 Products of complete combustion (PCC) are useful fo simple calculations.
 Can be considered an upper bound on the extent of reaction.
 Useful for defining heating values and providing a combustion basis.
 At high temperatures products dissociate to form $H_2$, $CO$, $H$, $O$, $OH$, etc.
Criteria for equilibrium

Consider a fixed mass at equilibrium

$dU = dQ + dW$
 $dW = PdV$
 at equilibrium differential changes are reversible, so $dQ = TdS$.

$dU = TdS  PdV$

Assume const $T$ and $P$, then we can move $T$ and $P$ inside the derivatives:
 $dU = d(TS)  d(PV)$.

Collect terms:
 $d(U+PVTS) = d(HTS)= dG = 0$.

At equilibrium $$dG_{T,P} = 0$$

Hence $G$ is a critical point.
 When not at equilibrium changes are not reversible and $dS > dQ/T$ so $dQ < TdS \rightarrow dG>0$,
 Hence $G$ is a minimum at equilibrium.
 That is, for a reversible process, we have $dS=dQ/T$. For an irreversible process we have $dS>dQ/T$ or $dS+\Delta = dQ/T\rightarrow dQ = TdS + T\Delta$. Inserting into the first expression for $dU$ gives $dG=T\Delta$ or $dG>0$. So, at the equilibrium point irreversible changes give an increase in $G$, implying $G$ is a minimum at the equilibrium point.

Note equilibrium is a state not a process.
 The $T,P$ in $dG_{T,P}=0$ does not mean that $T$ and $P$ are somehow constant.
 Rather, if you look in directions of constant $T$ and $P$, then the $G$ curve will be minimum at the equilibrium point.
Other equivalent equilibrium conditions
 Above, we had
$$dU = TdS  PdV$$
 For constant $T$, $P$, we have $dG_{T,P}=0$.
 For constant $S$, $V$, we have $dU_{S,V}=0$.
 For constant $S$, $P$, we have $dU=d(PV)\rightarrow dH_{S,P}=0$.
 For constant $T$, $V$, we have $dU=d(TS)\rightarrow dA_{T,V}=0$, where $A$ is Helmholz free energy.
 All of these four expressions are true of an equilibrium state at a given temperature and pressure.
 Similarly, all of these four expressions are true at a given $S$ and $V$, etc.
 Again, this is because equilibrium is a state, not a process.
Water gas shift (WGS) equilibrium

For some rich mixtures, augmenting products of complete combustion with water gas shift equilibrium gives an accurate approximation to full equilibrium. $$CO + H_2O \Leftrightarrow CO_2 + H_2$$

Following Turns, we can write:
$$C_xH_y + aO_2 + 3.76aN_2 \rightarrow bCO_2 + cCO + dH_2O + eH_2 + 3.76aN_2$$ *
 Here, $a=(x+y/4)/\Phi$, where $\Phi$ is equivalence ratio.
 We have 4 unknowns: $b$, $c$, $d$, $e$.
 We have 4 constraints: three elemental balances (C, H, O) and the WGS reaction equilibrium constraint.
 C: $b+c = x$
 H: $2d + 2e = y$
 O: $2b + c + d = 2a$
 WGS reaction: $K_{eq} = (b\cdot e)/(c\cdot d)$
 $K_{eq} = \Pi_i(P_i/P_0)^{\nu_i} = \Pi_i(x_iP/P_0)^{\nu_i}$.
 For WGS reaction, $\sum_i\nu_i=0$ so $P$, $P_0$ and total moles cancel giving $K_{eq}=(b\cdot e)/(c\cdot d)$
 Also, $K_{eq} = \exp(\Delta G_{rxn}^o(T)/RT)$
 The first three equations can be combined with the last equation to give a quadratic expression in a single variable $b$.
 This is then solved analytically for $b$, from which $c$, $d$, and $e$ are recovered from the first three expressions.
 The total moles is then found, followed by the mole fraction, and finally mass fractions.
 To find $K_{eq}$, temperature must be known. This can be found using $h=h(T,y_i)$, where $H$ is a known function of the stoichiometry, i.e., $h=(1\xi)h_{\xi=0} + (\xi)h_{\xi=1}$.
 This will require iteration to converge both $T$ and $y_i$.
 Note, this only works up to $\xi_{st,CO}$, which is the stoichiometric mixture fraction for the reaction
$$C_xH_y + \frac{x}{2}O_2 + 3.76\frac{x}{2}N_2 \rightarrow xCO + \frac{y}{2}H_2 + 3.76\frac{x}{2}N_2$$
 For methaneair, $\xi_{st,CO}=0.189$.
Example
 A comparison of full equilibrium, products of complete combustion, and water gas equilibrium is given here .
 The WGS equilibrium gives very good results up to $\xi<0.15$ for methaneair combustion.
 Note the very large difference in the adiabatic temperature between PCC and full equilibrium. Errors as high as 800 K occur at $\xi=0.16$.
 Soot formation is active in this region, and is sensitive to temperature.
 Radiation depends on $T^4$, and on soot, hence accurate representations of the temperature profile are of importance in many practical applications.