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