ME470 Chemical and Phase Equilibrium Inst: Shoeleh Di Julio
Chapter 16, Solution 4C.
(a) Kp1, (b) 1/Kp1, (c) Kp1, (d) Kp1, (e) Kp12.
Chapter 16, Solution 19.
The temperature at which 5 percent of diatomic oxygen dissociates into monatomic oxygen at a specified
pressure is to be determined.
Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are
ideal gases.
Analysis The stoichiometric and actual reactions can be written as
Stoichiometric:
O 2  2O (thus  O2  1 and  O  2)
Actual:
O 2  0.95O 2  0
.1O



react.
O2 ↔ 2O
5%
3 atm
prod.
The equilibrium constant Kp can be determined from
 O  O 2
N O  P 

K p  O 

N OO 2  N total 
2

0.12 
3



0.95  0.95  0.1 
2 1
 0.03008
From Table A-28, the temperature corresponding to this Kp value is
T = 3133 K
Chapter 16, Solution 26.
Hydrogen is heated to a high temperature at a constant pressure. The percentage of H2 that will dissociate
into H is to be determined.
Assumptions 1 The equilibrium composition consists of H2 and H. 2 The constituents of the mixture are
ideal gases.
Analysis The stoichiometric and actual reactions can be written as
Stoichiometric: H 2  2H (thus  H2  1 and  H  2)
Actual:
H2 
 
xH 2  
yH
react.
prod.
H balance:
2 = 2x + y or y = 2 - 2x
Total number of moles: Ntotal = x + y = x + 2 - 2x = 2 - x
The equilibrium constant relation can be expressed as
 H  H 2
N H  P 

K p  H 

N HH 2  N total 
2
From Table A-28, ln Kp = -2.534 at 3200 K. Thus Kp = 0.07934. Substituting,
(2  2 x) 2
0.07934 
x
 8 


2 x
2 1
Solving for x,
x = 0.95
Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is
1 - 0.95 = 0.05 or 5.0%
H2
3200 K
8 atm
Chapter 16, Solution 32.
Liquid propane enters a combustion chamber. The equilibrium composition of product gases and the rate of
heat transfer from the combustion chamber are to be determined.
Assumptions 1 The equilibrium composition consists of CO2, H2O, CO, N2, and O2. 2 The constituents of
the mixture are ideal gases.
Analysis (a) Considering 1 kmol of C3H8, the stoichiometric combustion equation can be written as
C 3 H 8 ()  a th (O 2  3.76 N 2 ) 
 3CO 2  4H 2 O + 3.76 a th N 2
C3H8
where ath is the stoichiometric coefficient and is determined
from the O2 balance,
25C
2.5ath  3  2  1.5ath


ath  5
Air
Then the actual combustion equation with 150% excess air
and some CO in the products can be written as
C3H8 ( )  12.5( O2  376
. N2 )


Combustion
chamber
2 atm
CO
1200 K CO2
H2O
O2
N2
12C
xCO2  (3  x )CO + (9  0.5x )O2  4H2O + 47N 2
After combustion, there will be no C3 H8 present in the combustion chamber, and H2O will act like an inert
gas. The equilibrium equation among CO2, CO, and O2 can be expressed as
CO 2  CO + 12 O 2 (thus  CO 2  1,  CO  1, and  O2  12 )
and


CO
N CO
N OO 2  P  ( CO  O 2  CO2 )
2


Kp 
N

 CO2
 total 
N CO
2
where
N total  x  (3  x )  (9  0.5x )  4  47  63  0.5x
From Table A-28, ln K p  17 .871 at 1200 K. Thus K p  1.73  10 8 . Substituting,
1.73  10 8 
(3  x)(9  0.5x) 0.5
x
1.51
2




63

0
.
5
x


Solving for x,
x  2.9999999  3.0
Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss in
accuracy. Then the combustion equation and the equilibrium composition can be expressed as
C3 H 8 ()  12.5(O 2  3.76N 2 ) 
 3CO 2  7.5O 2  4H 2 O + 47N 2
and
3CO 2  7.5O 2  4H 2O +47N 2
(b) The heat transfer for this combustion process is determined from the steady-flow energy balance
Ein  Eout  Esystem on the combustion chamber with W = 0,
 Qout 
 N h
P

f
h h
   N h
P
R

f
 h h

R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The
h f of liquid propane is obtained by adding the hfg at 25°C to h f of gaseous propane).
h f
kJ/kmol
-118,910
0
0
-241,820
-393,520
Substance
C3H8 ()
O2
N2
H2O (g)
CO2
h 285 K
h 298 K
h1200 K
kJ/kmol
--8696.5
8286.5
-----
kJ/kmol
--8682
8669
9904
9364
kJ/kmol
--38,447
36,777
44,380
53,848
Substituting,
 Qout  3( 393,520  53,848  9364)  4( 241,820  44,380  9904)
 7.5( 0  38,447  8682)  47( 0  36,777  8669)
 1( 118,910  h298  h298 )  12.5( 0  8296.5  8682)
 47( 0  8186.5  8669)
 185,764 kJ / kmol of C3H 8
or
Qout  185,764 kJ / kmol of C3H8
The mass flow rate of C3H8 can be expressed in terms of the mole numbers as

m
12
. kg / min
N 

 0.02727 kmol / min
M 44 kg / kmol
Thus the rate of heat transfer is
Q out  N  Qout  (0.02727 kmol/min)(185,746kJ/kmol)  5066 kJ/min
N 2  12 O 2  NO is ln Kp = -7.569, which is very small. This
indicates that the amount of NO formed during this process will be very small, and can be disregarded.
The equilibrium constant for the reaction
1
2
Chapter 16, Solution 39C.
The number of Kp relations needed to determine the equilibrium composition of a reacting mixture is equal
to the difference between the number of species present in the equilibrium mixture and the number of
elements.
Chapter 16, Solution 40.
Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature
is to be determined.
Assumptions 1 The equilibrium composition consists of H2O, OH,
O2, and H2. 2 The constituents of the mixture are ideal gases.
H2O, OH
Analysis The reaction equation during this process can be expressed as
H 2O


x H 2O  y H2  z O2 + w OH
Mass balances for hydrogen and oxygen yield
H balance:
(1)
2  2x  2 y  w
H2O 
O2 , H 2
3400 K
1 atm
O balance:
(2)
1  x  2z  w
The mass balances provide us with only two equations with four unknowns, and thus we need to have two
more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the
mixture. They are
H 2 O  H 2  12 O 2
(reaction 1)
H 2 O  12 H 2  OH
(reaction 2)
The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be
ln K P1  1891
.


K P1  015092
.
ln K P2  1576
.


K P2  0.20680
The Kp relations for these two simultaneous reactions are

K P1

N HH 2 N OO 2  P  ( H 2  O 2  H 2O )
2
2



N

 H 2O
 total 
NH O

and
K P2
2
where

(  
)
OH
N HH 2 N OH
 P  H 2 OH H 2O
2






N HH O2O  N total 
2
N total  N H2O  N H2  NO2  NOH  x  y  z  w
Substituting,
( y)( z )1 / 2
0.15092 
x
1/ 2


1


 x  y  z  w
(3)
1/ 2


1


(4)
 x  y  z  w
Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields
x = 0.574
y = 0.308
z = 0.095
w = 0.236
Therefore, the equilibrium composition becomes
0.574H 2O  0.308H 2  0.095O 2  0.236OH
0.20680 
( w)( y)1/ 2
x
Chapter 16, Solution 54.
The hR value for the dissociation process O2  2O at a specified temperature is to be determined using
enthalpy and KP data.
Assumptions Both the reactants and products are ideal gases.
Analysis (a) The dissociation equation of O2 can be expressed as
O 2  2O
The hR of the dissociation process of O2 at 3100 K is the amount of energy absorbed or released as one
kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 3100 K, and can be
determined from
hR 
 N h
P

f
 h h
   N h
P
R

f
 h h

R
Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables,
Substance
O
O2
hf
kJ/kmol
249,190
0
h 298 K
h 2900 K
kJ/kmol
6852
8682
kJ/kmol
65,520
110,784
Substituting,
hR  2(249 ,190  65,520  6852 )  1(0  110 ,784  8682 )
 513,614 kJ/kmol
(b) The hR value at 3100 K can be estimated by using KP values at 3000 K and 3200 K (the closest two
temperatures to 3100 K for which KP data are available) from Table A-28,
ln
K P 2 hR  1
h 1
1 
1 
   or ln K P 2  ln K P1  R   

K P1 Ru  T1 T2 
Ru  T1 T2 
 3.072  (4.357 ) 
hR
1 
 1



8.314 kJ/kmol  K  3000 K 3200 K 
hR  512,808 kJ/kmol
Chapter 16, Solution 66.
A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The
composition of the liquid phase is given. The composition of the vapor phase is to be determined.
Assumptions The mixture is ideal and thus Raoult’s law is applicable.
Properties At 25C, Psat,H2O  3.170 kPa and Psat,NH3  1003.5 kPa .
Analysis The vapor pressures are
PH 2O  y f ,H 2O Psat,H 2O (T )  0.50 (3.170 kPa)  1.585 kPa
PNH3  y f , NH3 Psat,NH3 (T )  0.50 (1003 .5 kPa)  501 .74 kPa
Thus the total pressure of the mixture is
Ptotal  PH2O  PNH3  (1.585  501.74) kPa  503.33 kPa
Then the mole fractions in the vapor phase become
y g ,H 2O 
y g , NH3 
PH 2O
Ptotal
PNH3
Ptotal

1.585 kPa
 0.0031 or 0.31%
503.33 kPa

501 .74 kPa
 0.9969 or 99.69%
503.33 kPa
H2O + NH3
25C