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ITK-234 Termodinamika Teknik Kimia 2
Chemical Equilibria
Dicky Dermawanwww.dickydermawan.net78.net
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Reaction Coordinate e
......AA......AA 33332211
3 1 1 1
H 3COOHCH
224 HCOOHCH
224
ddn
.....dndndndn
i
i
4
4
3
3
2
2
1
1
ddn
.....3
dn
1
dn
1
dn
1
dn
i
iHCOOHCH 224
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Reaction Coordinate e
i0i0i
0
0
i0iii
i0ii
nn nn
nn
n
n
n
ny
nn
If, initially there are 2 mol CH4, 1 mol H2O, 1 mol CO, and 4 mol H2:
28
34y
28
1y
28
1y
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
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Reaction Coordinate
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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
224
H 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
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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
11,i
1,i
1,4
1,4
1,3
1,3
1,2
1,2
1,1
1,1 ddn
.....dndndndn
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
1
dn
1
dn
1
dn
2224
224
jj,i
j,i
j,4
j,4
j,3
j,3
j,2
j,2
j,1
j,1d
dn.....
dndndndn
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Reaction Coordinate e
ij,ij0i0i
jjj,10
jjj,10
jjj,i0i
ii
jjj,i0ii
nn nn
nn
n
n
n
ny
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
228
434y
228y
228
1y
228
21y
228
2y
22
24
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Reaction Coordinate
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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-1
T/K
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Reaction Coordinate & Equilibrium Condition (Cont’)
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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
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Method of Equilibrium
Constant
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Problems
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Problem Hint: Different Way Leads to The Same Results
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Problems
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Calculation of Equilibrium Constant
DD oK298,fi
oK298 GG
DD oK298,fi
oK298 HH
2T
TR
H
dT
Klnd
D
298o
K298 Kln298RG D
dTCHHT
298
op
o298T DDD
D opi
op CC
dTTR
HKlnd
2T
Kln
Kln
T
K298
D
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)K298(o
fGD
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)K298(o
fGD
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)K298(o
fGD
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)K298(o
fGD
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)K298(o
fHD
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)K298(o
fHD
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)K298(o
fHD
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)K298(o
fHD
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Heat Capacity Data - Gases
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Heat Capacity Data - Gases
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Heat Capacity Data - Solid
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Heat Capacity Data - Liquid
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Problems
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Problems
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Problems
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Problems
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Problems
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Problems
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Problems
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Problems
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Problems
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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-1
T/K
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Multiple Reactions
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Problems
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Problems
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Problems
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Lagrange Multiplier for Complex Reaction
0aPˆn
nlnTRG
kikki
iofi
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.
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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