KESEIMBANGAN REAKSI KIMIA

1. KESEIMBANGAN REAKSI KIMIA (3)

2. PHASE RULE FOR REACTING SYSTEMS

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 rindependent 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 CaCO3 into an evacuated space. Solution A single chemical reaction occurs: CaCO3(s) CaO(s) + CO2(g) r = 1 N = 3  = 3 (solid CaCO3, solid CaO, and gaseous CO2) F = 2 –  + N – r =2 – 3 + 3 – 1 = 1

9. REACTION IN HETEROGENEOUS SYSTEM

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. Method 1 • 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.

12. Method 2 • 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.

13. Method 3 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: Henry’s Law (44)

14. KESEIMBANGAN REAKSI MULTI

15. Jikakeadaankeseimbangandalamsuatusistemreaksitergantungpadaduaataulebihreaksikimiaindependen, makalangkah-langkahuntukmenentukankomposisikeseimbanganadalah: • Tentukanreaksikimia yang terjadi. • Komposisikeseimbangandihitungseperti yang telahdibahas. • Konstantakeseimbanganuntuksetiapreaksidievaluasidengancaraseperti yang telahdibahas. • Untukreaksitunggal: (12)

16. Untukreaksi multi: (45) dengan j adalahnomorreaksikimia. Untukreaksifasa gas, persamaan (45) menjadi: (46) Jikacampurankeseimbanganberupa gas ideal: (47)

17. CONTOH Terhadapsenyawa n-butanadilakukanreaksicrackingpada 750 K dan 1,2 bar sehinggadihasilkan olefin. Hanyaduareaksi yang memilikikonversikeseimbangan yang signifikan, yaitu: C4H10 C2H4 + C2H6 (I) C4H10  C3H6 + CH4 (II) Jikakeduareaksiinimencapaikeseimbangan, bagaimanakomposisiproduk? Data: K(I) = 3,856 K(II) = 268,4

18. Penyelesaian: Basis: 1 mol umpan C4H10

20. Keseimbangankimia: (A) (B)

21. (C) Per. (C) dimasukkanke pers. (A):

24. Contoh Setumpukanbatubara (dianggapterdiridarikarbonmurni) dalamsebuahgasifierdialiri steam danudarasehinggaterjadireaksi yang menghasilkan gas yang terdiridari H2, CO, O2, CO2, dan N2. Jika gas yang diumpankankedalamgasifierterdiridari 1 mol steam dan 2,38 mol udara, hitungkomposisikeseimbangandarialiran gas yang keluardarigasifierpada P = 20 bar dantemperatur1500 K. DiketahuinilaiG0f,1500untuk: H2O = – 164.310 J/mol CO = – 243.740 J/mol CO2 = – 396.160 J/mol Karenatemperaturcukuptinggi, makacampuran gas dapatdianggapsebagai gas ideal.

25. Penyelesaian Kemungkinanreaksi yang terjadi: C + O2 CO2 (a) C + CO2  2 CO (b) H2O + C  H2 + CO (c) Harga K untukmasing-masingreaksipada1500K adalah:

26. Dengancara yang samadiperoleh: Harga Kasangatbesar, yang berartibahwareaksi (a) merupakanreaksiirreversibeldansemua O2habisbereaksidengan C menjadi CO2 [reaksi (a)].

27. Jumlah mol gas mula-mula: H2O = 1 mol N2 = 0,79  2,38 = 1,88 mol n0 =3,38 O2 = 0,21  2,38 = 0,5 mol Jumlah mol gas setelahreaksi: nH2O = 1 – c nN2 = 1,88 nCO2 = 0,5 – b nCO = 2 b + c H2 = c

28. Fraksi mol masing-masingkomponensetelahreaksi:

29. Untukreaksi (b): (A)

30. Untukreaksi (c): (B)

31. Persamaan (A) dan (B) diselesaikansecarasimultandenganmenggunakan Excel Solver denganbatasan: dan Hasilperhitungan: b = 0,4895 dan c = 0,9863 Jikanilaibdancinidimasukkankepersamaanuntukfraksi mol masing-masingkomponenmakaakandiperoleh: