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Hubungan Tetapan Kesetimbangan Dengan Energi Bebas Gibbs

Hubungan Tetapan Kesetimbangan Dengan Energi Bebas Gibbs. Riza Gustia (A1C109020). α A + β B ↔ γ C + δ D. Jika reaksi mencapai kesetimbangan , ΔG= 0 Menurut persamaan reaksi isotherm van’t hoff : ΔG= ΔG⁰ + RT ln Kes = 0 ΔG⁰ = - RT ln Kes ΔG⁰ = - RT ln K

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Hubungan Tetapan Kesetimbangan Dengan Energi Bebas Gibbs

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  1. HubunganTetapanKesetimbanganDenganEnergiBebas Gibbs RizaGustia (A1C109020)

  2. αA + βB ↔ γC + δD Jikareaksimencapaikesetimbangan, ΔG= 0 Menurutpersamaanreaksi isotherm van’thoff: ΔG= ΔG⁰ + RT lnKes = 0 • ΔG⁰ = - RT lnKes • ΔG⁰ = - RT ln K • K= tetapankesetimbangan =

  3. Untukreaksi gas ideal : ΔG⁰ = - RT lnKp Kp = kes Untukreaksidalamlarutan ideal : ΔG⁰ = - RT lnKc Kc = kes

  4. Hubunganantara K, KpdanKc K = Kγ . Kp Untuk gas ideal Kγ =1, sehingga K = Kp K = Kγ . Kc Kγ = Koefisienkeaktifan

  5. Contohsoal 1 Dari reaksi : ½ N2 + 3/2H2 ↔ NH3 . Pada 10 atmdan 450o : N2 = 25 %, H2 = 75 %, dan NH3 = 2,04 % denganKγ = 0,0995. Tentukan K danKppadakeadaaanini :

  6. Penyelesaian Diket : Kγ = 0,0995 T = 273 + 450 = 723 K Ptotal = 10 atm Ditanya : K danKp :…….. ? Jawab : Ptotal = PNH3 + ( PN2+H2) PN2+H2 = Ptotal - PNH3

  7. PNH3 = 2,04% x 10 atm = 0,204 atm PN2 = 25% x 9,796 atm = 2,449 atm PH2 = 75% x 9,796 atm = 7,347 atm Kp = = Kp = 6,55 .10-3 K = Kγ . Kp K = 0,995 x 0,00655 K= 6,52 . 10-3

  8. Contohsoal 2 N2O4 (g) ↔ 2NO2 (g) Pada 25oC, tekanan total 0,2118 atm, tekananparsial NO2adalah 0,1168 atm. Hitunglah ∆Gountuk N2O4(g) menjadi NO2 (g)

  9. Penyelesaian Diket : Ptotal = 0,2118 atm PNO2 = 0,1168 atm Dit : ∆Go = ? Jawab : PN2O4 = Ptotal – PNO2 = 0,2118 atm – 0,1168 atm = 0,0950 atm

  10. Kp = = = 0,144 ∆GO = - RT lnKp ∆GO = - 2,303 RT log Kp = - 2,303 (8,314 J. mol-1. K-1) (298 K) (-0.842) = 4,80 x 103 J.mol-1 = 4,80 kJ . mol-1

  11. Latihan • Persamaanreaksi : NO2(g) ↔ NO(g) + O2(g) .Pada 25oC, tekananparsial NO2, NO,dan O2berturut-turutadalah 0,3 atm; 0,2 atm; dan 0,3 atm. TentukanlahKpdan ∆Goreaksitersebut. • Dari reaksi : N2O4 (g) ↔ 2NO2 (g) ,padasuhutertentu, tekananparsial 2NO2sebesar 0,15 atm, dan Kγ = 0,998 dan K = 0,03. tentukantekanan total darireaksitersebut.

  12. Jawaban • 2 NO2(g) ↔ 2NO(g) + O2(g) Kp = = = 0,133 ∆GO = - RT lnKp = - (8,314 J. mol-1. K-1) (298 K) (-2,074 ) = 4998,2 J. mol-1 = 4,9982 kJ. mol-1

  13. N2O4 (g) ↔ 2NO2 (g) K = Kγ . Kp Kp = = = 0,03 Kp = P N2O4 = = = 0,75 Ptotal = P NO2 + P N2O4 = 0,15 + 0, 75 Ptotal = 0,9 atm

  14. TERIMAKASIH

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