Calculting equilibrium composition of a mixture of gases - Exercise 16-89 from Çengel’s Thermodynamics book (7th ed)

This is exercise 16-89 from [1]: a mixture of 1 mol of ${\mathrm{H}}_2\mathrm{O}$, 2 mols of ${\mathrm{O}}_2$ and 5 mols of ${\mathrm{N}}_2$ is heated up to 2200 K and 5 atm of pressure. The final composition will have the formation of ${\mathrm{H}}_2$ as well; what is the equilibrium composition, and is it reasonable to assume there will be no $\mathrm{OH}$ in the final composition?

I teach several classes on heat engines, steam generators and internal combustion engines, and for me the prime application of chemical thermodynamics and equilibrium composition is emissions. For a given combustion reaction, will pollutants such as $\mathrm{NO}$ and $\mathrm{CO}$ be formed? How much of them? The exercise explained above is a nice example of such calculations.

All calculations below assume all components are ideal gases.

First, we need to write the reaction:

$${\mathrm{H}}_2\mathrm{O}+2{\mathrm{O}}_2+5{\mathrm{N}}_2\to x{\mathrm{H}}_2\mathrm{O}+y{\mathrm{O}}_2+z{\mathrm{N}}_2+w{\mathrm{H}}_2$$

where the unknowns $x,y,v,w$ form a vector to be found with an appropriate system of equations. Since we have three elements ($\mathrm{N},\mathrm{O},\mathrm{H}$), or equivalently $({\mathrm{N}}_2,{\mathrm{O}}_2,{\mathrm{H}}_2)$), we can write three mass balances:

For hydrogen gas:

$$x+w-1=0$$

For oxygen gas:

$$2.5-0.5x-y=0$$

For nitrogen gas:

$$5-z=0$$

(Do you have any doubts on why these equations are as such?)

One missing equation has to be determined from equilibrium considerations. Looking at tables of equilibrium constants, I posit that the free hydrogen gas is formed by dissociation:

$${\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_2+\frac{1}{2}{\mathrm{O}}_2$$ whose equilibrium constant at 2200 K is $\ln K_p=-6.768$.

The definition of equilibrium coefficient for this latter equation is [1]:

$$K_p=\frac{N_{{\mathrm{H}}_2}^{{\nu }_{{\mathrm{H}}_2}}{N}_{{\mathrm{O}}_2}^{{\nu }_{{\mathrm{O}}_2}}}{N_{{\mathrm{H}}_2\mathrm{O}}^{{\nu }_{{\mathrm{H}}_2\mathrm{O}}}}{\left(\frac{P}{N_{\mathrm{total}}}\right)}^{{\nu }_{{\mathrm{H}}_2}+{\nu }_{{\mathrm{O}}_2}-{\nu }_{{\mathrm{H}}_2\mathrm{O}}}$$ where the ${\nu }_i$ are the stoichiometric coefficients in the dissociation equation, the $N_i$ are the real molar contents in the actual system where the dissociation occur, and $P$ is the pressure in atm:

$$K_p=\frac{w^1{y}^{0.5}}{x^1}{\left(\frac{5}{x+y+z+w}\right)}^{0.5}$$

Take a minute and see if you can understand this equation. It took me almost an hour to properly understand it.

The vector of unknowns then is the solution of a 4-dimensional function, and this problem can be solved numerically with R:

library(rootSolve)
model <- function(a) c(F1 = 1-a[1] - a[4], 
                       F2 = 2.5-0.5*a[1]-a[2],
                       F3 = 5-a[3],
                       F4 = ((a[4]*a[2]**0.5)/(a[1])*(5/(sum(a)))^0.5) -exp(-6.768))

ss <- multiroot(f = model, start = c(1,1,1,1), positive=TRUE)
print(ss)

$root
[1] 0.998972571 2.000513714 5.000000000 0.001027429

$f.root
           F1            F2            F3            F4 
 1.692006e-15 -1.332268e-15  0.000000e+00  1.414678e-13 

$iter
[1] 4

$estim.precis
[1] 3.612301e-14

Keep in mind that I had to tweak this code until it worked. For instance, if you write the fourth equation in terms of logarithms, and not with exponentials as I did, you might have some numerical problems (try!). Also, the solution is sensitive to initial conditions; the final solution makes sense, as oxygen and nitrogen are practically preserved and some of the water vapor in fact dissociates.

As for the underlying assumption: will there be $\mathrm{OH}$ in the final composition? The equilibrium constant for ${\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{OH}+\frac{1}{2}{\mathrm{H}}_2$ at 2200 K is $\ln K_p=-7.148$. The dissociation constants from water vapor to ${\mathrm{H}}_2$ and $\mathrm{OH}$ are very similar, and hence the reactions will occur in parallel, giving some amount of $\mathrm{OH}$ in the products, contrary to our assumptions. If you actually perform an experiment similar to this problem and encounter some errors in the final composition, this is a likely source of deviations.

The smaller the value of $K_p$, the harder is for the reaction to occur; you can see from the above equations that this coefficient is a measure of how much products form from the reactants. Hence, for similar $K_p$, the amount of ${\mathrm{H}}_2$ and $\mathrm{OH}$ formed will also be similar.

References

[1]: Çengel, Y. A., & Boles, M. A. Termodinâmica (7 ed.). Porto Alegre: AMGH, 2013.