Chapter 15
Chemical Reactions
Study Guide in PowerPoint
to accompany
Thermodynamics: An Engineering Approach, 7th edition
by Yunus A. Çengel and Michael A. Boles
The combustion process is a chemical reaction whereby fuel is oxidized and energy
is released.
Fuels are usually composed of some compound or mixture containing carbon, C, and
hydrogen, H2.
CH4
Methane
Examples of hydrocarbon fuels are:
C8H18
Octane
Coal
Mixture of C, H2, S,
O2, N2 and non-combustibles
Initially, we shall consider only those reactions that go to completion. The
components prior to the reaction are called reactants and the components after the
reaction are called products.
For more information and animations illustrating this topic visit the Animation Library
developed by Professor S. Bhattacharjee, San Diego State University, at this link.
2
test.sdsu.edu/testhome/vtAnimations/index.html
Reactants  Products
For complete or stoichiometric combustion, all carbon is burned to carbon dioxide
(CO2) and all hydrogen is converted into water (H2O). These two complete
combustion reactions are as follows:
C  O2  CO2
1
H2  O2  H2 O
2
Example 15-1
A complete combustion of octane in oxygen is represented by the balanced combustion
equation. The balanced combustion equation is obtained by making sure we have the
same number of atoms of each element on both sides of the equation. That is, we
make sure the mass is conserved.
C8 H18  A O2  B CO2  D H2 O
Note we often can balance the C and H for complete combustion by inspection.
C8 H18  A O2  8 CO2  9 H2 O
3
The amount of oxygen is found from the oxygen balance. It is better to conserve
species on a monatomic basis as shown for the oxygen balance.
O: A(2)  8(2)  9(1)
A  12.5
C8 H18  12.5 O2  8 CO2  9 H2 O
Note: Mole numbers are not conserved, but we have conserved the mass on a total
basis as well as a specie basis. Also, 12.5 kmol of O2 are used to form the CO2 and
H2O. If 15 kmol of O2 were supplied to the process, 2.5 kmol of O2 would be excess
oxygen in the products.
The complete combustion process is also called the stoichiometric combustion, and
all coefficients are called the stoichiometric coefficients.
4
In most combustion processes, oxygen is supplied in the form of air rather
than pure oxygen.
Air is assumed to be 21 percent oxygen and 79 percent nitrogen on a
volume basis. For ideal gas mixtures, percent by volume is equal to
percent by moles. Thus, for each mole of oxygen in air, there exists 79/21
= 3.76 moles of nitrogen. Therefore, complete or theoretical combustion of
octane with air can be written as
C8 H18  12.5 ( O2  3. 76 N 2 ) 
8 CO2  9 H2 O  47 N 2
5
Air-Fuel Ratio
Since the total moles of a mixture are equal to the sum of moles of each component,
there are 12.5(1 + 3.76) = 59.5 moles of air required for each mole of fuel for the
complete combustion process.
Often complete combustion of the fuel will not occur unless there is an excess of air
present greater than just the theoretical air required for complete combustion.
To determine the amount of excess air supplied for a combustion process, let us define
the air-fuel ratio AF as
kmol air
AF 
kmol fuel
Thus, for the above example, the theoretical air-fuel ratio is
12.5(1  3.76)
kmol air
AFth 
 59.5
1
kmol fuel
6
On a mass basis, the theoretical air-fuel ratio is
kg air
kmol air
kmol air
AFth  59.5
kmol fuel [8(12)  18(1)] kg fuel
kmol fuel
kg air
 1512
.
kg fuel
28.97
Percent Theoretical and Percent Excess Air
In most cases, more than theoretical air is supplied to ensure complete combustion and
to reduce or eliminate carbon monoxide (CO) from the products of combustion. The
amount of excess air is usually expressed as percent theoretical air and percent excess
air.
AFactual
100%
AFth
AFactual  AFth
Percent excess air 
100%
AFth
Percent theoretical air 
7
Show that these results may be expressed in terms of the moles of oxygen only as
Percent theoretical air 
Percent excess air 
N O2 actual
N O2 th
100%
N O2 actual  N O2 th
N O2 th
100%
Example 15-2
Write the combustion equation for complete combustion of octane with 120 percent
theoretical air (20 percent excess air).
C8 H18  12
. (12.5) (O2  3.76 N 2 ) 
8 CO2  9 H2 O  (0.2)(12.5) O2  12
. (47) N 2
Note that (1)(12.5)O2 is required for complete combustion to produce 8 kmol of
carbon dioxide and 9 kmol of water; therefore, (0.2)(12.5)O2 is found as excess
oxygen in the products.
C8 H18  12
. (12.5) (O2  3.76 N 2 ) 
8 CO2  9 H2 O  2.5 O2  12
. (47) N 2
8
Second method to balance the equation for excess air (see the explanation of this
technique in the text) is:
C8 H18  12
. Ath (O2  3.76 N 2 ) 
8 CO2  9 H2 O  0.2 Ath O2  12
. Ath (3.76) N 2
O:
12
. Ath (2)  8(2)  9(1)  0.2 Ath (2)
Ath  12.5
NOTE: THIS METHOD WORKS ONLY FOR COMPLETE COMBUSTION!!!
WHEN PERCENT OR EXCESS THEROETICAL AIR IS SUPPLIED TO AN
INCOMPLETE COMBUSTION PROCESS, THE AMOUNT OF SUPPLIED O2
(1.2ATH IN THIS EXAMPLE) NOT USED FOR FORMING CO2 MUST BE
ACCOUNTED FOR IN THE EXCESS O2 IN THE PODUCTS.
9
Incomplete Combustion with Known Percent Theoretical Air
Example 15-3
Consider combustion of C8H18 with 120 % theoretical air where 80 % C in the fuel goes
into CO2.
Theoretical
Combustion
C8 H18  Ath O2  8 CO2  9 H 2O
theoretical O :
Ath (2)  8(2)  9(1)
Ath  12.5
The chemical reaction equation for incomplete Combustion with excess
theoretical air and X amount of O2 in products is
C8 H18  12
. (12.5) (O2  3.76 N 2 ) 
0.8(8) CO2  0.2(8) CO  9 H2 O  X O2  12
. (47) N 2
10
O balance gives
O:
12
. (12.5)(2)  0.8(8)(2)  0.2(8)(1)  9(1)  X (2)
X  3.3
Why is X > 2.5, the amount of O2 in the previous complete combustion example?
Then the balanced equation is
C8 H18  12
. (12.5) (O2  3.76 N 2 ) 
6.4 CO2  16
. CO  9 H2 O  3.3 O2  12
. (47) N 2
Combustion Equation When Product Gas Analysis Is Known
Example 15-4
Propane gas C3H8 is reacted with air such that the dry product gases are 11.5 percent
CO2, 2.7 percent O2, and 0.7 percent CO by volume. What percent theoretical air was
supplied? What is the dew point temperature of the products if the product pressure is
100 kPa?
We assume 100 kmol of dry product gases; then the percent by volume can be
interpreted to be mole numbers. But we do not know how much fuel and air were
11
supplied or water formed to get the 100 kmol of dry product gases.
X C3 H8  A (O2  3.76 N 2 ) 
115
. CO2  0.7 CO  2.7 O2  B H2 O  A(3.76) N 2
The unknown coefficients A, B, and X are found by conservation of mass for each
species.
C: X (3)  115
. (1)  0.7(1)
X  4.07
H: X (8)  B(2)
O: A(2)  115
. (2)  0.7(1)
 2.7(2)  B(1)
N 2 : A(3.76)  85.31
B  16.28
A  22.69
The balanced equation is
4.07 C3 H8  22.69 (O2  3.76 N 2 ) 
115
. CO2  0.7 CO  2.7 O2  16.28 H2 O  85.31 N 2
Second method to find the coefficient A:
Assume the remainder of the 100 kmol of dry product gases is N2.
kmol N 2  100  (115
.  0.7  2.7)  851
.
12
Then A is
851
.
A
 22.65
3.76
( fairly good check )
These two methods don’t give the same results for A, but they are close.
What would be the units on the coefficients in the balanced combustion equation?
Later in the chapter we will determine the energy released by the combustion process
in the form of heat transfer to the surroundings. To simplify this calculation it is
generally better to write the combustion equation per kmol of fuel. To write the
combustion equation per unit kmol of fuel, divide by 4.07:
C3 H8  557
. (O2  3.76 N 2 ) 
2.83 CO2  017
. CO  0.66 O2  4.0 H2 O  20.96 N 2
The actual air-fuel ratio is
kg air
kmol air

kg fuel
1kmol fuel[3(12)  8(1)]
kmol fuel
kg air
 17.45
kg fuel
(5.57)(1  3.76) kmol air 28.97
AFactual
13
The theoretical combustion equation is
C3 H8  5 (O2  3.76 N 2 ) 
3 CO2  4.0 H2 O  18.80 N 2
The theoretical air-fuel ratio is
kg air
(5)(1  3.76) kmol air 28.97
kmol air
AFth 
kg fuel
1kmol fuel[3(12)  8(1)]
kmol fuel
kg air
 15.66
kg fuel
The percent theoretical air is
AFactual
Percent theoretical air 
100%
AFth
17.45

100  111%
15.66
14
or
Percent theoretical air 

N O2 actual
N O2 th
100%
557
.
100  111%
5
The percent excess air is
AFactual  AFth
Percent excess air 
100%
AFth
17.45  15.66

100  11%
15.66
Dew Point Temperature
The dew point temperature for the product gases is the temperature at which the
water in the product gases would begin to condense when the products are cooled at
constant pressure. The dew point temperature is equal to the saturation temperature
of the water at its partial pressure in the products.
15
Tdp  Tsat at Pv  yv Pproducts
N water
yv 
 Ne
products
Example 15-5
Determine dew point temperature of the products for Example 15-4.
4
yv 
 0.1398
2.83  0.17  0.66  4  20.96
Pv  yv Pproducts  0.1398(100 kPa)
 13.98 kPa
Tdp  Tsat at13.98 kPa
=52.31oC
What would happen if the product gases are cooled to 100oC or to 30oC?
16
Example 15-6
An unknown hydrocarbon fuel, CXHY is reacted with air such that the dry product
gases are 12.1 percent CO2, 3.8 percent O2, and 0.9 percent CO by volume. What is
the average makeup of the fuel?
We assume 100 kmol (do you have to always assume 100 kmol?) of dry product
gases; then the percent by volume can be interpreted to be mole numbers. We do
not know how much air was supplied or water formed to get the 100 kmol of dry
product gases, but we assume 1 kmol of unknown fuel.
CX HY  A (O2  3.76 N 2 ) 
12.1 CO2  0.9 CO  38
. O2  B H2 O  D N 2
The five unknown coefficients A, B, D, X, and Y are found by conservation of mass
for each species, C, H, O, and N plus one other equation. Here we use the
subtraction method for the nitrogen to generate the fifth independent equation for the
unknowns.
CX HY  A (O2  3.76 N 2 ) 
12.1 CO2  0.9 CO  38
. O2  B H2 O  D N 2
17
The unknown coefficients A, B, D, X, and Y are found by conservation of mass for
each species. Here we assume the remainder of the dry product gases is nitrogen.
N 2 : D  100  (12.1  0.9  38
. )  83.2
D
83.2

 22.13
3.76 3.76
O: A(2)  (12.1)(2)  (0.9)(1)  (38
. )(2)  B (1)
O2 : A 
B  1154
.
C: 1( X )  12.1(1)  (0.9)(1)
X  13.0
H: 1(Y )  B(2)
Y  23.08
The balanced equation is
C13 H23.08  22.13 (O2  3.76 N 2 ) 
12.1 CO2  0.9 CO  38
. O2  1154
. H2 O  83.2 N 2
18
Enthalpy of Formation
Consider the chemical reaction that occurs during the
decomposition process in the compost pile shown below. On the
day this photo was taken, the ambient temperature was 69°F. At
two feet inside the pile, the temperature was 135°F. This process
is an exothermic reaction and heat energy is given off during the
decomposition.
19
When a compound is formed from its elements (e.g., methane, CH4, from C and H2),
heat transfer occurs. When heat is given off, the reaction is called exothermic.
When heat is required, the reaction is called endothermic. Consider the following.
The reaction equation is
C  2 H2  CH4
The conservation of energy for a steady-flow combustion process is
Ein  Eout
Qnet  HReactants  HProducts
Qnet  HProducts  HReactants
20
Qnet 
N h
e e
Products

Nh
i i
Reactants
Qnet  1hCH4  (1hC  2hH2 )
A common reference state for the enthalpies of all reacting components is
established as
The enthalpy of the elements or their stable compounds is defined to
be ZERO at 25oC (298 K) and 1 atm (or 0.1 MPa).
Qnet  1hCH4  (1(0)  2(0))
 hCH4
o
This heat transfer is called the enthalpy of formation for methane, h f . The
superscript (o) implies the 1 atm pressure value and the subscript (f) implies 25°C
o
data, h f is given in Table A-26.
During the formation of methane from the elements at 298 K, 0.1 MPa, heat is given
off (an exothermic reaction) such that
Qnet  h
o
f CH
4
kJ
 74,850
kmolCH4
21
o
The enthalpy of formation h f is tabulated for typical compounds. The enthalpy of
formation of the elements in their stable form is taken as zero. The enthalpy of
formation of the elements found naturally as diatomic elements, such as nitrogen,
oxygen, and hydrogen, is defined to be zero. The enthalpies of formation for several
combustion components are given in the following table.
Substance
Formula M
h fo kJ/kmol
Air
28.97
0
Oxygen
O2
32
0
Nitrogen
N2
28
0
Carbon dioxide
CO2
44
-393,520
Carbon monoxide
CO
28
-110,530
Water (vapor)
H2Ovap
18
-241,820
Water (liquid)
H2Oliq
18
-285,830
Methane
CH4
16
-74,850
Acetylene
C2H2
26
+226,730
Ethane
C2H6
30
-84,680
Propane
C3H8
44
-103,850
Butane
C4H10
58
-126,150
Octane (vapor)
C8H18
114
-208,450
Dodecane
C12H26
170
-291,010
22
The enthalpies are calculated relative to a common base or reference called the
enthalpy of formation. The enthalpy of formation is the heat transfer required to form
the compound from its elements at 25oC (77 F) or 298 K (537 R), 1 atm. The
enthalpy at any other temperature is given as
h  hfo  (hT  h o )
o
Here the term h is the enthalpy of any component at 298 K. The enthalpies at the
temperatures T and 298 K can be found in Tables A-18 through A-25. If tables are
not available, the enthalpy difference due to the temperature difference can be
calculated from
Based on the classical sign convention, the net heat transfer to the reacting system is
Qnet  H P  H R

o
o
N
[
h

(
h

h
 e f T )]e 
Products
o
o
N
[
h

(
h

h
 i f T )]i
Reactants
In an actual combustion process, is the value of Qnet positive or negative?
23
Example 15-7
Butane gas C4H10 is burned in theoretical air as shown below. Find the net heat
transfer per kmol of fuel.
Balanced combustion equation:
C4 H10  6.5 (O2  3.76 N 2 ) 
4 CO2  5 H2 O  24.44 N 2
The steady-flow heat transfer is
Qnet  H P  H R

o
o
N
[
h

(
h

h
 e f T )]e 
Products
o
o
N
[
h

(
h

h
 i f T )]i
Reactants
24
Reactants: TR = 298 K
Comp
h fo
Ni
kmol/kmol
fuel
kJ/kmol
kJ/kmol
Ni [hfo  (hT  h o )]i
kJ/kmol fuel
ho
hT
kJ/kmol
C4H10
1
-126,150
--
--
-126,150
O2
6.5
0
8,682
8,682
0
N2
24.44
0
8,669
8,669
0
 N [h
HR 
i
o
f
 (hT  h o )]i
Reactants
 126,150
kJ
kmol C4 H10
hT
ho
Ne [h fo  (hT  h o )]e
kJ/kmol
kJ/kmol fuel
Products: TP = 1000 K
Comp
h fo
Ne
kmol/kmol
fuel
kJ/kmol
kJ/kmol
CO2
4
-393,520
42,769
9,364
-1,440,460
H2O
5
-241,820
35,882
9,904
-1,079,210
N2
24.44
0
30,129
8,669
+524,482
25
HP 
 N [h
e
o
f
 (hT  h o )]e
Products
kJ
 1,995,188
kmol C4 H10
Qnet  H P  H R
 1,869,038
kJ
kmol C4 H10
Adiabatic Flame Temperature
The temperature of the products of
combustion when the process takes
place adiabatically is called the
adiabatic flame temperature.
Shown here is the view of the riser
tubes during the combustion process
inside a steam power plant boiler.
Photo courtesy of Progress Energy
Carolinas, Inc.
26
Example 15-8
Liquid octane C8H18(liq) is burned with 400 percent theoretical air. Find the adiabatic
flame temperature when the reactants enter at 298 K, 0.1 MPa, and the products
leave at 0.1MPa.
The combustion equation is
C8 H18  4(12.5) (O2  3.76 N 2 ) 
8 CO2  37.5 O2  9 H2 O  188 N 2
27
The steady-flow heat transfer is
Qnet  H P  H R

o
o
N
[
h

(
h

h
 e f T )]e 
Products
o
o
N
[
h

(
h

h
 i f T )]i
Reactants
 0 ( Adiabatic Combustion)
Thus, HP = HR for adiabatic combustion. We need to solve this equation for TP.
Since the temperature of the reactants is 298 K, (
HR 
Nh
i
hT  h o )i = 0,
o
f i
Reactants
 1( 249,950)  4(12.5)(0)  4(12.5)(3.76)(0)
kJ
 249,950
kmol C4 H10
28
Since the products are at the adiabatic flame temperature, TP > 298 K
HP 

N e [h fo  (hTP  h o )]e
Products
 8(393,520  hTP  9364)CO2
 9(241,820  hTP  9904) H 2O
 37.5(0  hTP  8682)O2
 188(0  hTP  8669) N2
 (7, 443,845  8hTP , CO2  9hTP , H 2O
 37.5hTP , O2
Thus, setting HP = HR yields
N h
e TP , e
kJ
 188hTP , N 2 )
kmol C4 H10
 8hTP , CO2  9hTP , H2O  37.5hTP , O2  188hTP , N 2
Pr oducts
 7,193,895
29
To estimate TP, assume all products behave like N2 and estimate the adiabatic flame
temperature from the nitrogen data, Table A-18.
242.5hTP , N 2  7,193,895
hTP , N 2  29,6655
.
kJ
kmol N 2
Tp  985 K
Because CO2 and H2O are triatomic gases and have specific heats greater than
diatomic gases, the actual temperature will be somewhat less than 985 K. Try TP =
960 K and 970K.
Ne
Interpolation gives: TP = 962 K.
CO2
h8960 K
h40,607
970 K
41,145
H2O
9
34,274
34,653
O2
37.5
29,991
30,345
N2
188
28,826
29,151
7,177,572
7,259,362
N h
e TP , e
Produts
30
Example 15-9
Liquid octane C8H18(liq) is burned with excess air. The adiabatic flame temperature is
960 K when the reactants enter at 298 K, 0.1 MPa, and the products leave at
0.1MPa. What percent excess air is supplied?
Let A be the excess air; then combustion equation is
C8 H18  (1  A)(12.5) (O2  3.76 N 2 ) 
8 CO2  12.5 A O2  9 H2 O  (1  A)(12.5)(3.76) N 2
31
The steady-flow heat transfer is
Qnet  H P  H R

 N [h
e
o
f
 N [h
 (hT  h o )]e 
Products
i
o
f
 (hT  h o )]i
Reactants
 0 ( Adiabatic combustion)
Here, since the temperatures are known, the values of hT are known. The product
P
gas mole numbers are unknown but are functions of the amount of excess air, A.
The energy balance can be solved for A.
A3
Thus, 300 percent excess, or 400 percent theoretical, air is supplied.
Example 15-10
Tabulate the adiabatic flame temperature as a function of excess air for the complete
combustion of C3H8 when the fuel enters the steady-flow reaction chamber at 298 K
and the air enters at 400 K.
The combustion equation is
C3 H8  (1  A)(5) (O2  3.76 N 2 ) 
3 CO2  5 A O2  4 H2 O  (1  A)(5)(3.76) N 2
32
where A is the value of excess air in decimal form.
The steady-flow heat transfer is
Qnet  H P  H R

 N [h
e
o
f
 N [h
 (hT  h o )]e 
Products
i
o
f
 (hT  h o )]i
Reactants
 0 ( Adiabatic combustion)
Percent Excess Air
Adiabatic Flame Temp. K
0
2459.3
20
2191.9
50
1902.5
100
1587.1
217
1200
33
Enthalpy of Reaction and Enthalpy of Combustion
When the products and reactants are at the same temperature, the enthalpy of
reaction hR, is the difference in their enthalpies. When the combustion is assumed
to be complete with theoretical air supplied the enthalpy of reaction is called the
enthalpy of combustion hC. The enthalpy of combustion can be calculated at any
value of the temperature, but it is usually determined at 25oC or 298 K.
hC  H P  H R when TP  TR  25o C  298 K

Heating Value
o
N
h
 e fe
Products
o
N
h
 i fi
Reactants
The heating value, HV, of a fuel is the absolute value of the enthalpy of combustion
or just the negative of the enthalpy of combustion.
HV  hC
The lower heating value, LHV, is the heating value when water appears as a gas in
the products.
LHV  hC   hC with H2 Ogas in products
34
The lower heating value is often used as the amount of energy per kmol of fuel
supplied to the gas turbine engine.
The higher heating value, HHV, is the heating value when water appears as a liquid
in the products.
HHV  hC  hC with H2 Oliquid in products
The higher heating value is often used as the amount of energy per kmol of fuel
supplied to the steam power cycle.
See Table A-27 for the heating values of fuels at 25oC. Note that the heating values
are listed with units of kJ/kg of fuel. We multiply the listed heating value by the molar
mass of the fuel to determine the heating value in units of kJ/kmol of fuel.
The higher and lower heating values are related by the amount of water formed
during the combustion process and the enthalpy of vaporization of water at the
temperature.
HHV  LHV  N H2Ohfg H2O
35
Example 15-11
The enthalpy of combustion of gaseous octane C8H18 at 25oC with liquid water in the
products is -5,500,842 kJ/kmol. Find the lower heating value of liquid octane.
LHVC8 H18 gas  HHVC8 H18 gas  N H 2O h fg H 2O
 5,500,842
kJ
kmol H 2O
kJ
9
(44, 010)
kmol C8 H18
kmol C8 H18
kmol H 2O
 5,104, 752
kJ
kmol C8 H18
LHVC8 H18 liq  LHVC8 H18 gas  h fg C8 H18
kJ
 (5,104752  41,382)
kmol C8 H18
kJ
 5, 063,370
kmol C8 H18 liq
Can you explain why LHVliq< LHVgas?
36
Closed System Analysis
Example 15-12
A mixture of 1 kmol C8H18 gas and 200 percent excess air at 25oC, 1 atm, is burned
completely in a closed system (a bomb) and is cooled to 1200 K. Find the heat
transfer from the system and the system final pressure.
Apply the first law closed system:
37
Assume that the reactants and products are ideal gases; then
PV  NRu T
The balanced combustion equation for 200 percent excess (300 percent theoretical)
air is
C8 H18  (3)(12.5) (O2  3.76 N 2 ) 
8 CO2  25 O2  9 H2 O  141 N 2
38
Qnet  8( 393,520  53,848  9364  8.314(1200)) CO2
 9( 241,820  44,380  9904  8.314(1200)) H2 O
 25(0  38,447  8682  8.314(1200)) O2
 141(0  36,777  8669  8.314(1200)) N 2
 1( 208,450  h298 K  h o  8.314(298)) C8 H18
 37.5(0  8682  8682  8.314(298)) O2
 141(0  8669  8669  8.314(298)) N 2
kJ
 112
.  10
kmol C8 H18
6
To find the final pressure, we assume that the reactants and the products are idealgas mixtures.
PV
1 1  N1 Ru T1
PV
2 2  N 2 Ru T2
39
where state 1 is the state of the mixture of the reactants before the combustion
process and state 2 is the state of the mixture of the products after the combustion
process takes place. Note that the total moles of reactants are not equal to the total
moles of products.
PV
N 2 Ru T2
2 2

PV
N 1 Ru T1
1 1
but V2 = V1.
40
Second Law Analysis of Reacting Systems
Second law for the open system
The entropy balance relations developed in Chapter 7 are equally applicable to both
reacting and nonreacting systems provided that the entropies of individual
constituents are evaluated properly using a common basis.
Taking the positive direction of heat transfer to be to the system, the entropy balance
relation can be expressed for a steady-flow combustion chamber as
Qk
 T  SReact  SProd  S gen  SCV
k
( kJ / k )
41
For an adiabatic, steady-flow process, the entropy balance relation reduces to
S gen , adiabatic  SProd  SReact  0
The third law of thermodynamics states that the entropy of a pure crystalline
substance at absolute zero temperature is zero. The third law provides a common
base for the entropy of all substances, and the entropy values relative to this base
are called the absolute entropy.
The ideal-gas tables list the absolute entropy values over a wide range of
temperatures but at a fixed pressure of Po = 1 atm. Absolute entropy values at other
pressures P for any temperature T are determined from
s (T , P)  s o (T , Po )  Ru ln
P
Po
[ kJ / ( kmol  K )]
For component i of an ideal-gas mixture, the absolute entropy can be written as
si (T , Pi )  sio (T , Po )  Ru ln
yi Pm
Po
[ kJ / ( kmol  K )]
where Pi is the partial pressure, yi is the mole fraction of the component, and Pm is
the total pressure of the mixture in atmospheres.
42
Example 15-13
A mixture of ethane gas C2H6 and oxygen enters a combustion chamber at 1 atm,
25oC. The products leave at 1 atm, 900 K. Assuming complete combustion, does the
process violate the second law?
The balanced combustion equation is
C2 H6  35
. O2  2 CO2  3 H2 O
43
The mole fractions for the reactants and the products are
1
1
yC2 H6 

1  35
.
4.5
35
.
35
.
yO2 

1  35
.
4.5
2
2
yCO2 

23 5
3
3
y H2 O 

23 5
Now calculate the individual component entropies.
For the reactant gases:
44
S React 
N s
i i
Reactants
 1(242.0)  35
. (207.1)
kJ
 966.9
kmolC2 H6  K
45
For the product gases:
46
S Prod 
N s
e e
Products
 2(2712
. )  3(232.6)
 1240.2
kJ
kmolC2 H6  K
The entropy change for the combustion process is
SProd  SReact  (1240.2  966.9)
kJ
kmolC2 H6  K
kJ
 273.3
kmolC2 H6  K
Now to find the entropy change due to heat transfer with the surroundings. The
steady-flow conservation of energy for the control volume is
Qnet sys  H P  H R

o
o
N
[
h

(
h

h
 e f T )]e 
Products
o
o
N
[
h

(
h

h
 i f T )]i
Reactants
47
Qnet sys  2( 393,520  37,405  9364) CO2
 3( 241,820  31,828  9904) H2 O
 1( 214,820  h298 K  h o ) C2 H6
 35
. (0  8682  8682) O2
kJ
 1306
.
 10
kmol C2 H6
6
Qk Qnet sys
T  T
k
o
kJ
1306
.
 10
kmol C2 H6

(25  273) K
kJ
 4,383
kmol C2 H6  K
6
48
The entropy generated by this combustion process is
Since Sgen, or Snet , is ≥ 0, the second law is not violated.
49