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

Exercise from Çengel’s Thermodynamics

Author

Fábio P. Fortkamp

Published

March 17, 2022

This is exercise 16-89 from [1]: 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.

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 [1]:

\[
K_p = \frac{N_{\mathrm{H_2}}^{\nu_{\mathrm{H_2}}} N_{\mathrm{O_2}}^{\nu_{\mathrm{O_2}}}}{N_{\mathrm{H_2O}}^{\nu_{\mathrm{H_2O}}}}\left(\frac{P}{N_{\mathrm{total}}} \right)^{\nu_{\mathrm{H_2}} + \nu_{\mathrm{O_2}} - \nu_{\mathrm{H_2O}}}
\] 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:

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.

References

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