a. The representation of K should be consistent with the chemical reaction, and designated with the temperature Example: N2O4(g) 2NO2(9) 11.01 b. In a heterogeneous(multi-phase)equilibrium, K only relates to the gas pressures and solution concentrations Example: CaCO3(s) CaO(s)+ Co2(g) K =P EXample: Zn(s)+ 2 H+(aq)# Zn+(aq)+ H2(g) K=[zn2+]·P2/[H]2 6
a. The representation of K should be consistent with the a. The representation of K should be consistent with the chemical reaction, and designated with the temperature. Example: N 2 O 4 (g) ' 2 NO 2 (g) K 0 373K = 11.01 b. In a heterogeneous (multi-phase) equilibrium, K only rel t t th d l ti t ti la tes to the gas pressures an d solution concen trations. Example: CaCO 3 (s) ' CaO (s) + CO 2 (g) K = PCO2 Example: Zn (s) + 2 H + (aq ) ' Zn2+ Example: Zn (s) + 2 H (aq)+H (g) + (aq ) ' Zn2+ (aq ) + H 2 (g) K = [Zn2+] • PH2 / [H + ] 2 6
C. In(diluted)water solution, the [water] is 1 and unchanged EXample: Cr2 O,2(aq)+ H2o (# 2 Cro42(aq)+2 H*(aq) K=(CrO42]2·叶H]2)/[Cr2O2] d. For different chemical reactions K has different values N2(g)+3H2(g)s2NH3(g) 佐N2(9)+3/2H2(9)sNH3(9)Kxr=(K)2 2NH3(g)sN2(g)+3H2(g) K =1/KI 7
c. I (dil t d) t l ti th [ t ] i 1 d h d In (dil u t ed) wa ter solution, the [wa ter ] is 1 an d unc hange d. Exam ple: C r 2 O 7 2- ( a q ) + H 2O (l) ' 2 CrO 4 2- ( a q ) + 2 H + p ( a q ) 2 7 ( q ) 2 ( ) 4 ( q ) ( q ) K = ([CrO 4 2- ] 2 • [H + ] 2) / [Cr 2 O 7 2- ] d. For different chemical reactions, K has different values. N 2 (g) + 3 H 2 (g) ' 2 NH 3 N 2 (g) K T (g) + 3 H 2 (g) ' 2 NH 3 (g) K T ½ N 2 (g) + 3/2 H 2 (g) ' NH 3 (g) K T ’ = (K T )1/2 2 NH3 (g) ' N 2 (g) + 3 H 2 (g) K T ’’ = 1 / K T 7
Practice Q: For a reaction: COCI2(g)+ Co(g)+ Cl2(g) In a fixed volume container at 900 K there was some CoCl2(g)with initial pressure of 101.3 kPa. When this dissociation reaction reached equilibrium, the total pressure of the container is 189.6 kPa Calculate the equilibrium constant Solution COCl2(g) Co(g)+ CI2(g) itial 101.3 0 0 Change X X Equilibriu 101.3 (101.3-X) 1896 X=88.3 kPa K=(883)*(88.3)/(101.3-88.3)=600 8
Practice Q: For a reaction: COCl2 (g) ' CO (g) + Cl2 (g) In a fixed volume container at 900 K there was some In a fixed volume container at 900 K, there was some COCl2 (g) with initial pressure of 101.3 kPa. When this dissociation reaction reached eq, p uilibrium, the total pressure of the container is 189.6 kPa. Calculate the equilibrium constant. Solution: COCl2 (g) ' CO (g) + Cl2 (g) I iti l Initial: 101 3 0 0 101.3 0 0 Change: –x x x Equilibrium: 101.3 – x x x (101 3 – x) + x + x = 189 6 x = 88 3 kPa 8 (101.3 – x) + x + x = 189.6 x = 88.3 kPa K = (88.3) * (88.3) / (101.3 – 88.3) = 600
Equilibrium Constant&△G For an arbitrary gas reaction aa g bb(g CC(g)+ dd(g △G=c△Gc+d△G-(a△GA+b△G) △G=c△G°c+d△G°-(a△Ga+b△G°g) rt[(c In Pc+dIn pd)-(a In PA+ b In PB) sInce:c△G°c+d△G-(a△G°a+b△G°g)=△G° Therefore △G=△G+RTln[(Pc·P/(Pa·PBb)]
Equilibrium Constant & Δ G For an arbitrary gas reaction: For an arbitrary gas reaction: a A (g) + b B (g) ' c C (g) + d D (g) ΔG = c Δ G C + d Δ G D – (a Δ G A + b Δ G B ) Δ G Δ G o + d Δ G o ( Δ G o + b Δ G o Δ G = c Δ G ) o C + d Δ G o D – ( a Δ G o a + b Δ G o B ) + RT [(c ln P C + d ln P D ) – (a ln PA+ b ln P B [( )] C D ) ( A B)] since: c Δ G o C + d Δ G o D – (a Δ G o a + b Δ G o B) = Δ G o Therefore: Δ G = Δ G o + RT ln [ ( P C c • P D d)/( P a a • P B b) ] 9 Δ G Δ G RT ln [ ( P C P D ) / ( P a P B ) ]
Equilibrium Constant&△G Or in a solution reaction △G=△G°+RTin[cc·C/Ca·Cgb)] Qp Q aB AGr=AG°+ RTIn g △Gr<0 Reactant side→ Product side AGT =0 No net reaction (System in equilibrium) AG >0 Product side- Reactant side 10
Equilibrium Constant & Δ G Or in a solution reaction: ΔG = Δ G o + RT ln [ (C C c • C D d) / (C a a • C B b) ] C d C d b P B a A d D C C Q P P P P = ⋅ ⋅ b C B a A d D C C Q C C C C = ⋅ ⋅ Δ G T = Δ G T ° + RT ln Q A B A B G T G T Q Δ G T < 0 Reactant side Æ Product side Δ G T = 0 No net reaction (System in equilibrium) Δ G > 0 Product side Æ Reactant side 10 Δ G T > 0 Product side Æ Reactant side