Chem 170 1st Edition Lecture 21Outline of Last Lecture I. Chapter 14 – Entropy and Free EnergyII. Ch. 15 – Chemical equilibriuma. Characteristics of equilibrium stateb. Empirical law of mass actionOutline of Current Lecture III. Equilibrium Constant (K) examplesIV. Rules of KV. Reaction Coefficient Quotient (Q)VI.∆ G=∆ G0+RTln(Q)Current LectureCh. 15 – Chemical equilibriumEquilibrium Constant (K) examplesex. −¿(aq)−¿(aq)+Cl O3¿−¿(aq)↔2 C l¿3 C l O¿−¿C l¿¿−¿Cl O3¿¿¿¿K=¿note that pure liquids and pure solids do not appear in K, because their activity is 1ex. H2O(l)↔ H2O(g)K=PH2O(g), at 25°C PH2O(g)=0.03126 atmthen amount of H2O(l ) does not matterex. I2(s)↔ I2(aq)K=[I2]These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.if the reaction is at equilibrium (aka. the solution is saturated, where someI2(s) is present) then adding more or taking away I2(s) does not matter. Therefore the amount of pure solvent does not matter.ex. CaCO3(s)↔ CaO(s)+C O2(g)K=PC O2Rules of K: 1. Gasses use partial pressures (atm)2. Dissolved solvents use concentration (M)3. Pure liquids and pure solids do not appear in K even if the solvent participates in the reaction.ex. +¿(aq)−¿(aq)+H¿CH3COOH(aq)↔CH3COO¿is the same as +¿(aq)−¿(aq)+H3O¿CH3COOH(aq)+H2O(l)↔ CH3COO¿4. When there is the reaction aA+bB ↔cC +dD, thenK=[C ]eqc[D]eqd[ A ]eqa[B]eqb for solutionsPPPP(¿¿ A)a(¿ ¿B)b¿(¿¿C)c(¿¿ D)d¿¿K =¿ for gases (use partial pressures)ex. 2Cl2(g)+2 HgO(s)+H2O(l)↔ HgO ∙ HgCl(s)+2 HOCl(aq)K=[HOcl]2[PCl]2when both a dissolved solute and a gas coexist when the reaction is at equilibrium it is a heterogeneous equilibrium. K is unitless so this combination of dissolved solute and gas is alright.2 A+B ↔ CK1=[C][A]2[B]if the reaction is reversedC ↔ 2 A+B K2=1K1=[A]2[B][C]if the reaction is multiplied by some factor4 A +2 B ↔2 CK3=K12=[C]2[A]4[B]2if the reaction is divided by some factorA +(12) B↔(12)CK3=K112=[C]12[ A ][B]12A +B↔ 2 CK=[C]2[ A][B]2C ↔2 D+EK=[D][E][C]2A +B↔ 2 D+EK=[D][E][A][B]Reaction Coefficient Quotient (Q)Q has the same form as the equilibrium constant (K), but Q is used for conditions away from equilibrium. If the reaction is at equilibrium then Q=K.ex. 2 A+B ↔ C, where A=1M, B=1M, C=10M and K=10M.Q=[10][1]2[1]=1 0 =K, therefore the reaction is at equilibriumex. 2 A+B ↔ C, where A=1M, B=1M, C=0M and K=10Q=[0][1]2[1]=0 < KWhen Q<K the reaction goes to the right (toward the products).ex. 2 A+B ↔ C, where A=0M, B=0M, C=10M and K=10.Q=[10][0]2[0]=∞>KWhen Q>K the reaction goes to the left (toward the reactants).∆ G=∆ G0+RTln(Q)If the reaction is at equilibrium (ΔG=0 and Q=K)0=∆ G0+RTln(K)∆ G0=−RTln(K )If the reaction is spontaneous ∆ G0<0then the reaction would go to the products side (Q<K)K=e∆ G−RTIf the reaction is not spontaneous ∆ G0>0then the reaction would go to the reactants side
View Full Document
Unlocking...