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Solution Manual for Combustion 4th Edition

Irvin Glassman
Department of Mechanical and Aerospace Engineering
Princeton University
Princeton, New Jersey
Richard Yetter
Department of Mechanical and Nuclear Engineering
The Pennsylvania State University
University Park, PA
This solution manual is a revision to the solution manual for "Combustion, 3rd. Ed." that
was written in its final form by Prof. Queiroz and Dr. Black of Brigham Young
University.

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A Solution Manual for:

COMBUSTION. 4th Ed.
by
I. Glassman, and R.A. Yetter
CHAPTER 1
Problem 1
Calculate the heat of reaction when methane and air in stoichiometric proportions are brought
into a calorimeter at 500 K. The product composition is brought to the ambient temperature (298
K) by the cooling water. The pressure in the calorimeter is assumed to remain at 1 atm, but the
water formed has condensed.
Solution: The heat of reaction can be calculated using Eqn. (10) in the text. However, as also
suggested in the textbook, this equation can be simplified for the general case when H 0T0 = H 0298 .
The modified equation becomes:
Q = H =

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

n j H 0T H 0298
j react 1

298 i

298

+ ( H 0f )

298

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. The
general equation for the complete combustion of one mole of methane in air is:
CH 4 + a th ( O 2 + 3.76N 2 ) a CO 2 + b H 2 O + c N 2

where ath is a coefficient describing how much air is needed for stoichiometric combustion.
Balancing the above reaction one has:
CH 4 + 2 ( O 2 + 3.76N 2 ) CO 2 + 2 H 2 O + 7.52 N 2

In order to apply Eqn. (1) to the above reaction, the following properties from Tables in the
textbook are needed (note that for this problem T2 = 298 K and T1 = 500 K):
gases

CH4
O2
N2
H2O(liq)
CO2

H 0f (kJ/mol)

H 0T1 H 0298 (kJ/mol)

H 0T2 H 0298 (kJ/mol)

74.90
0.0
0.0
286.04
393.77

8.21
6.09
5.92
0.0
0.0

0.0
0.0
0.0
0.0
0.0

Equation (1) then becomes:

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Q = 1( 393.77 + 0.0 )CO + 2 ( 286.04 + 0.0 )H O + 7.52 ( 0.0 + 0.0 ) N 1( 74.90 + 8.21)CH
2

2 ( 0.0 + 6.09 )O 7.52 ( 0.0 + 5.92 ) N

2

Q=955.85 kJ

or

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Problem 2
Calculate the flame temperature of normal octane (liquid) burning in air at an equivalence ratio
of 0.5. For this problem assume there is no dissociation of the stable products formed. All
reactants are at 298 K and the system operates at 1-atm pressure. Compare your results with
those given in the graphs in the text. Explain any differences.
Solution: We will assume that adiabatic conditions exist and that the water in the products is in
the vapor phase. In this case, Eqn. (10) in the text can be used in an iterative approach to solve
for the adiabatic flame temperature. However, this equation can be simplified for the general
case when H 0T0 = H 0298 . For Q = 0, the modified equation becomes:

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

= n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since
the conditions for the reactants are specified, the solution approach is to find T2 such that the
above equality will be satisfied.
The general equation for the complete combustion of one mole of octane with excess air is:
C8 H18( liq ) + ( m )( a th )( O 2 + 3.76N 2 ) a CO 2 + b H 2 O + c O 2 + d N 2
where ath is the coefficient describing how much air is needed for stoichiometric combustion and
m is the coefficient used to represent the amount of excess air, respectively. For the case when c
= 0 and m = 1 (no excess air), a simple mass balance yields ath = 12.5. Therefore, for an excess
air of 100% (m = 2) or an equivalence ratio of 0.5, the chemical reaction becomes:
C8 H18( liq ) + 25 ( O2 + 3.76N 2 ) 8 CO 2 + 9 H 2 O + 94 N 2 + 12.5 O 2
Since the conditions of the reactants are specified (T1 = 298 K), Table 2 in the text and Appendix
A provide the following thermodynamic properties:
gases

C8H18(liq)
O2
N2

H 0f (kJ/mol)

H 0T1 H 0298 (kJ/mol)

250.12
0.0
0.0

0.0
0.0
0.0

The enthalpy of the reactants (Hr, given by the right-hand side of Eqn (1)) can be calculated as:
H r = 1( 250.12 + 0.0 )C H + 11( 0.0 + 0.0 )O + 41.36 ( 0.0 + 0.0 ) N = 250.12kJ
8

18

For the products it follows from thermodynamic properties in the text that:
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H 0f (kJ/mol)
0.0
0.0
242.00
393.77

gases

O2
N2
H2O(vap)
CO2

Therefore, the expression for the enthalpy of the products (Hp, given by the left-hand side of
Eqn. (1) in terms of the enthalpy of the individual gases in the products becomes
let H = H 0T2 H 0298 :

) (

H p = 8 393.77 + H CO2 + 9 242.00 + H H2O + 12.5 0.0 + H O2 + 94 0.0 + H N2

Simplifying the above equation yields:

8 H CO2 + 9 H H2O + 12.5H O2 + 94H N2 = 5, 078.04
A first guess for the products temperature can be obtained by assuming that all product gases are
N2. This approach gives the following as an initial guess for T2 (Note that 123.5 = 8 + 9 + 12.5 +
94):
123.5 H N2 = 5,078.04

so

H N2 = 41.12 kJ/mol

therefore T2 = 1577

We will then assume that T2 is between 1500 and 1600 K. If that is so, we can develop the
following table (all Hs are in kJ/mol and (Hp - Hr)'s are in kJ):
T2
1500
1550
1506

H CO2 (kJ/mol)
61.76
64.69
62.11

Interpolate

H H2O
48.13
50.51
48.41

H O2
40.64
42.48
40.86

1500
T2
1550

H N2
38.43
40.18
38.64

Hp Hr (kJ)
-30.37
202.00
-2.55

-30.37
00
202.00

T2 1500

50

30.37

232.37

so

T2 = 1506 K

The enthalpy difference for this temperature was also calculated in the above table to show that
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once you have a good idea of the temperature interval where the adiabatic temperature should be
in, linear interpolation does a good job of predicting the temperature.
Comparing the results for the adiabatic flame temperature found above to that found using Fig. 3
on page 22 of the textbook is done as follows:
Enthalpy per gram C8H18

250.12
96 + 18

2.194 kJ/gm

H/C =

18
= 2.25
8

With these values Fig. 3 yields T2 1505 K, which is very close to the calculated adiabatic
temperature.

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Problem 3

Carbon monoxide is oxidized to carbon dioxide in an excess of air (1 atm) in an afterburner so

that the final temperature is 1300 K. Under the assumption of no dissociation determine the airfuel ratio required. Report the results on both a molar and mass basis. For the purposes of this
problem assume that air has the composition of 1 mole of oxygen to 4 moles of nitrogen. The
carbon monoxide and air enter the system at 298 K.
Solution: We will assume that adiabatic conditions exist. In this case, Eqn. (1.10) in the
textbook can be used to calculate the initial composition of the mixture which will give an
adiabatic flame temperature of T2 = 1300 K. This equation can be simplified for the general case
when H 0T0 = H 0298 For Q = 0, the modified equation becomes:

n ( H
i

i prod

0
T2

H 0298 + ( H 0f )

= n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively.
Since the temperature for the reactants and products have been given (T1 = 298 K and T2 = 1300
K), the problem simplifies to finding x in the reaction shown below, which will satisfy Eqn. (1)
above.
CO + x ( O 2 + 4 N 2 ) = CO 2 + ( x 0.5 ) O 2 + 4 x N 2

The following table can be developed from information in the textbook.

Gases
CO2
CO
O2
N2

H 0f (kJ/mol)

H 0T1300 H 0298 (kJ/mol)

393.77
110.62
0.0
0.0

50.19
Not needed
33.37
31.52

Substituting these values in Eqn. (1) one has:

2x 1
( 0.0 + 33.37 )O2 + 4 x ( 0.0 + 31.52 ) N2
2
2
1( 110.62 + 0.0 )CO x ( 0.0 + 0.0 )O + 4 x ( 0.0 + 0.0 ) N = 0.0

1( 393.77 + 50.19 )CO +

343.58 + 33.37 x 16.69 + 126.08 x + 110.62 = 0.0

159.45 x = 249.65
x = 1.566
Therefore the total number of moles of air needed to react with one mole of CO is 1.566 * (1 +
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4) = 7.83 moles (which is the A/F ratio on a molar basis). The air/fuel ratio on a mass basis is:
AF =

m air 7.83 moles of air 29 kg / mol

=
m fuel
1 mole of CO 28 kg / mol

AF = 8.11

so

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kg of air

kg of CO

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Problem 4
The exhaust of a carbureted engine, which is operated slightly fuel rich, has an efflux of
unburned hydrocarbons entering the exhaust manifold. One can assume that all the hydrocarbons
are equivalent to ethylene (C2H4) and all the remaining gases are equivalent to inert nitrogen
(N2). On a molar basis there are 40 moles of nitrogen for every mole of ethylene. The
hydrocarbons are to be burned over an oxidative catalyst and converted to carbon dioxide and
water vapor only. In order to accomplish this objective, ambient (298 K) air must be injected
into the manifold before the catalyst. If the catalyst is to be maintained at 1000 K, how many
moles of air per mole of ethylene must be added? Take the temperature of the manifold gases
before air injection as 400 K. Assume the composition of air to be 1 mole of oxygen to 4 moles
of nitrogen.
Solution: We will assume that adiabatic conditions exist and that no dissociation of the stable
products will take place. In this case, Eqn. (1.10) in the textbook can be used to calculate the
initial composition of the mixture which will give an adiabatic flame temperature of T2 = 1000
K. This equation can be simplified for the general case when H 0T0 = H 0298 . For Q = 0, the

modified equation becomes:

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

= n j H T0 H 0298 + ( H 0f )
298 j
j react 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants, respectively. Since
the temperature for the reactants and products have been given (Tl = 400 K and T2 = 1000 K), the
problem simplifies to finding x in the reaction shown below, which will satisfy equation (1)
above.
C2H4 + 40N2 + x(O2 + 4N2) 2 CO2 + 2 H2O + (x - 3) O2 + (40 + 4x) N2

(2)

The following table can be developed from information in the text.

Gases

C2H4
N2
O2
CO2
H2O(vap)

H 0f (kJ/mol

H 0T400 H 0298 (kJ/mol)

H 0T1000 H 0298 (kJ/mol)

52.34
0.0
0.0
393.77
242.00

4.89
2.97
3.03
Not needed
Not needed

Not needed
21.47
22.72
33.43
26.00

Substituting these values in Eqn. (1) with the coefficients from Eqn. (2) one has:

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1( 52.34 + 4.89 )C H + 40 ( 0.0 + 2.97 ) N + x ( 0.0 + 0.0 )O + 4 x ( 0.0 + 0.0 ) N

2

2 ( 393.77 + 33.43)CO 2 ( 242.00 + 26.00 )H O ( x 3)( 0.0 + 22.72 )O

2

( 40 + 4 x )( 0.0 + 21.47 ) N

= 0.0
2

57.22 + 118.91 + 720.68 + 432.00 22.72 x + 68.17 858.96 85.90 x = 0.0

108.62 x = 538.01 and
x = 4.953
Therefore, the total number of moles of air needed to react with one mole of C2H4 is
4.953
x (1 + 4) = 24.75 moles.

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Problem 5
A combustion test was performed at 20-atm pressure in a hydrogen-oxygen system. Analysis of
the combustion products, which were considered to be in equilibrium, was as follows:
Compound
H2O
H2
O2
O
H
OH

Mole fraction
0.4855
0.4858
0.0
0.0
0.0216
0.0069

What must the combustion temperature have been in the test?

Solution: Since all the products are in equilibrium, any equilibrium reaction between the
products can be used to solve for the temperature. The simplest reaction to choose (and the only
one that is an equilibrium reaction of formation) is:

1
H2 H
2
The expression for the equilibrium constant of formation is then given by
1/ 2

K P,f ( H )

n P
= H1/ 2

n H2 n

Since values are given in mole fraction

K P,f ( H ) =

n = 1 . Therefore,

0.0216
1/ 2
20 ) = 0.138593
1/ 2 (
0.4858

log10 K P,f ( H ) = 0.85826

Looking up the temperature in Table 2 in the appendix for H which corresponds to this value
yields
T ~ 2959 K
If one desires to use the other possible equilibrium reaction given by
1
H 2 + OH H 2 O
2
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The equilibrium constant is given by

KP =

n H 2O
1/ 2

n H2 n OH

1/ 2

0.4855

( 0.0069 ) ( 0.4858)

1/ 2

1
= 22.573349
20

or
log10 KP = 1.353596
Then, in order to solve for the final temperature of the mixture one would have to calculate the
value for the equilibrium constant for the above reaction and find the corresponding temperature
as shown below.
KP =

K P,f ( H2O )
K P,f ( OH )

Temp.

K P,f ( OH )

K P,f ( H2O )

KP

Log10 KP

2900 K
3000 K

1.1066
1.1614

31.2608
22.0293

28.2494
18.9679

1.4510
1.2780

Interpolating for the temperature between 2900 and 3000 K that would give log10 KP
= 1.353596 yields:
T ~ 2956 K

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Problem 6
Whenever carbon monoxide is present in a reacting system it is possible for it to
disproportionate into carbon dioxide according to the equilibrium 2CO Cs + CO2. Assume
that such an equilibrium can exist in some crevice in an automotive cylinder or manifold.
Determine whether raising the temperature decreases or increases the amount of carbon
present. Determine the KP for the reaction equilibrium and the effect of raising the
pressure on the amount of carbon formed.
Solution: One simply considers how the equilibrium is shifted by the temperature and pressure
to solve the problem.
(i) Temperature effect

2 CO Cs + CO2
KP =

(1)

pCO2

(2)

2
pCO

in terms of equilibrium constants of formation for C + O2 CO2 one has

K P,f ( CO2 ) =

p CO2

(3)

p O2

1
and for C + O 2 CO
2
K P,f ( CO ) =

pCO
p1O22

(4)

For Eqn. (1) it follows that

KP =

K P,f ( CO2 )
K 2P,f ( CO )

or log10 K P = log10 K P,f ( CO2 ) 2 log10 K P,f ( CO )

Evaluating this log equation in two different temperatures one has
(a) At 1000 K

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log10 K P,f ( CO2 ) = 20.68 and 2 log10 K P,f ( CO ) = 2 (10.46 ) = 20.92 so

log Kp = 0.24
or KP = 0.58
(b) At 2000 K
log10 K P,f ( CO2 ) = 10.35 and 2 log10 K P,f ( CO ) = 2 ( 7.47 ) = 14.94 so

log KP = 4.59
or KP = 2.57 105
Therefore, KP drops as temperature increases.
Equation (1) and (2) show that as KP decreases, the reaction given by Eqn. (1) is shifting to the
left, thus less carbon forms as the temperature increases.
(ii) Pressure effect
KP =

p CO2
2
p CO

n CO2 P
=

n CO 2 n

Thus, as the pressure increases more CO2 forms, shifting the reaction from left to right and thus
more carbon forms as the pressure is increased.

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Problem 7
Determine the equilibrium constant KP at 1000 K for the following reaction

2 CH4 2 H2 + C2H4
Solution: As explained on page 14 of the text, the equilibrium constant (given in terms of the
equilibrium constant of formation) is:

KP =

K 2P,f ( H2 ) K P,f ( C2 H4 )
K 2P,f ( CH4 )

At 1000 K it follows from Table 2 in the appendix that (note Table 2 provides the logl0 of K):
K P,f ( H2 ) = 100 = 1
K P,f ( C2 H4 ) = 106.213 = 6.1235 107
K P,f ( CH4 ) = 101.011 = 0.0975
so evaluating Eqn. (1) for KP one has
KP =

1 ( 6.1235 107 )

( 0.0975)

= 6.442 105
K P = 6.442 10-5

so

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Problem 8
The atmosphere of Venus is said to contain 5% carbon dioxide and 95% nitrogen by volume. It
is possible to simulate this atmosphere for Venus reentry studies by burning gaseous cyanogen
(C2N2) and oxygen and diluting with nitrogen in the stagnation chamber of a continuously
operating wind tunnel. If the stagnation pressure is 20 atm, what is the maximum stagnation
temperature that could be reached and still have Venus atmosphere conditions? If the stagnation
pressure were 1 atm, what would be the maximum temperature? Assume all gases enter the
chamber at 298 K. Take the heat of formation of cyanogen as ( H 0f ) = 381.84 k/mol.
298

Solution: In order to maintain Venus atmosphere, the composition of 100 moles of products
would have to be 5 CO2 + 95 N2. Eqn. (10) in the textbook can be used to calculate the
stagnation temperature of this system. This equation can be simplified for the general case when
H 0T0 = H 0298 . For Q = 0, the modified equation becomes:

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

= n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(1)

In this Equation, T2 and T1 are the temperatures of the products and reactants respectively. Since
the conditions for the reactants are specified the solution approach is to find T2 such that the
above equality will be satisfied.
(a) for Venus atmosphere the general combustion reaction would be
a C2N2 + b O2 + c N2 5 CO2 + 95 N2
which can be balanced for one mole of C2N2 as follows:
C2N2 + 2 O2 + 36 N2 2 CO2 + 38 N2,
In preparation to apply Eqn. (1) the following tables can be generated.
gases

H 0f ( kJ mol )

H 0T1 H 0298 ( kJ mol )

C2N2
O2
N2

381.84
0.0
0.0

0.0
0.0
0.0

An expression for the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (1)) in
terms of the enthalpy of the individual gases in the reactants becomes:
H r = 1( 381.84 + 0.0 )C N + 2 ( 0.0 + 0.0 )O + 36 ( 0.0 + 0.0 ) N = 381.84 kJ
2

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For the products it follows that:

gases

H 0f ( kJ mol )

N2
CO2

0.0
393.77

Thus, an expression for the enthalpy of the products (Hp, given by the left-hand side of Eqn. (1))
in terms of the enthalpy of the individual gases in the products becomes let H = H 0T2 H 0298 :

H p = 2 393.77 + H CO2 + 38 0.0 + H N2

Equating Hr and Hp yields:

2 H CO2 + 38 H N 2 = 1,169.37
A first guess for the products temperature can be obtained by assuming that all product gases are
N2. This approach gives the following as an initial guess for T2 (Note that 40 = 2 + 38):
40 H N2 = 1,169.37 so
H N 2 = 29.23kJ mol , therefore, T2 1232 K
We will then assume that T2 is between 1200 and 1300 K. If that is so, we can develop the
following table (all Hs are in kJ/mol and (Hp - Hr)'s are in kJ):
T2

H CO2 ( kJ mol )

H N2

Hp Hr (kJ)

44.51
50.19

28.13
31.52

11.51
128.87

1200
1300
Interpolate

1200
T2
1300

11.51
00
128.31

T2 1200 11.51
=
100
139.88
T 2 = 1208 K

so

Since K P,f ( CO2 ) = 1.75 1017 at 1200 K, no significant dissociation of the products occurs.
Therefore, pressure has no effect and T2 would remain the same at all pressures.
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Problem 9
A mixture of 1 mole of N2 and 0.5 mole O2 is heated to 4000 K at 0.5 atm pressure and results in
an equilibrium mixture of N2, O2, and NO only. If the O2 and N2 were initially at 298 K and the
process is one of steady heating, how much heat was required to bring the final mixture to 4000
K on the basis of one initial mole of N2?
Solution: The first step is to find the equilibrium concentration of products. Once this is done,
one can use Eqn. (1.10) to calculate the amount of heat necessary to bring the final mixture to
4000 K. Consider the following reaction at equilibrium.

1
1 N 2 + O 2 a N 2 + b O 2 + c NO
2

(1)

balancing this reaction yields the following relationship between a, b and c.

N: 2 = 2 a + c
0: 1 = 2 b + c
Solving these two equations for b and c in terms of a it follows that
b=

2a 1
2

and c = 2 2a

Therefore, Eqn. ( 1) becomes:

1N 2 + 0.5O 2 a N 2 +

2a 1
O 2 + ( 2 2a ) NO
2
1
1
N 2 + O 2 NO , the expression for the equilibrium
2
2

For the equilibrium reaction given by

constant of formation is given by:
11 2 1 2

K P,f ( NO )

P
n NO
=

1/ 2
1/ 2
n N2 n O2 n

2a 1
+ 2 2a = 3 . At 4000 K, 1og10 KP,f(NO) =
2
- 0.524 or KP,f(NO) = 0.29923. Squaring Eqn. (3) would give the following equation for the
variable a:
The total number of moles is given by

n = a +

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( 2 2a )

2a 1
a
2

K 2P,f ( NO ) = 0,

Solving for a in the above equation and then using the mass balance equations to solve for b and
c yields: a = 0.909, b = 0.409 and c = 0.182. Equation (2) now becomes
1 N2 + 0.5 O2 0.909 N2 + 0.409 O2 + 0.182NO
The amount of heat required to bring the above mixture to 4000 K is calculated using Eqn. (1.10)
in the text by taking T2 = 298 K and T1 = 4000 K:
Q =

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(4)

from tables in the text one has:

gases

O2
N2
NO

H 0f ( kJ mol )

H 0T1 H 0298 ( kJ mol )

H 0T2 H 0298 ( kJ mol )

0.0
0.0
90.43

139.01
130.16
132.76

0.0
0.0
0.0

so the equation for Q becomes

Q = 0.182 ( 90.43 + 132.76 ) NO + 0.409 ( 0.0 + 139.01)O + 0.909 ( 0.0 + 130.16 ) N
2

1( 0.0 + 0.0 ) N 1/ 2 ( 0.0 + 0.0 )O

2

Q = - 215.79 kJ

so
if no dissociation occurred the value of Q would be
Q = 1/ 2 ( 0.0 + 139.01)O + 1( 0.0 + 130.16 ) N
2

1( 0.0 + 0.0 ) N 1/ 2 ( 0.0 + 0.0 )O

2

so

Q
= - 199.67 kJ

Therefore, more energy is required to raise the temperature of the mixture to 4000 K due to
dissociation.
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Problem 10
Calculate the adiabatic decomposition temperature of benzene under the constant pressure
condition of 20 atm. Assume that benzene enters the decomposition chamber in the liquid state at
298 K and decomposes into the following products: carbon (graphite), hydrogen, and methane.
Solution: The solution for this problem is an iterative one. In order to calculate the final
temperature of the mixture, one needs the composition of the mixture at equilibrium. But the
composition of the mixture can only be found if the final temperature is known. Therefore, one
will guess the final temperature and calculate the mixture composition using the equilibrium
concept. Once the composition is know at the guessed temperature, the adiabatic flame
temperature is calculated and checked against the guessed value. This process is repeated until
the guessed and calculated temperatures are obtained. Consider the following reaction at
equilibrium:
C6H6 (1) a C (s) + b H2 + c CH4

(1)

balancing this reaction yields the following relationship between a, b and c.

C: 6 = a + c
H: 6 = 2 b + 4 c
solving these two equations for a and b in terms of c it follows that
b=

6 4c
= 3 2c
2

and a = 6 c

Therefore, Eqn. (1) becomes:

C6 H 6(1) ( 6 c ) C(s ) +

6 4c
H 2 + c CH 4
2

(2)

Now consider the equilibrium reaction given by

C(s) + 2 H2 CH4
For this equilibrium reaction, the expression for the equilibrium constant of formation of CH4 is
1 2

K P,f ( CH4 )

n CH P
= KP = 2 4

n H2 n

(3)

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1/25/20119:27AM

6 4c
+ c = 3 c (note that the number of moles of
2
graphite are not included in the calculation for the total number of moles at equilibrium).
Equation (3) then becomes:

The total number of moles is given by

n =

c b+c
KP = 0
b 2 20
expressing b in terms of c gives:

( 3 2c )

3+ c
KP = 0
20

(4)

In order to calculate the adiabatic flame temperature one has to apply Eqn. (1.11) in the text
taking T1 = 298 K.
Q = 0 =

n ( H
i

0
T2

H 0298 + ( H f0 )

i prod

n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants
are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties:
gases

C6H6 (1)

H 0f ( kJ mol )

H 0T1 H 0298 ( kJ mol )

49.03

0.0

And the enthalpy of the reactants (Hr, given by the right-hand side of Eqn. (5)) can be calculated
as:
Hr = 1 (49.03)C6H6 (1) = 49.03 kJ
Let us first guess that T2 = 1000 K; therefore, logl0 KP = - 1.011 or KP = 0.097499. Solving Eqn.
(4) for c using this value of KP and then solving the mass balance equations for a and b results
in:
c = 0.9943, b = 1.0114, and a = 5.0057
For a product temperature of 1000 K, application of Eqn. (5) using the values of a, b, and c found
above yields:
T2
1000

H C(s) ( kJ mol )

H H2

H CH4

Hp

Hp Hr (kJ)

8.47

20.70

38.20

26.87

22.16

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2008 Glassman and Yetter.

1/25/20119:27AM

If T2 = 1100 K, logl0 Kp. = - 1.44 or KP = 0.03631. Solving Eqn. (4) for c using this value of KP
and then solving the mass balance equations for a and b gives:
c = 0.7409, b = 1.5182, and a = 5.2591
Following the same procedure as previously outlined yields:
T2
1100

H C(s) ( kJ mol )

H H2

H CH4

Hp

Hp Hr (kJ)

14.01

23.74

45.58

88.01

38.98

Interpolation for the value of temperature where Hp Hr = 0 yields T2 ~ 1036 K.

At T2 = 1036K, logl0 Kp = -1.1654 or Kp = 0.06832, c = 0.9101, b = 1.1798, a = 5.0899 and
T2
1036

H C(s) ( kJ mol )

H H2

H CH4

Hp

Hp Hr (kJ)

10.47

21.79

40.86

48.02

1.01

Since Hp Hr ~0

T 2 ~ 1036 K

23

2008 Glassman and Yetter.

1/25/20119:27AM

Problem 11
Calculate the flame temperature and the product composition of liquid ethylene oxide
decomposing at 20-atm pressure by the irreversible reaction:

C2H4O(1) a CO + b CH4 + c H2 + d C2H4

The four products are as specified. The equilibrium known to exist is
2 CH4 2 H2 + C2H4
The heat of formation H 0f of liquid ethylene oxide is = - 76.62 kJ/mol. It enters the
decomposition chamber at 298 K.
Solution: The solution for this problem is an iterative one. In order to calculate the final
temperature of the mixture, one needs the composition of the mixture at equilibrium. But the
composition of the mixture can only be found if the final temperature is known. Therefore, one
will guess the final temperature and calculate the mixture composition using the equilibrium
concept. Once the composition is know at the guessed temperature, the adiabatic flame
temperature is calculated and checked against the guessed value. This process is repeated until
the guessed and calculated temperatures are the same. Consider the following reaction at
equilibrium:

C2H4O(1) a CO + b CH4 + c H2 + d C2H4

balancing this reaction yields the following equations for a, b, c and d.
C: 2 = a + b +2 d
O: 1 = a
H: 4 = 4 b + 2 c + 4 d
simplifying it follows that:
b+2d=1

(1)

4b+2c+4d=4

(2)

The other equation for the solution of b, c, and d will come from the Kp expression for the given
equilibrium reaction.
2 CH4 2 H2 + C2H4
For this equilibrium reaction, the expression for the equilibrium constant in terms of the
equilibrium constant of formation of each individual species is,
24

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1/25/20119:27AM

KP =

K 2P,f ( H2 ) K P,f ( C2 H4 )
K 2P,f ( CH4 )

n 2H2 n C2 H4 P
=

2
n CH
n
4

2 +1 2

(3)

or in terms of b, c, and d ( with P = 20 atm.) one has

KP =

c2d
20
2
b 1+ b + c + d

(4)

By combining the equation above with the mass balance equations the following expression can
be written for KP:
KP

(1 b )
=
b

20
5b

(5)

The approach now is to guess T2, look up KP and solve for the three unknowns in Eqns. (1), (2),
and (5). The correct guess will be the one that satisfies the adiabatic flame temperature equation.
In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by
taking T1 = 298 K.
Q = 0 =

n ( H
i

0
T2

H 0298 + ( H f0 )

i prod

n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(6)

With Eqns. (5) and (6) the iterative process may now start. Since the conditions for the reactants
are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties:
gases

C2H4O(1)
CO
CH4
C2H4

H 0f ( kJ mol )

76.62
110.62
74.90
52.34

Applying Eqn. (6) yields let H = H 0T2 H 0298 :

) (

) (

76.62 = a ( 110.62 + H CO ) + b 74.90 + H CH4 + c 0.0 + H H2 + d 52.34 + H C2 H4

First guess that T2 = 1400 K, from Tables in the text one has:

25

2008 Glassman and Yetter.

(7)

1/25/20119:27AM

log10 K P,f ( H2 ) = 1
log10 K P,f ( C2 H4 ) = 5.664
log10 K P,f ( CH4 ) = 2.372
so from Eqn. (3), log10 KP = 0.12023. Solving Eqn. (5) for b and then Eqns. (1) and (2) for c and
d gives:
a = 1.0, b = 0.7557, c = 0.2443, and d = 0.1221
Application of Eqn. (7) using T2 = 1400 K and the values of a, b, c, and d just calculated yields:
T2
1400

HCO (kJ/mol)
35.36

H CH4
69.66

H H2
33.08

H C2 H4
91.26

Hp (kJ)
-53.60

If we now take T2 = 1300 K, from Tables in the text one has:

log10 K P,f ( H2 ) = 1
log10 K P,f ( C2 H4 ) = 5.766
log10 K P,f ( CH4 ) = 2.107
so from Eqn. (3), log10 KP = - 1.552 or KP = 0.028054. Solving Eqn. (5) for b and then Eqns. (1)
and (2) for c and d gives:
a = 1.0, b = 0.8398, c = 0.1603, and d = 0.0801
Application of Eqn. (7) using T2 = 1300 K and the values of a, b, c, and d just calculated yields:
T2
1300

HCO (kJ/mole)

H CH4

H H2

H C2 H4

Hp (kJ)

31.89

61.34

29.93

80.63

-74.65

Extrapolation to the value of T2 where Hp ~ Hr (76.62) yields:

2 ~ 1291 K
T

26

2008 Glassman and Yetter.

1/25/20119:27AM

Problem 12
Liquid hydrazine (N2H4) decomposes exothermically in a monopropellant rocket operating at
100-atm chamber pressure. The products formed in the chamber are N2, H2, and ammonia (NH3)
according to the irreversible reaction

N2H4(1) a N2 + b H2 + c NH3
Determine the adiabatic decomposition temperature and the product composition a, b, and c.
Take the standard heat of formation of liquid hydrazine as 50.03 kJ/mole. The hydrazine enters
the system at 298 K.
Solution: The solution for this problem is an iterative one. In order to calculate the final
temperature of the mixture, one needs the composition of the mixture at equilibrium. But the
composition of the mixture can only be found if the final temperature is known. Therefore, one
will guess the final temperature and calculate the mixture composition using the equilibrium
concept. Once the composition is known at the guessed temperature, the adiabatic flame
temperature is calculated and checked against the guessed value. This process is repeated until
the guessed and calculated temperatures are the same. Consider the following reaction at
equilibrium:

N2H4(1) a N2 + b H2 + c NH3
balancing this reaction yields the following equations for a, b, and c.
N: 2 = 2 a + c

(1)

H: 4 = 2 b + 3 c

(2)

The other equation necessary for the solution of a, b, and c will come from the Kp for the
following equilibrium reaction.
1
3
N 2 + H 2 NH 3
2
2
For this equilibrium reaction, the equilibrium constant of formation of NH3 can be expressed in
terms of the mixture composition at equilibrium as shown:
11 2 3 2

K P,f ( NH3 )

n NH P
= K p = 1 2 33 2

n H2 n N2 n

(3)

or in terms of a, b, and c (with P = 100 atm) one has:

27

2008 Glassman and Yetter.

1/25/20119:27AM

Kp =

c
a b

1/ 2 3/ 2

a+b+c
100

(4)

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2),
and (4). The correct guess will be the one that satisfies the adiabatic flame temperature equation.
In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by
taking T1 = 298 K.
Q = 0 =

n ( H
i

H 0298 + ( H f0 )

0
T2

i prod

n j H 0T1 H 0298 + ( H 0f )
298 j
j

react

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants
are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic
properties:
H 0f ( kJ mol )

gases

N2H4(1)
N2
H2
NH3

50.03
0.0
0.0
46.22

Applying Eqn. (5) yields let H = H 0T2 H 0298 :

) (

) (

50.03 = a 0.0 + H N2 + b 0.0 + H H2 + c 46.22 + H NH3

(6)

Let us first guess that T2 = 900 K, from Tables in the text it follows that:
log10 K P,f ( NH3 ) = 2.915

Solving Eqns. (l), (2) and (4) results in:

a = 0.9487, b = 1.8462, and c = 0.1026
For a product temperature of 900 K, application of Eqn. (6) using the values of a, b, and c, found
above, yields (all H's are in kJ/mol and Hp is in kJ):
T2
900 K

H N2 ( kJ mol )

H H2

H NH3

Hp (kJ)

18.23

17.69

27.09

47.98

First guess that T2 = 1000 K, from Tables in the text it follows that:
28

2008 Glassman and Yetter.

1/25/20119:27AM

log10 K P,f ( NH3 ) = 3.233

Solving Eqns. (1), (2) and (4) one has:

a = 0.9739, b = 1.9218, and c = 0.05216
For a product temperature of 1000 K, application of Eqn. (6) using values of a, b, and c, found
above yields:
T2
1000 K

H N2 ( kJ mole )

H H2

H NH3

Hp (kJ)

21.47

20.70

32.60

60.00

Interpolation to the value of T2 where Hp ~ Hr (50.03) yields:

T 2 ~ 917 K

29

2008 Glassman and Yetter.

1/25/20119:27AM

Problem 13
Gaseous hydrogen and oxygen are burned at l-atm pressure under the rich conditions designated
by the following combustion reaction:

O2 + 5 H2 a H2O + b H2 + c H
The gases enter at 298 K. Calculate the adiabatic flame temperature and the product composition
a, b, and c.
Solution: The solution for this problem is an iterative one. In order to calculate the final
temperature of the mixture, one needs the composition of the mixture at equilibrium. But the
composition of the mixture can only be found if the final temperature is known.

Therefore, one will guess the final temperature and calculate the mixture composition using the
equilibrium concept. Once the composition is known at the guessed temperature, the adiabatic
flame temperature is calculated and checked against the guessed value. This process is repeated
until the guessed and calculated temperatures are the same. Consider the following reaction at
equilibrium:
O2 + 5 H2 a H2O + b H2 + c H
balancing this reaction yields the following equations for a, b, and c.
O: 2 = a

(1)

H: 10 = 2 a + 2 b + c

(2)

The other equation necessary for the solution of a, b, and c will come from the KP expression for
the following equilibrium reaction.
1
H2 H
2
For this equilibrium reaction, the equilibrium constant of formation of H can be expressed in
terms of the mixture composition at equilibrium as shown:
11/ 2

K P,f ( H )

n P
= K P = 1/H2

n H2 n

(3)

or in terms of a, b, and c (with P = 1 atm) one has

KP =

c
1
b a+b+c

(4)

1/ 2

30

2008 Glassman and Yetter.

1/25/20119:27AM

The approach now is to guess T2, look up Kp and solve for the three unknowns in Eqns. (1), (2),
and (4). The correct guess will be the one that also satisfies the adiabatic flame temperature
equation.
In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by
taking T1 = 298 K.
Q = 0 =

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants
are specified (T1 = 298 K), Table 1.1 in the text provides the following thermodynamic
properties:
gases

H 0f ( kJ mol )

O2
H
H2
H2O

0.0
218.09
0.0
242.00

Applying Eqn. (5) yields let H = H 0T2 H 0298 :

) (

0.0 = a 242.00 + H H2O + b 0.0 + H H2 + c ( 218.09 + H H )

Let us first guess that T2 = 2500 K, from Table 2 it follows that,
log10 Kp,f(H) = 1.601
Solving Eqns. (1), (2) and (4) one has:
a = 2.0, b = 2.9516, and c = 0.0967
For a product temperature of 2500 K application of Eqn. (6) using the values of a, b, and c,
yields (all H's are in kJ/mole and Hp is in kJ):
T2
2500 K

H H2O ( kJ mol )

H H2

HH

Hp (kJ)

99.03

70.54

45.80

52.21

Let us first guess that T2 = 2700 K, from Table A2 it follows that

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2008 Glassman and Yetter.

(6)

1/25/20119:27AM

log10 Kp,f(H) = 1.247

Solving Eqns. (1), (2) and (4) one has:
a = 2.0, b = 2.8912, and c = 0.2176
For a product temperature of 2700 K application of Eqn. (6) using the values of a, b, and c, found
above yields (all H's are in kJ/mol and Hp is in kJ):
T2
2700 K

H H2O ( kJ mol )

H H2

HH

Hp (kJ)

109.89

77.77

49.96

18.96

Interpolation to the value of T2 where Hp ~ Hr (Hr = 0.0) yields:

T2 ~ 2647 K

32

2008 Glassman and Yetter.

1/25/20119:27AM

Problem 14
The liquid propellant rocket combination nitrogen tetroxide (N2O4) and UDMH (unsymmetrical
dimethyl hydrazine) has optimum performance at an oxidizer to fuel weight ratio of 2 at a
chamber pressure of 67 atm. Assume that the products of combustion of this mixture are N2,
CO2, H2O, CO, H2, O H, OH, and NO. Set down the equations necessary to calculate the
adiabatic combustion temperature and the actual product composition under these conditions.
These equations should contain all the numerical data in the description of the problem and in
the tables in the appendices. The heats of formation of the reactants are:
N 2 O 4(1)

H f0 298 = 2.90 kJ mol

UDMH (1)

0f 298 = +53.17 kJ mol

The propellants enter the combustion chamber at 298 K.

Solution: The solution for this problem consists of finding the correct number of equations
necessary to solve for the unknown quantities in this problem. There are 9 unknown species
coefficients and the temperature is unknown. Therefore, 10 equations must be found to solve the
problem. These equations consist of the energy, 4 mass balance, and 5 equilibrium equations.

To specify the general reaction equation, the mole proportion of reactants must be found first.
This can be done as follows.
The molecular weights of the reactants are:
for N2O4,

MW = 92

for C2N2H8,

MW = 60

From the problem statement one has:

mass UDMH
=2
mass N 2 O 4
By assuming 1 mole N2O4 this equation becomes
mass UDMH
=2
60
or

nUDMH (92) = 120

33

2008 Glassman and Yetter.

1/25/20119:27AM

So
nUDMH = 1.304
The general equation for the reaction then becomes:
C2N2H8 + 1.304 N2O4 a N2 + b CO2 + c H2O + d CO + e H2 + f O + g H + h OH + i NO
The mass balance equations can be written as:
C: 2 = b + d
N: 4.609 = 2a + i
H: 8 = 2c + 2e + g + h
O: 5.216 = 2b + c + d + f + h + i

(1)
(2)
(3)
(4)

The energy equation would be the equation for adiabatic flame temperature (Eq. (11)) found in
the text.

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

= n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

The following information can be used to simplify the equation

H fN2O4 = 2903 kJ mol
H f UDMH = 53.172 kJ mol
The right side of the equation is:
1(53.172) + 1.304 (-2.093) = 50.44 kJ
Therefore, the energy equation becomes:
50.44 =

n ( H
i

0
T2

H 0298 + ( H f0 )

i prod

(5)

298 i

Some of the possible equilbrium reactions include:

H2 2 H
CO + O CO2
N2 + 2 O 2 NO
H2 + 2 OH 2 H2O
34

2008 Glassman and Yetter.

1/25/20119:27AM

CO + 2 OH H2O + CO2
These reactions give the following equilibrium equations:
K p1 =

n 2H
n H2

n 2NO P
K p2 =

n N2 n O2 n

n CO2 P
K p3 =

n CO n O n

n 2H2O P
K p4 =
2
n
n H2 n OH

g 2 67

e n i

b
=
df

ni

67

(6)

(7)

(8)

c2 n i

e h 2 67

(9)

c b ni

d h 2 67

(10)

n H O n CO P
K p5 = 2 2 2

n CO n OH n

where

i2 ni
= 2

a f 67
=

= a + b + c + d + e + f + g + h + i, and the Kp's are evaluated at the adiabatic flame

temperature.

35

2008 Glassman and Yetter.

1/25/20119:27AM

Problem 15
Consider a fuel burning in inert airs and oxygen where the combustion requirement is only 0.21
moles of oxygen. Order the following mixtures as to their adiabatic flame temperatures with the
given fuel.

a)
b)
c)
d)

pure O2
0.21 O2 + 0.79 N2
0.21 O2+0.79 Ar
0.21 O2 + 0.79 CO2

(air)

Solution: The temperatures order with respect to the Cp of the inert and excess oxygen. The
lower the Cp of the inert gas the higher the temperature. Thus, Ar being monoatomic has the
lowest Cp, O2 and N2 are about the same, but higher than Ar since they are diatomic and have
vibrational and rotational degrees of freedom, and CO2 being triatomic has the highest Cp. Thus
the order from highest to lowest is

c, a and/or b, d

36

2008 Glassman and Yetter.

1/25/20119:27AM

Problem 16
Propellant chemists have proposed a new high energy liquid oxidizer, penta-oxygen O5, which is
also a monopropellant. Calculate the monopropellant decomposition temperature at a chamber
pressure of 10 atm if it assumed the only products are O atoms and O2 molecules. The heat of
formation of the new oxidizer is estimated to be very high, + 1025 kJ / mol. Obviously the
amounts of O2 and O must be calculated for one mole of O5 decomposing. The O5 enters the
system at 298 K. Hint: The answer will lie somewhere between 4000 and 5000K.
Solution: The solution for this problem is an iterative one. In order to calculate the final
temperature of the mixture, one needs the composition of the mixture at equilibrium. But the
composition of the mixture can only be found if the final temperature is known. Therefore, one
will guess the final temperature and calculate the mixture composition using the equilibrium
concept. Once the composition is known at the guessed temperature, the adiabatic flame
temperature is calculated and checked against the guessed value. This process is repeated until
the guessed and calculated temperatures are the same. Consider the following reaction at
equilibrium:

O5 a O2 + b O
balancing this reaction yields the following equation for a.
O: 5 = 2a + b or b = 5 - 2a

(1)

The other equation necessary for the solution of a and b will come from the Kp expression for
the following equilibrium reaction.
1
O2 O
2
For this equilibrium reaction, the expression for the equilibrium constant in terms of the
equilibrium constant of formation of each individual species is,
KP =

K P,f ( O )
K1/P,f2( O2 )

n P
= 1/O2

n O2 n

2 1

(2)

or in terms of a and b ( with P = 10 atm.) one has

1/ 2

b 10
K p = 1/ 2
a a + b

(3)

By combining the equation above with the mass balance equation the following expression can
be written for Kp:
37

2008 Glassman and Yetter.

1/25/20119:27AM

5 2a 10
a1/ 2 5 1

1/ 2

Kp =

(4)

The approach now is to guess T2 look up Kp and solve for the two unknowns in Eqns. (1) and
(4). The correct guess will be the one that satisfies the adiabatic flame temperature equation.
In order to calculate the adiabatic flame temperature, Eqn. (1.11) in the text can be used by
taking T1 = 298 K.

n ( H
i

0
T2

H 0298 + ( H 0f )

i prod

n j H 0T H 0298 + ( H 0f )
298 j
j react 1

298 i

(5)

With Eqns. (4) and (5) the iterative process may now start. Since the conditions for the reactants
are specified (T1 = 298 K), Table 1 in the text provides the following thermodynamic properties:
H 0f ( kJ mol )

gases

O
O2
O5

249.36
0
1025

Applying Eqn. (5) yields let H = H 0T2 H 0298 :

1025 = a 0.0 + H O2 + b ( 249.36 + H O )

(6)

First guess that T2 = 4500 K, from Tables in the text one has:
log10 K P,f ( O2 ) = 1
log10 K P,f ( O ) = 0.543
so from Eqn. (2), log10 Kp = 0.543 or Kp = 3.491. Solving Eqn. (4) for a and then Eqn. (1) for b
gives:
a = 1.29, b = 2.42
Application of Eqn. (6) using T2 = 4500 K and the values of a and b just calculated yields:
T2

H O2 ( kJ mol )

HO

Hp (kJ/mol)

4500

159.9

88.45

1023.77

38

2008 Glassman and Yetter.

1/25/20119:27AM

Since the guessed value of T2 give Hp ~ H (1025 kJ/mol) the decomposition temperature is
T 2 ~ 4500 K

39

2008 Glassman and Yetter.