1

KESEIMBANGANREAKSI KIMIA

(1)

The transformation of raw materials into products of greater value by means of chemical reaction is a

major industry, and a vast array of commercial products is obtained by chemical synthesis.

Sulfuric acid, ammonia, ethylene, propylene, phosphoric acid, chlorine, nitric acid, urea,

benzene, methanol, ethanol, and ethylene glycol

fibers, paints, detergents, plastics, rubber, paper, fertilizers, insecticides, etc. 2

3

Clearly, the chemical engineer must be familiar with chemical-reactor design and operation.

4

• Both the rate and the equilibrium conversion of a chemical reaction depend on the temperature, pressure, and composition of reactants.

• Consider the oxidation of SO2 to SO3.

2 SO2 (g) + O2 (g) 2 SO3 (g)

• The rate of reaction increases as temperature increases.

• The equilibrium conversion to SO3 falls as temperature rises, decreasing from about 90% at 520°C to 50% at about 680°C.

V2O5

300C

5

• The equilibrium conversions represent maximum possible conversions regardless of catalyst or reaction rate.

• The evident conclusion is that both equilibrium and rate must be considered in the exploitation of chemical reactions for commercial purposes.

• Although reaction rates are not susceptible to thermodynamic treatment, equilibrium conversions are.

• Therefore, the purpose of this chapter is to determine the effect of temperature, pressure, and initial composition on the equilibrium conversions of chemical reactions.

6

• Many industrial reactions are not carried to equilibrium; reactor design is then based primarily on reaction rate.

• However, the choice of operating conditions may still be influenced by equilibrium considerations.

• Moreover, the equilibrium conversion of a reaction provides a goal by which to measure improvements in a process.

7

• Similarly, it may determine whether or not an experimental investigation of a new process is worthwhile.

• For example, if thermodynamic analysis indicates that a yield of only 20% is possible at equilibrium and if a 50% yield is necessary for the process to be economically attractive, there is no purpose to an experimental study.

• On the other hand, if the equilibrium yield is 80%, an experimental program to determine the reaction rate for various conditions of operation (catalyst, temperature, pressure, etc.) may be warranted.

8

KOORDINAT REAKSI

9

Reaksi secara umum:

...AA...AA 44332211

i adalah koefisien stoikiometri reaksi

Konvensi tanda untuk i:

• Positif (+) untuk produk • Negatif (–) untuk reaktan

CONTOH:CH4 + H2O CO + 3 H2

3111224 HCOOHCH

(1)

10

Perubahan mol spesies yang ada dalam reaksi berbanding lurus dengan bilangan stoikiometrinya.

Jika 3 mol CH4 berkurang karena bereaksi, maka H2O juga berkurang 3 mol, sementara itu 3 mol CO dan 9 mol H2 terbentuk.

.dstdndndndn

1

1

3

3

1

1

2

2

ddndndndndn

i

i

4

4

3

3

2

2

1

1

ddn ii ( i = 1, 2, 3, ..., N)

Koordinat reaksi

(2)

(3)

CH4 + H2O CO + 3 H2

(1) (2) (3) (4)

11

313dn

1

1

313dn

2

2

313dn

3

3

339dn

4

4

CH4

H2O

H2

CO

12

(4)

Pers. (2) dan (3) menyatakan perubahan akibat perubahan jumlah mol spesies yang bereaksi.

Definisi dari dilengkapi dengan pernyataan = 0 untuk kondisi awal sistem, sebelum reaksi.

Jadi integrasi pers. (3) dari kondisi awal sebelum reaksi dengan = 0 dan ni = ni0 ke kondisi setelah reaksi:

0i

n

ni ddn

i

0i

iii 0nn ( i = 1, 2, 3, ..., N)

iii 0nn

13

Penjumlahan untuk semua spesies:

i

ii

ii

i 0nnn

0nn

i

inn i

i0 0nn

iiDengan:

Jadi fraksi mol yi dari satu spesies jika dihubungkan dengan :

0

iiii n

nnn

y 0 (5)

14

CONTOH

Untuk sistem dengan reaksi:

CH4 + H2O CO + 3 H2

Mula-mula ada 2 mol CH4, 1 mol H2O, 1 mol CO, dan 4 mol H2. Tentukan pernyataan untuk yi sebagai fungsi .

PENYELESAIAN

21113i

i

84112nni

i0 0

15

0

iiii n

nnn

y 0

28

2y

4CH

28

1y OH2

28

1yCO

2834

y2H

16

CONTOH

Sebuah tangki berisi hanya n0 uap air. Jika dekomposisi terjadi menurut reaksi

H2O H2 + ½ O2

Tentukan pernyataan yang menghubungkan jumlah mol dan fraksi mol tiap spesies dengan koordinat reaksi.

PENYELESAIAN

5,015,01i

i

Jumlah mol masing-masing spesies:

iii 0nn

17

0OH nn2

2Hn

5,0n2O

5,0nn 0

5,0n

ny

0

0OH2

5,0n

y0

H2

5,0n

5,0y

0O2

iii 0nn

18

Stoikiometri Reaksi MultiJika ada dua atau lebih reaksi independen yang berlangsung bersamaan, maka digunakan subskrip j sebagai indeks untuk reaksi.

j : koordinat untuk reaksi j.

i,j : bilangan stoikiometri untuk spesies i dalam reaksi j

Karena jumlah mol satu spesies ni dapat berubah karena beberapa reaksi, maka persamaan umum yang analog dengan persamaan (3) adalah:

j

jj,ii ddn ( i = 1, 2, 3, ..., N)

ddn ii

19

Integrasi dari ni = ni0 dan j = 0:

j

jj,iii 0nn ( i = 1, 2, 3, ..., N) (6)

Jika semua spesies dijumlahkan:

jj

ij,i0

i jjj,i

ii

ii nnnn

0

Definisi dari bilangan stoikiometri total ( i i):

i

j,ij j

jj0nn

jjj

jjj,ii

ii n

n

nn

y0

( i = 1, 2, 3, ..., N) (7)

20

j

jj,111 0nn

j

jj,222 0nn

(+)

jjj,22

jjj,1121 00

nnnnn

i jjj,i

ii

i jjj,ii

ii 00

nnnn

21

CONTOH

Untuk sistem dengan reaksi:

CH4 + H2O CO + 3 H2 (1)

CH4 + 2H2O CO2 + 4 H2 (2)

Mula-mula ada 2 mol CH4 dan 3 mol H2O, tentukan pernyataan untuk yi sebagai fungsi 1 dan 2.

22

PENYELESAIAN

Tabel bilangan stoikiometri:

CH4 H2O CO CO2 H2 j

1 – 1 – 1 1 0 3 2

2 – 1 – 2 0 1 4 2

ij

j

jj,iii 0nn

21j

jj,111CH 2nnn04

21j

jj,222OH 23nnn02

CH4 + H2O CO + 3 H2 (1)

CH4 + 2H2O CO2 + 4 H2 (2)

23

Pers. (7):

1j

jj,333CO 0nnn

2j

jj,444CO 02nnn

21j

jj,455H 43nnn02

21j

jj0 225nn

jjj

jjj,ii

ii n

n

nn

y0

24

21

21CHCH 225

2n

ny 4

4

21

21OHOH 225

23n

ny 2

2

21

1COCO 225n

ny

21

2COCO 225n

ny 2

2

21

21HH 225

43n

ny 2

2

25

APLIKASI KRITERIA KESEIMBANGAN

PADA REAKSI KIMIA

26

The total Gibbs energy of a closed system at constant T and P must decrease during an irreversible process and that the condition for equilibrium is reached when Gt attains its minimum value.

At this equilibrium state,

0dG P,Tt

Thus if a mixture of chemical species is not in chemical equilibrium, any reaction that occurs at constant T and P must lead to a decrease in the total Gibbs energy of the system.

The significance of this for a single chemical reaction is seen in the following figure.

(1)

27Figure 1. The total Gibbs energy in relation to the reaction coordinate

28

Since is the single variable that characterizes the progress of the reaction, and therefore the composition of the system, the total Gibbs energy at constant T and P is determined by .

The arrows along the curve Figure 1 indicate the directions of changes in (G )T,P that are possible on account of reaction.

The reaction coordinate has its equilibrium value , at the minimum of the curve.

29

Figure 1 indicates the two distinctive features of the equilibrium state for given temperature and pressure:

• The total Gibbs energy Gt is a minimum.

• Its differential is zero.

Each of these may serve as a criterion of equilibrium.

Thus, we may write an expression for Gt as a function of and seek the value of which minimizes Gt , or we may differentiate the expression, equate it to zero, and solve for .

30

THE STANDARD GIBBS-ENERGY CHANGE

AND THE EQUILIBRIUM CONSTANT

31

the fundamental property relation for single-phase systems, provides an expression for the total differential of the Gibbs energy:

i

ii dndTnSdPnVnGd

If changes in the mole numbers ni occur as the result of a single chemical reaction in a closed system, then by Eq. (2) each dni may be replaced by the product i d.

Equation (1) then becomes:

(2)

i

ii ddTnSdPnVnGd (3)

32

i

ii ddTnSdPnVnGd

i

iiddT

nSddP

nVdnGd

P konstan dP = 0T konstan dT = 0

iiid

nGd (T dan P konstan)

iii

P,T

nG

33

Since nG is a state function, the right side of this equation is an exact differential expression; whence,

P,T

t

P,Tiii

GnG

Thus the quantity iii represents, in general, the rate of change of the total Gibbs energy of the system with the reaction coordinate at constant T and P.

Figure 1 shows that this quantity is zero at the equilibrium state.

A criterion of chemical-reaction equilibrium is therefore:

0i

ii

(4)

(5)

34

Recall the definition of the fugacity of a species in solution

iii f̂lnRTT (6)

Gibbs free energy for pure species i in its standard state at the same temperature:

0ii

0i flnRTTG (7)

The difference between these two equations is:

0i

i0ii f

f̂lnRTG (8)

35

0i

i0ii f

f̂lnRTG

0i

i0ii f

f̂lnRTG

36

Combining Eq. (5) with Eq. (8) to eliminate i gives for the equilibrium state of a chemical reaction:

0ff̂

lnRTGi

0i

i0ii

0ff̂lnRTGi

0ii

i

0ii

i

RT

Gff̂ln i

0ii

0ii

i

i

(9)

(11)

(10)

37

i

0ii

i

0iii

i

0iii

iff̂lnRTff̂lnRTff̂lnRT

321

03

30

2

20

1

1

ff̂

lnff̂

lnff̂

lnRT

321

03

30

2

20

1

1

ff̂

ff̂

ff̂

lnRT

i

0i

i

i ff̂

lnRT

38

0ff̂

lnRTGi

0i

i

ii

0ii

0ff̂lnRTGi

0ii

i

0ii

i

RT

G

ff̂

ln i

0ii

0i

i

i

i

i

0ii0

i

i

iG

ff̂

lnRTi

39

RT

GKlnff̂ln i

0ii

0ii

i

i

RT

GKln i

0ii

RT

GexpK i

0ii

40

In exponential form, this equation becomes

(12) Kff̂ i0ii

i

RT

GexpK

0

This equation defines K; it is given alternative expression by

RTG

Kln0

Also by definition

(13)

(14)

i

0ii

0 GG (15)

41

Since Gi0 is a property of pure species i in its

standard state at fixed pressure, it depends only on temperature.

By Eq. (15) it follows that G0 and hence K, are also functions of temperature only.

In spite of its dependence on temperature, K is called the equilibrium constant for the reaction; iiGi

0, represented by G0, is called the standard Gibbs-energy change of reaction.

42

The fugacity ratios in Eq. (12) provide the connection between the equilibrium state of interest and the standard states of the individual species, for which data are presumed available.

The standard states are arbitrary, but must always be at the equilibrium temperature T.

The standard states selected need not be the same for all species taking part in a reaction.

However, for a particular species the standard state represented by Gi

0 must be the same state as for the fugacity fi

0.

43

The function G0 = i i Gi0 in Eq. (15) is the difference

between the Gibbs energies of the products and reactants (weighted by their stoichiometric coefficients) when each is in its standard state as a pure substance at the standard-state pressure, but at the system temperature.

Thus the value of G0 is fixed for a given reaction once the temperature is established, and is independent of the equilibrium pressure and composition.

Other standard property changes of reaction are similarly defined.

Thus, for the general property M:

0i

0 MM (16)

44

For example, the relation between the standard heat of reaction and the standard Gibbs energy change of reaction may be written for species i in its standardstate:

dT

RTGdRTH

0i20

i

Total derivatives are appropriate here because the properties in the standard state are functions of temperature only.

(17)

45

Multiplication of both sides of Eq. (17) by i and summation over all species gives:

dT

RTGdRTH i

0ii

2

i

0ii

(18)

dT

RTGdRTH

020

(19)

Or:

46

EFFECT OF TEMPERATURE ONTHE EQUILIBRIUM CONSTANT

47

Since the standard-state temperature is that of the equilibrium mixture, the standard property changes of reaction, such as G0 and H0, vary with the equilibrium temperature.

The dependence of G0 on T is given by Eq. (17), which may be rewritten:

2

00

RTH

dTRTGd

In view of Eq. (14), this becomes

2

0

RTH

dTKlnd (20)

RTG

Kln0

(14)

48

dTRT

HKlnd 2

0

2

1

2

1

T

T2

0Kln

KlndT

RTH

Klnd

49

Equation (20) gives the effect of temperature on the equilibrium constant, and hence on the equilibrium conversion.

If H0 < 0 (exothermic) K decreases as T increases

If H0 > 0 (endothermic) K increases with T

If H0 is assumed independent of T, integration of Eq. (20) from a particular temperature T' to an arbitrary temperature T leads to the simple result:

'T1

T1

RH

'KK

ln0

(21)

50

51

However, if H0 is T dependent, as defined in Eq. (11) of previous Chapter:

2

0

RTH

dTKlnd

T

T

P0

0

0

dTRC

RHH

dTdTRC

T1

RTH

KlndT

T

P20

0

52

dTdTRC

T1

RTH

KlndT

T

P20

0

T

T

T

T

P20

Kln

Kln 0 00

dTdTRC

T1

RTH

Klnd

T

T

T

T

P20

00 0

dTdTRC

T1

RTH

KlnKln

T

T

T

T

P20

0 0 0

dTdTRC

T1

RTH

KK

ln

53

T

T

T

T

P20

0 0 0

dTdTRC

T1

RTH

KK

ln

T

T

T

T

P20

0 0 0

dTdTRC

T1

RTH

expKK

T

T

T

T

P20

00 0

dTdTRC

T1

RTH

expKK

54

T

T

T

T

P20

00 0

dTdTRC

T1

RTH

expKK

T

T

T

T

PT

T20

00 00

dTdTRC

T1

expdTRT

HexpKK

K1K2

T

T

T

T

T

T

P20

00 0 0

dTdTRC

T1

dTRT

HexpKK

55

Integration from T0 to T results in:

210 KKKK

0

00

0 RTG

expK

11

RTH

expTT

1RT

HexpK

0

000

0

00

1

2

02

1TB

211

lnAexpK

2

2

20

220

1T

D2121

TC61

(22)

(23)

(24)

(25)

56

EXAMPLECalculate the equilibrium constant for the vapor-phase hydration of ethylene at 418.15 to 593.15K.

SOLUTIONReaction:

C2H4 (g) + H2O (g) C2H5OH (g)

57

C2H4 H2O C2H5OH – 1 – 1 + 1A 1,424 3,470 3,518B 14,394 10-3 4,450 10-3 20,001 10-3

C – 4,392 10-6 0 – 6,002 10-6

D 0 0,121 105 0H0

f,298 52.510 – 241.818 – 235.100G0

f,298 68.460 – 228.572 – 168.490

58

376,1470,3424,1518,3A

33 10157,410450,1394,14001,20B

66 10610,110000,0392,4002,6C

55 10121,010121,0000,0000,0D

10298 molJ792.45818.241510.52100.235H

10298 molJ378.8572.228490.68490.168G

366,29

15,298314,8378.8

expRT

GexpK

0

00

0

59

For T = 418.15K

TT

1RT

HexpK 0

0

00

1

310985,4

15,41815,298

115,298314,8

792.45exp

4025,115,29815,418

TT

0

9860,0K2

13210 10443,19860,010985,4366,29KKKK

60

For T = 593.15K

TT

1RT

HexpK 0

0

00

1

410023,1

15,59315,298

115,298314,8

792.45exp

9894,115,29815,593

TT

0

9794,0K2

34210 10942,29794,010023,1366,29KKKK