Criteria in Terms of Chemical Force Potential

5. Criteria in Terms of Chemical Force Potential

The reaction CO+ 1/2O 2 → CO 2 , must satisfy the criterion provided by Eq. (14b) to pro- ceed. In the context of the above discussion, dividing Eq. (15) by 0.002 or the degree of reac-

tion,

(14c) Recall that µ κ = ˆg k , i.e.,

dG T,P /0.002 = (–1) µ CO + (–1/2) µ O 2 + µ CO 2 < 0.

(14d) where k and ν k represent the reacting species and its stoichiometric coefficient for a reaction.

dG T,P = Σν k ˆg k = Σν k µ k ≤ 0,

In case of the reaction CO+ 1/2O 2 → CO 2 , ν CO = -1, ν O2 = -1/2, ν CO2 = 1. Equation (14c) can

be alternately expressed in the form

µ CO + ν O2 µ O 2 > µ CO 2 .

Defining the chemical force for the reactants and products as

F R = µ CO + (1/2) µ O 2 , and F P = µ CO 2 for reaction CO+ 1/2O2 →CO2 (15) The criterion dG T,P < 0 leads to the relation F R >F P . This criterion is equally valid for an adia-

batic closed rigid system (U, V, m specified), and adiabatic and isobaric system (H, P, m speci- fied), isentropic rigid closed system (S, V, m specified), isentropic and isobaric systems (S, P, m specified), and, finally, isothermal and isovolume systems (T, V, m specified).

From Eq. (2),

d ξ = dN k / ν k = dN CO /(-1) = dN CO2 /(+1) = 0.002.

Replacing 0.002 in Eq. (14c) by d ξ, the stoichiometric coefficients by ν k , and generalizing for any reaction

( ∂G/∂ξ) T,P = Σµ k ν k < 0. (16) The Gibbs energy decreases as the reaction progresses and eventually reaches a minimum

value at equilibrium. Defining the chemical affinity as

F = –( ∂G/∂ξ) T,P , (17) Equation (16) assumes the form

(18a) Similarly, following the relations for dA, dU and dS, we can show that

(-( ∂G/∂ξ) T,P =F=- Σµ k ν ) > 0. k

(-( ∂A/∂ξ) T,V =F=- Σµ k ν ) > 0, k

(18b)

(-( ∂U/∂ξ) S,V =F=- Σµ k ν ) > 0, k

(19a)

(-( ∂H/∂ξ) S,P =F=- Σµ k ν ) > 0, and k

(19b)

(20) The last expression shows that the entropy increases in an isolated system as chemical reaction

(T( ∂S/∂ξ) U,V = T( ∂S/∂ξ) H,P =F = - Σµ k ν ) > 0. k

proceeds. For a reaction to proceed under any of these constraints, the affinity F > 0.

In the CO oxidation example, the values of F for the reactants and products are

(21) Since (F R –F P ) > 0 for oxidation to proceed,

F R = µ CO + (1/2) µ O 2 , and F P = µ CO 2 .

F R >F P , (22) which is similar to the inequality T hot >T cold that allows heat transfer to occur from a hotter to a

colder body. In a manner similar to the temperature (thermal potential), F R and F P are analo- gous intensive properties called chemical force potentials. The chemical potential µ

is the

) same as partial molal Gibb’s function g k , (= h k -T s k ), which is a species property. Each spe-

cies has a unique way of distributing its energy and, thus, fixing the entropy. A species distrib- uting energy to a larger number of states has a low chemical potential and is relatively more stable. During chemical reactions, the reacting species proceed in a direction to form more stable products (i.e., towards lower chemical potentials). The physical meaning of the reaction potential is as follows: For a specified temperature, if the population of the reacting species

(e.g., CO and O 2 ) is higher (i.e., higher value of F R ) than the product molecules (i.e., CO 2 at lower F P ), then there is a high probability of collisions amongst CO and O 2 resulting in a reac- tion that produces CO 2 . On the other hand, if the population of the product molecules (e.g., CO 2 ) is higher (larger F P value) as compared to the reactant molecules CO and O 2 (i.,e., lower

F R ), there is a higher probability of collisions amongst CO 2 molecules which will break into CO and O 2 . If the temperature is lowered, the molecular velocities are reduced and the transla- tional energy may be insufficient to overcome bond energy among the atoms in the molecules that is required to the potential F(T, P X i ).

a. Example 1 Five kmole of CO, three of O 2 , and four of CO 2 are instantaneously mixed at 3000 K and 101 kPa at the entrance to a reactor. Determine the reaction direction and the val- a. Example 1 Five kmole of CO, three of O 2 , and four of CO 2 are instantaneously mixed at 3000 K and 101 kPa at the entrance to a reactor. Determine the reaction direction and the val-

Solution We assume that if the following reaction occurs in the reactor:

CO+ 1/2 O 2 → CO 2 , then

(A)

F R >F P (B) so that the criterion dG T,P < 0 is satisfied. The reaction potential for this reaction is

F R = (1) µ CO + (1/2) µ O 2 , and

(C)

(D) For ideal gas mixtures,

F P = (1) µ CO 2 .

(E) The larger the CO mole fraction, the higher the value of µ CO and, hence, F.

µ CO = ˆg CO = g CO (T,P) + R T ln X CO = g CO (T,p CO ).

g CO (T,P) = h CO (T,P) – T s CO (T,P)

h f,CO 0 =( + ( h t,3000K – h t,298K o ) CO )– 3000 ×( s CO (3000) – 8.314(ln ×P/1)) = (–110530+93541) –3000 ×273.508–8.314×ln 1)

g CO = –837513 kJ per kmole of CO. (F) Similarly, at 3000K and 1 bar,

g O 2 = –755099 kJ kmole –1 , and g = –1242910 kJ kmole CO –1 2 . (G) The species mole fractions

X CO =5 ÷(5+3+4) = 0.417, X O 2 =3 ÷(5+4+3) = 0.25, and X CO 2 = 0.333. (H) Further,

µ CO = ˆg CO (3000K, 1 bar, X CO = 0.417)

g = CO (3000K, 1 bar) + 8.314 ×3000×ln(0.417) = –837513 + 8.314 × 3000 × ln 0.467

(I) Similarly, µ O 2 = (3000K, 1 bar, X O 2 =0.25) = –789675 kJ per kmole of O 2 .

= –856504 kJ kmole –1 of CO in the mixture.

(J) µ CO 2 = (3000K, 1 bar, X CO 2 =0.333) = –1270312 kJ per kmole of CO 2 .

(K) Therefore, based on the oxidation of 1 kmole of CO,

F R = –856504 + 1/2(–789675) = –1254190 kJ, and (L)

F P =–1270312 kJ, i.e., (M)

-1260 F, MJ

N CO2 , kmole

Figure 1: The reaction potential with respect to the number of moles of CO 2 produced.

F R >F P , (N) which implies that assumed direction is correct and hence CO will oxidize to CO 2 .

The oxidation of CO occurs gradually. As more and more moles of CO 2 are produced, its molecular population increases, increasing the potential F P . Simultaneously, the CO and O 2 populations decrease, thereby decreasing the reaction potential F R until the reaction ceases when chemical equilibrium is attained. Thus chemical equilibrium is achieved when F R =F P , i.e., dG T,P =0 . This is illustrated in Figure 1 . The correspond- ing species concentrations are

N CO 2 = 5.25 kmole, N CO = 3.75 kmole, and N O 2 = 2.375 kmole.

(Recall the evaporation example discussed in Chapter 7 where A reaches a min imum value at specified values of T, V and G. From a thermodynamic perspective, this problem is similar to placing a cup of cold water in bone dry air. Evaporation will oc- cur when dG T,P < 0, but after a finite amount of water is transformed into the vapor, evaporation will cease at which g H2O(l) =g H2O(g) and dG T,P = 0.)

The Gibbs energy at any section

G= Σµ k N k = µ CO N CO + µ O 2 N O 2 + µ CO 2 N CO 2 , i.e.,

G = -856504 ×5 -789675×3-1270312×4 = -11,732,793 kJ. Figure 2 plots values of G vs N CO2 . The plot in Figure 2 shows that G reaches a mini-

mum value when F R =F P . Nitrogen does not participate in the reaction. Therefore, dN N 2 = 0 and, so, the expres-

sions for F R and F P are unaffected. However, the mole fractions of the reactants change so that the values of F R and F P are different, as is the equilibrium composition. The G expression for this case is

G= Σµ k N k = µ CO N CO + µ O 2 N O 2 + µ CO 2 N CO 2 + µ CO 2 N CO 2 + µ N2 N N2

T,P < 0, Possible

G, MJ

dG T,P > 0, impossible

-11760 dG T,P = 0, Equilibrium

N CO2 , kmoles

Figure 2: Illustration of the minimization of the Gibbs energy at equilibrium with respect to the number of moles of carbon dioxide produced.

Remarks The overall reaction has the form

5 CO + 3 O 2 + 4 CO 2 → 3.75 CO + 2.375 O 2 + 5.25 CO 2 . The assumed direction (i.e., CO + 1/2 O 2 → CO 2 ) is possible if dG T,P < 0 or F R >F P .

The mixture is at equilibrium if dG T,P = 0 (as illustrated in Figure 2 ) or F R =F P . If F R <F P , the reverse reaction CO 2 → CO + 1/2 O 2 becomes possible.