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 \(\mathrm{H_2O}\)
, 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 [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:
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 \(\mathrm{H_2O} \rightleftharpoons \mathrm{OH} + \frac{1}{2} \mathrm{H_2}\)
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.