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.

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2 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. 3

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 SO 2 to SO 3. 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) The rate of reaction increases as temperature increases. The equilibrium conversion to SO 3 falls as temperature rises, decreasing from about 90% at 520°C to 50% at about 680°C. V2O5V2O5 300  C 4

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. 5

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. 6

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. 7

8 8

9 Reaksi secara umum: adalah koefisien stoikiometri reaksi Konvensi tanda untuk i : Positif (+) untuk produk Negatif (–) untuk reaktan CONTOH: CH 4 + H 2 O  CO + 3 H 2 (1) 9

10 Perubahan mol spesies yang ada dalam reaksi berbanding lurus dengan bilangan stoikiometrinya. Jika 3 mol CH 4 berkurang karena bereaksi, maka H 2 O juga berkurang 3 mol, sementara itu 3 mol CO dan 9 mol H 2 terbentuk. ( i = 1, 2, 3,..., N) Koordinat reaksi (2) (3) 10 CH 4 + H 2 O  CO + 3 H 2 (1)(2) (3) (4)

11 CH 4 H2OH2O H2H2 CO 11

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 n i = n i0 ke kondisi setelah reaksi: ( i = 1, 2, 3,..., N) 12

13 Penjumlahan untuk semua spesies: Dengan: Jadi fraksi mol y i dari satu spesies jika dihubungkan dengan  : (5) 13

14 CONTOH Untuk sistem dengan reaksi: CH 4 + H 2 O  CO + 3 H 2 Mula-mula ada 2 mol CH 4, 1 mol H 2 O, 1 mol CO, dan 4 mol H 2. Tentukan pernyataan untuk y i sebagai fungsi . PENYELESAIAN 14

15 15

16 CONTOH Sebuah tangki berisi hanya n 0 uap air. Jika dekomposisi terjadi menurut reaksi H 2 O  H 2 + ½ O 2 Tentukan pernyataan yang menghubungkan jumlah mol dan fraksi mol tiap spesies dengan koordinat reaksi. PENYELESAIAN Jumlah mol masing-masing spesies: 16

17 17

18 Jika 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 n i dapat berubah karena beberapa reaksi, maka persamaan umum yang analog dengan persamaan (3) adalah: ( i = 1, 2, 3,..., N) 18

19 Integrasi dari n i = n i0 dan  j = 0: ( i = 1, 2, 3,..., N)(6) Jika semua spesies dijumlahkan: Definisi dari bilangan stoikiometri total (   i i ): ( i = 1, 2, 3,..., N)(7) 19

20 20 (+)

21 CONTOH Untuk sistem dengan reaksi: CH 4 + H 2 O  CO + 3 H 2 (1) CH 4 + 2H 2 O  CO 2 + 4 H 2 (2) Mula-mula ada 2 mol CH 4 dan 3 mol H 2 O, tentukan pernyataan untuk y i sebagai fungsi  1 dan  2. 21

22 PENYELESAIAN Tabel bilangan stoikiometri: CH 4 H2OH2OCOCO 2 H2H2 j 1– 1 1032 2 – 20142 i j 22 CH 4 + H 2 O  CO + 3 H 2 (1) CH 4 + 2H 2 O  CO 2 + 4 H 2 (2)

23 Pers. (7): 23

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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 G t attains its minimum value. At this equilibrium state, 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) 26

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

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. 28

29 Figure 1 indicates the two distinctive features of the equilibrium state for given temperature and pressure: The total Gibbs energy G t is a minimum. Its differential is zero. Each of these may serve as a criterion of equilibrium. Thus, we may write an expression for G t as a function of  and seek the value of  which minimizes G t, or we may differentiate the expression, equate it to zero, and solve for . 29

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31 the fundamental property relation for single-phase systems, provides an expression for the total differential of the Gibbs energy: If changes in the mole numbers n i occur as the result of a single chemical reaction in a closed system, then by Eq. (2) each dn i may be replaced by the product i d . Equation (1) then becomes: (2) (3) 31

32 32 P konstan  dP = 0 T konstan  dT = 0 (T dan P konstan)

33 Since nG is a state function, the right side of this equation is an exact differential expression; whence, Thus the quantity  i i  i 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: (4) (5) 33

34 Recall the definition of the fugacity of a species in solution (6) Gibbs free energy for pure species i in its standard state at the same temperature: (7) The difference between these two equations is: (8) 34

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36 Combining Eq. (5) with Eq. (8) to eliminate  i gives for the equilibrium state of a chemical reaction: (9) (11) (10) 36

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40 In exponential form, this equation becomes (12) This equation defines K; it is given alternative expression by Also by definition (13) (14) (15) 40

41 Since G i 0 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  G 0 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;  i i G i 0, represented by  G 0, is called the standard Gibbs-energy change of reaction. 41

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 G i 0 must be the same state as for the fugacity f i 0. 42

43 The function  G 0 =  i i G i 0 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  G 0 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: (16) 43

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 standard state: Total derivatives are appropriate here because the properties in the standard state are functions of temperature only. (17) 44

45 Multiplication of both sides of Eq. (17) by i and summation over all species gives: (18) (19) Or: 45

46 46

47 Since the standard-state temperature is that of the equilibrium mixture, the standard property changes of reaction, such as  G 0 and  H 0, vary with the equilibrium temperature. The dependence of  G 0 on T is given by Eq. (17), which may be rewritten: In view of Eq. (14), this becomes (20) (14) 47

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49 Equation (20) gives the effect of temperature on the equilibrium constant, and hence on the equilibrium conversion. If  H 0 < 0 (exothermic)  K decreases as T increases If  H 0 > 0 (endothermic)  K increases with T If  H 0 is assumed independent of T, integration of Eq. (20) from a particular temperature T' to an arbitrary temperature T leads to the simple result: (21) 49

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51 However, if  H 0 is T dependent, as defined in Eq. (11) of previous Chapter: 51

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53 53

54 54 K1K1 K2K2

55 Integration from T 0 to T results in: (22) (23) (24) (25) 55

56 EXAMPLE Calculate the equilibrium constant for the vapor-phase hydration of ethylene at 418.15 to 593.15K. SOLUTION Reaction: C 2 H 4 (g) + H 2 O (g)  C 2 H 5 OH (g) 56

57 C2H4C2H4 H2OH2OC 2 H 5 OH – 1 + 1 A1,4243,4703,518 B 14,394  10 -3 4,450  10 -3 20,001  10 -3 C – 4,392  10 -6 0 – 6,002  10 -6 D0 0,121  10 5 0  H 0 f,298 52.510– 241.818– 235.100  G 0 f,298 68.460– 228.572– 168.490 57

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59 For T = 418.15K 59

60 For T = 593.15K 60