Multireaction equilibria

Independent reactions, combining equilibrium constants, the multi-extent method and Gibbs-energy minimisation, with a methane reforming-plus-shift example.

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Why it matters

Real reactors rarely run a single reaction. Steam reforming couples methane reforming with the water-gas shift; combustion, gasification, cracking and isomerisation all involve several simultaneous equilibria. Getting the product slate right (for example the H₂/CO ratio for synthesis gas) needs all the reactions solved together, either by one extent per independent reaction or by minimising the total Gibbs energy directly.

Key ideas

Independent reactions. Only reactions that cannot be written as combinations of others count. For a set of species built from a set of elements, the number of independent reactions is usually the number of species minus the number of elements (strictly, minus the rank of the element-species matrix). For CH₄, H₂O, CO, CO₂ and H₂ (elements C, H, O): 5 − 3 = 2 independent reactions, for example reforming and shift. Any third reaction (such as CH₄ + 2H₂O ⇌ CO₂ + 4H₂) is their sum and gives no new information.

Combining reactions. If reaction 3 = reaction 1 + reaction 2, then ΔG°₃ = ΔG°₁ + ΔG°₂ and K₃ = K₁·K₂. Reversing a reaction inverts K; multiplying a reaction by n raises K to the power n.

Extent method. Assign one extent εⱼ to each independent reaction j. Then nᵢ = nᵢ₀ + Σⱼ νᵢ,ⱼ εⱼ and n = n₀ + Σⱼ νⱼ εⱼ. Each reaction must satisfy its own equilibrium condition, Πᵢ (f̂ᵢ/fᵢ°)^νᵢ,ⱼ = Kⱼ, so there are as many nonlinear equations as unknown extents. For ideal gases this becomes Πᵢ yᵢ^νᵢ,ⱼ = Kⱼ(P/P°)^(−νⱼ). The equations are solved simultaneously, usually by Newton's method or by nesting one-variable searches; every mole number must stay non-negative.

Gibbs-energy minimisation. Alternatively, minimise G^t = Σnᵢμᵢ subject to element balances, using Lagrange multipliers λₖ: for each species, ΔG°f,ᵢ + RT ln(yᵢφ̂ᵢP/P°) + Σₖ λₖaᵢₖ = 0, with aᵢₖ the number of atoms of element k in species i. No reactions need to be written, which is convenient for many species (combustion, gasification) and is how simulator "Gibbs reactors" work.

What changes the outcome. Each Kⱼ responds to temperature through its own ΔH°ⱼ (van't Hoff), so raising T can favour one reaction and suppress another: reforming (endothermic) is favoured at high T while the shift (exothermic) is favoured at low T, which is why reformers run hot and shift converters cool. Pressure acts on each reaction according to its νⱼ.

Conservation checks. After solving, verify element balances (C, H, O atoms in equal out) and that each Kⱼ is reproduced. These two checks catch most algebra errors.

Formulas

Number of independent reactions = N_species − rank(element matrix) (usually N_species − N_elements) nᵢ = nᵢ₀ + Σⱼ νᵢ,ⱼ·εⱼ, n = n₀ + Σⱼ νⱼ·εⱼ, νⱼ = Σᵢ νᵢ,ⱼ Πᵢ yᵢ^νᵢ,ⱼ = Kⱼ·(P/P°)^(−νⱼ) (ideal gases, each reaction j) Kⱼ = exp(−ΔG°ⱼ/(R·T)) K₃ = K₁·K₂ when reaction 3 = reaction 1 + reaction 2 ΔG°f,ᵢ + R·T·ln(yᵢ·φ̂ᵢ·P/P°) + Σₖ λₖ·aᵢₖ = 0 (Gibbs minimisation, each species) Σᵢ nᵢ·aᵢₖ = Aₖ (element balance, each element k)

Symbols: nᵢ moles of species i (mol); εⱼ extent of reaction j (mol); νᵢ,ⱼ stoichiometric number of i in reaction j; yᵢ mole fraction; Kⱼ equilibrium constant of reaction j (dimensionless); P (bar), P° = 1 bar; ΔG°ⱼ standard Gibbs energy change (J/mol); λₖ Lagrange multiplier for element k (J/mol); aᵢₖ atoms of element k per molecule of i; Aₖ total atoms of element k (mol); R = 8.314 J/(mol·K); T (K).

Worked examples

Example 1 (standard): parallel isomerisation. Given: in the liquid phase, A ⇌ B (K₁ = 1.5) and A ⇌ C (K₂ = 4.0) reach equilibrium from pure A; treat the solution as ideal. Find the equilibrium mole fractions.

  1. Ideal solution, ν = 0 for both: K₁ = x_B/x_A and K₂ = x_C/x_A.
  2. x_A + x_B + x_C = 1 gives x_A(1 + K₁ + K₂) = 1, so x_A = 1/6.5 = 0.1538.
  3. x_B = 1.5 × 0.1538 = 0.2308; x_C = 4.0 × 0.1538 = 0.6154.
  4. Note that B and C are in the ratio K₁:K₂ regardless of the starting point. Answer: x_A = 0.154, x_B = 0.231, x_C = 0.615

Example 2 (GATE level): methane steam reforming with shift. Given: feed 1 mol CH₄ + 2 mol H₂O at 1 bar and a temperature where (1) CH₄ + H₂O ⇌ CO + 3H₂ has K₁ = 26.6 and (2) CO + H₂O ⇌ CO₂ + H₂ has K₂ = 1.37. Ideal gases. Find the equilibrium composition.

  1. Species 5, elements 3: two independent reactions, extents ε₁ and ε₂.
  2. Moles: CH₄ = 1 − ε₁; H₂O = 2 − ε₁ − ε₂; CO = ε₁ − ε₂; CO₂ = ε₂; H₂ = 3ε₁ + ε₂; total n = 3 + 2ε₁ (ν₁ = +2, ν₂ = 0).
  3. Equations: y_CO·y_H₂³/(y_CH₄·y_H₂O) = K₁(P/P°)^(−2) = 26.6 and y_CO₂·y_H₂/(y_CO·y_H₂O) = 1.37.
  4. Solve simultaneously (for example, for a trial ε₁ find ε₂ from the shift equation, then adjust ε₁ until the reforming equation holds): ε₁ = 0.9577, ε₂ = 0.2477.
  5. Moles: CH₄ 0.0423, H₂O 0.7946, CO 0.7101, CO₂ 0.2477, H₂ 3.1209; total 4.9155.
  6. Mole fractions: CH₄ 0.0086, H₂O 0.1617, CO 0.1445, CO₂ 0.0504, H₂ 0.6349.
  7. Checks: C atoms 0.0423 + 0.7101 + 0.2477 = 1.000; O atoms 0.7946 + 0.7101 + 2 × 0.2477 = 2.000; substituting the y values reproduces K₁ = 26.6 and K₂ = 1.37. Answer: CH₄ conversion 95.8%; H₂/CO molar ratio = 3.1209/0.7101 = 4.40

Common mistakes

  • Writing a dependent reaction as an extra equation and getting an inconsistent or singular system.
  • Forgetting that a species appearing in two reactions takes contributions from both extents (CO in Example 2 is ε₁ − ε₂).
  • Using n₀ instead of n₀ + Σνⱼεⱼ as the total moles.
  • Accepting a root with a negative mole number.
  • Multiplying K values when reactions are added but forgetting to raise K to a power when a reaction is scaled.
  • Skipping the element-balance check at the end.

For GATE CH

  • NAT problems: overall K from component reactions, number of independent reactions for a species list, compositions for simple parallel or series equilibria in closed form.
  • Two-reaction gas-phase problems with simple numbers, where one reaction has ν = 0 so pressure cancels.
  • Conceptual MCQs on Gibbs minimisation, element balances and how temperature affects coupled endothermic and exothermic reactions.
  • Practise writing the mole table with two extents and checking it against element balances.

Quick check

  1. How many independent reactions relate CH₄, H₂O, CO, CO₂ and H₂?
  2. If reaction 3 = 2 × reaction 1 − reaction 2, how is K₃ related to K₁ and K₂?
  3. For A ⇌ B (K₁ = 2) and B ⇌ C (K₂ = 3) in an ideal liquid starting from pure A, what is x_A at equilibrium?
  4. What constraints are used in the Gibbs-minimisation method instead of reaction equations?
  5. Why is steam reforming run at high temperature but the shift reaction at lower temperature?

Answers: 1. Two. 2. K₃ = K₁²/K₂. 3. 1/(1 + 2 + 6) = 1/9 = 0.111. 4. Element (atom) balances. 5. Reforming is endothermic (K rises with T); the shift is exothermic (K falls with T).

Interview questions

All Chemical Engineering Thermodynamics interview questions

Try answering each one aloud before you open it.

  1. 1.What is multireaction equilibria in chemical engineering thermodynamics?Concept

    Multireaction equilibria refer to the state where multiple chemical reactions occur simultaneously and reach equilibrium. In this state, the rate of the forward reactions equals the rate of the reverse reactions for all reactions involved. This concept is crucial in processes where multiple reactions are interdependent, such as in catalytic converters or complex reaction networks.

  2. 2.Explain the significance of the Gibbs free energy in determining multireaction equilibria.Concept

    At constant T and P a closed reacting system reaches equilibrium at the minimum of its total Gibbs energy, subject to conservation of each element. That minimum is equivalent to Σᵢνᵢ,ⱼμᵢ = 0 for every independent reaction j, i.e. each reaction satisfies its own K = exp(−ΔG°ⱼ/RT). Simulators often minimise G^t directly with Lagrange multipliers for the element balances, which avoids having to choose a reaction set.

  3. 3.How do you apply the equilibrium constant to multireaction equilibria?Concept

    The equilibrium constant (K) is used to express the ratio of product concentrations to reactant concentrations at equilibrium. In multireaction equilibria, each reaction has its own equilibrium constant. These constants help determine the concentrations of species at equilibrium and are used in conjunction with mass balance equations to solve for unknowns in complex reaction systems.

  4. 4.Why is the stoichiometric matrix important in analyzing multireaction equilibria?Application

    The stoichiometric matrix is a mathematical representation of the stoichiometry of a set of reactions. It is important because it helps in organizing and solving the mass balance equations for multireaction systems. By using the stoichiometric matrix, engineers can systematically analyze the relationships between different species and reactions, making it easier to solve for equilibrium concentrations.

  5. 5.What happens if the temperature of a multireaction system is increased?Application

    Increasing the temperature of a multireaction system generally affects the equilibrium position of the reactions involved. According to Le Chatelier's principle, the system will adjust to counteract the change. For endothermic reactions, the equilibrium will shift towards the products, while for exothermic reactions, it will shift towards the reactants. The exact effect depends on the enthalpy changes of the individual reactions.

  6. 6.How does pressure affect multireaction equilibria involving gases?Application

    Pressure changes can significantly affect multireaction equilibria involving gases. According to Le Chatelier's principle, increasing pressure will shift the equilibrium towards the side with fewer moles of gas, while decreasing pressure will favor the side with more moles of gas. This is because the system seeks to minimize the change in pressure by adjusting the number of gas molecules.

  7. 7.Explain how catalysts influence multireaction equilibria.Application

    Catalysts speed up the rate of chemical reactions without being consumed in the process. In multireaction equilibria, catalysts lower the activation energy for both forward and reverse reactions equally, allowing the system to reach equilibrium faster. However, they do not change the position of the equilibrium; they only affect the rate at which equilibrium is achieved.

  8. 8.Calculate the equilibrium concentrations for a system with the following reactions: A ⇌ B (K1 = 2) and B ⇌ C (K2 = 3), given initial concentrations [A] = 1 mol/L, [B] = 0 mol/L, [C] = 0 mol/L.Numerical

    At equilibrium [B] = 2[A] and [C] = 3[B] = 6[A]. The total is conserved: [A] + [B] + [C] = 1 mol/L, so 9[A] = 1. Hence [A] = 0.111 mol/L, [B] = 0.222 mol/L and [C] = 0.667 mol/L. The initial 1 mol/L is the total, not the equilibrium value of [A].

  9. 9.For the reaction system: 2A + B ⇌ C (K = 4, concentration basis in mol/L), if the initial concentrations are [A] = 2 mol/L, [B] = 1 mol/L, and [C] = 0 mol/L, find the equilibrium concentration of C.Numerical

    With x mol/L of C formed: [A] = 2 − 2x, [B] = 1 − x, [C] = x. Then 4 = x/[(2 − 2x)²(1 − x)] = x/[4(1 − x)³], i.e. 16(1 − x)³ = x. Solving numerically gives x = 0.655, so [C] = 0.655 mol/L, [A] = 0.689 mol/L and [B] = 0.345 mol/L; substituting back reproduces K = 4.

  10. 10.What role does the reaction quotient play in multireaction equilibria?Concept

    The reaction quotient (Q) is a measure of the relative amounts of products and reactants present during a reaction at any point in time. In multireaction equilibria, Q is compared to the equilibrium constant (K) to predict the direction in which the reaction will proceed to reach equilibrium. If Q < K, the reaction will proceed forward, producing more products. If Q > K, the reaction will proceed in reverse, producing more reactants.

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