CH₄ and 300 mol/s H₂O. Methane conversion is 80 % and the outlet contains 30 mol/s CO₂. Find the outlet flows and verify with element balances.

1. ξ₁ = 0.80 × 100 = 80 mol/s (CH₄ appears only in reaction 1).
2. ξ₂ = 30 mol/s (CO₂ appears only in reaction 2).
3. CH₄ = 100 − 80 = 20; H₂O = 300 − 80 − 30 = 190; CO = 80 − 30 = 50; H₂ = 3 × 80 + 30 = 270; CO₂ = 30 mol/s.
4. Total = 20 + 190 + 50 + 270 + 30 = 560 mol/s (= 400 + 2ξ₁ + 0·ξ₂ ✓).
5. Element check – C: in 100; out 20 + 50 + 30 = 100 ✓. O: in 300; out 190 + 50 + 2 × 30 = 300 ✓. H: in 4 × 100 + 2 × 300 = 1000; out 4 × 20 + 2 × 190 + 2 × 270 = 1000 ✓.
6. H₂ mole fraction = 270 / 560 = 0.482.

Answer: Outlet 20 CH₄, 190 H₂O, 50 CO, 270 H₂, 30 CO₂ mol/s; y_H₂ = 0.482

Common mistakes

• Using a positive coefficient for a reactant in n_i = n_i,0 + ν_i·ξ.
• Taking ξ equal to the moles of reactant consumed when its coefficient is not 1.
• Assuming total moles are conserved in a reacting system.
• Writing one extent per species, or using dependent reactions as if they were independent.
• Forgetting inerts in the outlet total, which changes all mole fractions.
• Calling the reactant with fewer moles "limiting" without dividing by its coefficient.

For GATE CH

Expect NAT questions giving a feed and a conversion (or one outlet flow) and asking for an outlet mole fraction, two-reaction systems with yield and selectivity, and equilibrium-conversion problems where mole fractions are written in terms of ξ. Practise building a species table with ξ and checking with element balances.

Quick check

1. For A + 3B → 2D with 5 mol A, 9 mol B and ξ = 2 mol, how much B remains?
2. For 2X + Y → Z with 6 mol X and 4 mol Y, which reactant is limiting and what is ξ_max?
3. In Example 1, why does the total molar flow fall?
4. How many extents are needed for three reactions of which one is the sum of the other two?
5. Which balances need no generation or consumption terms in a reacting system?

Answers: 1. 3 mol; 2. X, ξ_max = 3 mol; 3. Σν = −1, so each unit of extent removes 1 mol; 4. two; 5. element (atomic) balances and the total mass balance.