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The intensive state of a PVT system containing N chemical species and  phases in equilibrium is characterized by the intensive variables, temperature.

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Presentasi berjudul: "The intensive state of a PVT system containing N chemical species and  phases in equilibrium is characterized by the intensive variables, temperature."— Transcript presentasi:

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3 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.

4 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: or (42)

5 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.

6 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: or (43) This is the phase rule for reacting systems

7 Example Determine the number of degrees of freedom F for a system of two miscible nonreacting species which exists in vapor/liquid equilibrium Solution

8 Example Determine the number of degrees of freedom F for a system prepared by partially decomposing CaCO 3 into an evacuated space. Solution A single chemical reaction occurs: CaCO 3(s)  CaO (s) + CO 2(g) r = 1 N = 3  = 3 (solid CaCO 3, solid CaO, and gaseous CO 2 ) F = 2 –  + N – r =2 – – 1 = 1

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10 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 liquid water B to form an aqueous solution C. Several choices in the method of treatment exist.

11 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  G 0 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

12 The reaction may be assumed to occur in the liquid phase. In this case, the equilibrium constant is evaluated from  G 0 data based on standard states for the species as liquids. Method 2

13 Alternatively, the reaction may be written: A (g) + B (l)  C (aq) in which case the  G 0 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) Henry’s Law

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15 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)

16 (45) Untuk reaksi multi: dengan j adalah nomor reaksi kimia. Untuk reaksi fasa gas, persamaan (45) menjadi: (46) Jika campuran keseimbangan berupa gas ideal: (47)

17 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: C 4 H 10  C 2 H 4 + C 2 H 6 (I) C 4 H 10  C 3 H 6 + CH 4 (II) Jika kedua reaksi ini mencapai keseimbangan, bagaimana komposisi produk? Data: K (I) = 3,856 K (II) = 268,4 CONTOH

18 Penyelesaian: j i,j j C 4 H 10 C2H4C2H4 C2H6C2H6 C3H6C3H6 CH 4 I – II – Basis: 1 mol umpan C 4 H 10

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20 Keseimbangan kimia: (A) (B)

21 (C) Per. (C) dimasukkan ke pers. (A):

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24 Contoh Setumpukan batubara (dianggap terdiri dari karbon murni) dalam sebuah gasifier dialiri steam dan udara sehingga terjadi reaksi yang menghasilkan gas yang terdiri dari H 2, CO, O 2, CO 2, dan N 2. 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  G 0 f,1500 untuk: H 2 O = – J/mol CO= – J/mol CO 2 = – J/mol Karena temperatur cukup tinggi, maka campuran gas dapat dianggap sebagai gas ideal.

25 Penyelesaian Kemungkinan reaksi yang terjadi: C + O 2  CO 2 (a) C + CO 2  2 CO(b) H 2 O + C  H 2 + CO(c) Harga K untuk masing-masing reaksi pada 1500K adalah:

26 Dengan cara yang sama diperoleh: Harga K a sangat besar, yang berarti bahwa reaksi (a) merupakan reaksi irreversibel dan semua O 2 habis bereaksi dengan C menjadi CO 2 [reaksi (a)].

27 Jumlah mol gas mula-mula: H 2 O= 1 mol N 2 = 0,79  2,38 = 1,88 mol n 0 =3,38 O 2 = 0,21  2,38 = 0,5 mol Jumlah mol gas setelah reaksi: n H2O = 1 –  c n N2 = 1,88 n CO2 = 0,5 –  b n CO = 2  b +  c H 2 =  c

28 Fraksi mol masing-masing komponen setelah reaksi:

29 Untuk reaksi (b): (A)

30 Untuk reaksi (c): (B)

31 Persamaan (A) dan (B) diselesaikan secara simultan dengan menggunakan Excel Solver dengan batasan: 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: dan


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