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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 values of FR , FP, and G. What is the equilibrium composition of the gas leaving the reactor? How is the process altered if seven kmole of inert N 2 is injected into the reactor? Solution We assume that if the following reaction occurs in the reactor: CO+ 1/2 O 2 → CO 2 , then (A) FR > FP (B) so that the criterion dGT,P < 0 is satisfied. The reaction potential for this reaction is FR = (1) μCO + (1/2) μO 2 , and (C)
FP = (1) μCO 2. (D)
For ideal gas mixtures, μCO = gˆ (^) CO = g (T,P) + R T ln XCO = g (^) CO(T,pCO). (E)
The larger the CO mole fraction, the higher the value of μCO and, hence, F. g (^) CO(T,P) = h (^) CO(T,P) – T s (^) CO(T,P)
= [ h (^) f,CO^0 + ( h (^) t,3000K – h (^) t,298K)CO ]– 3000×( s (^) COo^ (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, gO 2 = –755099 kJ kmole–1^ , and gCO 2 = –1242910 kJ kmole–1^. (G)
The species mole fractions XCO = 5÷(5+3+4) = 0.417, X (^) O 2 = 3÷(5+4+3) = 0.25, and X (^) CO 2 = 0.333. (H)
Further, μCO = (^) gˆ CO (3000K, 1 bar, XCO = 0.417) = g (^) CO(3000K, 1 bar) + 8.314× 3000 ×ln(0.417) = –837513 + 8.314 × 3000 × ln 0. = –856504 kJ kmole–1^ of CO in the mixture. (I) Similarly, μO 2 = (3000K, 1 bar, X (^) O 2 =0.25) = –789675 kJ per kmole of O 2. (J)
μCO 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, FR = –856504 + 1/2(–789675) = –1254190 kJ, and (L)
FP =–1270312 kJ, i.e., (M) FR > FP, (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 FP. Simultaneously, the CO and O 2 populations decrease, thereby decreasing the reaction potential FR until the reaction ceases when chemical equilibrium is attained. Thus chemical equilibrium is achieved when FR = FP, i.e., dGT,P=0. This is illustrated in Figure 1. The corresponding species concentrations are N (^) CO 2 = 5.25 kmole, NCO = 3.75 kmole, and N (^) O 2 = 2.375 kmole. (Recall evaporation example discussed in Chapter 07 where A reaches a minimum at given T and V and G reaches a minimum at given T and P. From a thermodynamic perspective, chemical reaction is a similar problem. In evaporation of water from a cup into bone dry air, evaporation occurs due to
gH2O(!) > gH2O(g). The evaporation will continue to occur with dGT,P < 0; but after a finite amount of water is transformed into the vapor, evaporation will cease(From a thermodynamic perspective, this problem is similar to placing a cup of cold water in bone dry air. Evaporation will occur when dGT,P < 0, but after a finite amount of water is transformed into the vapor, evaporation will cease at which g
H2O(!) = gH2O(g) and dGT,P = 0.) The Gibbs energy at any section
**-
-**
4 4.5 5 5.5 6 N (^) CO2 , kmole
F, MJ
FR
F (^) P
Equilibrium B C D E
Figure 1: The reaction potential with respect to the number of moles of CO 2 produced.
C (s) + 1/2 O 2 → CO, and (I) C(s) + O 2 → CO 2 (II) 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 |(FR –FP )|I > |(FR – FP)|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 reaction proceeds. Using Eq.(23), FR = g (^) C (T,P) + R T ln αˆ (^) k. (A)
Since solid carbon (C(s)) is a pure component and hence the activity aˆ (^) C(s) = 1. Further,
hC = h (^) f ,Co^ + (^) 298Kc (^) p,C
T
∫ dT^.
where h (^) f ,Co^ = 0 kJ kmole–1^ , and (B)
s (^) C = s (^) Co^ (298K) + (^) 298K(c (^) p,C / T)
T
∫ dT^.
Now, s (^) Co^ (298K) = 5.74 kJ kmole–1^ K–1^. (C)
Hence, using Eqs. (B) and (C), gCo^ = g (^) C (298K, 1 bar) = h (^) 298K – 298×s (^) C (298K), i.e., gCo^ = 0 – 298×5.74 = –1711 kJ kmole–1^. (D)
For solids and liquids, g (T, P)k ≈ g (T). Assume that 0.001 moles of C(s) reactok 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. bar. Therefore, s (^) O 2 = 205.03 – 8.314 × ln 0.9999 × 205.03 kJ K–1^ kmole–1^ , i.e.,
g (^) O 2 (298K, 1 bar) = 0 – 298 × 205.03 = –61099 kJ kmole–1^. (E)
Similarly, XCO = 0.001÷(0.001 + 49.995) ≈ 0.00002, and s (^) CO(T, pCO) = 197.54 – 8.314×ln (0.00002) = 287.5 kJ K–1^ kmole–1^ , so that g (^) CO(298K, 1 bar) = –110530 – 298 × 287.5 = –196205 kJ kmole–1^ .(F) Employing Eqs. (D) and (E), FR = (^) g (^) C + 1/2 (^) g (^) O 2 = –1710 + 0.5 × (61099) = –32260 kJ, and FP = FCO = –196205kJ kmole–1^ , i.e., FR – FP = –32260 + 196205 = 163945 kJ. For reaction I,
(dG/|dNC |)I = (dG/|dξ|)I = –(FR – FP)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
-6 .0 E + 0 5
-5 .0 E + 0 5
-4 .0 E + 0 5
-3 .0 E + 0 5
-2 .0 E + 0 5
-1 .0 E + 0 5
0 .0 E + 0 0
0 .0 0 .1 0 .2 0 .3 0 .4 0 .5 0 .6 0 .7 0 .8 0 .9 1. N (^) c , k m o le s c o n s u m e d
F,kJ
F (^) P , II
F (^) P , I
F (^) R ,II
F (^) R , I
Figure 3: The reaction potentials for reactions I and II with respect to the number of moles of carbon that are consumed.
FR = –1710+(–61099) = –62810 kJ, and FP = = –484048 kJ kmole –1^ , i.e., (G) FR – FP = –62810 + 484048 = 421238 kJ.
Hence, (dG/|dNC |)II = (dG/dξ)II = (FR – FP)II = –394390 kJ.
The variations in the reaction potentials for reactions I and II with respect to the number of moles of carbon that are consumed at a reactant temperature of 298 K are presented in Figure 3, and the corresponding variation in GI and GII in Figure 4At 298 K CO 2 production dominates. The analogous variations in GI and GII at 3500 K are presented in Figure 5 At the higher temperature CO formation is favored.
Remarks Since g (^) k (T, P, Xk ) = h (^) k – T s (^) k = h (^) k – T ( s (^) ko^ – R ln P Xk /1) = h (^) k – T ( s (^) ko^ – R /1)) + R T ln Xk = g (^) k (T,P)+ R T ln Xk , in general, the values of g (^) k (T, P, Xk ) are a function of the species mole fractions. If we assume that | g (^) k (T,P)| » | R T ln Xk ,|, then g (^) k (T, P, Xk ) ≈ 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, ∆GI = g (^) CO
- ( g (^) C + 1/2 g (^) O 2 ). Likewise, we can evaluate ∆GII to determine whether |∆GII| > |∆GI| ;if so, the CO 2 production reaction is favored. Values of ∆G(T,P) at 1 bar, i.e., ∆Go(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) H 2 + 1/2 O 2 → H 2 O. (V)
For instance, for reaction III, ∆Go^ (298 K) = 2 g (^) CO – ( g (^) C + g (^) CO 2 ) = 120080 kJ,
which a positive number or FR = g (^) CO2 + g (^) C < FP =2 g (^) CO. This implies that the reaction cannot proceed in the indicated direction. In case of reaction III, the reaction potential of the products (FP) is initially low and the value of FR is higher. However, the equilibrium state is reached at a very low CO concentration when dGT,P = 0, i.e., FR = FP. Thereafter, FP > FR or dGT,P > 0, and the reaction does not proceed. In other words, ∆Go>0 implies that
C + CO 2 → large amounts of leftover C and CO + small amounts of CO. (H)
On the other hand for reaction II, ∆Go^ < 0 implies that
C + O 2 → small amounts of leftover C and O 2 + large amounts of CO 2. (I)
Generally, the value of ∆Go^ for a reaction indicates the extent of completion of that reaction. A relatively large negative value of ∆Go^ implies that FR » FP, and this requires the largest decrease in the reactant population (or extent of completion of reaction) before chemical equilibrium is reached. Normally, a positive value for ∆G implies that the reaction will produce an insignificant amount of products (reaction III). We will now show that the value of ∆Go^ 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 oI^ (298 K) = g (^) CO – ( g (^) C + 1/2 g (^) O 2 ), (J)
∆Go (^) II^ (298 K) = g (^) CO 2 – ( g (^) C + g (^) O 2 ),and (K)
∆G oIV^ (298 K) = g (^) CO 2 – ( g (^) CO + 1/2 g (^) O 2 ), (L)
-5.0E+
-4.0E+
-3.0E+
-2.0E+
-1.0E+
0.0E+
1.0E+
2.0E+
0 500 1000 1500 2000 2500 3000 3500 4000 T, K
G ∆∆∆∆ 0 , kJ
C+CO2=2CO
C+O2=CO
C+1/2 O2=CO
CO+1/2 O2=CO H2+1/2 O2=H2O
Figure 6. The variation in the value of ∆Go^ with respect to the temperature for several reactions.
(pCO/1)^1 (p (^) O 2 /1)1/2^ /(p (^) CO 2 /1)^1 = (6/1)^1 (1/1)1/2^ /(2/1)^1 = 3, and the criterion Ko(T) = 0.0002 ≥ (pCO/1)^1 (p (^) O 2 /1)1/2^ /(p (^) CO 2 /1)^1
or (pCO/1)^1 (p (^) O 2 /1)1/2^ /(p (^) CO 2 /1)^1 ≤Ko(T)=0. is violated. Hence, CO will oxidize to CO2 , i.e., the reverse path is favored.
Example 5 5 kmole of CO, 3 of O2 , 4 of CO 2 , and 7 of N 2 are introduced into a reactor at 3000 K and 2000 kPa. Determine the equilibrium composition of gas leaving reactor, assuming that the outlet (product) stream contains CO, O 2 , N 2 , and CO 2. Will the equilibrium composition change if the feed is altered to 6 kmole of CO, 3 kmole of CO 2 , 3.5 kmole of O 2 , and 7 kmole of N 2 enter the reactor? Assume that the outlet stream contains the same species. Will the CO concentration at the outlet change if the pressure changes, say to 101 kPa? Solution Assume that the chemical reaction proceeds according to the reaction CO 2 → CO + 1/2 O so that Ko(T) = ((p (^) CO/1)(p (^) O 2 /1)1/2^ )/(p (^) CO 2 /1), where p (^) CO = XCO P = (NCO/N) P, and N = NCO + N (^) O 2 + N (^) CO 2. Therefore, Ko(T) = (NCON (^) O1/2 2 )(P/(1×N))1/2^ /N (^) CO 2. (A)
The conservation of C and O atoms provide two additional equations. The overall balance equation in terms of the three unknown concentrations is 5CO + 3O 2 + 4CO 2 + 7N 2 → NCOCO + N (^) O 2 O 2 + N (^) CO 2 CO 2 + NN2 N 2 (B
There are 4 unknowns N (^) CO, NCO 2 and NO 2 ) and we have three atom balance equations.
Carbon atoms: 5 + 4 = N (^) CO + NCO2 (C) Oxygen atoms: 5 1 + 32 + 4 *2 = N (^) CO 1 + N (^) CO 2 * 2 + N (^) O 2 * 2 (D) Nitrogen atoms : 7 2 = NN2 * 2 (D’)
The fourth equation is given by equilibrium condition at 3000 K: CO 2 ⇔ CO + 1/2 O 2 reaction, log 10 K = –0.48. Using this value in Eq. (A)
0.327 = (N (^) CON (^) O1/2 2 )(P/(1×N))1/2^ /N (^) CO 2. (E)
Using Eq. (C), NCO = NC – N (^) CO 2 , i.e., NCO = 9 – N (^) CO 2. (F)
Further, using Eqs. (D) and (F), N (^) O 2 = (NO – NC – N (^) CO 2 )/2, i.e., (G)
N (^) O 2 = (19 – 9 – N (^) CO 2 )/2. (H)
Therefore, the number of moles at the exit N = NCO + N (^) O 2 + N (^) CO 2 + NN2 = (NC – N (^) CO 2 ) + (NO – NC )/2 + N (^) CO (^2)
= NN2 + (NO +NC – N (^) CO 2 )/2 = 21 – N (^) CO 2 /2 (I)
Applying Eqs. (A) and (G)–(I), at 20 bar, at the exit N (^) CO 2 = 6.96 kmole, and NCO = 2.04 kmole, N (^) O 2 = 1.52 kmole, and N = 17.52 kmole. When the feed stream is altered to react 6 kmole of CO, 3 kmole of CO 2 , 3. kmole of O 2 , and 7 kmole of N 2 , the respective inputs of C, O and N atoms remain unaltered at 9, 19 and 14 respectively. Therefore, the equilibrium composition is unchanged. This indicates that it does not matter in which form the atoms of the reacting species enter the system. The same composition, for instance, could be achieved by reacting a feed stream containing 9 kmole of C(s) (solid carbon, such as charcoal), 9.5 kmole of O 2 and 7 kmole of N 2 (which is treated as an inert in this problem). From Eq. (A) we note that for a specified temperature, the value of K o(T) is unique. Therefore, if the pressure changes, but the temperature does not.. Eq. (E) dictates that the composition is altered and more CO 2 is produced as the pressure is increased.
Example 6 Consider a PCW assembly that is immersed in an isothermal bath at 3000 K. It initially consists of 9 kmole of C atoms and 19 kmole of O atoms (total mass = 912.01+ 1916 = 412 kg) is allowed to reach chemical equilibrium at 3000 K and 1 bar. What is the equilibrium composition? What is the value of the Gibbs energy? If we keep placing sand particles one at a time on the piston so that one achieves a final pressure of 4 bar; i.e. we allowed sufficient time to reach chemical equilibrium at that pressure, what is the resulting equilibrium composition and Gibbs energy? Solution We leave it to the reader to show that at equilibrium N (^) CO 2 = 5.25 kmole, NCO = 3.7 kmole, and N (^) O 2 = 2.37 kmole. (A)
Therefore, N = ΣNk = 11.37 kmole. (B) The Gibbs energy, G = N (^) CO 2 gˆ (^) CO 2 + NCO gˆ (^) CO + N (^) O 2 gˆ (^) O 2 , where (C)
gˆ (^) CO 2 =^ gCOo^2 (T)+^ R^ T ln(p^ CO 2 /1) =^ hf ,COo^2 +(^ h^ t,T–^ h^ t,298K)+^ R^ T ln(X^ CO 2 P/1), and (D)
Similar phenomenon occurs when reacting gases flow at slowest possible velocity through a diffuser where pressure at exit of diffuser is say 4 bars. If we follow the 412 kg mass when it flows through diffuser, it will reach equilibrium composition given by Eq. (D). However if the same mass flows at high velocity, the composition at exit of diffuser may be almost same as those at inlet!
Example 7 Determine the relations between the partial pressures and temperature for the following scenarios: pure H 2 SO 4 (l) dissociating upon evaporation, H 2 SO 4 (l) → H 2 O(g) + SO 3 (g), and an ideal mixture of 40% volatile H 2 SO 4 (l) and 60% nonvolatile liquid or solid participating in the same reaction. Assume that g (^) H 2 SO (^4) = –690013 kJ kmole–1^ , g (^) H 2 O(g ) = –228572 kJ kmole–1^ , g (^) SO 3 (g ) = –371060 kJ kmole– at 298 K and 1 bar. Estimate (p (^) H 2 O(g ) )(p (^) SO 3 (g ) ) at 298 K for pure H 2 SO 4 (l) and for an ideal mixture of 40% volatile H 2 SO 4 (l) and 60% nonvolatile liquid. Solution Pure H 2 SO 4 (l) The problem involves a mixture of phases. We will select the standard state to be the liquid state for H 2 SO 4 (l). Then from Eq. (35a), Ko(T) = Π(f (^) k (T,P) ˆα (^) k /f (^) k (T, 1 bar)) νk^. The gaseous species are assumed to be ideal so that f (^) k(g)(T, P) = P, f (^) k(g)(T, 1 bar) = 1, and ˆα (^) k(g)(T, P, Xk ) = Xk. In the liquid phase at 1 bar f (^) H 2 SO 4 (T, P)/f (^) H 2 SO 4 (T, 1 bar) ≈ 1. Further, since the liquid is pure, αˆ (^) H 2 SO 4 = 1, and Ko(T) =(p (^) H 2 O(g ) /1)^1 (p (^) SO 3 (g ) /1)^1. At 298 K, when X (^) H 2 SO 4 = 1 (p (^) H 2 O(g ) )(p (^) SO 3 (g ) ) = 1.44× 10 –16^ , which indicates that the partial pressures are very low, i.e., there is negligible dissociation. Ideal Liquid Mixture Since the liquid phase is in an ideal mixture, the activity of H 2 SO 4 (l) = X (^) H 2 SO 4 , and Ko(T) = (P X (^) H 2 O(g ) /1)^1 (P X (^) SO 3 (g ) /1)^1 /X (^) H 2 SO (^4) = (p (^) H 2 O(g ) /1)(p (^) SO 3 (g ) /1)/X (^) H 2 SO 4 , i.e., (p (^) H 2 O(g ) )(p (^) SO 3 (g ) ) = X (^) H 2 SO 4 Ko(T). The Gibbs energy change ∆Go(T) = (^) g (^) H 2 O(g ) + (^) g (^) SO 3 (g ) – (^) g (^) H 2 SO (^4) = –228572 –371060 + 690013 = 90381 kJ kmole –1^ , and ln Ko(T) = – ∆Go(T)/T = –36.48, i.e., Ko(T) = 1.44× 10 –^. At 298 K, when X = 0.4, (p (^) H 2 O(g ) )(p (^) SO 3 (g ) ) = 0.4×1.44× 10 –16^ = 0.576× 10 –^.
Remarks: During the vaporization of H 2 SO 4 (l) it is possible to produce H 2 O, SO 2 , SO 3 , and O 2 , rather than H 2 SO 4 (g). The pertinent reactions are
H 2 SO 4 (l) → H 2 O(g) + SO 2 (g) + 1/2 O 2 (g), and H 2 SO 4 (l) → H 2 O(g) + SO 3 (g) At equilibrium, can you determine the SO 2 and SO 3 concentrations?
Example 8 One kmole each of C(s) and O 2 enter a reactor at 298 K. The species CO, CO 2 , and O 2 leave the reactor at 3000 K and 1 bar at equilibrium. Find value of the equilibrium composition at the exit. What is the heat transfer across the boundary? What will happen if the inlet stream is altered to contain 1/2 kmole of oxygen and one kmole of CO. Also explain what happens if the outlet contains C(s), CO, and O 2. Solution The overall chemical reaction is C(s) + O 2 → a CO 2 + b CO + c O 2. (A) The species leaving the reactor are in an equilibrium state so that the following reaction must be in equilibrium, namely, CO 2 → CO + 1/2 O 2. (B) From an atom balance, C atoms 1 = a + b (C) O atoms: 2 = 2a + b + 2c. (D) Therefore, b = 1 – a, and c = (1 – a)/2. (E,F) The total moles leaving the reactor are N = a + b+ c = (3 – a)/2. (G) The exit equilibrium condition requires that Ko(T) = p (^) CO pCO1/ 2^2 /p (^) O 2. (H) For the carbon dioxide dissociation reaction at 3000 K, Ko(3000 K) = 0.327. Since, XCO = b/N, X (^) CO 2 = a/N, X (^) O 2 = c/N, and pk = Xk×1 bar, Solving the three unknowns a,b and c from Eqs. (B), (C) and (H) a = 0.563, b = 0.437, and c = 0.219. Applying the First Law
dEcv/dt = Q "^ cv – W "^ cv + Σ,k N "^ ik h
i,k –^ Σk N "^ e,k h
e,k ,^ (I)
Remarks Since, Ko(T) = (p (^) NH 3 /1 (p (^) H 2 O /1) (p (^) SO 3 /1), (D)
and Ko^ increases with temperature, decomposition is also favored at higher temperatures.
Example 10 Derive the Clausius–Clapeyron Relation from the van’t Hoff equation by considering the equilibrium of liquid and vapor. Solution Consider an isothermal and isobarically maintained air duct into which water droplets are injected. The chemical potentials of the liquid drops and the vapor are different, which cause a transfer of species from one phase into the other (from liquid to vapor during evaporation). The liquid droplets will eventually reach equilibrium with the vapor. At the equilibrium condition
H 2 O(l) → H 2 O(g), and (A) Ko(T) = p (^) H 2 O (g)/1 (B)
If the equilibrium constant is known at a reference temperature Tref, Koref(T) = p (^) H 2 O ,ref(g)/1, and (C)
from the van’t Hoff equation ln {Ko/ K oref^ } = – (∆H (^) Ro^ / R ) (1/T – 1/Tref). (D) For the vaporization process, ∆H (^) Ro^ = h (^) g – h (^) f, or ∆H (^) Ro^ = h (^) fg, i.e., (E) ln (p (^) H 2 O (g)/p (^) H 2 O (g),ref) = – ( h (^) fg/ R ) (1/T –1/T (^) ref), (F)
which is a relation for the change in the partial pressure of the vapor as the temperature changes. This is almost same as the Clausius–Clapeyron Relation
Example 11 Consider the reaction CO 2 → CO + 1/2 O 2. (A) That occurs in an isobaric and isothermal reactor. Discuss the effect on the equilibrium composition when the temperature is increased at a specified pressure and vice versa. Solution Recall that d(ln Ko)/dT = ∆H (^) Ro^ / R T^2 , (B)
where ∆H (^) Ro^ > 0 for the reaction. Hence, d(ln K o)/dT > 0, and the value of K increases with an increase in the pressure. Consequently, since
Ko(T) = (p (^) CO/1) (p (^) O 2 /1).0.5^ /(p (^) CO 2 /1), (C)
The value of p (^) CO 2 decreases (as does that of X (^) CO 2 ). The effect of increasing the temperature is to dissociate more CO 2. Simplifying Eq. (C), Ko(T) = XCO(X (^) O 2 )0.5^ (P/1)0.5^ /X (^) CO 2. (D)
The value of K o^ is a function of temperature alone. Therefore, increasing the pressure should cause the value of X (^) CO 2 to increase, i.e., a relatively lower amount of dissociation will occur. Since each mole of CO 2 that dissociates produces 1. moles of the other two species, a lower dissociation results in a smaller amount of product (in terms of moles), thereby lowering the pressure (which counteracts the pressure increase). This is an example of the Le Chatelier principle which states that any inhomogeneity or disturbance that is introduced into a system must result in a process which counteracts that inhomogeneity or disturbance.
Example 12 Consider the stoichiometric combustion of one kmole of CH 4 with air. The products are at 2250 K and 1 bar stream and contain CO 2 , CO, H 2 O, H 2 , O 2 , N 2 , and OH. Determine the equilibrium composition. Solution The overall chemical reaction is CH 4 + 2(O 2 + 3.76 N 2 ) → N (^) CO 2 CO 2 + NCO CO + N (^) H 2 O H 2 O + N (^) H 2 H 2 + N (^) O 2 O 2 + N (^) N 2 N 2 + NOH OH. (A) There are seven species of unknown composition. The four atom conservation equations for C, H, N, and O atoms are:
C atoms: 1 = N (^) CO 2 + NCO, (B)
H atoms: 4 = 2×N (^) H 2 O + 2×N (^) H 2 + NOH, (C)
N atoms: 7.52×2 = 2×N (^) N 2 , and (D)
O atoms: 2×2 = 2×N (^) CO 2 + N (^) CO + N (^) H 2 O + 2×N (^) O 2 + NOH. (E) We, therefore, require three additional relations. At equilibrium, for the reactions CO 2 → CO + 1/2 O 2 , Ko^ CO 2 = pCO(p (^) O 2 )0.5^ /(p (^) CO 2 ), (F)
H 2 O → H 2 + 1/2 O2, K (^) H 2 O = (p (^) H 2 )(p (^) O 2 )0.5/(p (^) H 2 O ), and (G)
OH → 1/2H 2 + 1/2 O 2 , KOH = (p (^) H 2 )0.5^ (p (^) O 2 )0.5/(pOH). (H)
Assume for example N (^) CO, NO2 , NCO2. Solve for other species from Eqs. (B) to (E). Then check whether Eqs. (F) to (H) are satisfied. If not iterate. One of the authors had developed spreadsheet based program which presents solution for all
( gOo^2 + R T ln (p (^) O 2 /1)) + 2 λ = 0, and (J)
( gOo^ + R T ln (p (^) O/1)) + λ = 0. (K) Multiplying Eq. (K) by 2 and subtracting it from Eq. (J), gOo^2 – 2 gOo^ + R T ln (p (^) O 2 /1)– 2 R T ln (p (^) O/1) = 0, i.e.,
(pO/1)^2 /(p (^) O 2 /1) = exp (–(2 gOo^ – gOo 2 )/( R T)) or (NO)^2 (P/N)(2–1)/N (^) O 2 = Ko, where (L) N = N (^) O 2 + NO. (M)
The equilibrium constant K = exp (–∆Go/ R T), where ∆Go^ = 2 gOo^ – gOo^2. (N)
With the values NO = 4 – 2N (^) O 2 and N = 4 – 2N (^) O 2 + NO2 = 4 – N (^) O 2 ,
(4 – 2N (^) O 2 )^2 (P/(1(4 – N (^) O 2 )))/N (^) O 2 = Ko. (O) the pressure P = 1 bar, (N (^) O 2 )^2 – 4 N (^) O 2 + 4^2 /(4+Ko^ ) = 0. (P) Now, gOo^2 = –755102 kJ kmole–1^ , and gOo^ = –323359 kJ kmole–^. We can solve for N (^) O 2 and selecting the root, such that N (^) O 2 > 0 and NO > 0, i.e., N (^) O 2 = 1.8875, NO = 0.225, and Gmin = –1.52× 106.
Example 14 A steady flow reactor is fired with 1 kmole of C 10 H 20 , with 10 % excess air. The species (1 to 5) leaving are CO, CO 2 , H 2 , H2O, OH, O2, NO, and N2 at T= 2500 K , 1 bar. Determine equilibrium composition of species leaving the reactor. Assume idea gas behavior., Solution The stoichiometric O2 can be determined to be :
(C 10 H 20 ) + 15 {O 2 + 3.76 N 2 } = 10 CO 2 + 10 H 2 O + 56.4 N (^2)
With 10 % excess air, O2 supplied = 16.5 kmoles , N2 supplied 62.04 kmoles
(C 10 H 20 ) + 16.5 {O 2 + 3.76 N 2 } → CO, CO 2 , H2, H2O, OH, O2, NO,N
This system is an open system. Now we follow a fixed mass ( 140+528+ 1737= 2405 kg) as it travels the reactor. Suppose this mixture is instantaneously heated to 2500 K at 1 bar and then calculate assuming various values for the moles of 8 species CO, CO2 etc subject to satisfaction of atom conservation for C, H, N, and O and select the composition at which G is minimum at same T and P. We use Larange multiplier method to arrive at the composition.
The four elements C,H,N and O are denoted by subscript j and eight species denoted by subscript k. The coefficients djk , i.e., d 11 = 1, d 12 = 1, … are provided in the following table: Coefficients djk
Element j → C H N O Species k ↓ CO 1 1 CO2 1 2 H2 2 H2O 2 1 NO 1 1 N2 2 OH 1 1 O2 2
The atom conservation equations (Σk djk Nk – Aj ) ( see Eq. (81)) yield the relations:
j = 1 (C atoms): 1NCO + 1N (^) CO 2 + 0N (^) H 2 + 0N (^) H 2 O + 0NNO + 0NN2 +0NOH + 0NO2 – 10 = 0,
(A)
j = 2 (H atoms): 0 NCO + 0 N (^) CO 2 + 2 N (^) H 2 + 2 N (^) H 2 O + 0NNO + 0NN2 +1NOH + 0NO2 – 20 = 0,
(B)
j = 3 (O atoms): 1NCO + 2 N (^) CO 2 + 0 N (^) H 2 + 1 N (^) H 2 O + 1NNO + 0NN2 +1NOH + 2NO2 - 31.6 =
0, (C)
j = 4 (N atoms): 0NCO + 0 N (^) CO 2 + 0N (^) H 2 + 0 N (^) H 2 O + 1NNO + 2NN2 +0NOH + 0NO2 - 118.6=
0, (D)
Dividing these equations by the total moles (N= ΣNk ), we obtain the relations
1XCO + 1X CO 2 + 0X H 2 + 0X H 2 O + 0XXO + 0XN2 +0XOH + 0X O2 – 10/N = 0, (E)
0 XCO + 0 X CO 2 + 2 X H 2 + 2 X H 2 O + 0XNO + 0XN2 +1XOH + 0XO2 – 20/N = 0, (F)
1XCO + 2 X CO 2 + 0 X H 2 + 1 X H 2 O + 1XNO + 0XN2 +1XOH + 2XO2 - 31.6/N= 0, (G)
0XCO + 0 X CO 2 + 0X H 2 + 0 X H 2 O + 1XNO + 2XN2 +0XOH + 0XO2 - 118.6/N = 0,(H)
where N is solved from the identity
ΣXk = 1 (I) G = G(T,P,N 1 , N 2 , …, NK). (J)
Multiplying Eqs. (A)–(D ), respectively by λC , λH, λN and λO, and adding with Eq. (J) we form a function
