Consider the water-gas-shift reaction:
$$
\mathrm{H}_2(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})+\mathrm{CO}(g)
$$
At high temperatures and low to moderate pressures the reacting species form an idealgas mixture. Application of the summability equation to Eq. (11.26) yields:
$$
G=\sum_i y_i G_i+R T \sum_i y_i \ln y_i
$$
When the Gibbs energies of the elements in their standard states are set equal to zero, $G_i=\Delta G_{f_i}^{\circ}$ for each species, and then:
$$
G=\sum_i y_i \Delta G_{f_i}^{\circ}+R T \sum_i y_i \ln y_i
$$
At the beginning of Sec. 13.2 we noted that Eq. (14.64) is a criterion of equilibrium. Applied to the water-gas-shiftreaction with the understandingthat $T$ and $P$ are constant, this equation becomes:
$$
d G^t=d(n G)=n d G+G d n=0 \quad n \frac{d G}{d \varepsilon}+G \frac{d n}{d \varepsilon}=0
$$
Here, however, $d n / d \varepsilon=0$. The equilibrium criterion therefore becomes:
$$
\frac{d G}{d \varepsilon}=0
$$
Once the $y_i$ are eliminated in favor of $\varepsilon$, Eq. (A) relates G to $E$. Data for $\Delta G_{f_i}^{\circ}$ for the compounds of interest are given with Ex. 13.13. For a temperature of $1000 \mathrm{~K}$ (the reaction is unaffected by $P$ ) and for a feed of $l \mathrm{~mol} \mathrm{H}_2$ and $l \mathrm{~mol} \mathrm{CO}_2$ :
(a) Determine the equilibrium value of $\varepsilon$ by application of Eq. (B).
(b) Plot G vs. $\varepsilon$, indicating the location of the equilibrium value of $\varepsilon$ determined in (a).