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Equilibrium I
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:
$d(U+PV-TS) = d(H-TS)= dG = 0$.
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.
\(CO + H_2O \Leftrightarrow CO_2 + H_2\)
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
\(C_xH_y + \frac{x}{2}O_2 + 3.76\frac{x}{2}N_2 \rightarrow xCO + \frac{y}{2}H_2 + 3.76\frac{x}{2}N_2\)