Lesson 14 of 15

Chemical & Phase Equilibrium

The Gibbs function as the arbiter of equilibrium: equilibrium constants and how they shift with temperature and pressure, dissociation in hot flames, nitric oxide, and the same principle deciding how phases share a substance.

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The question

Lesson 13 burned methane completely: every carbon atom to CO₂, every hydrogen to water. Real flames never quite manage it. Measure the gases in a stoichiometric flame and you find carbon monoxide, hydrogen, hydroxyl radicals, free oxygen atoms and a few thousand parts per million of nitric oxide. Nobody made a mistake. Why doesn't burning ever quite finish?

What does equilibrium minimise?

An isolated system heads toward maximum entropy (Lesson 6). But a flame, a chemical reactor or a glass of iced water is rarely isolated: it is held at a fixed temperature and pressure by its surroundings. For such a system, combining the first and second laws gives a new criterion. With TT and pp fixed,

dGT,p≤0,dG_{T,p} \le 0,

where G=H−TSG = H - TS is the Gibbs function GG = H − TS. At fixed temperature and pressure, a system reaches equilibrium where G is a minimum.. Every spontaneous change at constant temperature and pressure lowers GG, and equilibrium is the state of minimum GG. The two terms pull in opposite directions: a low enthalpy favours strong chemical bonds (CO₂ and water), a high entropy favours many, scattered molecules (CO, O₂, H, O). Temperature sets the exchange rate between them. At room temperature enthalpy wins outright. In a flame at 2500 K, entropy claims a share.

ExploreThe bottom of the Gibbs valley
00.20.40.60.81extent of dissociation ξG − G(pure CO₂) [RT]equilibrium ξ = 0.295
2800 K
1.00 atm
Equilibrium constant K0.1499log₁₀K = -0.824
Mole fractionsCO₂ 61.5%CO 25.7% · O₂ 12.8%
Challenges0/2
  • Find conditions where half the CO₂ has dissociated.
  • At 3000 K, use pressure alone to cut the dissociation below 10%.
One mole of CO₂ held at temperature T and pressure p, allowed to dissociate: CO₂ ⇌ CO + ½O₂. The curve is the Gibbs function of the mixture against the extent of reaction ξ (the fraction of CO₂ dissociated), relative to pure CO₂, in units of RT. Equilibrium is wherever G is lowest — the same answer the equilibrium constant gives.

The equilibrium constant

For a reaction between ideal gases, νAA+νBB⇌νCC+νDD\nu_A A + \nu_B B \rightleftharpoons \nu_C C + \nu_D D, setting dG=0dG = 0 at the minimum gives the Equilibrium constant KK(T) = exp(−ΔG°/RT): fixes the balance between reactants and products at equilibrium. It depends on temperature only.:

K(T)=yCνC yDνDyAνA yBνB(ppref)νC+νD−νA−νB,ln⁡K=−ΔG∘RˉT.K(T) = \frac{y_C^{\nu_C}\,y_D^{\nu_D}}{y_A^{\nu_A}\,y_B^{\nu_B}}\left(\frac{p}{p_\text{ref}}\right)^{\nu_C + \nu_D - \nu_A - \nu_B}, \qquad \ln K = -\frac{\Delta G^\circ}{\bar R T}.

Here ΔG∘\Delta G^\circ is the change in Gibbs function for the reaction with each gas pure at 1 atm. KK depends only on temperature, and it is tabulated — Moran and Shapiro's Table A-27 is the standard list:

ToolEquilibrium constants, read off a chart
1000150020002500300035004000-20-15-10-50temperature (K)log₁₀ KCO₂ ⇌ CO + ½O₂H₂ ⇌ 2HO₂ ⇌ 2OH₂O ⇌ H₂ + ½O₂½N₂ + ½O₂ ⇌ NON₂ ⇌ 2N
2500 K
log₁₀ K-1.438CO₂ ⇌ CO + ½O₂
Heat of reaction275 kJ/molpositive: absorbs heat
Challenges0/1
  • Find the temperature at which K for water dissociation reaches 0.01.
log₁₀K against temperature for six reactions important in flames, computed from the course’s ideal-gas Gibbs functions (they agree with Moran & Shapiro’s Table A-27 to within a few hundredths over most of the range). Every curve rises with temperature: dissociation reactions absorb heat, so heating drives them forward — van ’t Hoff’s rule.

Two rules tell you how the equilibrium shifts:

  • Temperature. For reactions that absorb heat, like every dissociation, KK rises steeply with temperature. The van 't Hoff equation makes it precise: dln⁡K/dT=ΔH/RˉT2d\ln K/dT = \Delta H/\bar R T^2.
  • Pressure. Raising the pressure pushes the equilibrium toward the side with fewer moles of gas. Dissociation increases the number of moles, so high pressure suppresses it. That is Le Chatelier's principle, falling straight out of the pΔνp^{\Delta\nu} factor.

Hot flames don't finish burning

Put these ideas inside a flame. At 2200 K, CO₂ and H₂O are partly dissociated into CO, H₂ and O₂. Dissociation absorbs energy, so the equilibrium flame temperature is lower than the complete-combustion one, and the products contain CO and H₂ even with enough oxygen to burn them. As the gas cools downstream, they mostly recombine, releasing their energy — the heat is delayed, not lost.

And the nitrogen, Lesson 13's inert ballast, is not quite inert. At flame temperatures, the equilibrium 12N2+12O2⇌NO\tfrac12\mathrm{N_2} + \tfrac12\mathrm{O_2} \rightleftharpoons \mathrm{NO} shifts appreciably to the right:

LabWhy burning never quite finishes
0.50.60.70.80.91140016001800200022002400equivalence ratio φflame temperature (K)0200040006000NO (ppm)
0.900
1.0 atm
Flame temperature2162 K24 K below complete combustion
Equilibrium CO0.26%
Equilibrium NO3563 ppmat 2162 K
Challenges0/2
  • Find a lean mixture that keeps equilibrium NO below 1000 ppm at the flame.
  • Find where dissociation costs the flame more than 75 K — and see what pressure does to it.
A lean-to-stoichiometric methane–air flame, with the products allowed to reach chemical equilibrium (CO₂ ⇌ CO + ½O₂ and H₂O ⇌ H₂ + ½O₂) while satisfying the energy balance. Yellow: complete-combustion flame temperature; red: equilibrium. Purple, right axis: equilibrium nitric oxide, ½N₂ + ½O₂ ⇌ NO, at the flame temperature.

That explains a puzzle about NOₓ emissions. At exhaust temperatures, equilibrium would leave almost no NO. But the reactions that make and destroy NO are slow; once the gas leaves the flame and cools, they freeze, and the NO formed in the hottest zone survives into the atmosphere, where it causes smog and acid rain. That is why low-NOₓ gas-turbine combustors burn lean, to keep the peak temperature down, and why their residence time at high temperature is kept short. Equilibrium thermodynamics tells you the destination; chemical kinetics tells you whether there is time to get there.

Check your intuition
A gas-turbine combustor is redesigned to run at 30 atm instead of 15 atm, at the same flame temperature. What happens to the equilibrium dissociation of CO₂?
Estimate first
At 3000 K and 1 atm, what fraction of pure water vapour is dissociated into H₂ and O₂, in percent?
0.0100.10110100
Drag along the line to place your guess — each tick is ten times the last.

Phase equilibrium

The same principle governs how a substance shares itself between phases. Each component in each phase has a chemical potential, μi=(∂G/∂ni)T,p\mu_i = (\partial G/\partial n_i)_{T,p}: the change in GG per mole added. Matter flows from high chemical potential to low until they are equal — exactly as heat flows from high temperature to low until temperatures are equal.

For a pure substance with liquid and vapour, equilibrium means gf=ggg_f = g_g: the Gibbs function is the same in both phases. That is what fixed the vapour pressure in Maxwell's equal-area construction (Lesson 11), and differentiating it along the saturation line gives the Clapeyron equation. It also explains why water evaporates from a glass in dry air, why salt lowers the freezing point of water, and how a distillation column separates crude oil into petrol, kerosene and diesel.

For mixtures with several components and phases, the Gibbs phase rule counts the freedom left over:

F=2+N−P,F = 2 + N - P,

where NN is the number of components and PP the number of phases. Pure water (N=1N = 1) as liquid and vapour (P=2P = 2) has F=1F = 1: fix the temperature and the pressure is fixed too — the saturation curve. At the triple point, with ice, liquid and vapour together, F=0F = 0: a single, unique state, 0.01 °C and 611.7 Pa, which is why it defines the Kelvin scale.

Going deeper
Going deeper: computing complex equilibria

Real combustion products involve a dozen species — CO₂, CO, H₂O, H₂, O₂, N₂, OH, H, O, N, NO and more — linked by many simultaneous reactions. Rather than writing an equilibrium constant for each, modern codes such as NASA's CEA or Cantera minimise the total Gibbs function directly, subject to conservation of each element, using Lagrange multipliers. The result is the same; the method scales to hundreds of species. The flame widget above uses two reactions plus NO, which captures the main effects for lean and stoichiometric methane flames; including OH, H and O would lower the stoichiometric flame temperature by roughly another 30 K.

Recap
What to carry forward
  • At fixed temperature and pressure, systems move toward minimum Gibbs function, G=H−TSG = H - TS.
  • For ideal-gas reactions, ln⁡K=−ΔG∘/RˉT\ln K = -\Delta G^\circ/\bar RT, with mole fractions and pressure combined in KK.
  • Temperature shifts equilibria according to the heat of reaction (van 't Hoff); pressure favours the side with fewer moles (Le Chatelier).
  • Hot flames are partly dissociated: lower flame temperatures, CO and H₂ in the products, and NO from nitrogen — which freezes in as the gas cools.
  • Phase equilibrium: equal chemical potentials in each phase; the Gibbs phase rule counts the degrees of freedom.

That completes the theory. Next: put it all to work. Can you design a power plant that beats the competition?

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