Chemical-Reaction Equilibra ERT 206: Thermodynamics Miss Anis Atikah Ahmad Tel: 04-9763245 Email: anis [email protected]
Jan 17, 2016
Chemical-Reaction Equilibra
ERT 206: ThermodynamicsMiss Anis Atikah Ahmad
Tel: 04-9763245 Email: anis [email protected]
OUTLINE
1. The Reaction Coordinate2. Application of Equilibrium Criteria to Chemical
Reactions3. The Standard Gibbs-Energy Change & the
Equilibrium Constant4. Effect of Temperature on the Equilibrium Constant5. Relation of Equilibrium Constants to Composition6. Multireaction Equilibria
1. The Reaction Coordinate
• The general chemical reaction is written as:
is the stoichiometric coefficient * positive (+) for product
* negative (-) for reactant stands for chemical formula
..... 44332211 AvAvAvAv
iv
iA
1. The Reaction Coordinate• For reaction:
the stoichiometric numbers are:
If 0.5 mol of CH4 and 0.5 mol of H2O dissappear in reaction;
Thus;0.5 mol of CO and 1.5 mol of H2 are formed.
224 3HCOOHCH
14
CHv 12
OHv 1COv 32Hv
dv
dn
v
dn
v
dn
v
dn ...
4
4
3
3
2
2
1
1Reaction coordinate (the extent
or degree to which a reaction has taken place)
The general relation connecting the differential change dni with dε:
Integrating;
Summation over all species;
where
1. The Reaction Coordinate
dvdn ii
0
0
dvdn i
n
n
i
i
i
iii vnn 0
vnn 0 i
inn i
inn00
iivv
The mole fraction yi of the species:
1. The Reaction Coordinate
vn
vn
n
ny iiii
0
0
Example 1
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 a function of ε
1. The Reaction Coordinate
224 3HCOOHCH
Recall:Example 1-SOLUTION
1. The Reaction Coordinate
vn
vn
n
ny iiii
0
0
224 3HCOOHCH
23111 i
ivv
8411200
iinn
28
24
CHy
28
1
COy28
12
OHy
28
342
Hy
14
CHv 12
OHv 1COv 32Hv
204CHn 1
02OHn 1
0COn 4
02Hn
Multireaction Stoichiometry:
When two or more independent reactions process simultaneously,
Where j serves as the reaction indexanda separate reaction coordinate εj applies to each reaction.
1. The Reaction Coordinate
j jj
j jjii
i vn
vny
0
,0
Example 2: Multireaction
Consider a system in which the following reaction occur:
If there are present initially 2 mol of CH4, and 3 mol H2O,
determine expressions for the mole fractions yi as a function of
ε1 and ε2.
1. The Reaction Coordinate
13 224 HCOOHCH
242 2224 HCOOHCH
Recall:
j jj
j jjii
i vn
vny
0
,0
Example 2: Multireaction- Solution
1. The Reaction Coordinate
13 224 HCOOHCH 242 2224 HCOOHCH
i CH4 H2O CO CO2 H2
j vj
1 -1 -1 1 0 3 2
2 -1 -2 0 1 4 2
j 204CHn 3
02OHn
Given:
21
21
225
24
CHy21
1
225
COy
21
21
225
232
OHy21
21
225
432
Hy
231111,1 i
ivv
53200
iinn
Recall:
j jj
j jjii
i vn
vny
0
,0
241212,2 i
ivv
21
2
2252
COy
• Criterion of equilibrium at constant T & P:– The total Gibbs
energy Gt is a minimum
– Its differential is zero
2. Application of Equilibrium Criteria to Chemical Reactions
Total Gibbs Energy vs Reaction Coordinate
0, PTtnGd
Any reaction that occurs at constant T & P must lead to decrease in the total Gibbs
energy of the system
From previous chapter,
Since ,
At constant T and P,
At equilibrium state,
Thus,
3. The Standard Gibbs-Energy Change & The Equilibrium Constant
ii
idndTnSdPnVnGd
dvdn ii
dvdTnSdPnVnGdi
ii
PT
t
PTiii
GnGv
,,
0
,
PT
tG
0i
iiv
Recall fugacity of a species in solution;
The fugacity of pure species in its standard state;
The difference between these two equations;
3. The Standard Gibbs-Energy Change & The Equilibrium Constant
iii fRTT ˆln
iii fRTTG ln
(1)
(2)
i
iii f
fRTG
ˆln
i
iii f
fRTG
ˆln (3)
Substituting into ;
OR
OR
In exponential form;
where ,
3. The Standard Gibbs-Energy Change & The Equilibrium Constant
i
iii f
fRTG
ˆln 0
iiiv
0ˆln
iiiii ffRTGv
0ˆln
i
v
iiii
i
ffRTGv
RT
Gvff i
ii
i
v
ii
i
ˆln
Kffi
v
ii
i ˆ
RT
GK exp
iiiGvG
Equilibrium constant
Standard Gibbs energy change of
reaction
Depends only on T
4. Effect of Temperature on the Equilibrium Constant
The relation between standard heat of reaction and the standardGibbs energy change of reaction;
Integrating;
dT
RTGdRTH
2
2RT
H
dT
RTGd
2
ln
RT
H
dT
Kd
RT
GK exp
TTR
H
K
K 11ln
For exothermic (-ΔH°): As T increases, K decreases.For endothermic (+ΔH°): As T increases, K increases.
Equilibrium constant as a function of temperature
5. Relation of Equilibrium Constants to Composition
• For Gas Phase Reaction:– The standard state for a gas is the ideal gas state of the
pure gas at the standard-state pressure P° of 1 bar.
– Since the fugacity of an ideal gas is equal to its pressure,
, thus becomes: Kff
i
v
ii
i ˆ
KPfi
v
i
i ˆ
Pfi
5. Relation of Equilibrium Constants to Composition
• Since , becomes;
where
• For pressures sufficiently low or temperature sufficiently high, the equilibrium mixture behaves essentially as an ideal gas. Thus,
KP
Py
v
i
v
ii
i
Pyf iii ˆ
KP
Py
v
i
vi
i
1ˆ i
KPfi
v
i
i ˆ
i
ivv
5. Relation of Equilibrium Constants to Composition
• For liquid phase reactions:
From previous chapter, . Thus,
Kffi
v
ii
i ˆ
iiii fxf ˆ
i
iii
i
iii
i
i
f
fx
f
fx
f
f ˆ
Can be substituted by Gibbs energy
expression
5. Relation of Equilibrium Constants to Composition
• For liquid phase reactions:
Integration of at constant T:
Thus, OR
or
i
iii f
fRTGG ln
PPVdPVGG i
P
P
iii
RT
PPV
f
f i
i
i
ln
iii fRTTG ln
iii fRTTG ln
i
iii f
fRTGG ln
SdTVdPdG
PPVf
fRT i
i
iln
RT
PPV
f
f i
i
i exp
5. Relation of Equilibrium Constants to Composition
Substituting into
Since
Thus,
iii
v
iii Vv
RT
PPKx i exp
i
iii
i
i
f
fx
f
f ˆ
RT
PPVx
f
f iii
i
i
exp
ˆ
RT
PPV
f
f i
i
i exp
Kffi
v
ii
i ˆ
K
RT
PPVx
i
v
iii
i
exp
5. Relation of Equilibrium Constants to Composition
For low to moderate pressure, the exponential term is close to unity and may be omitted. Thus
If the mixture is an ideal solution, is unity. Therefore,
Kx iv
iii
i
Kx iv
ii Law of mass action
5. Relation of Equilibrium Constants to Composition
Example 3The 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 behaves as an ideal gas.
(a) The reactants consist of 1 mol of H2O vapor & 1 mol of CO. The temperature is 1,100 K and the pressure is 1 bar.
(b) Same as (a) except 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 & 1 mol of CO. Other conditions are the same as in part (a)
(e) Same as (a) except that the temperature is 1,650 K.
gHgCOgOHgCO 222
5. Relation of Equilibrium Constants to Composition
Example 3-Solution(a) The reactants consist of 1 mol of H2O vapor & 1 mol of CO. The temperature is
1,100 K and the pressure is 1 bar.
At T=1100 K, 104/T=9.05.From the ln K vs 1/T graph,At 1/T=9.05, ln K =0Thus, K= 1
gHgCOgOHgCO 222
KP
Py
v
i
vi
i
01111 i
ivv
111
10
2
22
OHCO
COH
yy
yy
5. Relation of Equilibrium Constants to Composition
Example 3-Solution(a) The reactants consist of 1 mol of H2O vapor & 1 mol of CO. The temperature is
1,100 K and the pressure is 1 bar.
Thus, fraction of steam reacted = 0.5
gHgCOgOHgCO 222
2
1 eCOy
vn
vny iii
0
0
2110 n
2
12
eOHy
22
eCOy
22
eHy
12
22 OHCO
COH
yy
yy
Substitute into
1
1 2
2
e
e
5.0e
5. Relation of Equilibrium Constants to Composition
Example 3-Solution(b) Same as (a) except the pressure is 10 bar.
Because v=0, the increase in pressure has no effect on the ideal-gas reaction and εe is still 0.5.
gHgCOgOHgCO 222
KP
Py
i
vi
i
0
5. Relation of Equilibrium Constants to Composition
Example 3-Solution(c) Same as (a) except that 2 mol of N2 is included in the reactants.
N2 does not take part in the reaction, and serves only as diluent.
Thus, remain unchanged. Therefore, εe is still 0.5
gHgCOgOHgCO 222
12
22 OHCO
COH
yy
yy
5. Relation of Equilibrium Constants to Composition
Example 3-Solution(d) The reactants are 2 mol of H2O & 1 mol of CO. Other conditions are the same as
in part (a)
gHgCOgOHgCO 222
Thus,
Thus, fraction of steam reacted 0.667/2= 0.333
3
1 eCOy
3
22
eOHy
32
eCOy
32
eHy
12
22 OHCO
COH
yy
yy
Substitute into
3120 n
121
2
ee
e
667.0e
5. Relation of Equilibrium Constants to Composition
Example 3-Solution(e) Same as (a) except that the temperature is 1,650 K.
At T=1650 K, 104/T=6.06.From the ln K vs 1/T graph,At 1/T=6.06, ln K =-1.15Thus, K= 0.317
gHgCOgOHgCO 222
317.0
1 2
2
e
e
36.0e
Thus, fraction of steam reacted 0.36.Increasing the temperature reduce the conversion.
6. Multireaction Equilibra
• For liquid-phase reaction;
• For gas-phase reaction;
• If equilibrium mixture is an ideal gas;
ji
v
ii Kffji ,ˆ rj ,...,2,1
ji
v
i KPfji ,ˆ rj ,...,2,1
j
v
i
vi K
P
Py
j
ji
,