-
Advanced Thermodynamics
Note 12
Chemical-Reaction Equilibria
Lecturer:
-
Chemical reaction
Both the rate and equilibrium conversion of a chemical reaction
depend on the temperature, pressure, and composition of
reactants.Although reaction rates are not susceptible to
thermodynamic treatment, equilibrium conversions are.The purpose of
this lecture is to determine the effect of temperature, pressure,
and initial composition on the equilibrium conversions of chemical
reactions.
- A general chemical reaction:A differential change in the number
of moles of a reacting species:
Reaction coordinate, which characterizes the extent or degree to
which a reaction has taken place.
Reaction coordinate
-
For a system in which the following reaction occurs,
assume there are present initially 2 mol CH4, 1 mol H2O, 1 mol
CO and 4 mol H2. Determine expressions for the mole fractions yi as
functions of .
-
Consider a vessel which initially contains only n0 mol of water
vapor. If decomposition occurs according to the reaction,
find expressions which relate the number of moles and the mole
fraction of each chemical species to the reaction coordinate .
-
Multireaction
Two or more independent reactions proceed simultaneouslyi,j : the
stoichiometric number of species i in reaction j.the change of the
moles of a species ni:
total stoichiometric number:
integration
summation
-
Consider a system in which the following reactions occur,
if there are present initially 2 mol CH4 and 3 mol H2O,
determine expressions for the yi as functions of 1 and 2 .
i
CH4
H2O
CO
CO2
H2
j
j
1
-1
-1
1
0
3
2
2
-1
-2
0
1
4
2
-
Equilibrium criteria to chemical reactions
is the single variable that characterizes the progress of the
reactionthe total Gibbs energy at constant T and P is determined by
if not in chemical equilibrium, any reaction leads to a decrease in
the total Gibbs energy of the systemThe condition for
equilibrium:the total Gibbs energy is a minimumIts differential is
zero
-
Fig 13.1
-
Standard Gibbs energy change and the equilibrium constant
-
The fugacity of a species in solution:
For pure species i in its standard state:
-
The equilibrium constant for the reaction, f(T)
The standard Gibbs energy change of reaction, f(T)
Other standard property changes of reaction :
-
For a chemical species in its standard state:
summation
-
Readily calculated at any temperature from the standard heat of
reaction and the standard Gibbs energy change of reaction at a
reference temperature
-
Fig 13.2
-
Calculation the equilibrium constant for the vapor-phase
hydration of ethylene at 145 and at 320 C from data given in
App.c
The heat capacity data:
From Table C.4
TK0K1K2K298.1529.3661129.366418.1529.3664.985x10-30.9861.443x10-1593.1529.3661.023x10-40.97942.942x10-3
- Gas phase reaction:Ideal-gas state: Standard-state pressure Po
of 1 bar
An ideal solution:
An ideal gas:
f (T)
f (P)
f (composition)
- Liquid phase reaction:
Except at high pressures
Ideal solution
Law of mass action
-
Equilibrium conversions for single reactions
Single reaction in a homogeneous system:assuming an ideal gas and
the equilibrium constant is known:assuming an ideal solution and
the equilibrium constant is known:the phase composition at
equilibrium can be obtained
-
The water-gas-shift reaction, is carried out under the different
sets of conditions described below. Calculate the fraction of steam
reacted in each case. Assume the mixture behave as an ideal
gas.
(a) the reactant consists of 1 mol of H2O vapor and 1 mol of CO,
T = 1100 K, P = 1 bar
From Fig 13.2, K = 1
(b) the same as (a) except that the pressure is 10 bar
Since v = 0, the increase in pressure has no effect:
(c) the same as (a) except that 2 mol of N2 is included in the
reactants
Since N2 does not take part in the reaction and serves as a
diluent:
-
(d) the reactants are 2 mol of H2O and 1 mol of CO. other
conditions are the same as (a)
The fraction of steam that reacts is then 0.667 / 2 = 0.333
(e) the reactants are 1 mol of H2O and 2 mol of CO. other
conditions are the same as (a)
The fraction of steam that reacts is then 0.667
(f) the initial mixture consists of 1 mol of H2O, 1 mol of CO
and 1 mol of CO2 . other conditions are the same as (a)
The fraction of steam that reacts is then 0.333
-
(g) same as (a) except that the temperature is 1650 K
From Fig 13.2, K = 0.316
The reaction is exothermic, and conversion decreases with
increasing temperature.
-
Estimate the maximum conversion of ethylene to ethanol by vapor
phase hydration at 250C and 35 bars for an initial
steam-to-ethylene ratio of 5.
For a temperature of 250C, K = 10.02 x 10-3
Assuming the reaction mixture is an ideal solution.
Tc /K
Pc /bar
Tri
Pri
B0
B1
i
C2H4
282.3
50.40
0.087
1.853
0.694
-0.074
0.126
0.977
H2O
647.1
220.55
0.345
0.808
0.159
-0.511
-0.281
0.887
EtOH
513.9
61.48
0.645
1.018
0.569
-0.327
-0.021
0.827
-
The equilibrium conversion is a function of temperature,
pressure, and the steam-to-ethylene ratio in the feed:
Fig 13.4
-
In a laboratory investigation, acetylene is catalytically
hydrogenated to ethylene at 1120 C and 1 bar. If the feed is an
equilmolar mixture of acetylene and hydrogen, what is the
composition of the product stream at equilibrium?
Ideal gas
-
Acetic acid is esterified in the liquid phase with ethanol at
100 C and atmospheric pressure to produce ethyl acetate and water
according to the reaction:
If initially there is 1 mol of each of acetic acid and ethanol,
estimate the mole fraction of ethyl acetate in the reacting mixture
at equilibrium.
For the reaction at standard state (298 K):
Assuming ideal solution:
-
The gas phase oxidation of SO2 to SO3 is carried out at a
pressure of 1 bar with 20% excess air in an adiabatic reactor.
Assuming that the reactants enter at 25C and that equilibrium is
attained at the exit, determine the composition and temperature of
the product stream from the reactor
For the reaction at standard state (298 K):
Assuming 1 mol of SO2 entering the reactor,
O2: 0.5 x (1.2) = 0.6 mol entering
N2: 0.6 x (79/12) = 2.257 mol entering
In the product stream:
SO2: 1 - e
O2: 0.6 - 0.5e
SO3: e
N2: 2.257
total moles: 3.857-0.5e
Energy balance:
The enthalpy change of the products at T
-
Assuming ideal gas:
Assume T
ln K
T
Converges at T = 855.7 K, e = 0.77
...
- Reactions in heterogeneous systemsa criterion of vapor/liquid
equilibrium, must be satisfied along with the equation of
chemical-reaction equilibrium
Estimate the compositions of the liquid and vapor phases when
ethylene reacts with water to form ethanol at 200C and 34.5 bar,
conditions which assure the presence of both liquid and vapor
phases. The reaction vessel is maintained at 34.5 bar by connection
to a source of ethylene at this pressure. Assume no other
reactions.
According to phase rule (see later): F = 2 (say, P, T)
the material balance equations do not enter into the solution of
this problem.
Regarding the reaction occurring in the vapor phase:
Assuming ideal gas :
-
Phase equilibrium:
For vapor phase:
For liquid phase:
Ideal solution (vapor):
-
volatile
Known values:
P, T, , , , ,
-
Assume xEtOH
xH2O
yC2H4
K
check
Converged at:
Values of , are determined experimentally
xi
yi
EtOH
0.06
0.180
H2O
0.94
0.464
C2H4
0.00
0.356
=1
=1
-
Phase rule and Duhems theorem for reacting systems
For non-reacting systems phases and N chemical species:For reacting
systemsPhase rule variables in each phase: temperature, pressure
and N-1 mole fraction. Total number of these variable: 2 + (N-1)
Phase-equilibrium equations: (-1) Nr independent chemical reactions
at equilibrium within the system: r equationsF = [2+(N-1) ] - [(-1)
N] [r] :With special constraints s:
-
Determine the number of degree of freedom F for each of the
following systems:
A system of two miscible non-reacting species which exists as an
azeotrope in vapor/liquid equilibrium.A system prepared by
partially decomposing CaCO3 into an evacuated space.A system
prepared by partially decomposing NH4Cl into an evacuated space.A
system consisting of the gases CO, CO2, H2, H2O, and CH4 in
chemical equilibriumTwo non-reacting species in two phases
With no azeotrope:
With azeotrope (x1=y1, one constraint):
Single reaction:
Three chemical species and three phases:
Single reaction:
Three chemical species and two phases:
One constraint: gas phase is equimolar
-
(d) 5 species, a single gas phase, and no special
constraints:
???
The formation reactions:
Eliminating C
Eliminating C
Eliminating O2
Eliminating O2
r = 2
-
Duhems theorem
For non-reacting system :For any closed system formed initially
from given masses of particular chemical species, the equilibrium
state is completely determined by specification of any two
independent variables. For reacting system:A variable is introduced
for each independent reaction: j A equilibrium relation can be
written for each independent reaction
-
Multireaction equilibria
A separate equilibrium constant is evaluated for each reaction:gas
phase reaction
Ideal gas
-
A feed stock of pure-n-butane is cracked at 750K and 1.2 bar to
produce olefins. Only two reactions have favorable equilibrium
conversions at these conditions:
If these reactions reach equilibrium, what is the product
composition?
With a basis of 1 mol of n-butane feed:
Assuming ideal gas:
-
A bed of coal in a coal gasifier is fed with steam and air, and
produce a gas stream containing H2, CO, O2, H2O, CO2, and N2. If
the feed to the gasifier consists of 1 mol of steam and 2.38 mol of
air, calculate the equilibrium composition of the gas stream at P =
20 bar for temperature 1000, 1100, 1200, 1300, 1400, and 1500
K.
The feed stream:
1 mol C,O2 = (0.21)(2.38) = 0.5 mol,N2 = (0.79)(2.38) = 1.88
mol
The formation reactions for the compounds are:
K = ...
Gf J/mol
T (K)
H2O
CO
CO2
1000
-192424
-200240
-395790
1100
-187000
-209110
-395960
1200
-181380
-217830
-396020
1300
-175720
-226530
-396080
1400
-170020
-235130
-396130
1500
-164310
-243740
-396160
-
Assuming ideal gases for the remaining species:
With a basis of 1 mol of C feed:
Carbon is a pure solid phase:
Three equations and three unknowns can be solved.
-
Non- stoichiometric method
Alternative method to solve chemical reaction equilibrium
problemsbase on the fact that the total Gibbs energy of the system
has it minimum valuethe basis for a general scheme of computer
solution Find the set {ni} which minimizes Gt for specified T and
P, subject to the constraints of the material balances.
-
Lagranges undetermined multiplier method
The material balance on each element k :
multiply the Lagrange multipliers
summed over k
-
F (and Gt) minimum value?
gas phase reaction
pure ideal gas
-
Calculate the equilibrium compositions at 1000 K and 1 bar of a
gas-phase system containing the species CH4, H2O, CO, CO2, and H2.
In the initial unreacted state, there are present 2 mol of CH4 and
3 mol of H2O. Values of Gf at 1000 K of each species are given.
Ideal gas:
Element k
C
O
H
Ak = no. of atomic masses of k in the system
2
3
14
species
aik = no. of atoms of k per molecule of i
CH4
1
0
4
H2O
0
1
2
CO
1
1
0
CO2
1
2
0
H2
0
0
2
-
For the 5 species:
CH4 :
...
5 equations
For the 3 elements:
C :
H :
O :
3 equations
1 equations
Solve simultaneously
...
...
4
4
3
3
2
2
1
1
+
+
+
+
A
v
A
v
A
v
A
v
e
d
v
dn
v
dn
v
dn
v
dn
=
=
=
=
...
4
4
3
3
2
2
1
1
)
...,
,
2
,
1
(
N
i
d
v
dn
i
i
=
=
e
)
...,
,
2
,
1
(
0
0
N
i
d
v
dn
i
n
n
i
i
i
=
=
e
e
)
...,
,
2
,
1
(
0
N
i
v
n
n
i
i
i
=
+
=
e
e
v
n
n
n
i
i
+
=
=
0
e
e
v
n
v
n
n
n
y
i
i
i
i
+
+
=
=
0
0
2
3
1
1
1
=
+
+
-
-
=
=
i
i
n
n
2
2
4
3
H
CO
O
H
CH
+
+
8
4
1
1
2
0
0
=
+
+
+
=
=
i
i
n
n
e
e
2
8
2
4
+
-
=
CH
y
e
e
2
8
1
+
+
=
CO
y
e
e
2
8
1
2
+
-
=
O
H
y
e
e
2
8
3
4
2
+
+
=
H
y
2
1
2
1
1
1
=
+
+
-
=
=
i
i
n
n
2
2
2
2
1
O
H
O
H
+
0
0
0
n
n
n
i
i
=
=
e
e
2
1
0
0
2
+
-
=
n
n
y
O
H
e
e
2
1
0
2
+
=
n
y
H
e
e
2
1
2
1
0
2
+
=
n
y
O
e
-
=
0
2
n
n
O
H
e
2
1
2
=
O
n
e
=
2
H
n
j
j
j
i
i
d
v
dn
e
=
,
i
j
i
j
,
n
n
j
j
j
i
i
i
v
n
n
e
+
=
,
0
j
j
i
j
i
v
n
n
e
+
=
,
0
j
j
j
v
n
n
e
+
=
0
j
j
j
j
j
j
i
i
i
v
n
v
n
y
e
e
+
+
=
0
,
0
2
2
2
4
4
2
H
CO
O
H
CH
+
+
i
CH
4
H
2
O
CO
CO
2
H
2
j
j
1
-1
-1
1
0
3
2
2
-1
-2
0
1
4
2
2
1
2
1
2
2
5
2
4
e
e
e
e
+
+
-
-
=
CH
y
2
1
2
1
2
2
5
2
3
2
e
e
e
e
+
+
-
-
=
O
H
y
2
1
1
2
2
5
e
e
e
+
+
=
CO
y
2
1
2
2
2
5
2
e
e
e
+
+
=
CO
y
2
1
2
1
2
2
5
4
3
2
e
e
e
e
+
+
+
=
H
y
(
)
0
,
=
P
T
t
dG
)
(
e
f
G
t
=
+
-
=
i
i
i
dn
dT
nS
dP
nV
nG
d
m
)
(
)
(
)
(
e
d
v
dn
i
i
=
+
-
=
i
i
i
d
dT
nS
dP
nV
nG
d
e
m
n
)
(
)
(
)
(
0
)
(
)
(
,
,
=
=
=
m
equilibriu
at
P
T
t
P
T
i
i
i
G
nG
e
e
m
n
i
i
i
f
RT
T
ln
)
(
+
G
=
m
o
i
i
o
i
f
RT
T
G
ln
)
(
+
G
=
o
i
i
o
i
i
f
f
RT
G
ln
=
-
m
0
=
i
i
i
m
n
0
ln
=
+
i
o
i
i
o
i
i
f
f
RT
G
n
0
ln
=
+
i
i
v
o
i
i
o
i
i
i
f
f
RT
G
n
RT
G
f
f
i
o
i
i
i
v
o
i
i
i
-
=
n
ln
D
-
RT
G
K
o
exp
K
f
f
i
v
o
i
i
i
=
D
i
o
i
i
o
G
G
n
-
D
i
o
i
i
o
G
G
n
-
D
i
o
i
i
o
M
M
n
dT
RT
G
d
RT
H
o
i
o
i
-
=
2
dT
RT
G
d
RT
H
i
o
i
i
i
o
i
i
-
=
n
n
2
(
)
dT
RT
G
d
RT
H
o
o
D
-
=
D
2
dT
K
d
RT
H
o
ln
2
=
D
o
i
o
i
o
i
TS
H
G
-
=
-
=
i
o
i
i
i
o
i
i
i
o
i
i
S
T
H
G
n
n
n
o
o
o
S
T
H
G
D
-
D
=
D
D
+
D
=
D
T
T
o
P
o
o
dT
R
C
R
H
H
0
0
D
+
D
=
D
T
T
o
P
o
o
T
dT
R
C
R
S
S
0
0
0
0
0
0
T
G
H
S
o
o
o
D
-
D
=
D
(
)
D
-
D
+
D
-
D
+
D
=
D
T
T
o
P
T
T
o
P
o
o
o
o
T
dT
R
C
RT
dT
R
C
R
G
H
T
T
H
G
0
0
0
0
0
0
RT
G
K
o
D
-
=
ln
2
1
0
K
K
K
K
=
D
-
0
0
0
exp
RT
G
K
o
-
D
T
T
RT
H
K
o
0
0
0
1
1
exp
D
+
D
-
T
T
o
P
T
T
o
P
T
dT
R
C
dT
R
C
T
K
0
0
1
exp
2
OH
H
C
O
H
H
C
5
2
2
4
2
+
(
)
(
)
(
)
O
H
H
C
OH
H
C
2
4
2
5
2
-
-
=
D
mol
J
H
H
o
o
45792
298
0
-
=
D
=
D
mol
J
G
G
o
o
8378
298
0
-
=
D
=
D
o
o
i
P
f
=
K
P
f
i
v
o
i
i
=
P
y
f
i
i
i
f
=
(
)
K
P
P
y
o
i
v
i
i
i
n
f
-
=
i
i
n
n
i
i
f
f
=
(
)
K
P
P
y
o
i
v
i
i
i
n
f
-
=
1
=
i
f
(
)
K
P
P
y
o
i
v
i
i
n
-
=
i
i
i
i
f
x
f
g
=
K
f
f
x
i
v
o
i
i
i
i
i
=
g
o
i
i
P
P
i
o
i
i
f
f
RT
dP
V
G
G
o
ln
=
=
-
(
)
RT
P
P
V
f
f
o
i
o
i
i
-
=
ln
(
)
(
)
(
)
-
=
i
i
i
o
i
v
i
i
V
RT
P
P
K
x
i
n
g
exp
(
)
K
x
i
v
i
i
i
=
g
(
)
K
x
i
v
i
i
=
2
2
2
H
CO
O
H
CO
+
+
0
=
i
i
n
n
(
)
1
2
2
2
=
=
O
H
CO
CO
H
i
v
i
y
y
y
y
y
i
2
1
e
CO
y
e
-
=
2
2
e
CO
y
e
=
2
1
2
e
O
H
y
e
-
=
2
2
e
H
y
e
=
5
.
0
=
e
e
3
1
e
CO
y
e
-
=
3
2
e
CO
y
e
=
3
2
2
e
O
H
y
e
-
=
3
2
e
H
y
e
=
667
.
0
=
e
e
3
2
e
CO
y
e
-
=
3
1
2
e
O
H
y
e
-
=
3
1
2
e
CO
y
e
+
=
333
.
0
=
e
e
316
.
0
2
2
2
=
O
H
CO
CO
H
y
y
y
y
36
.
0
=
e
e
1
-
=
i
i
n
n
)
10
02
.
10
(
3
2
2
4
2
4
2
-
=
o
O
H
O
H
H
C
H
C
EtOH
EtOH
P
P
y
y
y
f
f
f
T
c
/K
P
c
/bar
T
ri
P
ri
B
0
B
1
i
C
2
H
4
282.3
50.40
0.087
1.853
0.694
-0.074
0.126
0.977
H
2
O
647.1
220.55
0.345
0.808
0.159
-0.511
-0.281
0.887
EtOH
513.9
61.48
0.645
1.018
0.569
-0.327
-0.021
0.827
)
10
02
.
10
(
1
35
)
887
.
0
(
)
977
.
0
(
)
827
.
0
(
3
2
4
2
2
2
4
2
4
2
-
=
=
O
H
H
C
EtOH
O
H
O
H
H
C
H
C
EtOH
EtOH
y
y
y
y
y
y
f
f
f
e
e
H
C
y
e
e
-
-
=
6
1
4
2
e
e
O
H
y
e
e
-
-
=
6
5
2
e
e
EtOH
y
e
e
-
=
6
233
.
0
=
e
e
2
2
2
2
H
C
H
C
+
4
2
2
2
2
H
C
H
C
+
4
2
2
2
2
H
C
H
H
C
+
o
II
o
I
o
G
G
G
D
+
D
=
D
II
I
K
RT
K
RT
K
RT
ln
ln
ln
-
-
=
-
(
)
(
)
1
10
5
.
2
10
4
6
5
=
=
=
-
II
I
K
K
K
1
2
2
2
4
2
=
H
C
H
H
C
y
y
y
e
e
H
y
e
e
-
-
=
2
1
2
e
e
H
C
y
e
e
-
-
=
2
1
2
2
e
e
H
C
y
e
e
-
=
2
4
2
293
.
0
=
e
e
414
.
0
2
2
2
=
=
H
C
H
y
y
172
.
0
4
2
=
H
C
y
O
H
H
COOC
CH
OH
H
C
COOH
CH
2
5
2
3
5
2
3
+
+
2
1
e
EtOH
AcH
x
x
e
-
=
=
e
e
v
n
v
n
n
n
x
i
i
i
i
+
+
=
=
0
0
2
2
e
O
H
EtAc
x
x
e
=
=
6879
.
0
=
e
e
344
.
0
2
/
6879
.
0
=
=
EtAc
x
J
G
J
H
o
o
4650
3640
298
298
-
=
D
-
=
D
8759
.
1
15
.
298
314
.
8
4650
ln
298
=
=
D
-
=
RT
G
K
o
8586
.
4
373
=
K
K
x
x
x
x
EtOH
AcH
O
H
EtAc
=
2
3
2
2
2
1
SO
O
SO
+
J
G
J
H
o
o
70866
98890
298
298
-
=
D
-
=
D
0
298
=
D
=
D
+
D
H
H
H
o
P
e
o
e
)
15
.
298
(
)
15
.
298
(
-
=
-
=
D
T
C
n
T
C
H
i
H
o
Pi
i
H
o
P
o
P
15
.
298
298
+
D
-
=
H
o
P
e
o
C
H
T
e
5
.
0
=
i
i
n
n
K
e
e
e
e
=
-
-
-
5
.
0
5
.
0
6
.
0
5
.
0
857
.
3
1
e
e
e
e
(
)
IDCPH
T
IDCPS
T
K
1
4
.
11894
3054
.
11
ln
-
+
+
-
=
e
e
0062
.
0
5
.
0
857
.
3
1
2
=
-
-
=
e
e
SO
y
e
e
)
(
5
2
)
(
2
)
(
2
2
g
g
g
OH
H
C
O
H
H
C
+
o
O
H
H
C
EtOH
P
f
f
f
K
2
4
2
=
bar
P
o
1
=
K
T
15
.
473
=
031
.
0
15
.
473
=
K
K
l
i
v
i
f
f
=
o
l
l
l
o
v
v
v
P
f
f
f
P
f
f
f
K
O
H
H
C
EtOH
O
H
H
C
EtOH
2
4
2
2
4
2
=
=
P
y
f
i
i
v
i
f
=
l
i
i
i
l
i
f
x
f
g
=
(
)
(
)
o
l
O
H
O
H
O
H
H
C
H
C
l
EtOH
EtOH
EtOH
P
f
x
P
y
f
x
K
2
2
2
4
2
4
2
g
f
g
=
sat
i
sat
i
sat
i
l
i
P
f
f
f
=
~
i
i
f
f
=
(
)
(
)
o
sat
O
H
sat
O
H
O
H
O
H
H
C
H
C
sat
EtOH
sat
EtOH
EtOH
EtOH
P
P
x
P
y
P
x
K
2
2
2
2
4
2
4
2
f
g
f
f
g
=
P
P
x
y
i
sat
i
sat
i
i
i
i
f
f
g
=
P
P
x
P
P
x
y
O
H
sat
O
H
sat
O
H
O
H
O
H
EtOH
sat
EtOH
sat
EtOH
EtOH
EtOH
H
C
2
2
2
2
2
4
2
1
f
f
g
f
f
g
-
-
=
O
H
EtOH
H
C
y
y
y
2
4
2
1
-
-
=
(
)
(
)
o
sat
O
H
sat
O
H
O
H
O
H
H
C
H
C
sat
EtOH
sat
EtOH
EtOH
EtOH
P
P
x
P
y
P
x
K
2
2
2
2
4
2
4
2
f
g
f
f
g
=
4
2
2
1
H
C
EtOH
O
H
x
x
x
-
-
=
(
)
(
)
O
H
O
H
H
C
EtOH
EtOH
x
y
x
K
2
2
4
2
0493
.
0
g
g
=
O
H
O
H
EtOH
EtOH
H
C
x
x
y
2
2
4
2
493
.
0
907
.
0
1
g
g
-
-
=
sat
O
H
2
f
sat
EtOH
f
4
2
H
C
f
sat
EtOH
P
sat
O
H
P
2
EtOH
O
H
x
x
-
=
1
2
031
.
0
15
.
473
=
K
K
x
i
y
i
EtOH
0.06
0.180
H
2
O
0.94
0.464
C
2
H
4
0.00
0.356
=1
=1
O
H
EtOH
2
g
g
N
F
+
-
=
p
2
r
N
F
-
+
-
=
p
2
s
r
N
F
-
-
+
-
=
p
2
2
0
2
2
2
2
=
-
+
-
=
-
+
-
=
r
N
F
p
1
2
=
-
-
+
-
=
s
r
N
F
p
)
(
2
)
(
)
(
3
g
s
s
CO
CaO
CaCO
+
)
(
2
)
(
)
(
3
g
s
s
CO
CaO
CaCO
1
1
3
3
2
2
=
-
+
-
=
-
+
-
=
r
N
F
p
)
(
)
(
3
)
(
4
g
g
s
HCl
NH
Cl
NH
+
1
1
1
3
2
2
2
=
-
-
+
-
=
-
-
+
-
=
s
r
N
F
p
)
(
)
(
3
)
(
4
g
g
s
HCl
NH
Cl
NH
+
s
r
N
F
-
-
+
-
=
p
2
CO
O
C
+
2
2
1
2
2
CO
O
C
+
O
H
O
H
2
2
2
2
1
+
4
2
2
CH
H
C
+
2
2
2
1
CO
O
CO
+
2
2
2
4
2
CO
H
O
CH
+
+
2
2
2
4
2
CO
H
O
CH
+
+
O
H
CO
H
CO
2
2
2
+
+
2
2
2
4
4
2
CO
H
O
H
CH
+
+
4
0
2
5
1
2
2
=
-
-
+
-
=
-
-
+
-
=
s
r
N
F
p
2
]
)
1
[(
]
)
1
(
2
[
=
+
-
-
+
-
+
=
N
N
N
F
p
p
p
2
1
1
]
)
1
[(
]
)
1
(
2
[
=
-
+
+
-
-
+
-
+
=
N
N
N
F
p
p
p
j
i
v
o
i
i
K
f
f
j
i
=
,
j
i
v
o
i
K
P
f
j
i
=
,
(
)
j
o
i
v
i
K
P
P
y
j
j
i
n
-
=
,
6
2
4
2
10
4
H
C
H
C
H
C
+
4
6
3
10
4
CH
H
C
H
C
+
II
I
II
I
H
C
y
e
e
e
e
+
+
-
-
=
1
1
10
4
II
I
I
H
C
H
C
y
y
e
e
e
+
+
=
=
1
6
2
4
2
II
I
II
CH
H
C
y
y
e
e
e
+
+
=
=
1
4
6
3
(
)
j
o
i
v
i
K
P
P
y
j
j
i
n
-
=
,
I
o
H
C
H
C
H
C
K
P
P
y
y
y
1
10
4
6
2
4
2
-
=
II
o
H
C
CH
H
C
K
P
P
y
y
y
1
10
4
4
6
3
-
=
1068
.
0
=
I
e
8914
.
0
=
II
e
001
.
0
10
4
=
H
C
y
0534
.
0
6
2
4
2
=
=
H
C
H
C
y
y
4461
.
0
4
6
3
=
=
CH
H
C
y
y
O
H
O
H
2
2
2
2
1
+
CO
O
C
+
2
2
1
2
2
CO
O
C
+
G
f
J/mol
T (K)
H
2
O
CO
CO
2
1000
-192424
-200240
-395790
1100
-187000
-209110
-395960
1200
-181380
-217830
-396020
1300
-175720
-226530
-396080
1400
-170020
-235130
-396130
1500
-164310
-243740
-396160
2
1
2
2
2
-
=
o
H
O
O
H
I
P
P
y
y
y
K
2
1
2
=
o
O
CO
II
P
P
y
y
K
2
2
O
CO
III
y
y
K
=
(
)
2
38
.
3
2
I
II
I
H
y
e
e
e
-
+
-
=
(
)
2
38
.
3
I
II
II
CO
y
e
e
e
-
+
=
(
)
(
)
2
38
.
3
1
2
1
2
I
II
III
II
I
O
y
e
e
e
e
e
-
+
-
-
-
=
(
)
2
38
.
3
1
2
I
II
I
O
H
y
e
e
e
-
+
+
=
(
)
2
38
.
3
2
I
II
III
CO
y
e
e
e
-
+
=
(
)
2
38
.
3
88
.
1
2
I
II
N
y
e
e
-
+
=
1
)
1
(@
)
20
(@
=
bar
f
bar
f
f
f
o
C
C
o
C
C
(
)
(
)
N
P
T
t
n
n
n
n
g
G
...
,
,
,
3
2
1
,
=
k
i
ik
i
A
a
n
=
0
=
-
k
i
ik
i
k
A
a
n
l
0
=
-
k
k
i
ik
i
k
A
a
n
l
-
+
=
k
k
i
ik
i
k
t
A
a
n
G
F
l
)
,...,
2
,
1
(
0
,
,
,
,
N
i
a
n
G
n
F
k
ik
k
n
P
T
i
t
n
P
T
i
j
j
=
=
+
=
l
)
,...,
2
,
1
(
0
N
i
a
k
ik
k
i
=
=
+
l
m
+
=
o
i
i
o
i
i
f
f
RT
G
ln
m
+
=
o
i
o
i
i
P
f
RT
G
ln
m
+
D
=
o
i
i
o
fi
i
P
P
y
RT
G
f
m
ln
)
,...,
2
,
1
(
0
ln
N
i
a
P
P
y
RT
G
k
ik
k
o
i
i
o
fi
=
=
+
+
D
l
f
Element
k
C
O
H
A
k
= no. of atomic masses of
k
in the system
2
3
14
species
a
ik
= no. of atoms of
k
per molecule of
i
CH
4
1
0
4
H
2
O
0
1
2
CO
1
1
0
CO
2
1
2
0
H
2
0
0
2
1
=
i
f
1
=
o
P
P
)
,...,
2
,
1
(
0
ln
N
i
a
RT
n
n
RT
G
k
ik
k
i
i
i
o
fi
=
=
+
+
D
l
0
4
ln
19720
4
=
+
+
+
RT
RT
n
n
RT
H
C
i
i
CH
l
l
2
2
4
=
+
+
CO
CO
CH
n
n
n
14
2
2
4
2
2
4
=
+
+
H
O
H
CH
n
n
n
3
2
2
2
=
+
+
CO
CO
O
H
n
n
n
2
2
2
4
H
CO
CO
O
H
CH
i
i
n
n
n
n
n
n
+
+
+
+
=
