Chapter 15

Chemical Reactions

Study Guide in PowerPoint

to accompany

Thermodynamics: An Engineering Approach, 5th editionby Yunus A. Çengel and Michael A. Boles

2

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.

Examples of hydrocarbon fuels are

CH4 MethaneC8H18 OctaneCoal 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.

3

Reactants Products

C O CO

H O H O

2 2

2 2 2

1

2

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:

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.

C H A O B CO D H O8 18 2 2 2 Note we often can balance the C and H for complete combustion by inspection.

C H A O CO H O8 18 2 2 28 9

4

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

A

: ( ) ( ) ( )

.

2 8 2 9 1

12 5

C H O CO H O8 18 2 2 212 5 8 9 .

Note: Mole numbers are not conserved, but we have conserved the mass on a total basis as well as a specie basis.

The complete combustion process is also called the stoichiometric combustion, and all coefficients are called the stoichiometric coefficients.

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

C H O N

CO H O N8 18 2 2

2 2 2

12 5 3 76

8 9 47

. ( . )

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

AFkmol air

kmol fuel

Thus, for the above example, the theoretical air-fuel ratio is

AFkmol air

kmol fuelth

12 5 1 3 76

159 5

. ( . ).

6

On a mass basis, the theoretical air-fuel ratio is

AFkmol air

kmol fuel

kg airkmol air

kg fuelkmol fuel

kg air

kg fuel

th

59 5

28 97

8 12 18 1

1512

.

.

[ ( ) ( )]

.

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.

Percent theoretical air AF

AFactual

th

100%

Percent excess air AF AF

AFactual th

th

100%

7

Show that these results may be expressed in terms of the moles of oxygen only as

Percent theoretical air N

NO actual

O th

2

2

100%

Percent excess air N N

N

O actual O th

O th

2 2

2

100%

Example 15-2

Write the combustion equation for complete combustion of octane with 120 percent theoretical air (20 percent excess air).

C H O N

CO H O O N8 18 2 2

2 2 2 2

12 12 5 376

8 9 0 2 12 5 12 47

. ( . ) ( . )

( . )( . ) . ( )

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.

C H O N

CO H O O N8 18 2 2

2 2 2 2

12 12 5 3 76

8 9 2 5 12 47

. ( . ) ( . )

. . ( )

8

Second method to balance the equation for excess air (see the explanation of this technique in the text) is:

C H A O N

CO H O A O A N

O A A

A

th

th th

th th

th

8 18 2 2

2 2 2 2

12 376

8 9 0 2 12 376

12 2 8 2 9 1 0 2 2

12 5

. ( . )

. . ( . )

: . ( ) ( ) ( ) . ( )

.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.

C H O N

CO CO H O X O N8 18 2 2

2 2 2 2

12 12 5 376

08 0 2 9 12 47

. ( . ) ( . )

. (8) . (8) . ( )

9

O balance gives

O X

X

: . ( . )( ) . (8)( ) . (8)( ) ( ) ( )

.

12 12 5 2 0 8 2 0 2 1 9 1 2

3 3

Why is X > 2.5?

Then the balanced equation is

C H O N

CO CO H O O N8 18 2 2

2 2 2 2

12 12 5 376

6 4 16 9 33 12 47

. ( . ) ( . )

. . . . ( )

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 supplied or water formed to get the 100 kmol of dry product gases.

10

X C H A O N

CO CO O B H O A N3 8 2 2

2 2 2 2

3 76

115 0 7 2 7 376

( . )

. . . ( . )

The unknown coefficients A, B, and X are found by conservation of mass for each species.

C X X

H X B B

O A

B A

N A

: ( ) . ( ) . ( ) .

: (8) ( ) .

: ( ) . ( ) . ( )

. ( ) ( ) .

: ( . ) .

3 115 1 0 7 1 4 07

2 16 28

2 115 2 0 7 1

2 7 2 1 22 69

376 85312

The balanced equation is

4 07 22 69 376

115 0 7 2 7 16 28 85313 8 2 2

2 2 2 2

. . ( . )

. . . . .

C H O N

CO CO O H O N

Second method to find the coefficient A:

Assume the remainder of the 100 kmol of dry product gases is N2.

kmol N2 100 115 0 7 2 7 851 ( . . . ) .

11

Then A is

A fairly good check 851

3 7622 65

.

.. ( )

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:

C H O N

CO CO O H O N3 8 2 2

2 2 2 2

557 3 76

2 83 017 0 66 4 0 20 96

. ( . )

. . . . .

The actual air-fuel ratio is

AF

kmol airkg air

kmol air

kmol fuelkg fuel

kmol fuel

kg air

kg fuel

actual

(5. )( . ) .

[ ( ) ( )]

.

57 1 3 76 28 97

1 3 12 8 1

17 45

12

The theoretical combustion equation is

C H O N

CO H O N3 8 2 2

2 2 2

5 376

3 4 0 1880

( . )

. .

The theoretical air-fuel ratio is

AF

kmol airkg air

kmol air

kmol fuelkg fuel

kmol fuel

kg air

kg fuel

th

(5)( . ) .

[ ( ) ( )]

.

1 3 76 28 97

1 3 12 8 1

15 66

The percent theoretical air is

Percent theoretical air

AF

AFactual

th

100%

17 45

1566100 111%

.

.

13

or

Percent theoretical air

N

NO actual

O th

2

2

100%

557

5100 111%

.

The percent excess air is

Percent excess air

AF AF

AFactual th

th

100%

17 45 15 66

15 66100 11%

. .

.

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.

14

T T P y P

yN

N

v

ve

dp sat v products

water

products

at

Example 15-5

Determine dew point temperature of the products for Example 15-4.

products

dp sat

40.1398

2.83 0.17 0.66 4 20.96 0.1398(100 )

13.98

at13.98kPa

=52.31

v

v v

o

y

P y P kPa

kPa

T T

C

What would happen if the product gases are cooled to 100oC or to 30oC?

15

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.

C H A O N

CO CO O B H O D NX Y

( . )

. . .2 2

2 2 2 2

376

121 0 9 38

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.

C H A O N

CO CO O B H O D NX Y

( . )

. . .2 2

2 2 2 2

3 76

121 0 9 38

16

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 D2 100 12 1 0 9 38 83 2: ( . . . ) .

O AD

O A B

B

C X

X

H Y B

Y

2 3 76

83 2

3 7622 13

2 12 1 2 0 9 1 38 2 1

1154

1 12 1 1 0 9 1

13 0

1 2

23 08

:.

.

..

: ( ) ( . )( ) ( . )( ) ( . )( ) ( )

.

: ( ) . ( ) ( . )( )

.

: ( ) ( )

.

The balanced equation is

C H O N

CO CO O H O N13 23 08 2 2

2 2 2 2

2213 376

121 0 9 38 1154 832. . ( . )

. . . . .

17

Enthalpy of Formation

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 H CH 2 2 4

The conservation of energy for a steady-flow combustion process is

E E

Q H H

Q H H

in out

net

net

Reactants Products

Products Reactants

18

Q N h N h

Q h h h

net e e i i

net CH C H

Products Reactants

1 1 24 2

( )

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).

Q h

h

net CH

CH

1 1 0 2 04

4

( ( ) ( ))

h fo

h fo

This heat transfer is called the enthalpy of formation for methane, . The superscript (o) implies the 1 atm pressure value and the subscript (f) implies 25oC data, 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

Q hkJ

kmolnet fo

CHCH

4

4

74 850,

19

The enthalpy of formation 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.

h fo

h foSubstance Formula M 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

20

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 h h hfo

To ( )

Here the term 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

h o

Based on the classical sign convention, the net heat transfer to the reacting system is

Q H H

N h h h N h h h

net P R

e fo

To

e i fo

To

i

[ ( )] [ ( )]Products Reactants

In an actual combustion process, is the value of Qnet positive or negative?

21

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:

C H O N

CO H O N4 10 2 2

2 2 2

65 376

4 5 24 44

. ( . )

.

The steady-flow heat transfer is

Q H H

N h h h N h h h

net P R

e fo

To

e i fo

To

i

[ ( )] [ ( )]Products Reactants

22

Reactants: TR = 298 K

h fo hT h o N h h hi f

oT

oi[ ( )] Comp Ni

kmol/kmol fuel

kJ/kmol kJ/kmol kJ/kmol kJ/kmol fuel

C4H10 1 -126,150 -- -- -126,150

O2 6.5 0 8,682 8,682 0

N2 24.44 0 8,669 8,669 0

H N h h h

kJ

kmol C H

R i fo

To

i

[ ( )]

,

Reactants

126 1504 10Products: TP = 1000 K

h fo hT h o N h h he f

oT

oe[ ( )] Comp Ne

kmol/kmol fuel

kJ/kmol kJ/kmol kJ/kmol kJ/kmol fuel

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

23

H N h h h

kJ

kmol C H

P e fo

To

e

[ ( )]

, ,

Products

1 995 1884 10

Q H H

kJ

kmol C H

net P R

1869 0384 10

, ,

Adiabatic Flame Temperature

The temperature the products have when a combustion process takes place adiabatically is called the adiabatic flame temperature.

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.

24

The combustion equation is

C H O N

CO O H O N8 18 2 2

2 2 2 2

4 12 5 3 76

8 37 5 9 188

( . ) ( . )

.The steady-flow heat transfer is

Q H H

N h h h N h h h

Adiabatic Combustion

net P R

e fo

To

e i fo

To

i

[ ( )] [ ( )]

( )Products Reactants

0

Thus, HP = HR for adiabatic combustion. We need to solve this equation for TP.

25

Since the temperature of the reactants is 298 K, ( )i = 0, h hTo

H N h

kJ

kmol C H

R i fo

i

Reactants

1 249 950 4 12 5 0 4 12 5 3 76 0

249 9504 10

( , ) ( . )( ) ( . )( . )( )

,

Since the products are at the adiabatic flame temperature, TP > 298 K

2

2

2

2

2 2

2 2

Products

, ,

, ,4 10

[ ( )]

8( 393,520 9364)

9( 241,820 9904)

37.5(0 8682)

188(0 8669)

( 7,443,845 8 9

37.5 188 )

P

P

P

P

P

P P

P P

o oP e f T e

T CO

T H O

T O

T N

T CO T H O

T O T N

H N h h h

h

h

h

h

h h

kJh h

kmol C H

26

Thus, setting HP = HR yields

N h h h h he T eoducts

T CO T H O T O T NP P P P P,Pr

, , , ,.

, ,

8 9 37 5 188

7 193 895

2 2 2 2

To estimate TP, assume all products behave like N2 and estimate the adiabatic flame temperature from the nitrogen data, Table A-18.

242 5 7 193 895

29 6655

985

2

2

2

. , ,

, .

,

,

h

hkJ

kmol N

T K

T N

T N

p

P

P

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.

h K960 h K970

N he T eP ,Produts

Ne

CO2 8 40,607 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

27

Interpolation gives: TP = 962 K.

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

C H A O N

CO A O H O A N8 18 2 2

2 2 2 2

1 12 5 376

8 12 5 9 1 12 5 376

( )( . ) ( . )

. ( )( . )( . )

28

The steady-flow heat transfer is

Q H H

N h h h N h h h

Adiabatic combustion

net P R

e fo

To

e i fo

To

i

[ ( )] [ ( )]

( )Products Reactants

0

Here, since the temperatures are known, the values of are known. The product gas mole numbers are unknown but are functions of the amount of excess air, A. The energy balance can be solved for A.

hTP

A 3Thus, 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

C H A O N

CO A O H O A N3 8 2 2

2 2 2 2

1 376

3 5 4 1 376

( )(5) ( . )

( )(5)( . )

29

where A is the value of excess air in decimal form.

The steady-flow heat transfer is Q H H

N h h h N h h h

Adiabatic combustion

net P R

e fo

To

e i fo

To

i

[ ( )] [ ( )]

( )Products Reactants

0

Percent Excess Air

Adiabatic Flame Temp. K

0 2459.3

20 2191.9

50 1902.5

100 1587.1

217 1200

30

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.

h H H when T T C K

N h N h

C P R P Ro

e fo

e i fo

i

25 298

Products Reactants

Heating Value

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 h h with H O in productsC C gas 2

31

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 h h with H O in productsC C liquid 2

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 hH O fg H O 2 2

32

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.

8 18 8 18 2 2

2

8 18 8 18 2

8 18

5,500,842 9 (44,010)

5,104,752

gas gasC H C H H O fg H OLHV HHV N h

kJ kmol H O kJ

kmol C H kmol C H kmol H O

kJ

kmol C H

8 18 8 18 8 18

8 18

8 18

(5,104752 41,382)

5,063,370

liq gasC H C H fg C H

liq

LHV LHV h

kJ

kmol C H

kJ

kmol C H

Can you explain why LHVliq< LHVgas?

33

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:

34

Assume that the reactants and products are ideal gases; then

PV NR Tu

The balanced combustion equation for 200 percent excess (300 percent theoretical) air is

C H O N

CO O H O N8 18 2 2

2 2 2 2

3 12 5 376

8 25 9 141

( )( . ) ( . )

35

Q

h h

kJ

kmol C H

net CO

H O

O

N

Ko

C H

O

N

8 393 520 53 848 9364 8 314 1200

9 241820 44 380 9904 8 314 1200

25 0 38 447 8682 8 314 1200

141 0 36 777 8669 8 314 1200

1 208 450 8 314 298

37 5 0 8682 8682 8 314 298

141 0 8669 8669 8 314 298

112 10

2

2

2

2

8 18

2

2

298

6

8 18

( , , . ( ))

( , , . ( ))

( , . ( ))

( , . ( ))

( , . ( ))

. ( . ( ))

( . ( ))

.

To find the final pressure, we assume that the reactants and the products are ideal-gas mixtures.

PV N R T

PV N R Tu

u

1 1 1 1

2 2 2 2

36

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

PV

N R T

N R Tu

u

2 2

1 1

2 2

1 1

but V2 = V1.

37

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

Q

TS S S S kJ kk

kgen CV React Prod ( / )

38

For an adiabatic, steady-flow process, the entropy balance relation reduces to

S S Sgen adiabatic, Prod React 0

The third law of thermodynamics states that the entropy of a pure crys talline 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 T P RP

PkJ kmol Ko

o uo

( , ) ( , ) ln [ / ( )]

For component i of an ideal-gas mixture, the absolute entropy can be written as

s T P s T P Ry P

PkJ kmol Ki i i

oo u

i m

o

( , ) ( , ) ln [ / ( )]

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.

39

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 isC H O CO H O2 6 2 2 235 2 3 .

40

The mole fractions for the reactants and the products are

y

y

y

y

C H

O

CO

H O

2 6

2

2

2

1

1 35

1

4 535

1 35

35

4 52

2 3

2

53

2 3

3

5

. ..

.

.

.

Now calculate the individual component entropies.

For the reactant gases:

41

S N s

kJ

kmol K

i i

C H

ReactReactants

1 242 0 35 207 1

966 92 6

( . ) . ( . )

.

42

For the product gases:

43

S N s

kJ

kmol K

e e

C H

ProdProducts

2 2712 3 232 6

1240 22 6

( . ) ( . )

.

The entropy change for the combustion process is

S SkJ

kmol K

kJ

kmol K

C H

C H

Prod React

( . . )

.

1240 2 966 9

273 3

2 6

2 6

Now to find the entropy change due to heat transfer with the surroundings. The steady-flow conservation of energy for the control volume is

Q H H

N h h h N h h h

net sys P R

e fo

To

e i fo

To

i

[ ( )] [ ( )]Products Reactants

44

Q

h h

kJ

kmol C H

net sys CO

H O

Ko

C H

O

2 393 520 37 405 9364

3 241820 31828 9904

1 214 820

35 0 8682 8682

1306 10

2

2

2 6

2

298

6

2 6

( , , )

( , , )

( , )

. ( )

.

Q

T

Q

T

kJkmol C H

K

kJ

kmol C H K

k

k

net sys

o

1306 10

25 273

4 383

6

2 6

2 6

.

( )

,

45

The entropy generated by this combustion process is

Since Sgen, or Snet , is ≥ 0, the second law is not violated.