复旦大学：《普通化学 General Chemistry》课程教学资源（课堂讲义）Chap 10 - Chemical Equilibrium

PART 10-Chemical Equilibrium Reference: Chapter 14, 17 in textbook

PART 10 PART 10 – Chemical Equilibrium Chemical Equilibrium Reference: Chapter 14 17 in textbook Reference: Chapter 14, 17 in textbook 1

Basic Properties 1. Dynamic equilibrium eg.N2O4(g)s2N○2(g) At equilibrium, Vforward = backward showing the system reaches an equilibrium state Q How to draw the graph by v-t? Xm0.6

Basic Properties z 1. Dynamic Equilibrium „ e.g. N 2 O 4 (g) ' 2 NO 2 (g) At ilib i At equilib rium, vforward = vbackward Æ showin gy q the s ystem reaches an e quilibrium state. „ Q: How to draw the graph by v ~ t? 2

Basic Properties 2. System reaches an equilibrium state spontaneously All the equilibrium states are under a set of conditions, and can be affected when conditions change 3. It does not matter whether the system starts from reactant side or product side e.g.N2O4(9)s2NO2(9) Case 1: Starting from 1 mol N2O4(g) Case 2: Starting from 2 mol NO2(g)

Basic Properties z 2. Sys e eac es a equ b u s a e t em reac hes an equili bri um s t a t e spontaneously „ All th ilib i t t d t f diti All the equilib rium s t a tes are un der a se t o f conditions, and can be affected when conditions change. z 3. It does NOT matter whether the system starts from reactant side or product side. „ eg N . . N O (g) ' 2 NO (g) 2 O 4 (g) ' 2 NO 2 (g) Case 1: Starting from 1 mol N 2 O 4 (g) Case 2: Starting from 2 mol NO 2 (g) 3

Basic Properties 4. Driving force The coexistence of two driving forces leads to an equilibrium e.g. n2O4 (g)# 2 NO2(g) Reduced energy( enthalpy):2N○2(g)→N2O4(g) Increased entropy: n2O4(9)>2 NO2(g) At Equilibrium:△G=0,(.e.△H=T·△S)

Basic Properties z 4 Dri ing Force 4. Dri ving Force „ The coexistence of two driving forces leads to an equilibrium. e.g. N 2 O 4 (g) ' 2 NO 2 (g) Reduced energy (enthalpy): 2 NO 2 (g) Æ N 2 O 4 Reduced energy (enthalpy): 2 NO (g) 2 (g) Æ N 2 O 4 (g) Increased entropy: N 2 O 4 (g) Æ 2 NO 2 (g) „ At Equilibrium: ΔG = 0, (i.e. ΔH = T • ΔS) 4

Equilibrium Constant Equilibrium Constant(K) Revisit: Kw, Ka, Kb, Ksp, Formally defined -Mass Law For a reaction aa tbb f cc+dd At equilibrium K=cc·CD)/(CAa·CB K=(Pc·P/(PAa·Pgb),(iA,B,C, D are gases) Equilibrium Constant K is only the function Of T

Equilibrium Constan t z E q () uilibrium Constant ( K ) „ Revisit: K w, K a, K b, Ksp, …… „ Formally defined – Mass Law For a reaction: a For a reaction: a A + b B ' c C + d D At equilibrium: K = (C C c • C D d) / (C A a • C B b ) K ( P c P d)/( P a P b K = ( P ) (if A B C D ) C c • P D d ) / ( PA a • P B b ), (if A, B, C, D are gases ) z Equilibrium Constant K is only the function Equilibrium Constant K is only the function of T 5

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 △Gr0 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 Product side Æ Reactant side 10 Δ G T > 0 Product side Æ Reactant side