Post on 17-Jan-2018
description
ITK-234 Termodinamika Teknik Kimia 2
Chemical Equilibria
Dicky Dermawanwww.dickydermawan.net78.net
dickydermawan@gmail.com
Reaction Coordinate e......AA......AA 33332211
3 1 1 1H 3COOHCH
224 HCOOHCH
224
e
ddn
.....dndndndn
i
i
4
4
3
3
2
2
1
1
e
ddn
.....3
dn
1dn
1
dn
1
dn
i
iHCOOHCH 224
Reaction Coordinate e
eee
e
i0i0i
0
0
i0iii
i0ii
nn nn
nnn
nnn
y
nn
If, initially there are 2 mol CH4, 1 mol H2O, 1 mol CO, and 4 mol H2:
ee
e
e
ee
e
e
e
eeee
2834
y 28
1y
281
y 28
2y
28n n n nn
23111
34n 1n 1n 2n
84112nn nn n
4n 1n 1n 2n
224
224
224
224
224
224
HCOOHCH
HCOOHCH
HCOOHCH
HCOOHCH
0,H0,CO0,OH0,CH0
0,H0,CO0,OH0,CH
Reaction Coordinate
Reaction Coordinate: Multiple Reactions
.......................................................................
......AA......AA 2,3232,3232,2222,121
......AA......AA j,3j3j,3j3j,2j2j,1j1
......AA......AA 1,3131,3131,2121,111
2224
224H 4COOH 2CH
H 3COOHCH
j v CH4 v H2O v CO v CO2 v H2 νj
1 -1 -1 1 0 3 22 -1 -2 0 1 4 2
Reaction Coordinate: Multiple Reactions
22,i
2,i
2,4
2,4
2,3
2,3
2,2
2,2
2,1
2,1 ddn
.....dndndndn
e
11,i
1,i
1,4
1,4
1,3
1,3
1,2
1,2
1,1
1,1 ddn
.....dndndndn
e
22,i
2,i2,H2,CO2,OH2,CH
11,i
1,i1,H1,CO1,OH1,CH
ddn
.....4
dn
1
dn
2
dn
1
dn
ddn
.....3
dn
1dn
1
dn
1
dn
2224
224
e
e
jj,i
j,i
j,4
j,4
j,3
j,3
j,2
j,2
j,1
j,1 ddn
.....dndndndn
e
Reaction Coordinate e
e
e
e
e
ij,ij0i0i
jjj,10
jjj,10
jjj,i0i
ii
jjj,i0ii
nn nn
nn
n
n
nn
y
nn
If, initially there are 2 mol CH4, 1 mol H2O, 1 mol CO, and 4 mol H2:
21
21H
21
2CO
21
1CO
21
21OH
21
21CH
228434
y 228
y
228
1y
22821
y 228
2y
22
24
eeee
ee
e
eee
ee
ee
eeee
Reaction Coordinate
Reaction Coordinate & Equilibrium Condition
0dG P,Tt
H2O CO CO2
1000 -192420 -200240 -3957901100 -187000 -209110 -3959601200 -181380 -217830 -2960201300 -175720 -226530 -3960801400 -170020 -235130 -3961301500 -164310 -243740 -396160
DGfo/J.mol-1T/K
Reaction Coordinate & Equilibrium Condition (Cont’)
Method of Equilibrium Constant
ffffK3
H1
CO1
OH1
CH 224
TR
Gexp)a(K
0ii
ii
i)f(K iFor gases:
224 H 3COOHCH
For ideal gases:
PKKK y
PKK y
Method of Equilibrium
Constant
Problems
Problem Hint: Different Way Leads to The Same Results
Problems
Calculation of Equilibrium Constant DD o
K298,fio
K298 GG
DD oK298,fi
oK298 HH
2T
TR
HdT
Klnd
D
298o
K298 Kln298RG D
dTCHHT
298
op
o298T DDD
D opi
op CC
dTTR
HKlnd
2T
Kln
Kln
T
K298
D
)K298(o
fGD
)K298(o
fGD
)K298(o
fGD
)K298(o
fGD
)K298(o
fHD
)K298(o
fHD
)K298(o
fHD
)K298(o
fHD
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 ReactionsA 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.
H2O CO CO2
1000 -192420 -200240 -3957901100 -187000 -209110 -3959601200 -181380 -217830 -2960201300 -175720 -226530 -3960801400 -170020 -235130 -3961301500 -164310 -243740 -396160
DGfo/J.mol-1T/K
Multiple Reactions
Problems
Problems
Problems
Lagrange Multiplier for Complex Reaction
0aPˆnn
lnTRGk
ikkiio
fi
D
i
kiki AanFirst Step: Atomic Balances
#2 Step:
No. of eqn = no. of involved atom
No. of eqn = no. of species involved
Use Computer to Solve All The Equation Simultaneously.
ProblemCalculate 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:
1-oCO f
1-oCO f
1-oOH f
-1oCH f
molJ 395790G
molJ 200240G
molJ 192420G
molJ 19720G
2
2
4
D
D
D
D