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

This is exercise 16-89 from : a mixture of 1 mol of $$\mathrm{H_2O}$$, 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:

$$\mathrm{H_2O} + 2\mathrm{O_2} + 5\mathrm{N_2} \to x\mathrm{H_2O} + y\mathrm{O_2} + z\mathrm{N_2} + w\mathrm{H_2}$$

where the unknowns $${x,y,v,w}$$ form a vector to be found with an appropriate system of equations. Since we have three elements ($\mathrm{N,O,H}$), or equivalently $$(\mathrm{N_2,O_2,H_2})$$), 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:

$$\mathrm{H_2O} \rightleftharpoons \mathrm{H_2} + \frac{1}{2} \mathrm{O_2}$$ whose equilibrium constant at 2200 K is $$\ln K_p = -6.768$$.

The definition of equilibrium coefficient for this latter equation is :

$$K_p = \frac{N_{\mathrm{H_2}}^{\nu_{\mathrm{H_2}}} N_{\mathrm{O_2}}^{\nu_{\mathrm{O_2}}}}{N_{\mathrm{H_2O}}^{\nu_{\mathrm{H_2O}}}}\left(\frac{P}{N_{\mathrm{total}}} \right)^{\nu_{\mathrm{H_2}} + \nu_{\mathrm{O_2}} - \nu_{\mathrm{H_2O}}}$$ 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 - a,
F2 = 2.5-0.5*a-a,
F3 = 5-a,
F4 = ((a*a**0.5)/(a)*(5/(sum(a)))^0.5) -exp(-6.768))

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

## $root ##  0.998972571 2.000513714 5.000000000 0.001027429 ## ##$f.root
##            F1            F2            F3            F4
##  1.692006e-15 -1.332268e-15  0.000000e+00  1.414678e-13
##
## $iter ##  4 ## ##$estim.precis
##  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 $$\mathrm{H_2O} \rightleftharpoons \mathrm{OH} + \frac{1}{2} \mathrm{H_2}$$ 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.

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