Gibbs Minimization Method

9. Gibbs Minimization Method

a. General Criteria for Equilibrium At a specified temperature and pressure, for any composite system, (dG T,P = Σµ k dN k )

≤ 0, where the equality holds at equilibrium. The Gibbs energy for a chemical reaction de- creases as the reaction progresses (cf. dG T,P <0) until it reaches a minimum value. Since, G =

G(T,P,N 1 ,N 2 ...), the criteria for the G minima are dG T,P = 0 and d 2 G T,P > 0. (80a,b) Consider methane–air combustion at specified temperature and pressure according to

the overall reaction CH 4 +a O 2 +b N 2 → c CO+d CO 2 +e O+f O 2 +g NO+h H 2 O+i H 2 +j N 2 +k C(s),

where a and b are known There are nine unknown species concentrations and four atom bal- ance equations for C, H, N, and O atoms. Therefore, five equilibrium relations are required, which may include the reactions

CH 4 → C(s) + 2 H 2 ,

(A)

CO 2 → CO + 1/2 O 2 ,

(B)

H 2 O → 1/2 O 2 +H 2 ,

(C) O 2 → 2 O, and (D)

N 2 +O 2 → 2 NO. (E) We may also select a linear combination of reactions. For instance, subtracting reaction (B)

from reaction (C) we obtain the reaction

H 2 O – CO 2 → 1/2 O 2 +H 2 – CO – 1/2 O 2 , i.e.,

(F) which is the familiar water gas shift reaction. Note that reaction (F) does not provide an inde-

H 2 O + CO → CO 2 +H 2 ,

pendent equilibrium relation that can be selected in addition to reactions (A)–(E). The solution procedure becomes far more complex as we encounter literally hundreds of species in a realis- tic applications. Therefore, it is useful to adopt a more general procedure during which the species concentrations are adjusted until the Gibbs energy reaches a minimum value at equilib- rium, i.e., dG

= 0 and d T,P 2 G T,P > 0 subject to the atom balance constraints and the specified temperature and pressure. The LaGrange multiplier method is useful in this regard.

t. Example 20

A piston–cylinder assembly contains two kmole of O 2 at 1 bar and 3000 K. The pres- sure and temperature are maintained constant. Chemical reaction proceeds and O at- oms are formed at the expense of O 2 . Determine the equilibrium composition starting with G= G(T, P, N O 2 ,N O ) and minimizing G subject to atom conservation.

Solution The O atom conservation equation is

2N O 2 +N O =4 (A) The Gibbs energy of the mixture must keep decreasing as the reaction proceeds and

equilibrium is achieved when it is at the minimum. G= Σµ k N k , i.e.,

(B)

(C) We will minimize Eq. (C) at the specified pressure and temperature subject to the

G = G(T, P, N O 2 ,N O )= µ O 2 N O 2 + µ O N O .

atom balance constraint Eq. (A). Using the LaGrange multiplier scheme

F = G(T,P,N O 2 ,N O )+ λ (2N O 2 +N O – 4) = 0, and ∂F/∂N O 2 = 0, ∂F/∂N O = 0. (D) From Eq. (D),

∂F/∂N O 2 = ∂G/∂N O 2 +2 λ = 0, i.e., µ O 2 +2 λ = 0 (and µ O + λ = 0). (E) Assuming the ideal mixture model to apply,

(F) Further, assuming ideal gas behavior,

µ O 2 = ˆg O 2 = g O 2 (T,P) + R T ln X O 2 .

(G) Therefore,

g O o 2 (T,P) = g O 2 + R T ln (P/1).

(H) µ

O 2 = g O 2 + R T ln (p O 2 /1), and

O = g O + R T ln (p O /1), so that (I)

(J) ( o g O + R T ln (p O /1)) + λ = 0.

( g O o 2 + R T ln (p O 2 /1)) + 2 λ = 0, and

(K) Multiplying Eq. (K) by 2 and subtracting it from Eq. (J),

O 2 –2 g O o + R T ln (p O 2 /1)– 2 R T ln (p O /1) = 0, i.e.,

(p o

O /1) /(p O 2 /1) = exp (–(2 g O – g O 2 )/( R T)) or (N O ) (P/N) /N O 2 =K , where (L) N=N O 2 +N O .

(M) The equilibrium constant

(N) With the values N O = 4 – 2N O 2 and N = 4 – 2N O 2 +N O2 =4–N O 2 , (4 – 2N

K = exp (– ∆G o / R T), where ∆G o =2 o g O – g O o 2 .

(O) the pressure P = 1 bar,

O 2 ) 2 (P/(1(4 – N O 2 )))/N O 2 =K .

(P) Now,

(N O 2 ) 2 –4N O 2 +4 2 /(4+K o ) = 0.

. We can solve for N O 2 and selecting the root, such that N O 2 > 0 and N O > 0, i.e., N

g o = –755102 kJ kmole –1 , and g o = –323359 kJ kmole –1 O 2 O

O 2 = 1.8875, N O = 0.225, and G min = –1.52 ×10 .

b. Multiple Components The solution can be explicitly obtained for two components. We will now generalize the methodology for multicomponent systems. The procedure to minimize G = G(T,P,N 1 , N 2 , …, N K ) subject to the atom balance equations is as follows. Formulate the atom balance equations for each element “j”

Σ k d jk N k =A j , j = 1, …, J, k = 1, …, K, (81) where d jk denotes the number of atoms of an element “j” in species k (e.g., for the element j=O

in species k =CO 2 , d = 2) and A j is the number of atoms of type j entering the reactor. This relation can be expressed using the La Grange multiplier method, i.e.,

λ j ( Σ k d jk N k –A j ), j = 1, …, J, k = 1, …, K. (82) The Gibbs energy at equilibrium must be minimized subject to this condition.

We create a function F=G+ Σ j λ j ( Σ k d jk N k –A j ), j = 1, …, J, k = 1, …, K, (83)

such that ∂F/∂N k =( ∂G/∂N k ) T,P +( Σ j λ j d jk ) T,P = 0, j = 1, …, J, k = 1, …, K, i.e., (84)

Next, we minimize F with respect to N k , k = 1, …, K, and for each species, µ k + Σ j λ j d jk = 0, j = 1, …, J, k = 1, …, K.

elemental potentials in species k. Example 21

A steady flow reactor is fired with 1 kmole of C 10 H 20 , with 5 % excess air. The spe- cies (1 to 5) leaving are CO, CO 2 ,H 2 H 2 O, OH, O 2 , NO, and N 2 at T= 2500 K and 1 bar. Determine equilibrium composition of species leaving the reactor. Assume ideal gas behavior.,

Solution The stoichiometric amount of O 2 can be determined from the stoichiometric relation (C 10 H 20 ) + 15 (O 2 + 3.76 N 2 ) = 10 CO 2 + 10 H 2 O + 56.4 N 2 With 5% excess air, the O 2 supplied = 16.5 kmole, the N 2 supplied is 62.04 kmole.

(C 10 H 20 ) + 15. 8 (O 2 + 3.76 N 2 ) → Products.

This system is an open system. Now we follow a fixed mass (140+528+ 1737= 2405 kg) as it travels the reactor. We will assume that this mixture is instantaneously heated to 2500 K at 1 bar and then calculate assuming various values for the moles of the 8 species subject to the atom conservation relations for C, H, N, and O. We will then select the composition at which G has a minimum value at this T and P using the LaGrange multiplier method to arrive at the composition. The four elements C, H, N and O are denoted by the subscript j and the eight species

denoted by subscript k. The coefficients d jk , i.e., d 11 = 1, d 12 = 1, … are provided in the following table: Coefficients d jk

Element j →

Species k ↓ CO

The atom conservation equations ( Σ k d jk N k –A j ) (see Eq. (81)) yield the relations: j = 1 (C atoms):

1N CO + 1N CO 2 + 0N H 2 + 0N HO 2 + 0N NO + 0N N2 +0N OH + 0N O2 – 10 = 0, (A) j = 2 (H atoms):

0N CO +0N CO 2 +2N H 2 +2N HO 2 + 0N NO + 0N N2 +1N OH + 0N O2 – 20 = 0, (B) j = 3 (O atoms):

1N CO +2N CO 2 +0N H 2 +1N HO 2 + 1N NO + 0N N2 +1N OH + 2N O2 - 31.6 = 0, (C) j = 4 (N atoms):

0N CO +0N CO 2 + 0N H 2 +0N HO 2 + 1N NO + 2N N2 +0N OH + 0N O2 - 118.6= 0, (D)

Dividing these equations by the total moles (N= ΣN k ), we obtain the relations

1X CO + 1X CO 2 + 0X H 2 + 0X HO 2 + 0X XO + 0X N2 +0X OH + 0X O2 – 10/N = 0, (E) 0X CO +0X CO 2 +2X H 2 +2X HO 2 + 0X NO + 0X N2 +1X OH + 0X O2 – 20/N = 0,

(F) 1X CO +2X CO 2 +0X H 2 +1X HO 2 + 1X NO + 0X N2 +1X OH + 2X O2 - 31.6/N= 0, (G) 0X CO +0X CO 2 + 0X H 2 +0X HO 2 + 1X NO + 2X N2 +0X OH + 0X O2 - 118.6/N = 0, (H)

and N is solved from the identity ΣX k = 1.

(I)

G = G(T,P,N 1 ,N 2 , …, N K ). (J) Multiplying Eqs. (A)–(D), respectively by λ C , λ H , λ N and λ O , and adding with Eq. (J)

we form a function

F=G+ λ C (N CO +N CO 2 – 10) + λ H (2N H 2 + 2N HO 2 +N OH

λ O (N CO + 2N CO 2 +N HO 2 +N OH +N NO +2N O2 – 31.6) +

λ N (N NO +2N N2 –118.6) (K) that is minimized at equilibrium, i.e.,

∂F/∂N CO =( ∂G/∂N CO ) T,P + λ C + λ O = 0,

(L)

∂F/∂N CO2 =( ∂G/∂N CO2 ) T,P + λ C +2 λ O = 0, and

(M)

(N) Rewrite Eqs. (L) to (N) in the form µ CO + λ C + λ O = 0,

∂F/∂N H 2 =( ∂G/∂N H 2 ) T,P + 2λ H = 0.

(O) µ CO2 + λ C +2 λ H = 0, and

(P) µ H 2 + 2λ H = 0,where µ k =( ∂G/∂N k ) T,P .

(Q) Recall that for an ideal mixture of gases or ideal mix of liquids or solids

µ k = ˆg k = g k (T,P) + R T ln (X k ). (R) Divide by R T, namely, µ k / R T=( g k (T,P)/ R T) + ln (X k 1).

(R´) Divide Eqs. (O) through (Q) by R T and using Eq. (R´)

( g CO (T,P)/ R T) + ln (X CO ).+ λ C ´+ λ O ´ = 0,

(S)

( g CO2 (T,P)/ R T) + ln (X CO2 ).+ λ C ´+2 λ H ´= 0, and

(T)

(U) where λ j ´= λ j / R T.

( g H2 (T,P)/ R T) + ln (X H2 ) + 2λ H ´ = 0,

The values of ln (X k ) can be determined from the linear equations Eqs. (S) to (U) for assumed values of modified LaGrange multipliers of λ C ´, λ H ´, λ N ´, and λ O ´. Then we

can check whether these four multipliers satisfy the four-element conservation equa- tions Eqs. (E) to (H). Thus, knowing the values of µ k , those of X k can be determined from linear equations such as Eq. (R). Irrespective of the number of unknown species,

the assumptions for the LaGrange multipliers are made only equal to number of atom balance equations. Good starting values can be obtained by assuming complete com- bustion, determining those mole fractions to obtain initial guesses for employing Eqs.

(S) to (U). For T = 2500 K and P = 1 bar, g k = h k –T s k , and

CO = –701455 kJ kmole –1 , g H 2 = - 419840 kJ kmole ,

g –1 CO 2 = –1078464 kJ kmole , and so on (S) Good starting values could be λ C ´, λ H ´, λ N ´, and λ O ´ = 19.7, 12.5, 14.0, 17.1, respec-

tively. The converged solutions are provided below. N CO = 2.04 kmole, N CO 2 = 7.96, N H 2 = 0.42, N HO 2 = 9.24,

N NO = 0.57, N N2 = 59, N OH = 0.69 , N O2 = 1.52, N= ΣN k = 81.4 kmole.

The NASA equilbrium code uses a descent Newton-Raphson method with a typical number of iterations equal to 8-12.