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