Answer to problem 2.1
Given : Air with a molar composition of 79%N2 and 21%O2
Find: The mass fraction of O2 and N2 in the air.
Approach: This is a simple conversion problem using:
MWmix=ΣxiMWi and Yi=xiMWi/MWmix
Given the molar composition:
x N 2  0.79
and
xO2  0.21
mixture molecular weight:
MWmix=ΣxiMWi= x N MW N  xO MWO =0.79(28.013)+0.21(32)=28.85 kg/kmol
2
2
2
2
Mass fraction of O2 and N2:
YO2  xO2 (
MWO2
MWmix
YN 2  x N 2 (
MW N 2
MWmix
)  0.21(
32
)  0.233
28.85
)  0.79(
28.013
)  0.767
28.85
Comments: Note that YO2  xO2 , since MWO2  MW mix and that ΣYi=1 as would be expected.
Answer to problem 2.2
Given: The following mixture:
Species
# Moles
CO
0.095
CO2
6
H 2O
7
N2
34
NO
0.005
Total
47.1
Find: a) The mole fraction, mole %, and ppm of NO in the mixture
b) Determine the MW of the mixture
c) Determine the mass fraction of each constituent.
a) xi 
Ni
N NO
0.005


 N i N CO  N CO2  N H 2O  N N2  N NO 0.095  6  7  34  0.005
x NO  106  10 6 kmol / kmol  mix =106 ppm
mole %= xi 100  0.0106 %
ppm= # NO/TOT#×106=
N NO A
 10 6  106 ppm
N
A
 i
Where A≡Avogadro’s Number=6.0223×1026 molecules/kmol.
b) MWmix=ΣxiMWi= xCO MWCO  xCO2 MWCO2  x H 2O MW H 2O  x N 2 MW N 2  x NO MW NO
=0.002(28.010)+0.127(44.011)+0.149(18.016)+0.722(28.013)+106×10-6(30.026)
=28.6 kg/kmol-mix
where x CO , x CO 2 , x H 2O and x N 2 are found in the same manner as x NO was found.
Species
MWi
MWmix
c) Yi  xi
CO:
CO2:
H2O:
N2:
NO:
xi
CO
0.002
CO2
0.127
H 2O
0.149
N2
0.722
NO
106×10-6
Total
1.0
(kgi/kgmix)
Y=0.002(28.01/28.6)=0.002
Y=0.127(44.011/28.6)=0.195
Y=0.149(18.016/28.6)=0.094
Y=0.722(28.013/28.6)=0.707
Y=106×10-6(30.006/28.6)=11110-6
Species
Yi
CO
0.002
CO2
0.195
H 2O
0.094
N2
0.707
NO
111×10-6
Total
1.0
Comments: Note that ppm=xi(1×106) and that
used to check your calculations.
Answer to problem 2.3
GIVEN:Mixture with 5 mole H2 and 3 mole O2.
FIND: x H 2 , x O2 , MWmix , YH 2 , YO2
SOLUTION:
a)
xi 
Ni
;
N tot
xH2 
5
 0.625
53
xO2  1  x H 2  0.375
x
i
 1 and
Y
i
 1 can often be
b) MWmix=ΣxiMWi= x H 2 MW H 2  xO2 MWO2 =0.625(2.016)+0.375(31.999)=13.260 kg/kmol
MWi
MWmix
Yi  xi
c)
YH 2  0.625
2.016
 0.095
13.26
YO2  1  YH 2  1  0.095  0.905
COMMENT: Even though the mole fraction of H2 is large, its low molecular weight results in its
having a small mass fraction.
Atp2.4
Given: O2-CH4 mixture at 300 K & 100 kPa, xO  0.2
2
FIND: YCH 4 , N CH 4 / 
Assumptions: ideal gas mixture.
SOLUTION:
a) YCH 4  xCH 4
= xCH 4
= 0.2
MWCH 4
MWmix
MWCH 4
xCH 4 MWCH 4  (1  xCH 4 ) MWO2
16.043
0.2(16.043)

 0.111
0.2(16.043)  0.8(31.999)
28.808
b) PCH 4   N CH 4 Ru T ; PCH 4  xCH 4 P
N CH 4 /  
xCH P
4
Ru T

(0.2)100  10 3
 8.018  10 3 kmol / m 3
8315(300)
COMMENT: Careful treatment of unit is required in part b.
Atp2.5
GIVEN: N2-Ar mixture with N N 2  3 N Ar , T=500 K, P=250 kPa
FIND: xi , MWmix , Yi , N N 2 / 
ASUMPTION: ideal gas mixture.
SOLUTION:
a)
x N2 
N N2
N mix

3N Ar
 3 / 4  0.75
3N Ar  N Ar
x Ar  1  0.75  0.25
b) MWmix=ΣxiMWi=0.75(28.014)+0.25(39.948)= 30.998
c) YN 2  x N 2 MW N 2 / MWmix  0.75(28.014) / 30.998  0.678
YAr  1  YN 2  0.322
d) PN 2   N N 2 Ru T
PN 2  x N 2 Ptot
N N2 /  
x N 2 Ptot
Ru T

(0.75)250  10 3
 0.0451kmol / m 3
8315(500)
COMMENT: Careful treatment of unit is required in part d.
Atp2-9
GIVEN: Propane (C3H8) burning at an air-fuel ratio (mass) of 18:1
FIND: The equivalence ratio, Φ
ASSUMPTIONS: Air is comprised of 79% N2 and 21% O2 by volume
APPROACH: Determine the stoichiometric A/F ratio and then the equivalence ratio
Stoichiometric relation: C3 H 8  aO2  3.76aN 2  3CO2  4 H 2 O  3.76aN 2
a
x Y /4

x=3, Y=8, Φ=1, a=5
A / F | Stoich  4.76a(
MWair
28.85
)  4.76(5)
 15.6
MW fuel
44.096
equivalence ratio:  
A / F | STOICH 15.6

 0.87
A / F | ACTUAL 18.0
COMMENT: Since Φ<1, this combustion process is fuel-lean. Also note thatΦ does not depend on
whether the A/F ratio are expressed in terms of moles or mass since Φ is also a ratio.
Atp2-10
GIVEN: An equivalence ratio of 0.6
FIND: The corresponding A/F ratios (mass) for methane (CH4), propane (C3H8) and decane
(C10H22)
ASSUMPTIONS: Air is comprised of 79% N2 and 21% O2 by volume
APPROACH: Use the relationships:
a
x Y /4
and

A / F | mass  4.76a(
MWair
)
MW fuel
methane (CH4): x=1,Y=4, MW=16.043 kg/kmole
a
1 4/ 4
 3.33
0 .6
so, A / F | mass  4.76(3.33)(
28.85
)  28.50kgair / kg fuel
16.043
propane (C3H8): x=3, Y=8,MW=44.096 kg/kmole
a
3 8/ 4
 8.33
0.6
so, A / F | mass  4.76(3.33)(
28.85
)  25.94kgair / kg fuel
44.096
decane (C10H22): x=10, Y=22, MW=142.284 kg/kmole
a
10  22 / 4
 25.83
0.6
so, A / F | Mass  4.76(25.83)(
28.85
)  24.93kgair / kg fuel
142.284
COMMENTS: Note how the A/F ratio (mass) changes only slightly from one hydrocarbon fuel to
another. While the A/F ratio (molar) varies from 15.9 (methane) to 123 (decane). This difference
in behavior is due to the MWfuel increasing as the molar A/F ratio increases.
Atp2-12
GIVEN: 1 mole of alcohol (CxHyOz) undergoing complete combustion
FIND: Stoichiometric balance equation and number of moles of air to burn 1 mole of alcohol.
ASSUMPTIONS: no dissociation and air is comprised of 79% N2 and 21% O2 by volume
APPROACH: Use conservation of elements for the combustion of one mole of alcohol.
Stoichiometric balance:
C x H y Oz  aO2  3.76aN 2  bCO2  cH 2O  3.76aN 2
conservation of Carbon :
x=b, so, b=x
conservation of H :
y=2c, so, c=y/2
conservation of O :
z+2a=2b+c=2x+y/2, so, a=x+y/4-z/2
number of moles of air to burn 1 mole of alcohol:
N air
a  3.76a

 4.76a
N fuel
1
N air
 4.76[ x  y / 4  z / 2]
N fuel
COMMENTS: Note that stoichiometric combustion of an alcohol (CxHyOz) requires less oxygen
than the combustion of a comparable hydrocarbon fuel (CxHy) due to the presence of oxygen in
the fuel.
Atp2-12
GIVEN: 1 mole of alcohol (CxHyOz) undergoing complete combustion
FIND: Stoichiometric balance equation and number of moles of air to burn 1 mole of alcohol.
ASSUMPTIONS: no dissociation and air is comprised of 79% N2 and 21% O2 by volume
APPROACH: Use conservation of elements for the combustion of one mole of alcohol.
Stoichiometric balance:
C x H y Oz  aO2  3.76aN 2  bCO2  cH 2O  3.76aN 2
conservation of Carbon :
x=b, so, b=x
conservation of H :
y=2c, so, c=y/2
conservation of O :
z+2a=2b+c=2x+y/2, so, a=x+y/4-z/2
number of moles of air to burn 1 mole of alcohol:
N air
a  3.76a

 4.76a
N fuel
1
N air
 4.76[ x  y / 4  z / 2]
N fuel
COMMENTS: Note that stoichiometric combustion of an alcohol (CxHyOz) requires less oxygen
than the combustion of a comparable hydrocarbon fuel (CxHy) due to the presence of oxygen in
the fuel.
Atp2-14
GIVEN: Stoichiometric mixture of isooctane & air
FIND: H per kmol of C8H18; hmix ; hmix
ASSUMPTIONS: Air is comprised of 79% N2 and 21% O2 by volume; ideal gas
APPROACH: We start by finding the stoichiometric proportions of each component:
C8 H18  a(O2  3.76 N 2 )  products (Eqn.2.30)
a=x+y/4=8+18/4=12.5
(Eqn.2.31)
so, (1)C8 H 18  12.5O2  47 N 2  products
as above is written for 1 kmol of C8H18;
a) H  (1)hC8 H18  12.5hO2  47hN2 ( J / kmolC8 H18 )
at 298 K, hisooctan e  224,109kJ / kmol (Evaluated from curve fit coefficients in table B.2)
hO2  h f ,O2  0
o
hN 2  h f , N 2  0
o
H  (1)( 224,109)  12.5(0)  47(0)  224,109( J / kmolC8 H18 )
b) hmix 
x h
i i
; xi  N i / N to t
i
xC8 H18  1 /(1  12.5  47)  0.0165
xO2  12.5 /(1  12.5  47)  0.2066
x N 2  47 /(1  12.5  47)  0.7769
hmix  0.0165(224,109)  0.2066(0)  0.7769(0)  3700kJ / kmol  mix
c) hmix 
Y h
i i
 hmix / MWmix
i
MWmix   xi MWi 0.0165(114.23)  0.2066(31.999)  0.7769(28.014)  30.260
hmix 
 3698
 122.2kJ / kg  mix
30.260
COMMENTS: We note that although both n-octane and isooctane are represented as C8H18, They
have different molecular structures as discussed in the Chapter 2 appendix. Because of these
structural differences, the enthalpy-of-formations of the two compounds have different values.
o
Table B.2 was used to calculate h f for isooctane as the value given in table B.1 is for n-octane.
Spread-sheet software simplifies calculating properties from the table B.2 curve fit coefficients.
Atp2-15
GIVEN: Isooctane-air, Φ=1, T=500 K
FIND: H(per kmol C3H8), hmix , hmix
ASSUMPTIONS: Air is 79% N2 and 21% O2; ideal gas
APPROACH: We need only evaluate the enthalpies of the constituents at 500K and then follow
the solution to prob.2-14.
Isooctane
O2
N2
hf
hs @ 500 K
h (500 K )
0
0
_
6097
5920
-175,807
6097
5920
Table B.2*
Table A.11
Table A.11
*Evaluated using spread sheet software
a) H  1(-175,807)  12.5(6097)  47(5920)  178,646 kJ/kmole
b) hmix 
C8 H18
H
178,646

 2953kJ / kmol  mix
N mix 1  12.5  47
c) hmix  hmix / MWmix 
2953
 97.59kJ / kg  mix
30.260
COMMENTS: Note that use of Table B.2 in combustion with A.7 and A.11
Atp2-16
GIVEN: Isooctane-air, Φ=0.7, T=500 K
FIND: H(per kmol C3H8), hmix , hmix
ASSUMPTIONS: see prob. 2-14
APPROACH: After calculating the properties of the constituents for   0.7 ,we follow the
same solution as for prob.2-17.
(1)C8 H 18 
12.5
(O2  3.76 N 2 )  products

C8H18
O2
N2
N
x
1
17.86
67.14
0.0116
0.2077
0.7807
h (500 K ) *
-175,807
6097
5920
*see Prob.2-15
N
i
 86.00
a) H=1(-175,807)+17.86(6097)+67.14(5920)=+330,554 kJ (for 1kmol C8H18)
b)
hmix 
H
330,554

 3844kJ / kmol  mix
N mix
86
c) MWmix 
hmix 
 x MW
i
i
 0.0116(114.230)  0.2077(31.999)  0.7807(28.014)  29.842
hmix
3844

 128.8kJ / kg  mix
MWmix 29.842
COMMENTS: Note how for non-stoichiometric combustion “ a /  ” is substituted for “a” in
Eqn.2.30. As expected, the mixture enthalpy increases with the addition of excess air.
Atp2-23
GIVEN: The lower heating value for methane, LHV=50,016 kJ/kg@298K
FIND: The enthalpy of formation of methane at 298 K
ASSUMPTIONS: complete combustion of methane to form CO2, H2O and N2
APPROACH: Use the stoichiometric relation to determine the proper A/F ratio and combustion
products for 1 kmole of methane. Then use the first law of thermodynamics to evaluate the
reactant enthalpy.
Q=LHV
Stoichiometric relation
CH 4  aO2  3.76aN 2  CO2  2H 2O  3.76aN 2
HR,298
a=x+y/4=2
Based on First Law analysis of control volume (CV) at steady-state
CV
HP,298
H R , 298  H P , 298  LHV 298
H P, 298  1[h f , 298  (h  h f , 298 )]CO2  2[h f , 298  (h  h f , 298 )] H 2O  7.52[h f , 298  (h  h f , 298 )] N 2
o
o
o
o
o
o
using appendix A-T=298 K
H P , 298  1[393546  0]  2[241847  0]  7.52[0  0]  877240kJ
H R, 298  1[h f , 298  (h  h f , 298 )]CH 4  2[h f , 298  (h  h f , 298 )]O2  7.52[h f , 298  (h  h f , 298 )] N 2
o
again using appendix A-T=298 K
o
o
o
o
o
H R, 298  1[h f , 298  0]CH 4  2[241847  0]  7.52[0  0]  1[h f , 298 ]CH 4
o
o
H R  1[h f , 298 ]CH 4  H P , 298  LHV 298  877240kJ  50,016kJ / kg(
o
16kg
)(1kmole)
kmole
[h f , 298 ]CH 4  76984kJ / kmole
o
Atp2-24
GIVEN: The mixture composition in problem 2-2 at T=1000 K
FIND: the absolute enthalpy of the mixture(kJ/kmol-mix)
APPROACH: Detremine the absolute enthalpy of each species in the mixture using appendix A
and then calculate the mixture absolute enthalpy from:
hmix   xi hi
mixture composition (from prob. 2-2) and species enthalpies (appendix A)
Species
#Moles
x
CO
0.095
CO2
h f , 298 (kJ / kmol)
(h1000  h f , 298 )( kJ / kmole)
0.002
-110541
21697
6
0.127
-393546
33425
H2O
7
0.149
-241847
25993
N2
34
0.722
0
21468
0.005
106ּ10-6
90297
22241
NO
o
o
hmix   xi hi ( xi hi ) CO  ( xi hi ) CO2  ( xi hi ) H 2O  ( xi hi ) N2  ( xi hi ) NO
h i  [h f , 298  (h1000  h f , 298 )]i
o
o
hmix  0.002[110541  21697]  0.127[393546  33425]  0.149[241847  225993]
 0.722[0  21468]  106  10 6 [90297  22241]  62563kJ / kmole
COMMENTS: Note how much the N2 contributes to the mixture specific enthalpy
(15500kJ/kmol-mix) despite having a relatively small absolute enthalpy itself (21468
kJ/kmole-N2). This is due to the large mole fraction of N2 present in the mixture.
Atp2-30
GIVEN: A stoichiometric propane (C3H8)-air mixture at 298 K
FIND: The adiabatic flame temperature, Tad
ASSUMPTIONS: no dissociation, constant specific heats evaluated at 298 K
APPROACH: Use element conservation to determine the correct fuel-air mixture and product
composition. Then use a first law analysis to evaluate Tad
Stoichiometric reaction:
C3 H 8  aO2  3.76aN 2  3CO2  4H 2 O  3.76aN 2
at   1 , a  x  y / 4  5
HR
T=298K
CV
HP
T=Tad
C3 H 8  5O2  18.8N 2  3CO2  4H 2 O  18.8N 2
Since the reactions are at T=298 K the O2 and N2 contributions
to the reactant enthalpy are zero
first law for adiabatic conditions: H R , 298  H P ,Tad
[ Nh ]C3H8  [ Nh ]O2  [ Nh ] N2  [ Nh ]CO2  [ Nh ] H 2O  [ Nh ] N2
where [ Nh ]O2  0 , [ Nh ] N 2  0 ,
hf
o
and c p properties from appendix A
h  h f , 298  (h  h f , 298 )  h f  c p (T  298)
o
o
o
(1)[ 103847  0]C3H8
 3[393546  37.198(Tad  298)]CO2  4[241847  33.448(Tad  298)] H 2O  18.8[0  29.071(Tad  298)] N 2
solving for Tad:
Tad=2879 K
COMMENTS: Note that this flame temperature is much greater than the adiabatic flame
temperature in appendix B.1. This is due to the assumption of no species dissociation and the
assumption of constant specific heats evaluated at 298 K. An examination of appendix A shows
that the specific heats can vary significantly from 298 K to 2879 K.
Atp2-31
GIVEN: A stoichiometric propane (C3H8)-air mixture at 298 K
FIND: The adiabatic flame temperature, Tad
ASSUMPTIONS: no dissociation, constant specific heats evaluated at 2000 K
APPROACH: Use element conservation to determine the correct fuel-air mixture and product
composition. Then use a first law analysis to evaluate Tad
Stoichiometric reaction:
C3 H 8  aO2  3.76aN 2  3CO2  4H 2 O  3.76aN 2
HR
at   1 , a  x  y / 4  5
C3 H 8  5O2  18.8N 2  3CO2  4H 2 O  18.8N 2
T=298K
Since the reactions are at T=298 K the O2 and N2 contributions
to the reactant enthalpy are zero
first law for adiabatic conditions: H R , 298  H P ,Tad
[ Nh ]C3H8  [ Nh ]O2  [ Nh ] N2  [ Nh ]CO2  [ Nh ] H 2O  [ Nh ] N2
CV
HP
T=Tad
where [ Nh ]O2  0 , [ Nh ] N 2  0 ,
hf
o
and c p properties from appendix A
h  h f , 298  (h  h f , 298 )  h f  c p (T  298)
o
o
o
(1)[ 103847  0]C3 H8
 3[393546  60.433(Tad  298)]CO2  4[241847  51.143(Tad  298)] H 2O  18.8[0  35.9881(Tad  298)] N
solving for Tad:
Tad=2222 K
COMMENTS: Note that this flame temperature is much closer to the value listed in appendix B.1
than the temperature calculated in prob.2-17. This is due to a more appropriate estimate of the
constant specific heats. The effects of dissociation on the flame temperature are still unaccounted
for.
frAtp2-32
GIVEN: A stoichiometric propane (C3H8)-air mixture
FIND: The adiabatic flame temperature, Tad
ASSUMPTIONS: no dissociation, species thermophysical propertied equal to those listed in
appendix A.
APPROACH: Use element conservation to determine the correct fuel-air mixture and product
composition. Then use appendix A to evaluate the species thermophysical peoperties and a first
law analysis to evaluate Tad. This is an iterative process in which a flame temperature is guessed,
the first law is checked, and if necessary a new flame temperature is chosen.
See problems 2-17 and 2-18 for correct fuel-air misture and product composition. A control
volume sketch for energy concentration is also shown.
First law for adiabatic conditions: H R , 298  H P ,Tad
→
HP-HR=0
[ Nh ]C3H8  [ Nh ]O2  [ Nh ] N2  [ Nh ]CO2  [ Nh ] H 2O  [ Nh ] N2
where [ Nh ]O2  0 , [ Nh ] N 2  0 ,
hf
o
and c p properties from appendix A
h  h f , 298  (h  h f , 298 )  h f  hS
o
o
o
values from appendix A
(1)[ 103847  0]C3 H 8  3[393546  hS ]CO2  4[241847  hS ] H 2O  18.8[0  hS ] N 2
rearranging in form : HP-HR=0:
3hS ,CO2  4hS , H 2O  18.83hS , N2  2.0442 10 6  0
T(K)
hS ,CO2 (kJ / kmol)
hS ,CO2 (kJ / kmol)
hS ,CO2 (kJ / kmol)
HP-HR*
2000
91420
72805
56130
-423476
2100
97477
2200
103562
2300
109670
2400
115798
* Linear interpolation using HP-HR=0
77952
83160
88426
93744
59738
63360
66997
70645
-316887
-209706
-101942
6296
Tad=2394 K
COMMENTS: Note that this flame temperature is slightly greater than that listed in appendix B.1
despite using accurate thermophysical properties form appendix A. This is due to neglecting
species dissociation.
Atp2-35
GIVEN: (C3H8)-air,   1 ,Ti=298, Pi =1 atm, constant-volume combustion
FIND: Tad, Pfinal
ASSUMPTIONS: no dissociation (given), constant c p ,i @298K, Air  21%O2  79%N 2
SOLUTION: Apply 1st law, Eqn.2.41: U R (Ti , Pi )  U Pr (Tad , Pf )
form Eqn.2.43, H R  H Pr  Ru ( N RTi  N PrTad )  0
to evaluate Eqn.2.43, we need the composition of both the reactants & products:
C3 H 8  5(O2  3.76 N 2 )  3CO2  4H 2 O  5(3.76) N 2
Reactant:
C3H8
O2
N2

N
h  (h f )
Nh
1
5
18.8
-103,847
0
0
-103,847
0
0
o
24.8 kmol
Table
B.1
-
-103,847 kJ
R
HR   N i hi  103,847kJ
R
Products:
N
CO2
H2O
N2
3
4
18.8

25.8
hf
o
-393,546
-241,845
0
o
c p @ 298 K
Nh f
Nc p
37.198
33.448
29.071
-1,180,638
-967,380
0
111.594
133.742
546.535
-2,148,018
791.921
Table
A.2
A.6
A.7
Pr
H Pr   N i hi   N i [h f ,i  c p ,i (Tad  Tref )]   Nih f ,i   N i c p ,i (Tad  Tref )
Pr
 2,148,018  791.921(Tad  Tref )
Substituting into Eqn.2.43 with Tref=Ti: H R  H Pr  Ru N R Ti  Ru N PrTad  0
 103,847  [2,148,018  791.921(Tad  Ti )]  8.3145(24.8)298.15  8.3145(25.8)Tad  0
Simplifying: 2218804-577.407Tad=0
Tad=3843 K

N R Ru Ti N Pr Ru Tad

Pi
Pf
Pf  1atm
 Pf  Pi
N Pr Tad
N r Ti
25.8 3843
 13.4atm
24.8 298.15
COMMENTS: As expected, the constant volume adiabatic flame temperature is significantlt
greater then the constant-P value of 2879 K from problem 2.30.
Atp2-38
GIVEN:
A /m
 F  18 , TA=800K (preheated), TF=450 K, P=1atm, hf,A=hf,Pr=0
A/ F  m
CP,F=3500 J/kg-K, cP,A=1200 J/kg-K, cP,Pr=1200J/kg-K, h f , F  1.16  10 J / kmole
9
FIND: Tad for P=1 atm
ASSUMPTIONS: Air≡79%N2/21%O2; properties as given.
SOLUTION: write 1st law (mass basis) recognizing that
 Pr  m
A m
F
m
m A hA
m F hF
 PrhPr
m
Q ,W  0
 A hA  m
 F hF  (m
A m
 F )hPr
m
 F and substitue properties (hi=hf,i+cp,i(Ti-Tref)):
Divided by m
(
m A
h
m
)(0  c P , A (TA  Tref ))  (1)[ F  c P , F (TF  Tref )]  ( A  1)[0  c P ,Pr (Tad T ref )]
m F
MWF
m F
1.16  10 9
18(1200(800  300))  (
 3500(450  300))  (18  1)(1200(Tad  300))
29
solve for Tad:
1.080  10 7  4.0525  10 7  2.28  10 4 (Tad  300)
Tad  2251  300  2551K
COMMENTS: (1) Use of simplified properties focuses attention on energy conservation. (2) use
of a mass-based 1st law simplifies the solution(HR=HPr)
Atp2-42
GIVEN: A closed vessel containing 1 kmole of O2 when there is no dissociation.
FIND: The mole fractions: xO and x O2 at the following conditions:
a) T=2500 K, P=1 atm;
b) T=2500 K, P=3 atm;
ASSUMPTIONS: ideal gas, system is in chemical equilibrium
APPROACH:There are two unknowns ( xO and x O2 ) so 2 equations must used. The first is the
definition of Kp and the second is
x
i
1
O2  2O
a) T=2500 K
Gr  2[ Ng f ,T ]O  [ Ng f ,T ]O2  2(88203)  0  176406kJ / kmole
o
o
K p  exp[
o
 G
176406
]  exp[ 
]  206.3  10 6
Ru T
8.315(2500)
2
x P
Kp  O
 206.3  10 6
xO2 Po
x
i
1st equation
 1  xO  xO2
xO2  1  xO
2nd equation
substituting the 2nd equation into the 1st equation and rearranging yields:
P 2
xO  K P xO  K P  0
Po
quadratic equation
 K P  K P  4( P / Po ) K P
2
solving for xO : x O 
2( P / Po )
note: only “+” yields physically realistic result
6
T=2500K, P=1atm: Kp  206.3  10 , P / Po  1 →
xO  0.0143 , xO2  1  xO  0.9857
6
T=2500K, P=3atm: Kp  206.3  10 , P / Po  3 →
xO  0.00826 , xO2  1  xO  0.9917
COMMENTS: Note how this system follows the principle of Le Chatelier. Increasing the system
pressure cause the system to shift towards more O2, thereby reducing the number of moles in the
system ( N O  N O2 )