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SOLUTION MANUAL 
CHAPTER 14 
Fundamentals of 
Thermodynamics 
Borgnakke Sonntag Borgnakke Sonntag 
 Borgnakke and Sonntag 
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 CONTENT CHAPTER 14 
 
 
 SUBSECTION PROB NO. 
 
 In-Text-Concept-Questions a-g 
 Concept-Study Guide Problems 1-17 
 Equilibrium and Phase Equilibrium 18-21 
 Chemical Equilibrium, Equilibrium Constant 22-74 
 Simultaneous Reaction 75-86 
 Gasification 87-93 
 Ionization 94-99 
 Applications 100-108 
 Review Problems 109-117 
 
 English Unit Problems 118-144 
 
 
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In-Text Concept Questions 
 
 
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14.a 
 For a mixture of O2 and O the pressure is increased at constant T; what happens 
to the composition? 
 
 An increase in pressure causes the reaction to go toward the side of 
smaller total number of moles, in this case toward the O2 . 
 
14.b 
 For a mixture of O2 and O the temperature is increased at constant P; what 
happens to the composition? 
 
 A temperature increase causes more O2 to dissociate to O. 
 
14.c 
 For a mixture of O2 and O I add some argon keeping constant T, P; what happens 
to the moles of O? 
 
 Diluting the mixture with a non-reacting gas has the same effect as 
decreasing the pressure, causing the reaction to shift toward the side of larger total 
number of moles, in this case the O. 
 
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14.d 
 When dissociations occur after combustion, does T go up or down? 
 
 Dissociation reactions of combustion products lower the temperature. 
 
14.e 
 For nearly all the dissociations and ionization reactions what happens with the 
composition when the pressure is raised? 
 
 The reactions move towards the side with fewer moles of particles, that is 
ions and electrons towards the monatomic side and monatomic species combine 
to form the molecules (diatomic or more) 
 
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14.f 
 How does the time scale for NO formation change when P is lower at same T? 
 
 Look at the expression for the time scale in Eq.14.40. As P is lowered the 
time scale becomes larger. The formation rates drops, and the effect is explained 
by the larger distance between the molecules/atoms (density lower), the same T 
essentially means that they have the same characteristic velocity. 
 
14.g 
 Which atom in air ionizes first as T increases? What is the explanation? 
 
 Using Fig. 14.11, we note that as temperature increases, atomic N ionizes 
to N+, becoming significant at about 6-8000 K. N has a lower ionization potential 
compared to O or Ar. 
 
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Concept-Study Guide Problems 
 
 
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14.1 
 Is the concept of equilibrium limited to thermodynamics? 
 
 Equilibrium is a condition in which the driving forces present are 
balanced, with no tendency for a change to occur spontaneously. This concept 
applies to many diverse fields of study – one no doubt familiar to the student 
being that of mechanical equilibrium in statics, or engineering mechanics. 
 
 
14.2 
 How does Gibbs function vary with quality as you move from liquid to vapor? 
 
 There is no change in Gibbs function between liquid and vapor. For 
equilibrium we have gg = gf. 
 
 
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14.3 
 How is a chemical equilibrium process different from a combustion process? 
 
 Chemical equilibrium occurs at a given state, T and P, following a 
chemical reaction process, possibly a combustion (which is a chemical reaction 
process) together with several reactions amongst the combustion products. 
Whereas the combustion is a one-way process (irreversible) the chemical 
equilibrium is a reversible process that can proceed in both directions. 
 
 
14.4 
 Must P and T be held fixed to obtain chemical equilibrium? 
 
 No, but we commonly evaluate the condition of chemical equilibrium at a 
state corresponding to a given temperature and pressure. If T and P changes in a 
process it means that the chemical composition adjusts towards equilibrium and 
the composition changes along with the process. 
 
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14.5 
 The change in Gibbs function ∆Go for a reaction is a function of which property? 
 
 The change in Gibbs function for a reaction ∆G is a function of T and P. 
The change in standard-state Gibbs function ∆Go is a function only of T. 
 
 
 
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14.6 
 In a steady flow burner T is not controlled, which properties are? 
 
 The pressure tends to be constant, only minor pressure changes due to 
acceleration of the products as density decreases velocity must increase to have 
the same mass flow rate. The product temperature depends on heat losses 
(radiation etc.) and any chemical reactions that may take place generally lowering 
the temperature below the standard adiabatic flame temperature. 
 
 
14.7 
 In a closed rigid combustion bomb which properties are held fixed? 
 
 The volume is constant. The number of atoms of each element is 
conserved, although the amounts of various chemical species change. As the 
products have more internal energy but cannot expand the pressure increases 
significantly. 
 
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14.8 
 Is the dissociation of water pressure sensitive? 
 
 Yes, since the total number of moles on the left and right sides of the 
reaction equation(s) is not the same. 
 
 
14.9 
 At 298 K, K = exp(−184) for the water dissociation, what does that imply? 
 
 This is an extremely small number, meaning that the reaction tends to go 
strongly from right to left – in other words, does not tend to go from left to right 
(dissociation of water) at all. 
 
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14.10 
 If a reaction is insensitive to pressure prove that it is also insensitive to dilution 
effects at a given T. 
 
 Assume the standard reaction we used to develop the expression for the 
equilibrium constant: νA A + νB B ⇔ νC C + νD D 
 let us assume we add an inert component E so the total moles become: 
 ntot = nA + nB + nC + nD + n E
 
This will now lower the mole fractions of A, B, C, and D. If the reaction is 
pressure insensitive then: νA + νB = νC + νD and the equilibrium equation 
becomes: 
 K = 
y
νC
C
 y
νD
D
y
νA
A
 y
νB
B
 





P
Po
νC + νD − νA − νB
 = C
 y
νD
D
y
νA
A
 y
νB
B
y
νC
 
 Since each yi = ni / ntot we get: 
 K = 
y
νC
C
 y
νD
D
y
νA
A
 y
νB
B
 = 
(nC/ntot)
νC (nD/ntot)
νD
 (nA/ntot)
νA (nB/ntot)
νB
 
 = 
n
νC
C
 n
νD
D
n
νA
A
 n
νB
B
 ntot
νA + νB − νC − νD
 = C
 n
νD
D
n
νA
A
 n
νB
B
n
νC
 
 
Now we see that the total number of moles that includes nE does not enter the 
equation and thus will not affect any progress of the reaction. 
 
 
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14.11 
 For a pressure sensitive reaction an inert gas is added (dilution), how does the 
reaction shift? 
 
 Assume the standard reaction we used to develop the expression for the 
equilibrium constant: νA A + νB B ⇔ νC C + νD D 
 let us assume we add an inert component E so the total moles become: 
 ntot = nA + nB + nC + nD + n E
 This will now lower the mole fractions of A, B, C, and D. If the reaction is 
pressure sensitive then: νA + νB ≠ νC + νD and the equilibrium equation 
becomes: 
 K = 
y
νC
C
 y
νD
D
y
νA
A
 y
νB
B
 





P
Po
νC + νD − νA − νB
 
 Since each yi = ni / ntot we get: 
 K = 
(nC/ntot)
νC (nD/ntot)
νD
 (nA/ntot)
νA (nB/ntot)
νB
 





P
Po
νC + νD − νA − νB
 
 = 
n
νC
C
 n
νD
D
n
νA
A
 n
νB
B
 





P
Po ntot
νC + νD − νA − νB
 
 
 As ntot is raised due to nE it acts as if the pressure P is lowered thus pushing the 
reaction towards the side with a larger number of moles. 
 
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14.12 
 In a combustion process is the adiabatic flame temperature affected by reactions? 
 
 The adiabatic flame temperature is lower due to dissociation reactions of 
the products as those products absorb some of the energy, i.e. 2 O atoms have 
more energy, h = 2(249 170 + ∆h) than a single O2 molecule, h = ∆h, see Table 
A.9. The temperature is also influenced by other reactions like the water gas 
reaction. 
 
 
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14.13 
 In equilibrium Gibbs function of the reactants and the products is the same; how 
about the energy? 
 
 The chemical equilibrium mixture at a given T, P has a certain total 
internal energy. There is no restriction on its division among the constituents. 
The conservation of energy from the reactants to the products will determine the 
temperature so if it takes place in a fixed volume (combustion bomb) then U is 
constant whereas if it is in a flow (like a steady flow burner) then H is constant. 
 When this is combined with chemical equilibrium it is actually a lengthy 
procedure to determine both the composition and the temperature in the actual 
process. 
 
 
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14.14 
 Does a dissociation process require energy or does it give out energy? 
 
 Dissociation reactions require energy and is thus endothermic. Notice 
from Table A.9 that all the atoms (N, O, H) has a much higher formation enthalpy 
than the diatomic molecules (which have formation enthalpy equal to zero). 
 
 
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14.15 
 If I consider the non-frozen (composition can vary) specific heat, but still assume 
all components are ideal gases, does that C become a function of temperature? of 
pressure? 
 
 The non-frozen mixture heat capacity will be a function of both T and P, 
because the mixture composition depends on T and P, while the individual 
component specific heat capacities depend only on T. 
 
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14.16 
 What is K for the water gas reaction in Example 14.4 at 1200 K? 
 
 Using the result of Example 14.4 and Table A.11 
 
 ln K = 
1
2 ( ln KI – ln KII ) 
 = 0.5 [–35.736 – (–36.363)] = + 0.3135 , 
 
 K = 1.3682 
 
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14.17 
 
 What would happen to the concentrations of the monatomic species like O, N if 
the pressure is higher in Fig. 14.11 
 
 Since those reaction are pressure sensitive (more moles on RHS than on 
LHS) the higher pressure will push these reactions to the left and reduce the 
concentrations of the monatomic species. 
 
 
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Equilibrium and Phase Equilibrium 
 
 
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14.18 
 Carbon dioxide at 15 MPa is injected into the top of a 5-km deep well in 
connection with an enhanced oil-recovery process. The fluid column standing in 
the well is at a uniform temperature of 40°C. What is the pressure at the bottom of 
the well assuming ideal gas behavior? 
 
 
Z 1 
Z 2 
CO 2 
cb 
 
(Z1 − Z2) = 5000 m, P1 = 15 MPa 
T = 40 oC = constant 
Equilibrium at constant T 
 −wREV = 0 = ∆g + ∆PE 
 = RT ln (P2/P1) + g(Z2 − Z1) = 0 
 
 ln (P2/P1) = 
9.807×5000
1000×0.188 92×313.2 = 0.8287 
 P2 = 15 MPa exp(0.8287) = 34.36 MPa 
 
 
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14.19 
 Consider a 2-km-deep gas well containing a gas mixture of methane and ethane at 
a uniform temperature of 30oC. The pressure at the top of the well is 14 MPa, and 
the composition on a mole basis is 90% methane, 10% ethane. Each component is 
in equilibrium (top to bottom) with dG + g dZ = 0 and assume ideal gas, so for 
each component Eq.14.10 applies. Determine the pressure and composition at the 
bottom of the well. 
 
 
Z 1 
Z 2 
mixture 
cb 
Gas 
A + B 
 
 
(Z1 − Z2) = 2000 m, Let A = CH4, B = C2H6 
P1 = 14 MPa, yA1 = 0.90, yB1 = 0.10 
T = 30 oC = constant 
From section 14.1, for A to be at equilibrium between 
 
1 and 2: WREV = 0 = nA(G
-
A1 − G
-
A2) + nA MA g (Z1 − Z2) 
 Similarly, for B: WREV = 0 = nB(G
-
B1 − G
-
B2) + nB MB g (Z1 − Z2) 
Using eq. 14.10 for A: R- T ln (PA2/PA1) = MA g (Z1 − Z2) 
with a similar expression for B. Now, ideal gas mixture, PA1 = yA1P, etc. 
Substituting: ln 
yA2P2
yA1P1
 = 
MAg(Z1-Z2)
R- T
 and ln 
yB2P2
yB1P1
 = 
MBg(Z1-Z2)
R- T
 
 ln (yA2 P2) = ln(0.9×14) + 
16.04×9.807(2000)
1000×8.3145×303.2 = 2.6585 
 => yA2 P2 = 14.2748 
 ln (yB2P2) = ln(0.1×14) + 
30.07×9.807(2000)
1000×8.3145×303.2 = 0.570 43 
 => yB2 P2 = (1 − yA2) P2 = 1.76903 
 Solving: P2 = 16.044 MPa & yA2 = 0.8897 
 
 
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14.20 
 A container has liquid water at 20oC , 100 kPa in equilibrium with a mixture of 
water vapor and dry air also at 20oC, 100 kPa. How much is the water vapor 
pressure and what is the saturated water vapor pressure? 
 
From the steam tables we have for saturated liquid: 
 Pg = 2.339 kPa, vf = 0.001002 m
3/kg 
 The liquid is at 100 kPa so it is compressed liquid still at 20oC so from 
Eq.14.15 at constant T 
 gliq – gf = ⌡⌠ v dP = vf (P – Pg) 
 The vapor in the moist air is at the partial pressure Pv also at 20
oC so we 
assume ideal gas for the vapor 
 gvap – gg = ⌡⌠ v dP = RT ln Pg
Pv 
 We have the two saturated phases so gf = gg ( q = hfg = Tsfg ) and now 
for equilibrium the two Gibbs function must be the same as 
 gvap = gliq = RT ln 
Pv
Pg
 + gg = vf (P – Pg) + g f
leaving us with 
 ln 
Pv
Pg
 = vf (P – Pg)/ RT = 
0.001002 (100 - 2.339)
0.4615 × 293.15 = 0.000723 
 Pv = Pg exp(0.000723) = 2.3407 kPa.This is only a minute amount above the saturation pressure. For the moist 
air applications in Chapter 11 we neglected such differences and assumed the 
partial water vapor pressure at equilibrium (100% relative humidity) is Pg. 
The pressure has to be much higher for this to be a significant difference. 
 
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14.21 
 Using the same assumptions as those in developing Eq. d in Example 14.1, 
develop an expression for pressure at the bottom of a deep column of liquid in 
terms of the isothermal compressibility, βT. For liquid water at 20
oC, βT = 0.0005 
[1/MPa]. Use the result of the first question to estimate the pressure in the Pacific 
ocean at the depth of 3 km. 
 
 d gT = v° (1 − βTP) dPT d gT + g dz = 0 
 v° (1 − β EEA⌡⌠v°(1-β AT AP) dP AT AEA = − g A⌡⌠dzE TP) dPT + g dz = 0 and integrate 
 EEEA
⌡
⌠
P A0 A
E
P (1-β ATAP) dP AT AEA = + A
g
v° EA A⌡
⌠
0
+HdzE A => P - P A0 EA - β ATE A A
1
2EA [P A
2 E
A − PA0 EAA
2 E
A] = A
g
v° EA H 
 P (1 − A
1
2EA β ATE A P) = P A0 EA − A
1
2EAβATE A P A0 EAA
2 E
A + A
g
v°EA H 
v° = vAf 20°CEA = 0.001002; H = 3000 m, g = 9.80665 m/s A
2 E
A; β ATE A = 0.0005 1/MPa 
 P (1 − A12 EA × 0.0005 P) = 0.101 − A
1
2 EA × 0.0005 × 0.101A
2 E
A 
 + [9.80665 × 3000/0.001002] × 10A-6E 
 = 29.462 MPa, which is close to P 
Solve by iteration or solve the quadratic equation 
 P = 29.682 MPa 
 
 
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Chemical Equilibrium, Equilibrium Constant 
 
 
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14.22 
 Which of the reactions listed in Table A.11 are pressure sensitive? 
 Check if: νAAE A + ν AB EA ≠ ν ACE A + νAD E 
 
 Reaction Check P sensitive? 
 HA2 EA ⇔ 2H 1 < 2 yes 
 OA2 EA ⇔ 2O 1 < 2 yes 
 NA2 EA ⇔ 2N 1 < 2 yes 
 2 HA2 EAO ⇔ 2 HA2 EA + 1 OA2 EA 2 < 3 yes 
 2 HA2 EAO ⇔ 2 HA2 EA + 2 OH 2 < 4 yes 
 2 COA2 EA ⇔ 2 CO + 1 OA2 EA 2 < 3 yes 
NA2 EA + OA2 EA ⇔ 2 NO 2 = 2 no 
NA2 EA + 2OA2 EA ⇔ 2 NOA2 EA 3 > 2 yes 
 
Most of them have more moles on RHS and thus will move towards the RHS if 
the pressure is lowered. Only the last one has the opposite and will move towards 
the LHS if the pressure is lowered. 
 
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14.23 
 Calculate the equilibrium constant for the reaction OA2 EA ⇔ 2O at temperatures of 
298 K and 6000 K. Verify the result with Table A.11. 
 Reaction OA2 EA ⇔ 2O 
At 25 Ao EAC (298.15 K): 
 ∆HA0 EA = 2Ah-E AA0f OE A − 1Ah
-E
AA
0
f O2 E A = 2(249 170) − 1(0) = 498 340 kJ/kmol 
 ∆S A0 EA = 2As-E AA0OEA − 1As
-E
AA
0
O2 EA = 2(161.059) − 1(205.148) = 116.97 kJ/kmol K 
 ∆GA0 EA = ∆HA0 EA − T∆S A0 EA = 498 340 − 298.15×116.97 = 463 465 kJ/kmol 
 ln K = − A
∆G0
ER- TE
A = − A
463 465
8.3145×298.15EA = −186.961 
At 6000 K: 
 ∆HA0 EA = 2(249 170 + 121 264) − (0 + 224 210) = 516 658 kJ/kmol 
 ∆S A0 EA = 2(224.597) −1(313.457) = 135.737 kJ/kmol K 
 ∆GA0 EA = 516 658 − 6000×135.737 = −297 764 kJ/kmol 
 ln K = A
+297 764
8.3145×6000E A = +5.969 
 
 
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14.24 
 Calculate the equilibrium constant for the reaction HA2 EA ⇔ 2H at a temperature of 
2000 K, using properties from Table A.9. Compare the result with the value listed 
in Table A.11. 
 From Table A.9 at 2000 K we find: 
 E∆h
-
HA2 A
 = 52 942 kJ/kmol; E s
-
HA2 A
 = 188.419 kJ/kmol K; Ah
−o
EfE A = 0 
 ∆h
-
H = 35 375 kJ/kmol; s
-
H = 154.279 kJ/kmol K; Ah
−o
EfE A = 217 999 kJ/kmol 
 ∆GA0 EA = ∆H − T∆S = HARHSE A − HALHSE A – T (S A
0 E
AARHSE A − S A
0 E
AALHSE A) 
 = 2 × (35 375 + 217 999) – 52943 – 2000(2×154.279 - 182.419) 
 = 213 528 kJ/kmol 
 ln K = −∆GA0 EA/AR
-E
AT = −213 528 / (8.3145 × 2000) = −12.8407 
 
 Table A.11 ln K = −12.841 OK 
 
 
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14.25 
For the dissociation of oxygen, OA2 EA ⇔ 2O, around 2000 K we want a 
mathematical expression for the equilibrium constant K(T). Assume constant heat 
capacity, at 2000 K, for OA2 EA and O from Table A.9 and develop the expression 
from Eqs. 14.12 and 14.15. 
 
From Eq.14.15 the equilibrium constant is 
 K = exp( − A
∆G0
ER−TE
A ) ; ∆GA0 EA = ∆HA0 EA – T ∆S A0 E 
and the shift is 
 ∆GA0 EA = 2 h-O - h
-
O2 - T(2s
-o
O – s
-o
O2) 
Substitute the first order approximation to the functions Ah-E A and As-o EA as 
 Ah-E A = h-2000 K + C
−
p (T – 2000) ; As
-o E
A = s-o2000 K + C
−
p ln A
T
2000E 
The properties are from Table A.9 and AR− E A = 8.3145 kJ/kmol K 
Oxygen OA2 EA: h
-
2000 K = 59 176 kJ/kmol, s
-o
2000 K = 268.748 kJ/kmol K 
 C−p = 
h-2200 K − h
-
2200 K
2200 - 1800 = A
66 770 − 51 674
400E A = 37.74 kJ/kmol K 
 
Oxygen O: h-2000 K = 35 713 + 249 170 = 284 883 kJ/kmol, 
s-o2000 K = 201.247 kJ/kmol K 
 C−p = 
h-2200 K − h
-
2200 K
2200 - 1800 = A
39 878 − 31 547
400E A = 20.8275 kJ/kmol K 
Substitute and collect terms 
 A
∆G0
ER−TE
A = A
∆Η0
ER−TE
A – A
∆S0
ER− E
A = A
∆Η
0
 2000
ER−TE
A + A
∆C−p 2000
ER− E
A [ AT − 2000TE A – ln A T2000EA] – A
∆S02000ER− E
 
 
Now we have 
 ∆HA02000E A/ AR
− E
A = (2 × 284 883 – 59 176)/8.3145 = 61 409.6 K 
 ∆AC−p 2000EA/ AR
− E
A = (2 × 20.8275 – 37.74)/8.3145 = 0.470864 
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∆S A02000EA/ AR
− E
A = (2 × 201.247 – 268.748)/8.3145 = 16.08587 
 
so we get 
 A
∆G0
ER−TE
A = A
61 409.6
TE A + 0.470864 [ A
T − 2000
TE A – ln A
T
2000EA ] – 16.08587 
 = A
60 467.9
TE A – 15.615 – 0.470864 ln A
T
2000E 
 Now the equilibrium constant K(T) is approximated as 
 
 K(T) = exp [ 15.615 – A60 467.9TE A + 0.470864 ln A
T
2000EA ] 
 
Remark: We could have chosen to expand the function ∆GA0 EA/ AR− E AT as a linear 
expression instead or even expand the whole exp(-∆GA0 EA/ AR− E AT) in a linear function. 
 
 
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14.26 
Find K for: COA2 EA ⇔ CO + 1/2OA2 EA at 3000 K using A.11 
 
The elementary reaction in A.11 is : 2COA2 EA ⇔ 2CO + OA2 E 
so the wanted reaction is (1/2) times that so 
 
K = K1/2A.11 = A exp(-2.217)E A = A 0.108935EA = 0.33 
or 
 
ln K = 0.5 ln KA.11 = 0.5 (−2.217) = −1.1085 
 K = exp(−1.1085) = 0.33 
 
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14.27 
 Plot to scale the values of ln K versus 1/T for the reaction 2 COA2 EA ⇔ 2 CO + OA2 EA. 
Write an equation for ln K as a function of temperature. 
 2 COA2 EA ⇔ 2 CO + 1 OA2 E 
 
 T(K) 
10A4 EA × A
1
TE 
 ln K T(K) 
10A4 EA × A
1
TE 
 ln K 
 2000 5.000 -13.266 4000 2.500 3.204 
 2400 4.167 -7.715 4500 2.222 4.985 
 2800 3.571 -3.781 5000 2.000 6.397 
 3200 3.125 -0.853 5500 1.818 7.542 
 3600 2.778 1.408 6000 1.667 8.488 
 
 For the range 
below ~ 5000 K, 
 
 ln K ≈ A + B/T 
 
Using values at 
 2000 K & 5000 K 
 
 A = 19.5056 
 B = -65 543 K 
 
 
8 
4 
0 
-4
-8
-12 
1 2 3 4 5 0 
1 
almost 
 linear 
10 T 
_ x 4 
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14.28 
Consider the reaction 2 COA2 EA ⇔ 2 CO + OA2 EA obtained after heating 1 kmol COA2 EA to 
3000 K. Find the equilibrium constant from the shift in Gibbs function and verify 
its value with the entry in Table A.11. What is the mole fraction of CO at 3000 K, 
100 kPa? 
From Table A.9 we get: 
∆h
-
CO = 93 504 Ah
-E
AA
0
f COE A = -110 527 s
-
CO = 273.607 
E∆h
-
COA2 A
 = 152 853 Ah-E AEA
0
f COA2 AEA = -393 522E s
-
COA2 A
 = 334.17 
E∆h
-
OA2 A
 = 98 013 E s
-
OA2 A
 = 284.466 
∆GA0 EA = ∆H - T∆S = 2 HACOEA + HEAOA2 AE
A – 2 HEACOA2 AE
A- T (2s
-
CO +E s
-
OA2 A
 - E2s
-
COA2 A
) 
 = 2 (93 504 – 110 527) + 98 013 + 0 – 2(152 853 - 393 522) 
 -3000(2×273.607 + 284.466 - 2×334.17) = 55 285 
 ln K = -∆GA0 EA/ AR
-E
AT = -55 285/ (8.31451×3000) = -2.2164 
 Table A.11 ln K = -2.217 OK 
 
 At 3000 K, 2 COA2 EA ⇔ 2 CO + 1 OA2 E 
 ln K = -2.217 Initial 1 0 0 
 K = 0.108935 Change -2z +2z +z 
 Equil. 1-2z 2z z 
 
We have P = P Ao EA = 0.1 MPa, and nAtotEA = 1 + z, so from Eq.14.29 
 K = A
yCOE
2 yO2
yCO2
2
E
A (A
P
P0 E
A) = A


2z
1 - 2z
2 E
A A


z
1 + z E A (1) = 0.108935 ; 
 4 zA3 EA = 0.108935 (1 – 2z)A2 EA(1 + z) => z = 0.22 
 
 yACOEA = 2z / (1 + z) = 0.36 
 
 
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14.29 
Carbon dioxide is heated at 100 kPa. What should the temperature be to see a 
mole fraction of CO as 0.25? For that temperature, what will the mole 
fraction of CO be if the pressure is 200 kPa? 
 
 2 COA2 EA ⇔ 2 CO + 1 OA2 E 
 Initial 1 0 0 
 Change -2z +2z +z 
 Equil. 1-2z 2z z 
 
We have P = P Ao EA = 0.1 MPa, and nAtotEA = 1 + z, so from Eq.14.29 
 yACOEA = A
2z
1 + z EA = 0.25 => z = 1/7 
K = A
yCOE
2 yO2
yCO2
2
E
A (A
P
P0 E
A) = A


2z
1 - 2z
2 E
A A


z
1 + z E A (1) = A


2
7 - 2
2 E
A A


1
7 + 1E A = 0.02 
 ln K = - 3.912 from A.11 T = 2785 K 
 K = 0.02 = A
yCOE
2 yO2
yCO2
2
E
A (A
P
P0 E
A) = A


2z
1 - 2z
2 E
A A


z
1 + z E A (2) => 
 zA3 EA = 0.0025 (1 – 2z)A2 EA(1 + z) => z = 0.11776 
 yACOEA = A
2z
1 + z EA = 0.2107 
 
 
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14.30 
 Assume a diatomic gas like OA2 EA or NA2 EA dissociate at a pressure different from P Ao EA. 
Find an expression for the fraction of the original gas that has dissociated at any T 
assuming equilibrium. 
 
Look at initially 1 mol Oxygen and shift reaction with x 
OA2 EA ⇔ 2 O 
 
Species OA2 E O 
Initial 1 0 
Change –x 2x 
Equil. 1–x 2x nAtotEA = 1 – x + 2x = 1 + x 
 yAOEA = A
2x
1 + xEA and yAO2 EA = A
1 – x
1 + xE 
Substitute this into the equilibrium equation as 
 KATE A = A
yOE
2
y02
E
A (A PPo E
A)A2-1EA = A 4x
2
E(1 + x)2EA A
1 + x
1 – x E A (A
P
Po E
A) = A 4x
2
E1 – x2E A (A
P
Po E
A) 
Now solve for x as 
 xA2 EA = (1 – x A2 EA) 
KTPo
4P 
 x = 
KT
4(P/Po) + KT
 
 
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14.31 
 Hydrogen gas is heated from room temperature to 4000 K, 500 kPa, at 
which state the diatomic species has partially dissociated to the monatomic 
form. Determine the equilibrium composition at this state. 
 
 HA2 EA ⇔ 2 H Equil. nAH2 EA = 1 - x 
 -x +2x nAH EA = 0 + 2x 
 n = 1 + x 
 K = A
(2x)2
E(1-x)(1+x)EA (A
P
P0 E
A) A2-1EA at 4000 K: ln K = 0.934 => K = 2.545 
 A
2.545
4×(500/100)EA = 0.127 25 = A
x2
E1-x2E
A Solving, x = 0.3360 
 nAH2 EA = 0.664, nAH EA = 0.672, ntot = 1.336 
 yAH2 EA = 0.497, yAH EA = 0.503 
 
 
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14.32 
Consider the dissociation of oxygen, OA2 EA ⇔ 2 O, starting with 1 kmol oxygen at 
298 K and heating it at constant pressure 100 kPa. At which temperature will we 
reach a concentration of monatomic oxygen of 10%? 
 
Look at initially 1 mol Oxygen and shift reaction with x 
OA2 EA ⇔ 2 O 
 
Species OA2 E O 
Initial 1 0 
Change -x 2x 
Equil. 1-x 2x nAtotEA = 1 - x + 2x = 1 + x 
 
 yAOEA = A
2x
1 + xEA = 0.1 ⇒ x = 0.1/(2 – 0.1) = 0.0526, yAO2 EA = 0.9 
 K = A
yOE
2
y02
E
A (A PPo E
A)A2-1EA = A0.1
2
E0.9E A 1 = 0.01111 ⇒ ln K = –4.4998 
 Now look in Table A.11: T = 2980 K 
 
 
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14.33 
Redo Problem 14.32 for a total pressure of 40 kPa. 
 
Look at initially 1 mol Oxygen and shift reaction with x 
OA2 EA ⇔ 2 O 
 
Species OA2 E O 
Initial 1 0 
Change -x 2x 
Equil. 1-x 2x nAtotEA = 1 - x + 2x = 1 + x 
 
 yAOEA = A
2x
1 + xEA = 0.1 ⇒ x = 0.1/(2 – 0.1) = 0.0526, yAO2 EA = 0.9 
 K = A
yOE
2
y02
E
A (A PPo E
A)A2-1EA = A0.1
2
E0.9E A 0.4 = 0.004444 ⇒ ln K = –5.4161 
 Now look in Table A.11: T = 2856 K 
 
 
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14.34 
Redo Problem 14.32, but start with 1 kmol oxygen and 1 kmol helium at 298 K, 
100 kPa. 
 
Look at initially 1 mol Oxygen and shift reaction with x 
 OA2 EA ⇔ 2 O 
 
Species OA2 E O He 
Initial 1 0 1 
Change -x 2x 
Equil. 1-x 2x 1 nAtotEA = 1 - x + 2x + 1 = 2 + x 
 
 yAOEA = A
2x
2 + xEA = 0.1 ⇒ x = 0.2/(2 – 0.1) = 0.10526, 
 yAO2 EA = (1 – x)/(2 + x) = 0.425 
 K = A
yOE
2
y02
E
A (A PPo E
A)A2-1EA = A 0.1
2
E0.425 EA 1 = 0.023529 ⇒ ln K = –3.7495 
 Now look in Table A.11: T = 3094 K 
 
 
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14.35 
 Calculate the equilibrium constant for the reaction:2COA2 EA ⇔ 2CO + OA2 EA at 3000 K 
using values from Table A.9 and compare the result to Table A.11. 
 
From Table A.9 we get: 
 kJ/kmol kJ/kmol kJ/kmol K 
∆h
-
CO = 93 504 Ah
-E
AA
o
f COE A = -110 527 s
-
CO = 273.607 
E∆h
-
COA2 A
 = 152 853 Ah-E AEA
o
f COA2 AEA = -393 522E s
-
COA2 A
 = 334.17 
E∆h
-
OA2 A
 = 98 013 Ah-E AEA
o
f OA2 AEA = 0 E s
-
OA2 A
 = 284.466 
∆GA0 EA = ∆H - T∆S = 2 HACOEA + HEAOA2 AE
A – 2 HEACOA2 AE
A- T (2s
-
CO +E s
-
OA2 A
 - E2s
-
COA2 A
) 
 = 2 (93 504 – 110 527) + 98 013 + 0 – 2(152 853 - 393 522) 
 -3000(2×273.607 + 284.466 - 2×334.17) = 55 285 kJ/kmol 
 ln K = -∆GA0 EA/ AR
-E
AT = -55 285/ (8.31451×3000) = -2.2164 
 Table A.11 ln K = -2.217 OK 
 
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14.36 
Find the equilibrium constant for: CO + 1/2OA2 EA ⇔ COA2 EA at 2200 K using Table 
A.11 
 
The elementary reaction in A.11 is: 2 COA2 EA ⇔ 2CO + OA2 E 
The wanted reaction is therefore (−0.5) times that so 
 
K = K−1/2A.11 = 1 / A exp(-10.232)E A = 1 / A 0.000036EA = 166.67 
or 
 
ln K = −0.5 ln KA.11 = −0.5 (−10.232) = 5.116 
 K = exp(5.116) = 166.67 
 
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14.37 
 Pure oxygen is heated from 25°C to 3200 K in an steady flow process at a 
constant pressure of 200 kPa. Find the exit composition and the heat transfer. 
 
The only reaction will be the dissociation of the oxygen 
 OA2 EA ⇔ 2O ; From A.11: K(3200) = exp(-3.069) = 0.046467 
Look at initially 1 mol Oxygen and shift reaction with x 
 nAO2 EA = 1 - x; nAOEA = 2x; nAtotEA = 1 + x; yAi EA = nAi EA/nAtotE 
 K = A
yOE
2
y02
E
A (A PPo E
A)A2-1EA = A 4x
2
E(1 + x)2 E
A A
1 + x
1 - xE A 2 = A
8x2
E1 - x2E
 
 xA2 EA = A
K/8
1 + K/8EA ⇒ x = 0.07599; 
 yA02
EA = A
1 – x
1 + xEA = 0.859; yA0 EA = 1– yA02
EA = 0.141 
 
 
 Aq-E A = nA02ex
EAAh-E AA02ex
EA + nA0exEAAh
-E
AAOex EA - Ah
-E
AA02in
EA = (1 + x)(yA02
EAAh-E AA02
EA + yA0 EAAh
-E
AAOEA) - 0 
 Ah-E AA02
EA = 106 022 kJ/kmol; Ah-E AAOEA = 249 170 + 60 767 = 309 937 kJ/kmol 
 ⇒ Aq-E A = 145 015 kJ/kmol OA2 E 
 q = Aq-E A/32 = 4532 kJ/kg ( = 3316.5 if no reaction) 
 
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14.38 
 Nitrogen gas, N2, is heated to 4000 K, 10 kPa. What fraction of the N2 is 
dissociated to N at this state? 
 NA2 EA <=> 2 N @ T = 4000 K, lnK = -12.671 
 
Initial 1 0 K = 3.14x10-6 
Change -x 2x 
Equil. 1-x 2x ntot = 1 - x + 2x = 1 + x 
 
 yAN2 EA = A
1 - x
1 + xE A , yAN EA = A
2x
1 + xEA 
K = 
y2N
yN2
 A





P
Po
2-1
AEE; => 3.14x10-6 = A
4x2
E1 - x2 E
A A


10
100 E A => x = 0.0028 
yN2 = A
1 - x
1 + xEA = 0.9944, yN = A
2x
1 + xEA = 0.0056 
 
 
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14.40 
One kilomole Ar and one kilomole O2 are heated up at a constant pressure of 
100 kPa to 3200 K, where it comes to equilibrium. Find the final mole 
fractions for Ar, OA2 EA, and O. 
The only equilibrium reaction listed in the book is dissociation of OA2 EA. 
So assuming that, we see in Table A.11: ln(K) = -3.072 
 Ar + OA2 EA ⇒ Ar + (1 - x) OA2 EA + 2x O 
The atom balance already shown in above equation can also be done as 
 Species Ar OA2 EA O 
 Start 1 1 0 
 Change 0 -x 2x 
 Total 1 1-x 2x 
The total number of moles is n AtotEA = 1 + 1-x + 2x = 2 + x so 
 yAArE A = 1/(2 + x); yAO2 EA = 1 - x/(2 + x); yAOEA = 2x/(2 + x) 
and the definition of the equilibrium constant (P AtotEA = P Ao EA) becomes 
 K = eA-3.072EA = 0.04633 = A
yOE
2
y02 E
A = A
4x2
E(2 + x)(1 - x)E 
The equation to solve becomes from the last expression 
 (K + 4)x A2 EA + Kx - 2K = 0 
If that is solved we get 
 x = -0.0057 ± 0.1514 = 0.1457; x must be positive 
 yAOEA = 0.1358; yA02EA = 0.3981; yAArE A = 0.4661 
 
 
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14.41 
Air (assumed to be 79% nitrogen and 21% oxygen) is heated in a steady state 
process at a constant pressure of 100 kPa, and some NO is formed. At what 
temperature will the mole fraction of NO be 0.001? 
 
 0.79 NA2 EA + 0.21 OA2 EA heated at 100 kPa, forms NO 
 
 NA2 EA + OA2 EA ⇔ 2 NO nAN2 EA = 0.79 - x 
 -x -x +2x nAO2 EA = 0.21 - x 
 nANOEA = 0 + 2x 
 ntot = 1.0 
 At exit, yANOEA = 0.001 = A
2x
1.0EA ⇒ x = 0.0005 
 ⇒ nAN2 EA = 0.7895, nAO2 EA = 0.2095 
 K = A
y2NO
EyN2yO2 E
A (A P
P0 E
A) A0 EA = A 10
-6
E0.7895×0.2095EA = 6.046×10A
-6E
A or ln K = -12.016 
 From Table A.11, T = 1444 K 
 
 
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14.42 
 Pure oxygen is heated from 25°C, 100 kPa to 3200 K in a constant volume 
container. Find the final pressure, composition, and the heat transfer. 
As oxygen is heated it dissociates 
 OA2 EA ⇔ 2O ln KAeq EA = -3.069 from table A.11 
C. V. Heater: UA2 EA - UA1 EA = A1 EAQA2 EA = HA2 EA - HA1 EA - P A2 EAv + P A1 EAv 
 Per mole OA2 EA: A1 EAAq
-E
AA2 EA = Ah
-E
AA2 EA - Ah
-E
AA1 EA + AR
-E
A[TA1 EA - (n A2 EA/nA1 EA)TA2 EA] 
 Shift x in reaction 1 to have final composition: (1 - x)OA2 EA + 2xO 
 n A1 EA = 1 nA2 EA = 1 - x + 2x = 1 + x 
 yAO2 E
A = (1 - x)/(1 + x) ; yAO EA = 2x/(1 + x) 
Ideal gas and VA2 EA = VA1 EA ⇒ P A2 EA = P A1 EAnA2 EATA2 EA/n A1 EATA1 EA ⇒ P A2 EA/PAo EA = (1 + x)TA2 EA/TA1 E 
Substitute the molefractions and the pressure into the equilibrium equation 
 KAeq EA = eA
-3.069E
A = A
yOE
2
y02 E
A (A
P2
EPoE
A) = (A 2x1 + xEA)A
2 E
A (A
1 + x
1 - xE A) (A
1 + x
1E A) (A
T2
ET1E
A) 
 ⇒ A
4x2
E1 - xEA = A
T1
ET2E
A eA-3.069EA = 0.00433 ⇒ x = 0.0324 
The final pressure is then 
 P A2 EA = P Ao EA(1 + x)A
T2
ET1E
A = 100 (1 + 0.0324) × A
3200
298.2EA = 1108 kPa 
 (nAO2 E
A) = 0.9676, (nAO EA) = 0.0648, n A2 EA = 1.0324 
 A1 EAAq
-E
AA2 EA = 0.9676 × 106022 + 0.0648 (249170 + 60767) - 0 
 + 8.3145 (298.15 - 1.0324 × 3200) = 97681 kJ/kmolO A2 E 
 yAO2 E
A = A
0.9676
1.0324 E A= 0.937; yAO EA = A
0.0648
1.0324E A = 0.0628 
 
 
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14.43 
 Find the equilibrium constant for the reaction 2NO + OA2 EA ⇔ 2NOA2 EA from the 
elementary reactions in Table A.11 to answer which of the nitrogen oxides, NO or 
NOA2 EA, is the more stable at ambient conditions? What about at 2000 K? 
 
 2 NO + OA2 EA ⇔ 2 NOA2 EA (1) 
 But NA2 EA + OA2 EA ⇔ 2 NO (2) 
 NA2 EA + 2 OA2 EA ⇔ 2 NOA2 EA (3) 
 Reaction 1 = Reaction 3 - Reaction 2 
 ⇒ ∆GA01 E A = ∆GA
0
3 E A - ∆GA
0
2 EA => ln KA1 EA = ln KA3 EA - ln KA2 E 
At 25 Ao EAC, from Table A.11: ln KA1 EA = -41.355 - (-69.868) = +28.513 
 or KA1 EA = 2.416×10A
12E 
an extremely large number, which means reaction 1 tends to go very strongly 
from left to right. 
At 2000 K: ln KA1 EA = -19.136 - (-7.825) = - 11.311 or KA1 EA = 1.224 × 10A
-5E 
meaning that reaction 1 tends to go quite strongly from right to left. 
 
 
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14.44 
 Assume the equilibrium mole fractions of oxygen and nitrogen are close to those 
in air, find the equilibrium mole fraction for NO at 3000 K, 500 kPa disregarding 
dissociations. 
 
 Assume the simple reaction to make NO as: 
 
 NA2 EA + OA2 EA ⇔ 2 NO Ar nAN2 EA = 0.78 - x 
 0.78 0.21 0 0.01 nAO2 EA = 0.21 - x 
 -x -x +2x nANOEA = 2x, nAarE A = 0.01 
 0.78–x 0.21–x 2x 0.01 ntot = 1.0From A.11 at 3000 K, ln K = -4.205, K = 0.014921 
 
 K = A
4x2
E(0.78 – x)(0.21 – x)EA ( A
P
P0 E
A) A0 EA  
 A
x2
E(0.78 – x)(0.21 – x)E A = A
0.014921
4E A = 0.00373 and 0 < x < 0.21 
 
 Solve for x: x = 0.0230  yANOEA = A
2x
1.0EA = 0.046 
 
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14.45 
The combustion products from burning pentane, C A5 EAHA12EA, with pure oxygen in a 
stoichiometric ratio exists at 2400 K, 100 kPa. Consider the dissociation of only 
COA2 EA and find the equilibrium mole fraction of CO. 
 C A5 EAHA12EA + 8 OA2 EA → 5 COA2 EA + 6 HA2 EAO 
 
 At 2400 K, 2 COA2 EA ⇔ 2 CO + 1 OA2 E 
 ln K = -7.715 Initial 5 0 0 
 K = 4.461 × 10A-4E Change -2z +2z +z 
 Equil. 5-2z 2z z 
 
Assuming P = P Ao EA = 0.1 MPa, and ntot = 5 + z + 6 = 11 + z 
 K = A
yCOE
2 yO2
yCO2
2
E
A (A
P
P0 E
A) = A


2z
5 - 2z
2 E
A A


z
11 + z E A (1) = 4.461 × 10 A
-4E
A ; 
Trial & Error (compute LHS for various values of z): z = 0.291 
 nACO2 E
A = 4.418; nACOEA = 0.582; nAO2 E
A = 0.291 => yACOEA = 0.0515 
 
 
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14.46 
 A mixture flows with 2 kmol/s COA2 EA, 1 kmol/s argon and 1 kmol/s CO at 298 K 
and it is heated to 3000 K at constant 100 kPa. Assume the dissociation of carbon 
dioxide is the only equilibrium process to be considered. Find the exit equilibrium 
composition and the heat transfer rate. 
 
 
 Reaction 2 COA2 E ⇔ 2 CO + OA2 E Ar 
 initial 2 1 0 1 
 change -2x +2x +x 
 equil. 2 - 2x 1 + 2x x 1 
 
 From Table A.11: 
K = exp( -2.217) = 0.108935 = A
y2CO yO2
Ey2CO2 E
A ( A PPo EA) A
3-2E
A 
 = A
(1 + 2x)2
E(4 + x)2 E
A A
x
4 + xEA A
(4 + x)2
E(2 - 2x)2 E
A (1)A1 EA = Ax4EA A
(1 + 2x)2
E4 + xE A A
1
(1 - x)2E
 
then 
0.43574 (4 + x) (1 – x)A2 EA = x (1 + 2x)A2 E 
trial and error solution gives x = 0.32136 
 
The outlet has: 1.35728 COA2 EA + 1.64272 CO + 1 Ar + 0.32136 OA2 E 
Energy equation gives from Table A.9 and A.5 (for argon C Ap EAΔT) 
 AQ
.E
A = ∑ n. ex h-ex - ∑ n. in h-in 
 = 1.35728 (152853 – 393522) + 1.64272 (93504 – 110527) 
+ 1 (39.948 ×0.52 × (3000 – 298)) + 0.32136 (98013) 
– 2 (-393522) – 1(-110527) – 1 (0) 
 = 630 578 kW 
 
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14.47 
 A mixture of 1 kmol carbon dioxide, 2 kmol carbon monoxide, and 2 kmol 
oxygen, at 25°C, 150 kPa, is heated in a constant pressure steady state process to 
3000 K. Assuming that only these same substances are present in the exiting 
chemical equilibrium mixture, determine the composition of that mixture. 
 
 initial mix: 
 1 COA2 EA, 2 CO, 
 2 OA2 E 
 
Constant 
pressure 
reactor 
Q 
Equil. mix: 
COA2 EA, CO, OA2 EA at 
T = 3000 K, 
P = 150 kPa 
 
 Reaction 2 COA2 E ⇔ 2 CO + OA2 E 
 initial 1 2 2 
 change -2x +2x +x 
 equil. (1 - 2x) (2 + 2x) (2 + x) 
 
nAtotEA = 1 – 2x + 2 + 2x + 2 + x = 5 + x so y = n/ n AtotE 
 
From A.11 at 3000 K: K = exp(−2.217) = 0.108935 
 For each n > 0 ⇒ −1 < x < +A12 E 
 K = A
y2COyO2
Ey2CO2 E
A (A P
P0 E
A) A1 EA = 4 ( A 1 + x1 - 2xEA) A
2 E
A ( A2 + x5 + xEA) (A
150
100EA) = 0.108935 
 or ( A 1 + x1 - 2xE A) A
2 E
A ( A2 + x5 + x EA) = 0.018 156, Trial & error: x = −0.521 
 A


nCO2 = 2.042
EnCO = 0.958
 
nO2 = 1.479
ntot = 4.479
 E A A



 
yCO2 = 0.4559
EyCO = 0.2139
yO2 = 0.3302
E 
 
 
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14.48 
Acetylene gas Acetylene gas, C A2 EAHA2 EA, is burned with stoichiometric air in a torch. 
The reactants are supplied at the reference conditions P A0 EA, TA0 EA. The products come 
out from the flame at 2800 K after a small heat loss by radiation. Consider the 
dissociation of COA2 EA into CO and OA2 EA and no others. Find the equilibrium 
composition of the products. Are there any other reactions that should be 
considered? 
 
Combustion: C A2 EAHA2 EA + 2.5 (OA2 EA + 3.76NA2 EA) → 2 COA2 EA + 1 HA2 EAO + 9.4 NA2 E 
 
 At 2800 K, 2 COA2 EA ⇔ 2 CO + 1 OA2 EA HA2 EAO NA2 E 
 ln K = -3.781 Initial 2 0 0 1 9.4 
 K = 0.0228 Change -2z +2z +z 0 0 
 Equil. 2-2z 2z z 1 9.4 
 
Assuming P = P Ao EA = 0.1 MPa, and ntot = 2 + z + 1 + 9.4 = 12.4 + z 
 K = A
yCOE
2 yO2
yCO2
2
E
A (A
P
P0 E
A) = A


2z
2 - 2z
2 E
A A


z
12.4 + z E A (1) = 0.0228; 
Solve: zA3 EA = 0.0228 (1 – z)A2 EA(12.4 + z) => z = 0.4472 
 yACOEA = A
2z
12.4 + z E A = 0.0696; yACO2 EA = A
2 - 2z
12.4 + z EA = 0.086; yAO2 EA = A
z
12.4 + z E A = 0.0348 
 yAH2OEA = A
1
12.4 + z EA = 0.0778; yAN2 EA = A
9.4
12.4 + z EA = 0.7317 
 
 Looking in A.11 there will be no dissociation of nitrogen and only a small 
dissociation of water and oxygen which could be included. 
 
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14.49 
 Consider combustion of methane with pure oxygen forming carbon dioxide and 
water as the products. Find the equilibrium constant for the reaction at 1000 K. 
Use an average heat capacity of Cp = 52 kJ/kmol K for the fuel and Table A.9 for 
the other components. 
 
For the reaction equation, 
 
CHA4 EA + 2 OA2 EA ⇔ COA2 EA + 2 HA2 EAO 
 
At 1000 K from Table A.9 and A.10 for the fuel at 298 K 
 
∆HA01000 KE A = 1(−393 522 + 33 397) + 2(−241 826 + 26 000) 
− 1[−74 873 + 52(1000 – 298.2)] − 2(0 + 22 703) 
= − 798 804 kJ/kmol 
 ∆S A01000 KEA = 1×269.299 + 2×232.739 – 1(186.251+ ln A
1000
298.2EA ) - 2×243.579 
 = 487.158 kJ/kmol K 
 ∆GA01000 KE A = ∆HA
0
1000 KEA - T ∆S A
0
1000 KE 
 = − 798 804 – 1000 × 487.158 = − 1 285 962 kJ/kmol 
 ln K = − A
∆G0
ER- TE
A = A
+ 1 285 962
8.3145×1000EA = + 154.665 , K = 1.4796 E 67 
 
 This means the reaction is shifted totally to the right. 
 
 
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14.50 
Water from the combustion of hydrogen and pure oxygen is at 3800 K and 50 
kPa. Assume we only have HA2 EAO, OA2 EA and HA2 EA as gases find the equilibrium 
composition. 
 
With only the given components we have the reaction 
 2 HA2 EAO ⇔ 2HA2 EA + OA2 E 
which at 3800 K has an equilibrium constant from A.11 as ln K = −1.906 
Assume we start with 2 kmol water and let it dissociate x to the left then 
 
 Species HA2 EAO HA2 EA OA2 EA 
 Initial 2 0 0 
 Change -2x 2x x 
 Final 2 − 2x 2x x Tot: 2 + x 
Then we have 
 K = exp(−1.906) = A
yH2
E
2 yO2
yH2O
2
E
A A





P
P0
2+1-2E
A = A



2x
2 + x
2
 x2 + x



2 - 2x
2 + x
2 
50
E100E 
which reduces to 
 0.148674 = A
1
(1- x)2
 
4x3
E2 + x 
1
4 
1
2EA or x A
3 E
A = 0.297348 (1 – x)A2 EA (2 + x) 
Trial and error to solve for x = 0.54 then the concentrations are 
 
 yAH2OEA = A
2 - 2x
2 + xE A = 0.362; yAO2 EA = A
x
2 + xEA = 0.213; yAH2 EA = A
2x
2 + xEA = 0.425 
 
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14.51 
 Repeat problem 14.47 for an initial mixture that also includes 2 kmol of 
nitrogen, which does not dissociate during the process. 
 
 This problem has a dilution of the reactant with nitrogen. 
 
 initial mix: 
 1 COA2 EA, 2 CO, 
 2 OA2 EA, 2 NA2 E 
 
Constant 
pressure 
reactor 
Q 
Equilibrium mix: 
 COA2 EA, CO, OA2 EA and NA2 EA 
at T = 3000 K, P = 150 kPa 
 
 Reaction 2 COA2 E ⇔ 2 CO + OA2 E 
 initial 1 2 2 
 change -2x +2x +x 
 equil. (1-2x) (2+2x) (2+x) 
 
 From A.11 at 3000 K: K = exp(−2.217) = 0.108935 
 For each n > 0 ⇒ −1 < x < +A12 E 
 Equilibrium: nACO2 EA = (1 – 2x), nACOEA = (2 + 2x), n AO2 EA = (2 + x), 
n AN2 EA = 2 so then n AtotEA = 7 + x 
 K = A
y2COyO2
Ey2CO2 E
A (A P
P0 E
A) A1 EA = 4 ( A 1 + x1 - 2xEA) A
2 E
A ( A2 + x7 + xEA) (A
150
100EA) = 0.108935 
 or ( A 1 + x1 - 2xEA) A
2 E
A ( A2 + x7 + xEA) = 0.018167 Trial & error: x = −0.464 
 
 A


nCO2 = 1.928
EnCO = 1.072
 
nO2 = 1.536
nN2 = 2.0
nTOT = 6.536
 E A A



 
yCO2 = 0.295
EyCO = 0.164
 
yO2 = 0.235
yN2 = 0.306
E 
 
 
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14.52 
Catalytic gas generators are frequently used to decompose a liquid, providing 
a desired gas mixture (spacecraft control systems, fuel cell gas supply, and so 
forth). Consider feeding pure liquid hydrazine, NA2 EAHA4 EA, to a gas generator, from 
which exits a gas mixture of NA2 EA, HA2 EA, and NHA3 EA in chemical equilibrium at 
100°C, 350 kPa. Calculate the mole fractions of the species in the equilibrium 
mixture. 
 Initially, 2 NA2 EAHA4 EA → 1 NA2 EA + 1 HA2 EA + 2 NHA3 E 
 
 Reaction: NA2 E + 3 HA2 E ⇔ 2 NHA3 E 
 initial 1 1 2 
 change -x -3x +2x 
 equil. (1-x) (1-3x) (2+2x) nATOTALE A = (4-2x) 
 K = A
y2NH3
EyN2y
3
H2 E
A ( A P
P0 E
A) A-2EA = A(2 + 2x)
2(4 - 2x)2
E(1 - x)(1 - 3x)3E
A ( A350100EA) A
-2E
 
At 100 Ao EAC = 373.2 K, for NHA3 EA use A.5 AC
-E
AAP0EA = 17.03×2.130 = 36.276 
 Ah-E AA0NH3 E A = −45 720 + 36.276(373.2 − 298.2) = −42 999 kJ/kmol 
 As-E AA0NH3 EA = 192.572 + 36.276 ln A
373.2
298.2EA = 200.71 kJ/kmol K 
Using A.9, 
 ∆HA0100 CE A = 2(−42 999) − 1(0+2188) − 3(0+2179) = −94 723 kJ 
 ∆S A0100 CEA = 2(200.711) − 1(198.155) − 3(137.196) = −208.321 kJ/K 
 ∆GA0100 CE A = ∆HA
0 E
A − T∆S A0 EA = −94 723 − 373.2(−208.321) = −16 978 kJ 
 ln K = − A
∆G0
ER−TE
A = A
+16 978
8.3145×373.2EA = 5.4716 => K = 237.84 
Therefore, 
 [A(1 + x)(2 - x)
E(1 - 3x)E A]A
2 E
A A
1
(1 - x)(1 - 3x)EA = A
237.84×3.52
E16E A = 182.096 
By trial and error, x = 0.226 
 A


nN2 = 0.774
EnH2 = 0.322
 
nNH3 = 2.452
nTOT = 3.518
 E A A



 
yN2 = 0.2181
EyH2 = 0.0908
yNH3 = 0.6911
E 
 
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14.53 
 Complete combustion of hydrogen and pure oxygen in a stoichiometric ratio at P Ao E
A, TAo EA to form water would result in a computed adiabatic flame temperature of 
4990 K for a steady state setup. How should the adiabatic flame temperature be 
found if the equilibrium reaction 2HA2 EA + OA2 EA ⇔ 2 HA2 EAO is considered? Disregard all 
other possible reactions (dissociations) and show the final equation(s) to be 
solved. 
 
 2HA2 EA + OA2 EA ⇔ 2HA2 EAO Species HA2 EA OA2 EA HA2 EAO 
 Initial 2 1 0 
 Shift −2x −x 2x 
 Final 2 − 2x 1 −x 2x 
 Keq = A
yH2O
E
2
yH2
2 yO2
E
A (A
P
P0 E
A)A-1EA, nAtotEA = 2 − 2x + 1 − x + 2x = 3 − x 
Energy Eq.: HAP EA = HAREA = HAPE
o
E A + ∆HAP EA = HARE
o
E A = 0 
 HAPEA = (1 − x)(2∆Ah
-E
AAH2
EA + ∆Ah-E AAO2
EA) + 2x(Ah-E AAfH2OE
o
E A + ∆Ah-E AAH2OEA) = 0 (1) 
Equilibrium constant: 
 KAeq EA = A
4x2
E(3-x)2 E
A A
(3 - x)2
E(2 - 2x)2 E
A A
3 - x
1 - xEA = A
x2(3 - x)
E(1 - x)3 E
A = KAT EA (2) 
A.10: Ah-E AAfH2OE
o
E A = −241 826 kJ/kmol; A.11: ln (KAT EA ) 
A.9: ∆Ah-E AAH2
EA(T), ∆Ah-E AAO2
EA(T), ∆Ah-E AAH2O
EA(T), 
Trial and error (solve for x, T) using Eqs. (1) and (2). 
Guess T = 3600 K: K = exp(1.996) = 7.35956, x = 0.53986 
 HAPEA = (1 − 0.53986)(2 × 111 367 + 122 245) 
+ 2 × 0.53986 × (−241 826 + 160 484) = 70 912 high 
Guess T = 3400 K: K = exp(3.128) = 22.8283, x = 0.648723 
 HAPEA = (1 − 0.648723)(2 × 103 736 + 114 101) 
+ 2 × 0.648723 × (−241 826 + 149 073) = −7381 low 
Now do a linear interpolationto get HAPEA = 0: 
T = 3400 + 200 × 7381/(70 912 + 7381) = 3419 K, K = 20.50 
 x = 0.63905, yAO2 E
A = 0.153; yAH2 E
A = 0.306; yAH2OE
A = 0.541 
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14.54 
 Consider the water gas reaction in Example 14.4. Find the equilibrium constant at 
500, 1000, 1200 and 1400 K. What can you infer from the result? 
 
 
As in Example 14.4, III HA2 EA + COA2 EA ⇔ HA2 EAO + CO 
 
 I 2 COA2 EA ⇔ 2 CO + OA2 E 
 
 II 2 HA2 EAO ⇔ 2 HA2 EA + OA2 EA 
 
Then, ln KAIIIEA = 0.5 (ln KAI EA − ln KAII EA ) 
 
At 500 K, ln KAIIIEA = 0.5 (−115.234 – (−105.385)) = − 4.9245 , 
K = 0.007 266 
 
At 1000 K, ln KAIIIEA = 0.5 (−47.052 – (− 46.321)) = − 0.3655 , 
K = 0.693 85 
 
At 1200 K, ln KAIIIEA = 0.5 (−35.736 – (−36.363)) = + 0.3135 , 
K = 1.3682 
 
At 1400 K, ln KAIIIEA = 0.5 (−27.679 – (−29.222)) = + 0.7715 , 
K = 2.163 
 
It is seen that at lower temperature, reaction III tends to go strongly from right to 
left, but as the temperature increases, the reaction tends to go more strongly from 
left to right. If the goal of the reaction is to produce more hydrogen, then it is 
desirable to operate at lower temperature. 
 
 
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14.55 
A piston/cylinder contains 0.1 kmol hydrogen and 0.1 kmol Ar gas at 25°C, 
200 kPa. It is heated up in a constant pressure process so the mole fraction of 
atomic hydrogen is 10%. Find the final temperature and the heat transfer 
needed. 
When gas is heated up HA2 EA splits partly into H as 
 HA2 EA ⇔ 2H and the gas is diluted with Ar 
 Component HA2 EA Ar H 
 Initial 0.1 0.1 0 
 Shift -x 0 2x 
 Final 0.1-x 0.1 2x Total = 0.2 + x 
 yAHEA = 0.1 = 2x /(0.2 + x) ⇒ 2x = 0.02 + 0.1x 
 ⇒ x = 0.010526 ⇒ nAtotEA = 0.21053 
 yAH2 EA = 0.425 = [(0.1 - x)/(0.2 + x)]; yAArE A = 1 – rest = 0.475 
Do the equilibrium constant: 
 K(T) = A
y2H
EyH2E
A ( A P
P0 E
A) A2-1EA = ( A 0.010.425EA) × (A
200
100EA) = 0.047059 
 ln (K) = −3.056 so from Table A.11 interpolate to get T = 3110 K 
To do the energy eq., we look up the enthalpies in Table A.9 at 3110 K 
 h AH2 EA = 92 829.1; hAHEA = hAfE A + ∆h = 217 999 + 58 447.4 = 276 445.4 
 hAArE A = 0 + C AP EA(3110 – 298.15) = 20.7863 × (3110-298.13) = 58 447.9 
 (same as ∆h for H) 
Now get the total number of moles to get 
 nAHEA = 0.021053; nAH2 EA = nAtotEA × A
1-x
2+xE A = 0.08947; n AArE A = 0.1 
Since pressure is constant W = P∆V and Q becomes differences in h 
 Q = n∆h = 0.08947 × 92 829.1 – 0 + 0.021053 × 276 446.4 
 – 0 + 0.1 × 58 447.9 
 = 19 970 kJ 
 
 Borgnakke and Sonntag 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or 
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other 
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States 
Copyright Act without the permission of the copyright owner is unlawful. 
 
14.56 
The van't Hoff equation 
 d ln K = A
∆Ho
R−T2E
A dTAPoE 
relates the chemical equilibrium constant K to the enthalpy of reaction ∆Ho. From 
the value of K in Table A.11 for the dissociation of hydrogen at 2000 K and the 
value of ∆Ho calculated from Table A.9 at 2000 K use van’t Hoff equation to 
predict the constant at 2400 K. 
 
 HA2 EA ⇔ 2H 
 ∆H° = 2 × (35 375 + 217 999) – 52 942 = 453 806 kJ/kmol 
 ln KA2000EA = −12.841; 
 Assume ∆H° is constant and integrate the Van’t Hoff equation 
 lnKA2400EA − lnKA2000EA = A
⌡
⌠
2400
2000
 
∆Ho
R−T2
 dTE A = − EA
∆H°
AR− AE
A (EA
1
TA2400AE
A − EA
1
TA2000AE
A) 
 lnKA2400EA = lnKA2000EA + ∆H° (EA
1
TA2400AE
A − EA
1
TA2000AE
A) / AR− E 
 = −12.841 + 453 806 (A
6-5
12000 EA) / 8.31451 = −12.841 + 4.548 
 = −8.293 
Table A.11 lists –8.280 (∆H° not exactly constant) 
 
 
 Borgnakke and Sonntag 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or 
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other 
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Copyright Act without the permission of the copyright owner is unlawful. 
 
14.57 
 A gas mixture of 1 kmol carbon monoxide, 1 kmol nitrogen, and 1 kmol 
oxygen at 25°C, 150 kPa, is heated in a constant pressure process. The exit 
mixture can be assumed to be in chemical equilibrium with COA2 EA, CO, OA2 EA, and NA2 EA 
present. The mole fraction of COA2 EA at this point is 0.176. Calculate the heat 
transfer for the process. 
 
 initial mix: 
 1 CO, 1 OA2 EA, 
 1 NA2 E 
 
Constant 
pressure 
reactor 
Q 
Equil. mix: 
 COA2 EA, CO, OA2 EA, NA2 E 
 yACO2 EA = 0.176 
 P = 150 kPa 
 
 reaction 2 COA2 E ⇔ 2 CO + OA2 E also, NA2 E 
 initial 0 1 1 1 
 change +2x -2x -x 0 
 equil. 2x (1-2x) (1-x) 1 
 
 yACO2 EA = 0.176 = A
2x
3-xEA ⇒ x = 0.242 65 
 A


nCO2 = 0.4853
EnCO = 0.5147
 
nO2 = 0.7574
nN2 = 1
E A A



 
yCO2 = 0.176
EyCO = 0.1867
 yO2 = 0.2747 E 
 K = A
y2COyO2
Ey2CO2 E
A ( A P
P0 E
A) A1 EA = A0.1867
2×0.2747
E0.1762 E
A ( A150100EA) = 0.4635 
 From A.11, TAPRODEA = 3213 K 
 From A.10, HAREA = -110 527 kJ 
 HAP EA = 0.4853(-393 522 + 166 134) + 0.5147(-110 527 + 101 447) 
 + 0.7574(0 + 106 545) + 1(0 + 100 617) 
 = +66 284 kJ 
 QACVEA = HAP EA - HAREA = 66 284 - (-110 527) = +176 811 kJ 
 
 
 Borgnakke and Sonntag 
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or 
instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other 
reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the b1976 United States 
Copyright Act without the permission of the copyright owner is unlawful. 
 
14.58 
A tank contains 0.1 kmol hydrogen and 0.1 kmol of argon gas at 25 Ao EAC, 200 kPa 
and the tank keeps constant volume. To what T should it be heated to have a mole 
fraction of atomic hydrogen, H, of 10%? 
 
For the reaction HA2 EA ⇔ 2H , K = A
yHE
2
yH2
E
A (A P
Po E
A)A2-1E 
 Assume the dissociation shifts right with an amount x then we get 
 
 reaction HA2 E ⇔ 2 H also, Ar 
 initial 0.1 0 0.1 
 change -x 2x 0 
 equil. 0.1 - x 2x 0.1 Tot: 0.2 + x 
 yAHEA = A
2x
0.2 + xEA = 0.10 ⇒ x = 0.010526 
We need to find T so K will take on the proper value, since K depends on P 
we need to evaluate P first. 
 P A1 EAV = nA1 EAAR
− E
ATA1 EA; P A2 EAV = nA2 EAAR
− E
ATA2 EA