A remarkably simple technique can be applied to solve for the equilibrium compositions of species. It is most effective when only a gas phase is present. This technique recognizes the simplicity of the fundamental problem of minimizing the Gibbs energy at equilibrium. By expressing the total Gibbs energy of the mixture in terms of its ideal solution components, we can simply request that the value of the Gibbs energy be minimized. The Gibbs energy of the mixture is calculated by Eqn. 10.42 and the needed chemical potential (partial molar Gibbs energy) is given by Eqn. 10.59:
where the last equality assumes all components are ideal gases. If we take the reference state as the elements in their natural form at the standard state, then, at the standard state pressure, 
. However, frequently the reactions are not at standard state pressure. The pressure effect on Gibbs energy is given by Eqn. 9.17. When the pressure effect is added,
Combining Eqns. 17.67 and 17.68 results in
To find equilibrium compositions, we just need to minimize Eqn. 17.69 by varying the mole numbers ni of each component while simultaneously satisfying the atom balance. Note that the mole fractions in the equation will also change as the mole numbers are varied. We do not need to explicitly write out the reactions. This method assumes that equilibrium is reached by whatever system of reactions is necessary. Most process simulators provide Gibbs minimization as a process unit.
Example 17.12. Butadiene by Gibbs minimization
Review Example 17.3(a) where steam is used to enhance conversion for 1-butene dehydrogenation. Gibbs energies of formation at 900 K for the hydrocarbons are summarized in that example.
The Gibbs energy of formation for water at 900 K is –198.204 kJ/mol. Vary conversion by selecting values of the reaction coordinate, calculating the Gibbs energy by Eqn. 17.69, and plotting the total Gibbs energy as a function of reaction coordinate. Demonstrate that Gibbs energy is minimized. Compare the equilibrium composition with that found in Example 17.3(a).
Solution
The initial moles of feed are 1 mol of 1-butene and 10 moles of steam. As an example calculation, select ξ = 0.1. Then the material balance provides, nC4H8 = 0.9, nC4H6 = nH2 = 0.1, nH2O = 10. The mole fractions are yC4H8 = 0.9/(0.9 + 2(0.1) + 10) = 0.08108, yC4H6 = yH2 = 0.0090, yH2O = 0.9009, and inserting the quantities into Eqn. 17.69, gives (inserting components in the order given above)
Repeating the calculation at various extents of reaction results in the following plot:
Careful analysis would show that the minimum is at ξ = 0.784 as in the earlier example. Note the changes in values are a small percentage of the absolute values of the total Gibbs energy, a numerical observation that is important in setting up convergence in Excel.
The previous example is somewhat contrived because the reaction was specified and the reaction coordinate was varied. The method is applicable without specifying reactions as long as the atom material balances are satisfied when selecting mole numbers.
Example 17.13. Direct minimization of the Gibbs energy with Excel
Apply the Gibbs minimization method to the problem of steam cracking of ethane at 1000 K and 1 bar where the ratio of steam to ethane in the feed is 4:1. Determine the distribution of C1 and C2 products, neglecting the possible formation of aldehydes, carboxylic acids, and higher hydrocarbons.
Solution
The solution is obtained using the worksheet GIBBSMIN contained in the workbook Rxns.xlsx (see Fig. 17.6).
Figure 17.6. Worksheet GIBBSMIN from the workbook Rxns.xlsx for Example 17.13.
Workbook, Rxns.xlsx, Worksheet GIBBSMIN, or MATLAB Gibbsmin.m.
One fundamental problem and one practical problem remain to be faced; there are several constraints that must be respected during the minimization process. These are the atom balances. We must keep in mind that we are not destroying matter, only rearranging it. So the number of carbon atoms, say, must be the same at the beginning and end of the process. Atom balance constraints must be written for every atom present. The atom balances are given straightforwardly by the stoichiometry of the species. For this example the balances are as follows.
O-balance: nOfeed – 2nCO2 –nCO –2nO2 – nH2O = 0
H-balance: nHfeed – 4nCH4– 4nC2H4 – 2nC2H2 – 6nC2H6 – 2nH2 – 2nH2O = 0
C-balance: nCfeed – nCH4 – 2nC2H4 – 2nC2H2 – nCO2 – nCO – 2nC2H6 = 0
The practical problem that remains is that the numerical solver often attempts to substitute negative values for the prospective species. This problem is easily treated by solving for the log(ni) during the iterations and determining the values of ni after the solution is obtained. Large negative values for the log(ni)s cause no difficulty. They simply mean that the concentrations of those species are small.
In order to apply Gibbs minimization, the Gibbs energy of formation is required for each component at the reaction temperature. This preliminary calculation is the same type of calculation as performed in Example 17.4 on page 653, but is not shown here. For example, the Gibbs energy of methane is simply the Gibbs energy of the formation reaction C(s) + 
at 1000 K. The values are embedded in the calculation of Gi shown in the worksheet in Fig. 17.6.
The primary product of this particular process is hydrogen. Fracturing hydrocarbons is a common problem in the petrochemical industry. This kind of process provides the raw materials for many downstream processes. The extension of this method to other reactions is straightforward. Some examples of interest would include several systems with environmental applications: carbon monoxide and NOx from a catalytic converter, or by-products from catalytic destruction of chlorinated hydrocarbons. Evaluating the equilibrium possibilities by this method is so easy that it should be a required preliminary calculation for any gas phase process reaction study.
Example 17.14. Pressure effects for Gibbs energy minimization
Apply the Gibbs energy minimization method to the methanol synthesis reaction using stoichio-metric feed at 50 bar and 500 K. The reaction has been discussed in Example 17.1 on page 643, and in Section 17.5 on page 649.
Solution
It is convenient to first find ΔGof,1 bar and then find Gi/RT for each species, and then apply these values in the Gibbs minimization.
For a basis of 1 mole CO: nC,feed = 1 = nCO + nMeOH; nH,feed = 4 = 2nH2 + 4nMeOH; nO,feed = nCO + nMeOH; the C-balance is redundant with the O-balance so only one of these should be included as a constraint to improve convergence. Minimizing the Gibbs energy gives yMeOH = 0.42 in agreement with the other method in Section 17.5 on page 649.
Note: The objective function changes weakly with mole numbers near the minimum, so tighten the convergence criteria or re-run the Solver after the first convergence. Convergence is sensitive to the initial guess. An initial guess which works is log(ni) = –0.1 for all i.
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