Generalized Relation for the Chemical Potential

6. Generalized Relation for the Chemical Potential

Recall from Chapter 8 that

(23) ˆa k = ˆf k /f k is the ratio of the fugacity of species k in a mixture to the fugacity of the same spe-

µ k = ˆg k = g k (T,P) + R T ln ˆ α k , where

cies in its pure state. Equation (15) can be generalized for any reaction in the form Σ ˆg k dN k = Σ g k (T,P) + R T ln ˆ α k ) dN k ≤ 0, (24)

where the activity coefficient ˆa k equals the species mole fraction for ideal mixtures and the equality applies to the equilibrium state.

b. Example 2 Consider the reactions

C (s) + 1/2 O 2 → CO, and (I)

C(s) + O 2 → CO 2

(II)

F R,II

F P, I

-2.0E+05

F,kJ -3.0E+05

N c , kmoles consumed

Figure 3: The reaction potentials for reactions I and II with respect to the number of moles of carbon that are consumed.

Which of the two reactions is more likely when 1 kmole of C reacts with 50 kmole of O 2 in a reactor at 1 bar and 298 K. Assume that c p,C / R = 1.771+0.000877 T–86700/T 2 in SI units and T is in K. Assume ideal mixture.

Solution If |(F R –F P )| I > |(F R –F P )| II , then the first reaction dominates and vice versa. Note that the reaction potentials are functions of the species populations and hence vary as a re- action proceeds. Using Eq. (23),

F R = g C (T,P) + R T ln ˆ α k . (A)

Since solid carbon (C(s)) is a pure component and hence the activity ˆa C(s) = 1. Further,

C = h fC , + ∫ 298 K pC , c dT .

where h fC o , = 0 kJ kmole –1 , and (B) s T

C = s o C (298K) + ∫ 298 K ( c pC , /) T dT .

Now,

s –1

C (298K) = 5.74 kJ kmole K . (C) Hence, using Eqs. (B) and (C), g o C = g C (298K, 1 bar) = h 298K – 298 × s C (298K), i.e.,

C = 0 – 298 ×5.74 = –1711 kJ kmole . (D) For solids and liquids, gTP k (,)≈ g o k (T) . Assume that 0.001 moles of C(s) react with

0.0005 moles of O 2 to produce 0.001 moles of CO. Hence, p O 2 =X O 2 P = (50– 0.0005) ÷(0.001+(50–0.0005)) = 0.9999 P = 0.9999 bar.

Therefore, s O 2 = 205.03 – 8.314 × ln 0.9999 = 205.03 kJ K –1 kmole –1 , i.e.,

-3.00E+06 -3.05E+06 -3.10E+06

-3.15E+06

-3.20E+06 -3.25E+06

G, kJ

G II

-3.30E+06 -3.35E+06 -3.40E+06

-3.45E+06 -3.50E+06

0 0.2 0.4 N , consumed 0.6 0.8 1

Figure 4: Variation in G I and G II with respect to the number of moles of carbon consumed for reactions I and II at 298 K.

g O 2 (298K, 1 bar) = 0 – 298 × 205.03 = –61099 kJ kmole –1 . (E) Similarly,

X CO = 0.001 ÷(0.001 + 49.9995) ≈ 0.00002, and s CO (T, p CO ) = 197.54 – 8.314 ×ln (0.00002) = 287.5 kJ K –1 kmole –1 , so that

(F) Employing Eqs. (D) and (E),

g CO (298K, 1 bar) = –110530 – 298 × 287.5 = –196205 kJ kmole –1 .

F R = g C + 1/2 g O 2 = –1710 + 0.5 × (61099) = –32260 kJ, and

F P = g = –196205kJ kmole CO –1 , i.e.,

F R –F P = –32260 + 196205 = 163945 kJ. For reaction I,

(dG/|dN C |) I = (dG/|d ξ|) I = –(F R –F P ) I = –163945 kJ.

For reaction (II), the corresponding amount of O 2 consumed is 0.0001 kmole while

0.0001 kmole of CO 2 is produced. Therefore, N O 2 = 50 – 0.001 = 49.999,

X O 2 = 0.999 × (0.0001 + 49.999) = 0.999,

X CO 2 = 0.001 × (0.001 + 49.999) ≈ 0.00002, and Consequently,

s O 2 = 205.03 – 8.314 × ln (0.999) ≈ 205.03 kJ K –1 kmole –1 , s CO2 = 213.74 – 8.314 × ln (0.0002) ≈ 303.70 kJ K –1 kmole –1 ,

g –1 O 2 = –61099 kJ kmole , and

g –1

CO 2 = -393546- 298 × 303.70 = –484048 kJ kmole .

For this reaction

F –1

R = –1710+(–61099) = –62810 kJ, and F P = = –484048 kJ kmole , i.e.,

(G)

-3.00E+07 -3.01E+07 -3.01E+07 -3.02E+07 -3.02E+07 -3.03E+07

G II

-3.03E+07 G, kJ -3.04E+07 -3.04E+07 -3.05E+07

-3.05E+07 -3.06E+07

N C , consumed, kmoles

Figure 5: Variation in G I and G II with respect to the number of moles of carbon consumed for reactions I and II at 3500 K.

Hence,

(dG/|dN C |) II = (dG/d ξ) II = - (F R –F P ) II = –394390 kJ.

The variations in the reaction potentials for reactions I and II with respect to the num- ber of moles of carbon that are consumed at a reactant temperature of 298 K are pre-

sented in Figure 3 , and the corresponding variation in G I and G II in Figure 4 . At 298 K CO 2 production dominates. The analogous variations in G I and G II at 3500 K are pre- sented in Figure 5 . At the higher temperature CO formation is favored.

Remarks Since,

g k (T, P, X k) = h k –T s k = h k –T( s o k – R ln P X k /1)

k –T{ s = k – R ln (P/1)} + R T ln X k

g k (T,P)+ = R T ln X k , in general, the values of g k (T, P, X k) are a function of the species mole fractions. If

we assume that | g k (T,P)| » | R T ln X k ,|, then g k (T, P, X k) ≈ g k (T,P). This offers an approximate method of determining whether reaction I or II is favored. For instance, if the reactions are assumed to go to completion, ∆G I = g CO –( g C + 1/2 g O 2 ). Likewise, we can evaluate ∆G II to determine whether | ∆G II |>| ∆G I | ;if so, the CO 2 production reaction is favored. Values of ∆G(T,P) at 1 bar, i.e., ∆G o (T) are tabulated. ( Tables 27A and 27B at T= 298 K)

In addition to reactions I and II, consider the following reactions: C(s) + CO 2 → 2 CO, (III)

CO + 1/2 O 2 → CO 2 , and

(IV)

(V) Figure 6 plots value of ∆G o with respect to the temperature for these five reactions.

H 2 + 1/2 O 2 →H 2 O.

For instance, for reaction III, ∆G o (298 K) = 2 g CO – ( g C + g CO 2 ) = 120080 kJ, For instance, for reaction III, ∆G o (298 K) = 2 g CO – ( g C + g CO 2 ) = 120080 kJ,

the products (F P ) is initially low and the value of F R is higher. However, the equilib- rium state is reached at a very low CO concentration when dG T,P = 0, i.e., F R =F P . Thereafter, F P > F R or dG T,P > 0, and the reaction does not proceed. In other words, ∆G o >0 implies that

C + CO 2 → large amounts of leftover C and CO 2 + small amounts of CO 2 . (H) On the other hand for reaction II, ∆G o < 0 implies that C+O 2 → small amounts of leftover C and O 2 + large amounts of CO 2 .

(I) Generally, the value of ∆G o for a reaction indicates the extent of completion of that

reaction. A relatively large negative value of ∆G o implies that F R »F P , and this re- quires the largest decrease in the reactant population (or extent of completion of reac- tion) before chemical equilibrium is reached. Normally, a positive value for ∆G im- plies that the reaction will produce an insignificant amount of products (reaction III).

We will now show that the value of ∆G o for reaction IV can be obtained in terms of the corresponding values for reactions I and II. For reactions I, II, and IV, respectively

∆G o I (298 K) = g CO –( g C + 1/2 g O 2 ),

(J)

∆G o II (298 K) = g CO 2 –( g C + g O 2 ),and

(K)

∆G o IV (298 K) = g CO 2 –( g CO + 1/2 g O 2 ),

(L) where the g k ’s are evaluated at 298 K, i.e., g k = g o k . Equation (L) assumes the form

∆G o

IV (298 K) = g CO 2 –( g C + g O 2 )–{ g CO –( g C +1/2 g O 2 )}= ∆G II – ∆G I , i.e., (M) ∆G o IV (298 K) = ∆G o II - ∆G o I = g CO 2 (298 K) – g CO (298 K)

oo

= –394390 + 137137 = –257253 kJ.

We can arbitrarily set g o k = 0 for the elemental species C and O 2 at T=298 K so that ∆G o

I (298 K)= g f,,CO (298 K), ∆G II (298 K)= g f,,CO2 (298 K). where 0 g f,,k is called Gibbs’ function of formation of species k from elements in natural form.