KESEIMBANGANREAKSI KIMIA
(3)
PHASE RULE FORREACTING SYSTEMS
The intensive state of a PVT system containing N chemical species and phases in equilibrium is characterized by the intensive variables, temperature T, pressure P, and N - 1 mole fractions for each phase.
These are the phase-rule variables, and their number is 2 + (N – 1)().
The masses of the phases are not phase-rule variables, because they have no influence on the intensive state of the system.
An independent phase-equilibrium equation may be written connecting intensive variables for each of the N species for each pair of phases present.
Thus, the number of independent phase-equilibrium equations is ( – 1 )(N).
The difference between the number of phase-rule variables and the number of independent equations connecting them is the number of variables that may be independently fixed.
Called the degrees of freedom of the system F, the number is:
N11N2F
N2F
or
(42)
For reacting system, the phase-rule variables are unchanged: temperature, pressure, and N – 1 mole fractions in each phase.
The total number of these variables is 2 + (N – 1)().
The same phase-equilibrium equations apply as before, and they number ( – 1)(N).
However, Eq. (12) provides for each independent reaction an additional relation that must be satisfied at equilibrium.
Since the pi's are functions of temperature, pressure, and the phase compositions, Eq. (12) represents a relation connecting phase-rule variables.
If there are r independent chemical reactions at equilibri-um within the system, then there is a total of ( – 1)(N) + r independent equations relating the phase-rule variables.
Taking the difference between the number of variables and the number of equations gives:
rN11N2F
rN2F
or
(43)
This is the phase rule for reacting systems
ExampleDetermine the number of degrees of freedom F for a system of two miscible nonreacting species which exists in vapor/liquid equilibrium
SolutionrN2F
20222
ExampleDetermine the number of degrees of freedom F for a system prepared by partially decomposing CaCO3 into an evacuated space.
Solution
A single chemical reaction occurs:CaCO3(s) CaO(s) + CO2(g)
r = 1N = 3 = 3 (solid CaCO3, solid CaO, and gaseous CO2)F = 2 – + N – r =2 – 3 + 3 – 1 = 1
REACTION INHETEROGENEOUS
SYSTEM
Li
Vi f̂f̂
When liquid and gas phases are both present in an equilibrium mixture of reacting species,
a criterion of vapor-liquid equilibrium, must be satisfied along with the equation of chemical-reaction equilibrium.
Consider, for example, the reaction of gas A with liquidwater B to form an aqueous solution C.
Several choices in the method of treatment exist.
• The reaction may be assumed to occur entirely in the gas phase with simultaneous transfer of material between phases to maintain phase equilibrium.
• In this case, the equilibrium constant is evaluated from G0 data based on standard states for the species as gases, i.e., the ideal-gas states at 1 bar and the reaction temperature.
Method 1
• The reaction may be assumed to occur in the liquid phase.
• In this case, the equilibrium constant is evaluated from G0 data based on standard states for the species as liquids.
Method 2
Alternatively, the reaction may be written:
A(g) + B(l) C(aq)
in which case the G0 value is for mixed standard states: C as a solute in an ideal 1-molal aqueous solution, B as a pure liquid at 1 bar, and A as a pure ideal gas at 1 bar.
For this choice of standard states, the equilibrium constant as given by Eq. (12) becomes:
Method 3
(44) Kff̂x
mff̂ff̂
ff̂0AABB
C0AA
0BB
0CC
Henry’s Law
KESEIMBANGANREAKSI MULTI
Jika keadaan keseimbangan dalam suatu sistem reaksi tergantung pada dua atau lebih reaksi kimia independen, maka langkah-langkah untuk menentukan komposisi keseimbangan adalah:• Tentukan reaksi kimia yang terjadi.• Komposisi keseimbangan dihitung seperti yang telah
dibahas.• Konstanta keseimbangan untuk setiap reaksi dievaluasi
dengan cara seperti yang telah dibahas.
Untuk reaksi tunggal:
(12)Kff̂ i
0i
i
i
(45)j0i
i
iK
ff̂ j,i
Untuk reaksi multi:
dengan j adalah nomor reaksi kimia.
Untuk reaksi fasa gas, persamaan (45) menjadi:
(46)j0i
iK
Pf̂ j,i
Jika campuran keseimbangan berupa gas ideal:
(47) j0ii
KPPy
jj,i
Terhadap senyawa n-butana dilakukan reaksi cracking pada 750 K dan 1,2 bar sehingga dihasilkan olefin. Hanya dua reaksi yang memiliki konversi keseimbangan yang signifikan, yaitu:
C4H10 C2H4 + C2H6 (I)
C4H10 C3H6 + CH4 (II)
Jika kedua reaksi ini mencapai keseimbangan, bagaimana komposisi produk?
Data: K(I) = 3,856
K(II) = 268,4
CONTOH
Penyelesaian:
ji,j
jC4H10 C2H4 C2H6 C3H6 CH4
I – 1 + 1 + 1 0 0 + 1II – 1 0 0 + 1 + 1 + 1
Basis: 1 mol umpan C4H10
1n0HC 104
1nn0HC0 104
IIIHC 1n104
IHC 42n
IHC 62n
IIHC 63n
IICH4n
III0 1n
III
IIIHC 1
1y104
III
IHC 1
y42
III
IHC 1
y62
III
IIHC 1
y63
III
IICH 1
y4
Keseimbangan kimia:
j0ii
KPPy
jj,i
I
1
0HC
HCHC KPP
yyy
104
6242
II
1
0HC
CHHC KPP
yyy
104
463
I
1
0IIIIII
2I K
PP
11
II
1
0IIIIII
2II K
PP
11
(A)
(B)
0144,04,268
856,3KK
BA
II
I2II
2I
12,00144,0II
I
III 12,0 (C)
I
1
0IIIIII
2I K
PP
11
Per. (C) dimasukkan ke pers. (A):
856,3
12,1
12,0112,0112,0 1
IIIIIIII
2II
856,3
12,1
12,0112,0112,0 1
IIIIIIII
2II
2133,3
12,1112,1112,0
IIII
2II
2II
2II 2544,112133,30144,0
2II
2II 0308,42133,30144,0
2133,30452,4 2II
8913,00452,42133,3
II 107,0I
0009,011y
III
IIIHC 104
0535,01
yIII
IHC 42
0535,01
yIII
IHC 62
4460,01
yIII
IIHC 63
4460,01
yIII
IICH4
ContohSetumpukan batubara (dianggap terdiri dari karbon murni) dalam sebuah gasifier dialiri steam dan udara sehingga terjadi reaksi yang menghasilkan gas yang terdiri dari H2, CO, O2, CO2, dan N2. Jika gas yang diumpankan ke dalam gasifier terdiri dari 1 mol steam dan 2,38 mol udara, hitung komposisi keseimbangan dari aliran gas yang keluar dari gasifier pada P = 20 bar dan temperatur 1500 K. Diketahui nilai G0
f,1500 untuk:H2O = – 164.310 J/molCO = – 243.740 J/molCO2 = – 396.160 J/mol
Karena temperatur cukup tinggi, maka campuran gas dapat dianggap sebagai gas ideal.
Penyelesaian
Kemungkinan reaksi yang terjadi:C + O2 CO2 (a)C + CO2 2 CO (b)H2O + C H2 + CO (c)
Harga K untuk masing-masing reaksi pada 1500K adalah:
RTGexpK
01500
a
1500314,8160.396exp
RT
Gexp 2CO
01500,f
131025,6
12,1514Kb
58,583Kc
Dengan cara yang sama diperoleh:
Harga Ka sangat besar, yang berarti bahwa reaksi (a) merupakan reaksi irreversibel dan semua O2 habis bereaksi dengan C menjadi CO2 [reaksi (a)].
Jumlah mol gas mula-mula:H2O = 1 molN2 = 0,79 2,38 = 1,88 mol n0 =3,38O2 = 0,21 2,38 = 0,5 mol
Jumlah mol gas setelah reaksi:nH2O = 1 – c
nN2 = 1,88nCO2 = 0,5 – b
nCO = 2 b + c
H2 = c
cb38,3n
Fraksi mol masing-masing komponen setelah reaksi:
cb
cOH 38,3
1y2
cbN 38,3
88,1y2
cb
bCO 38,3
5,0y2
cb
cbCO 38,3
2y
cb
cH 38,3
y2
Untuk reaksi (b):
12,15141
20yy 1
CO
2CO
2
121bi
ib
b0ii
KPPy
bi
706,75
38,35,02
bbc
2cb
(A) 038,35,0706,752 bbb2
cb
Untuk reaksi (c):
58,5831
20y
yy 1
OH
HCO
2
2
1111ci
ic
c0ii
KPPy
ci
179,29
38,312
cbc
ccb
(B) 038,31179,292 cbcccb
Persamaan (A) dan (B) diselesaikan secara simultan dengan menggunakan Excel Solver dengan batasan:
5,05,0 b 10 c
Hasil perhitungan:b = 0,4895 dan c = 0,9863
Jika nilai b dan c ini dimasukkan ke persamaan untuk fraksi mol masing-masing komponen maka akan diperoleh:
003,0y OH2
387,0y2N
002,0y2CO
405,0yCO
203,0y2H
dan