Questions: Chemical Equilibrium and Equilibrium Constant

Question 1

A combustion engineer assumes that burning methane at 2500 K produces only CO₂ and H₂O. A more accurate equilibrium analysis reveals significant CO and OH in the products. What is the primary reason the simplified assumption fails?

• A. Methane combustion is incomplete due to insufficient oxygen supply
• B. At high temperatures, K_p for dissociation reactions becomes large enough that CO₂ and H₂O partially reverse to CO, OH, and other species
• C. The reaction has not reached equilibrium at 2500 K
• D. High pressure prevents complete conversion to CO₂ and H₂O

Answer: (B)

The van't Hoff equation shows that K_p increases with temperature for endothermic reactions. The dissociation of CO₂ → CO + ½O₂ and H₂O → OH + ½H₂ are endothermic, so at high combustion temperatures their K_p values become significant — the equilibrium composition includes substantial minor species. Assuming only CO₂ and H₂O ignores these equilibrium shifts, causing serious error in energy and emissions calculations.

Question 2

For a reaction with ΔG° = −20 kJ/mol at 1000 K, which statement correctly describes K_p?

• A. K_p = 0, meaning the reaction does not proceed
• B. K_p < 1, meaning reactants are favored at equilibrium
• C. K_p > 1, meaning products are favored at equilibrium
• D. K_p = 1, meaning the mixture is at equal partial pressures

Answer: (C)

The relationship ΔG° = −RT ln K_p directly links standard Gibbs free energy to the equilibrium constant. A negative ΔG° means −RT ln K_p < 0, so ln K_p > 0, so K_p > 1 — products are thermodynamically favored. The quantitative relationship shows not just direction but how far toward products the equilibrium lies.

Question 3

For an exothermic combustion reaction, increasing the reaction temperature will shift the equilibrium toward greater product yield.

• T. True
• F. False

Answer: False

This is exactly backwards. The van't Hoff equation d(ln K_p)/dT = ΔH_rxn/(RT²) shows that for an exothermic reaction (ΔH_rxn < 0), K_p decreases as temperature rises. Le Chatelier's principle confirms this: adding heat to an exothermic reaction shifts equilibrium toward reactants. High combustion temperatures actually cause dissociation of CO₂ and H₂O back into intermediates — a critical consideration in combustor design.

Question 4

The equilibrium composition of a reacting gas mixture at constant T and P corresponds to the minimum of the Gibbs free energy over all possible compositions.

• T. True
• F. False

Answer: True

This is the fundamental thermodynamic principle underlying chemical equilibrium. At constant T and P, any spontaneous process decreases G, and the equilibrium state is the composition where G is minimized. Both the K_p approach and direct Gibbs minimization are computational implementations of this same physical principle — they yield identical equilibrium compositions.

Question 5

Why does finding the equilibrium composition of real combustion products at high temperature require iterative numerical solution rather than a simple stoichiometric calculation?

Model answer: Real combustion involves simultaneous equilibria among many species (CO₂, H₂O, CO, OH, H₂, O, NO, etc.). Each equilibrium is described by a nonlinear K_p expression, and all must be satisfied simultaneously along with atom-balance constraints. The resulting system of coupled nonlinear equations has no closed-form solution — compositions must be assumed, K_p equations and mass balances evaluated, adjustments made, and the process repeated until convergence.

Multiple equilibria are coupled: changing the amount of one species affects the partial pressures of all others, shifting every other equilibrium. Unlike a single-reaction problem solvable by a quadratic, real mixtures with many species require matrix-based or iterative methods such as Gibbs minimization with Lagrange multipliers.