ITK-234 Termodinamika Teknik Kimia 2
Chemical Equilibria
Dicky Dermawan
www.dickydermawan.net78.net
[email protected]
Reaction Coordinate e
1 A1   2 A 2  ......  3 A 3   3 A 3  ......
CH 4  H 2 O  CO  3 H 2
 CH 4  1
 H 2O  1
 CO  1
 H 2  3
dn1 dn 2 dn 3 dn 4
dn i



 ..... 
 de
1
2
3
4
i
dn CH 4
1

dn H2O
1
dn CO dn H2
dni


 ..... 
 de
1
3
i
Reaction Coordinate e
n i  n i0   i  e
yi 
n i n i0   i  e

n
n0    e
n  n0    e
n
 ni
n0 
 n i0

 i
If, initially there are 2 mol CH4, 1 mol H2O, 1 mol CO, and 4 mol H2:
n CH 4 ,0  2
n H 2 O, 0  1
n CO,0  1
n H 2 ,0  4
n 0  n CH 4 ,0  n H 2O,0  n CO,0  n H 2 ,0  2  1  1  4  8
n CH 4  2  e
n H 2O  1  e
n CO  1  e
n H2  4  3  e
   CH 4   H 2O   CO   H 2  1  1  1  3  2
n  n CH 4  n H 2O  n CO  n H 2  8  2  e
y CH 4 
2e
8 2e
y H 2O 
1 e
8 2e
y CO 
1 e
8 2e
y H2 
4  3 e
8 2e
Reaction Coordinate
Reaction Coordinate: Multiple Reactions
1 1 A1,1   2 1 A 2,1  ......   3 1 A3,1   3 1 A 3,1  ......
1 2 A1,2   2 2 A 2,2  ......   3 2 A3,2   3 2 A3,2  ......
.......... .......... .......... .....  .......... .......... .......... ......
1 j A1, j   2 j A 2, j  ...... 3 j A3, j  3 j A3, j  ......
CH 4  H 2 O  CO  3 H 2
CH 4  2 H 2 O  CO 2  4 H 2
j
1
2
v CH4 v H2O v CO v CO2
-1
-1
1
0
-1
-2
0
1
v H2
3
4
νj
2
2
Reaction Coordinate: Multiple Reactions
dn1,1
1,1
dn1,2
1,2
dn1, j
1, j
dn CH 4 ,1
1
dn CH 4 ,2
1

dn 2,1

dn 2,2

 2,1
 2,2
dn 2, j
 2, j

dn H 2O,1

dn H 2O,2
1
2

dn3,1

dn3,2



 3,1
 3,2
dn 3, j
 3, j
dn CO,1
1

dn 4,1

dn 4,2

 4,2
dn 4, j

dn CO 2 ,2
1
 4,1
 4, j
 ..... 
dni,1
 ..... 
dni,2
 ..... 
dn H 2 ,1
3

 i, 2
dn i, j
 i, j
 ..... 
dn H 2 ,2
4
 i,1
 de1
 de 2
 de j
dn i,1
 i,1
 ..... 
 de1
dn i,2
 i, 2
 de 2
Reaction Coordinate e
n i  n i0 
  i, j  e j
j
n i0 
  i, j  e j
n
j
yi  i 
n
n0 
1, j  e j

j
n  n0 
 1, j  e j
j
n
 ni
n0 
 n i0
j 
  i, j
i
If, initially there are 2 mol CH4, 1 mol H2O, 1 mol CO, and 4 mol H2:
y CH 4 
2  e1  e 2
8  2  e1  2  e 2
y H 2O 
y CO 2 
e2
8  2  e1  2  e 2
y H2 
1  e1  2  e 2
8  2  e1  2  e 2
4  3  e1  4  e 2
8  2  e1  2  e 2
y CO 
1  e1
8  2  e1  2  e 2
Reaction Coordinate
Reaction Coordinate & Equilibrium
Condition
DGf /J mol-1
CO
-200240
-209110
-217830
-226530
-235130
-243740
o
T/K
1000
1100
1200
1300
1400
1500
dG 
t
T, P
0
H2O
-192420
-187000
-181380
-175720
-170020
-164310
.
CO2
-395790
-395960
-296020
-396080
-396130
-396160
Reaction Coordinate & Equilibrium
Condition (Cont’)
Method of Equilibrium Constant
K
For gases:

K
(â i )


 exp 


i
 (f̂i )

 i  G i 0 

R T

i
CH 4  H 2O  CO  3 H 2

K  f̂ CH 4
  f̂ H O   f̂CO   f̂ H 
1
K  K y  K  P
For ideal gases:
K  K y  P
1
2
1
3
2
Method of
Equilibrium
Constant
Problems
Problem Hint: Different Way Leads to
The Same Results
Problems
Calculation of Equilibrium Constant
DG o298K 
DHo298K 
DCop 
  DH
i
o
f ,298K
  C 
i

  DG
i
o
f ,298K
DG o298K  R  298  ln K 298
o
p
T
DH T  DH o298 

DC op  dT
298
d ln K DH T

dT
R  T2

ln K T
DH T
 d ln K  R  T 2  dT
ln K 298 K
o
DG f ( 298K )
o
DG f (298K )
o
DG f (298K )
o
DG f (298K )
o
DH f ( 298K )
o
DH f (298K )
o
DH f (298K )
o
DH f (298K )
Heat Capacity Data - Gases
Heat Capacity Data - Gases
Heat
Capacity
Data Solid
Heat Capacity Data - Liquid
Problems
Problems
Problems
Problems
Problems
Problems
Problems
Problems
Problems
Multiple Reactions
A bed of coal (assume pure carbon) 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.
DGf /J mol-1
CO
-200240
-209110
-217830
-226530
-235130
-243740
o
T/K
1000
1100
1200
1300
1400
1500
H2O
-192420
-187000
-181380
-175720
-170020
-164310
.
CO2
-395790
-395960
-296020
-396080
-396130
-396160
Multiple Reactions
Problems
Problems
Problems
Lagrange Multiplier for Complex
Reaction
First Step: Atomic Balances

n i  a ik  A k
i
#2 Step:
DG ofi
No. of eqn = no. of
involved atom
 ni ˆ

 R  T  ln   i  P  
 k  a ik  0
n
 k

No. of eqn = no. of
species involved
Use Computer to Solve All The Equation Simultaneously.
Problem
Calculate the equilibrium composition 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 ther are present 2 mol of CH4
and 3 mol of H2O. Assume ideal gases.
At 1000 K: DG of CH  19720 J  mol -1
4
DG of H
2O
 192420 J  mol -1
DG of CO  200240 J  mol -1
DG of CO  395790 J  mol -1
2