KESETIMBANGAN KIMIA

A rections reactans reahes chemical equilbrium when there is no further change in concentrations of the reactants and products.

At equilbrium:The rate of the forward reaction is equal to the rate of the reverse reaction

Forward reaction is a reaction form products from reactants Reaktants Produk

Reverse reaction is a reaction form reactants from producsReaktants Producs

Reversible Reaction a reversible reaction consits of both a forward and a reverse reaction

Reactants Products

k1 rate raction not equal k2

k2k1

At equilibriumThe rate of the forward reaction is equal to the rate of the reverse reaction

Reaktan Produk

k 1 equal k2k1k2

Figure rate reaction vs Time rate reactionTime rate of forward reactionRate of reverse reactionequillibrium

Figure Consentration vs Timeconsentration (mol/L) time[Reactans] decrease[Producs] increase equilbrium

Equilibrium ConstantsaA + bB cC + dD

Equilibrium constants expression is :

Equilibrium contant eq contant expression Produk [C]c [ D]d Kc = ------------- = -------------- coofisients Reaktan [A]a [B]b

Ex : reactions H2 and I2

H2(g) + I2 2HI(g)

[ HI]2 Kc = ------------ [H2][I2]

Kinds of equilibrium1. Homogeneous equillibrium: reaction in which all the reactants and products are in the same state reaches

H2(g) + I2(g) 2HI(g)

2. Heterogeneous equillibrium : the reactan and the products are more states the equilibrium termed CaCO3(s) CaO(s) + CO2(g)

Calculating equilbrium constantsH2(g) + I2 2HI(g)

reactants Products [H2] = o,10 M [HI] = 1,04 M [I2] = 0,20 M

[HI]2 [ 1,04] 2Kc = --------- = ---------------- = 54 [H2][I2] [ 0,10] [ 0,20 ]

Kc for H2(g) + I2 2HI(g) at 427oC

Ekperimen [H2] [I2] [HI] Kc 1 0,10M 0,20M 1,04M 54 2 0,20M 0,20M 1,47M 54 3 0,30M 0,17M 1,66M 54

Using equilibrium constants[ Produk] P > R -----------------= Kc besar [Reactan]

[ Produk] P < R -----------------= Kc kecil [Reaktan]

B A B kc = ------ A

AAAB A > B Kc = kecilAABB A = B Kc = 1ABBB A < B Kc = besar

Le Chateliers prinsiple

when a stress (change in conditions) is placed on a reaction at equilibrium, the equilibrium shifts in the direction that realives the stress.

1.Effect of concentrasion changePCl5(g) PCl3(g) + Cl2(g)

[PCl5] increase forming of the producs [PCl5] decrese forming of the reactants [PCl3] increase forming of the reactants [PCl3] decrese forming of the producs [Cl2] increase forming of the reactants [Cl2] derease forming of the producs

2. Effect of Volume (Pressure) change on equilibrium

When a reaction has the same number of moles of reactants as producs, a volume changes does not effet the equillibrium

Exsample H2(g) + I2(g) 2HI(g)

2 NO2 2 NO + O2 pd T tetapP naik kesetimb bergeser kekiri (mengarah pd pembentukan NO2), V berkurang shg mengurangi efek kenaikan PP kurang kesetimb bergeser kekanan membentuk lebih banyak produk.

[NO]2[O2]Kc = ---------------- [NO2]2 (nNO/V) (nO2/V) n2NO x nO2 1 ------------------- = --------------- x --- (nNO2/V) n NO2 V nNO = jumlah mol NO nO2 = jumlah mol O2 nNO2 = jumlah mol NO2

2 NO2 2 NO + O2 pd T tetap

Bila P dinaikkan V mengecilBila V mengecil,agar nilai Kc tetap maka (nNO)2 (nO2) --------------- akan mengecil jumlah mol NO (nNO2)2 dan O2 berkurang dan jumlah mol NO2 meningkat artinya kesetimbangan bergeser kekiri

3. Efek Perubahan T N2(g) + O2(g) 2NO(g)

A Reaksi Endotermik N2(g) + O2(g) + panas 2NO(g)

Bila T dinaikkan maka Kc naik reaksi akan bergeser menuju reaktan Bila T diturunkan maka Kc turun reaksi bergeser menuju produk

B. Reaksi exotermik N2(g) + O2(g) 2NO(g) + panas

Bila T dinaikkan maka Kc turun shg reaksi menuju produk Bila T diturunkan maka Kc naik shg reaksi menuju reaktan

Kesetimbangan dalam larutan jenuh

Contoh kerusakan gigi dan batu ginjalBakteri dlm mulut bereaksi dg gula dan asam melarutkan enamel gigi yang dibuat mineral dinamakan hydroxyapatite, Ca5(PO4)3OHKerusakan gigi disusun dari garam kalsium seperti kalsium oksalat,CaC2O4 dan kalsium posphat, Ca3(PO4)2

Jika ion Ca2+ dan C2O42- saling larut dlm gigi membentuk endapan padatan CaC2O4

Ca2+(aq) + C2O42- (aq) CaC2O4(s)

3Ca2+(aq) + 2PO42-(aq) Ca3(PO4)2(s)

Konstanta hasil kelarutan (Ksp)

CaC2O4(s) Ca2+(aq) + C2O42- (aq) Pd saat kestmbangan [Ca2+] dan [C2O42- ] konstan shg kelarutan CaC2O4 dlm kontanta hasil kelarutan dapat dinamakan Ksp

Ksp = [Ca2+] [C2O42- ]

Perhitungan KspContoh dibuat larutan CaCO3 jenuh dengan melarutkan CaCO3 padat ditambah air dan distirer sampai pada kestimbangan dicapai CaCO3 Jika dihasilkan [Ca2+ ] = 7,1x10-5 M dan [CO32-] = 7,1x10-5 M maka untuk menghitung Ksp ada 4 langkah

Lankah 1 Tulis pers kesmtbngan kelarutan CaCO3(s) ) Ca2+(aq) + CO32- (aq)

Langkah 2 Tulis Ksp Ksp = [Ca2+] [CO32-]

Langkah 3 substitusi konsentrasi M ion2 Ksp = [7,1x10-5 M ] [7,1x10-5 M ] = 5,0 x 10-9

Perhitungan Kelarutan Molar (S) S adalah mol zat terlarut yang larut dalam satu liter larutanContoh kelarutan molar CdS adalah 1 x 10-12 mol per liter. Artinya bahwa 1 x 10-12 mol CdS masuk dalam ion Cd 2+ dan S2- untuk memberi [Cd 2+ ] = 1 x 10-12 M dan [S2- ] = 1 x 10-12 M

CdS(s) Cd 2+(aq) + S2- (aq) S= 1 x 10-12 mol/L [Cd 2+ ] = 1 x 10-12 M dan [S2- ] =1 x 10-12 MKami dapat menggunakan persamaan kesetibngan dan Ksp untuk menghitung molaritas masing2 ion dan menentukan kelarutan molar

Hitung kelarutan molar (S) PbSO4 jika Ksp adl 1,6x10-8Tahap 1 pers kestbangan larutan PbSO4(s) Pb 2+(aq) + SO42-(aq)

Tahap 2 tulis Ksp Ksp = [Pb 2+] [SO42-]

Tahap 3Tuunjukkan molaritas ion2 S dlm pers dg mengetahui harga Ksp Ksp = [S][S] = 1,6x10-8

Tahap 4 S x S = S2 = 1,6x10-8S = (1,6x10-8 ) = 1,3 x 10-4 mol/LS untuk PbSO4 adalah 1,3 x 10-4 mol/L artinya jika 1,3 x 10-4 mol PbSO4 dilarutkan dlm 1 liter larutan ada ion [Pb 2+] = 1,3 x 10-4 M dan [SO42-] = 1,3 x 10-4 M

Efek ion senamaKelarutan suatu elektrolit akan berkurang bila dilarutkan dlm larutan yang mengandung ion yang sama bila dibandingkan dg kelarutan dlm air murniContoh: AgCl(s) Ag+(aq) + Cl- (aq)[Cl- ]di+ kestmbngan bergeser kekiri

Efek GaramPenambahan ion yang tak senama cenderung untuk meningkatkan kelarutan

Daftar PustakaTimberlake.Karen C, 2005,Basic Chemistry,1st edition,Pearson Benjamin Cummings,San Francisco.Atkins.P.W, 1990,Physical Chemistry,4th edition,Oxford University Press,Tokyo.