CHAPTER 6- CHEMICAL REACTIAN EQUILIBRIA.PPT

Thermodynamics
Chapter 10
Chemical-reaction
Equilibria

In a chemical reaction, both the rate and the
equilibrium conversion are considered for
commercial purposes. Equilibrium conversion
represent the maximum possible conversion
regardless the reaction rate or catalyst.
This chapter focus on the effect of
temperature, pressure, and the initial
composition on the equilibrium conversions of
chemical reactions.

Chemical-reaction
Equilibria
10.1 The reaction
Coordinate
10.2 Application of
Equilibrium Criteria to
Chemical Reactions
10.3 The Standard
Gibbs-Energy change and
The Equilibrium Constant
10.4 Effect of Temperature
on The Equilibrium
Constant
10.5 Evaluation of
Equilibrium Constants
10.6 Relation of Equilibrium
Constant to Composition
10.7 Equilibrium Conversions
for Single Reactions
10.8 Phase Rule and Duhem’s
Theorem For Reacting System
10.9 Multireaction Equilibria

The general chemical reaction:
...
... 4
4
3
3
2
2
1
1 



 A
v
A
v
A
v
A
v
where |vi| is a stoichiometric coefficient
and Ai stands for a chemical formula.
For vi :
Positive ( + ) for product
Negative ( - ) for reactants
10.1 The Reaction Coordinate

For example,
CH4 + H2O → CO + 3H2
The stoichiometric numbers are :
1
4


CH
v 1
2


O
H
v 1

CO
v 3
2

H
v
The stoichiometric number for an inert
species is zero.
Since:

d
v
dn
v
dn
v
dn
v
dn




 ...
4
4
3
3
2
2
1
1

The differential change dni in the number of
moles of a reacting species and dε is :
dni = vi dε ( i = 1 , 2 ,….N)
This variable ε is called the reaction
coordinate, characterizes the extent or
degree to which a reaction has taken place.


 


0
0
d
v
dn i
n
n
i
i
i
gives

i
i
i v
n
n 
 0
( i = 1 , 2 ,….N)
Summation over all species yields :


 


i
i
i
i
i
i v
n
n
n 
0
or n = n0 + vε
where 

i
i
n
n 

i
i
n
n 0
0 

i
i
v
v

Thus, the mole fractions yi of the species
present are related to ε by :


v
n
v
n
n
n
y
i
i
i
i




0
0

Example 13.2
Consider a vessel which initially contains
only n0 mol of water vapor. If decomposition
occurs according to the reaction,
2
2
2
2
1
O
H
O
H 

find expressions which relate the number of
moles and the mole fraction of each chemical
species to the reaction coordinate.

Solution 13.2
For the given reaction
2
1
2
1
1
1 




v
Application of Eqs.(13.4) and (13.5) yields:


 0
2
n
n O
H


2
1
0
0
2



n
n
y O
H


2
H
n


2
1
0
2


n
yH

2
1
2

O
n


2
1
2
1
0
2


n
yO

The fractional decomposition of water vapor
is:
 
0
0
0
0
0
0 2
n
n
n
n
n
n
n O
H 






Thus when no = 1, ε is identified with the
fractional decomposition of the water vapor.



j
j
j
i
i d
v
dn 
,
After integration:
j
j
j
i
i
i v
n
n 


 ,
0
Summing over all species yields :
j
j i
j
i
i
j
j
j
i
i
i v
n
v
n
n 
  

 








 ,
0
,
0
For the mole fraction of
the species presents in
particular reaction: 




j
j
j
j
j
j
i
i
i
v
n
v
n
y


0
,
0
( i = 1 , 2 ,….N)

At the equilibrium state :
(dGt)T , P = 0
The total Gibbs energy, Gt is a minimum.
Its differential is zero.
6.2 Application of Equilibrium Criteria To
Chemical Reactions

CHAPTER 6- CHEMICAL REACTIAN EQUILIBRIA.PPT

The fundamental property relation for
single-phase systems, provides an
expression for the total differential of the
Gibbs energy:
      



i
i
i dn
dT
S
dP
nV
nG
d 
6.3 The Standard Gibbs-Energy Change
and The Equilibrium Constant

Since nG is a state function , the right side
of this an exact differential expression :
   
P
T
t
P
T
i
i
i
G
nG
v
,
,


















 


If changes in mole numbers ni occur as the
result of a single chemical reaction in a
closed system , then each dni may be
replaced by the product vi dε.
      



i
i
i d
v
dT
S
dP
nV
nG
d 


For chemical-reaction equilibrium :
0


i
i
i
v 
Recall the definition of the fugacity of a
species in solution :
  i
i
i f
RT
T ˆ
ln




For a pure species i in its standard start
at the same temperature :
  

i
i
i f
RT
T
G ln



The difference between these equations:


i
i
i
i
f
f
RT
G
ˆ
ln




For the equilibrium state of a chemical reaction:
 
  0
ˆ
ln 


i
i
i
i
i
f
f
RT
G
v 

  0
ˆ
ln 
 
 i
v
i
i
i
i
i
i
f
f
RT
G
v 

  RT
G
v
f
f i i
i
i
v
i
i
i 





ˆ
ln
 
 
i
v
i
i K
f
f
i

ˆ
ln
where 






 


RT
G
K

exp

Alternative expression for K :
RT
G
K




ln
where



i
i
iG
G 


K is dependence on temperature.
In spite of its dependence on temperature,
K is called the equilibrium constant for the
reaction.
Called as standard Gibbs-energy change
of reaction.

The dependence of ΔG˚ on T :
 
2
RT
H
dT
RT
G
d 





Then become:
2
ln
RT
H
dT
K
d 


Equation above gives the effect of
temperature on equilibrium constant and
hence on the equilibrium conversion.
6.4 Effect of Temperature on the
Equilibrium Constant

When ΔH˚
negative = exothermic reaction
and the equilibrium constant, K
decreases as the temp. increases.
positive = endothermic reaction
and the equilibrium constant, K
increases as the temp. increases.











 T
T
R
H
K
K 1
1
ln

If ΔH˚ is assumed independent of T,
integration from a reference T’ (at 298K)
to random T, then:

In Figure 13.2, a plot
of ln K vs 1/T for a
number of common
reactions, illustrates
this near linearity.

The effect of temperature on the equilibrium
constant is based on the definition of the
standard Gibbs energy:
G˚i = H˚i -TSi˚
Multiplication by vi and summation over all
species gives:


 

i
i
i
i
i
i
i
i
i S
v
T
H
v
G
v 


Hence, the standard property change of
reaction;
ΔG˚=ΔH˚ - TΔS˚
where the standard heat of reaction and
standard entropy change is related to
temperature:
dT
R
C
R
H
H
T
T
P






0
0









T
T
P
T
dT
R
C
R
S
S
0
0




However
0
0
0
0
T
G
H
S


 




whence












 T
T
P
T
T
P
T
dT
R
C
dT
R
C
T
RT
H
RT
H
G
RT
G
0
0
1
0
0
0






and recall,
RT
G
K




ln

As an alternative, the preceding equation
may be reorganized so as to factor K into
three terms , each representing a basic
contribution to its value :
K = K0 K1 K2
K0 represents the equilibrium constant at
reference temperature T0 :







 


0
0
0
RT
G
K
















T
T
RT
H
K 0
0
0
1 1
exp

      





 




















 


 2
2
2
0
2
2
0
2
0
2
1
2
1
2
1
6
1
1
2
1
1
ln
exp










T
D
CT
BT
A
K

Example 13.4
Calculate the equilibrium constant for the
vapor-phase hydration of ethylene at 145
and at 320°C from data given in App. C.
Solution 13.4
First determine values for ΔA, ΔB, ΔC, and
ΔD for the reaction:
C2H4 (g) + H2O (g) → C2H5OH (g)
6.5 Evaluation of Equilibrium Constant

The meaning of Δ is indicated by: A = (C2H5
OH) - (C2H4) - (H2O). Thus, from the heat-
capacity data of Table C.1:
ΔA = 3.518 - 1.424 - 3.470 = - 1.376
ΔB = (20.001 - 14.394 - 1.450) x 10-3
= 4.157 x l0-3
ΔC = (-6.002 + 4.392 - 0.000) x 10-6
= -1.610 x 10-6
ΔD = (-0.000 - 0.000 - 0.121) x 105
= -0.121 x 105

Values of ΔH°298 and ΔG°298 at 298.15K for the
hydration reaction are found from the heat-of-
formation and Gibbs-energy-of-formation data
of Table C.4:
ΔH°298 = -235,100 - 52,510 - (-241,818)
= -45,792 J mol-1
ΔG°298 = -168,490 - 68,460 - (-228,572)
= -8,378 J mol-1
For T = 145 + 273.15 = 418.15 K, values
of the integrals in Eq. (13.18) are:
121
.
23
0




T
T
P
dT
R
C
06924
.
0
0




T
T
P
T
dT
R
C

Substitution of values into Eq. (13.18) for a
reference temperature of 298.15 gives:
     
9356
.
1
06924
.
0
15
.
418
121
.
23
15
.
418
314
.
8
792
,
45
15
.
298
314
.
8
792
,
45
378
,
8
418










RT
G
     
8286
.
5
01731
.
0
15
.
593
632
.
22
15
.
593
314
.
8
792
,
45
15
.
298
314
.
8
792
,
45
378
,
8
593









RT
G
For T = 320 + 273.15 = 593.15 K, values
of the integrals in Eq. (13.18) are:
632
.
22
0



T
T
P
dT
R
C
01731
.
0
0



T
T
P
T
dT
R
C
Substitution of values into Eq. (13.18) for a
reference temperature of 298.15 gives:

At 418.15K,
: ln K = ln -1.9356 and K = 1.443 x 10-1
At 593.15K,
: ln K = ln -5.8286 and K = 2.942x 10-3
Application of Eqs. (13.21), (13.22), and (13.24)
provides an alternative solution to this example.
By Eq. (13.21),
  
366
.
29
15
.
298
314
.
8
378
,
8
exp
0 

K
  
473
.
18
15
.
298
314
.
8
792
,
45
0
0





RT
H 

The following results are obtained:
T/ K  K0 K1 K2 K
298.15 1 29.366 1 1 29.366
418.15 1.4025 29.366 4.985x10-3 0.9860 1.443x10-1
593.15 1.9894 29.366 1.023x10-4 0.9794 2.942x10-3
Clearly, the influence of K1, is far greater
than that of K2. This is a typical result, and
accounts for the fact that the lines on Fig.
13.2 are nearly linear.

The standard state for gas is the ideal
gas-state of the pure gas at the standard-
state pressure P˚ of 1 bar.
Since for ideal gas : f˚i = P˚
Thus :


P
f
f
f i
i
i
ˆ
ˆ
 and K
P
f
i
v
i
i









 
ˆ
where the constant K is a function of temp..
6.6 Relation of Equilibrium Constant
to Composition and Pressure

For a fixed temperature the composition at
equilibrium must change with pressure in
such a way that   K
P
f
i
v
i
i

 
ˆ
Hence an equilibrium expression displaying
pressure and composition;
  K
P
P
y
v
i
v
i
i
i








 

ˆ
P˚ is the standard –state pressure of 1 bar.
constant .

 i i
v
v
reactants
products
i
i
y
y
reactants
products
i
i



For pressure sufficiently low or temperature
sufficiently high, the equilibrium behaves
essentially as an ideal gas :
For equilibrium mixture assumed to be
in ideal solution, each is assumed to be
the fugacity of pure species i :
  K
P
P
y
v
i
v
i
i
i








 

i

ˆ
  K
P
P
y
v
i
v
i
i








 
1

i

i


 i i
v
v
reactants
products
i
i
y
y
reactants
products
i
i



Example 13.5
The water-gas-shift reaction,
CO(g) + H2O(g) → CO2(g) + H2(g)
is carried out under the different sets of
conditions described below. Calculate the
fraction of steam reacted in each case.
Assume the mixture behaves as an ideal gas.
6.7 Equilibrium Conversions for Single
Reactions

Calculate if the conditions:
(a)The reactants consist of 1 mol of H2O
vapor and 1 mol of CO. The temperature
is 1,100 K and the pressure is 1 bar.
(b) Same as (a) except that the pressure
is 10 bar.
(c) Same as (a) except that 2 mol of N2 is
included in the reactants.

(d)The reactants are 2 mol of H2O and 1 mol of
CO. Other conditions are the same as in (a).
(e) The reactants are 1 mol of H2O and 2 mol of
CO. Other conditions are the same as in (a).
(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 in (a)
(g) Same as (a) except that the temperature
is 1,650 K.

Solution 13.5
(a) For the given reaction at 1,100 K,
104/T = 9.05, and Fig. 13.2 provides the
value. ln K0 or K = 1. For this reaction
0
1
1
1
1 




 
i
i
v
v . Since the reaction mixture
)
(
1
2
2
2
A
K
y
y
y
y
O
H
CO
CO
H


By Eq. (13.5):
2
1 e
CO
y



2
1
2
e
O
H
y



2
2
e
CO
y


2
2
e
H
y


is an ideal gas, Eq. (13.28) applies, and
here becomes:
K
P
P
y
y
y
y v
v
O
H
CO
CO
H
i 
















2
2
2

Substitution of these values into Eq. (A) gives:
 
1
1
2
2

 e
e


Therefore the fraction of the steam that
reacts is 0.5.
(b) Since v = 0 , the increase in pressure
has no effect on the ideal-gas reaction,
and εe is still 0.5.
or εe = 0.5

(c) The N2 does not take part in the reaction,
and serves only as a diluent. It does increase
the initial number of moles no from 2 to 4, and
the mole fractions are all reduced by a factor
of 2. However, Eq. (A) is unchanged and
reduces to the same expression as before.
Therefore, εe is again 0.5.

(d)In this case, the mole fractions at
equilibrium are:
3
1 e
CO
y



3
2
2
e
O
H
y



3
2
e
CO
y


3
2
e
H
y


and Eq. (A) becomes:
  
1
2
1
2


 e
e
e



The fraction of steam that reacts is
then 0.667/2 = 0.333
or εe = 0.667

(e)Here the expressions for yCO and yH2O are
interchanged , but this leaves the equilibrium
equation the same as in (d). Therefore εe =
0.667, and the fraction of steam that reacts is
0.667.
(f) In this case Eq. (A) becomes:
 
 
1
1
1
2



e
e
e



The fraction of steam reacted is 0.333.
or εe = 0.333

(g) At 1,650 K, 104/T = 6.06, and from Fig.
13.2, in K = -1.15
or K = 0.316.
Therefore Eq. (A) becomes:
 
316
.
0
1
2

 e
e


The reaction is exothermic, and conversion
decreases with increasing temperature.
or εe = 0.36
(Must try: Examples 13.6, 13.7,13.8)

This is phase rule for reacting systems.
F = 2 – π + N –r
where π is number of phases , N number of
chemical species and r is number of
independent chemical reactions at
equilibrium within the system .
6.8 Phase Rule and Duhem’s Theorem
for Reacting System

j
i
v
i
i
K
f
f
j
i










,
ˆ

where j is the reaction index.
For gas phase reaction :
j
i
v
i
K
P
f
j
i










,
ˆ

For the equilibrium mixture is an ideal-gas,
  j
v
i
v
i K
P
P
y
j
j
i








 
,
6.9 Multireaction Equilibria
(Must try: Examples 13.12, 13.13)

CHAPTER 6- CHEMICAL REACTIAN EQUILIBRIA.PPT

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CHAPTER 6- CHEMICAL REACTIAN EQUILIBRIA.PPT