CHEM 106 1nd Edition Lecture 20Outline of Last Lecture I. EquilibriaII. Writing Equilibrium Constant ExpressionsIII. Values and Terminology for KOutline of Current Lecture II. Relationship between Kc and KpIII. Manipulating Equilibrium Constant ExpressionsIV. Combining ReactionsV. Reaction Quotient (Q)VI. heterogeneous EquilibriaCurrent LectureChapters 16: Chemical Equilibria (continued)16.3 Relationship between Kc and KpIdeal gas law, sections 6.4 pg. 264PV = nRT solve for PP = nvRTwhere nv is concentration (molarity) (m)P = MRT or P = [X] RTFor any gas x, Px = [X] RTEx: 2SO2 (g) + O2 (g) ↔ 2SO3 (g)PSO2 = [SO2] RT Po2 = [O2] RT PSO3 = [SO3] RTKp = PS O32PSO22PO2Kp = [S O3]RT ¿2¿[S O2]RT ¿2([O2]RT )¿¿¿= S O3¿2¿S O2¿2[O2]¿¿¿¿Kp = SORT ¿−1[¿¿2]2[O2]¿[ S O3]2¿More generally:aA + bB ↔ cC + dDKp = RT ¿∆ nPCcPDdPAaPBb=Kc(RT)=Kc ¿Ex: N2 + 3H2 ↔ 2NH3Kc = NHNH[¿¿2][¿¿2]3=2.8 e−9@ 300 K¿[¿¿ 3]2¿¿What is Kp?Kp=Kc(RT)∆ n∆ n=2−(3+1)=−224.86 ¿−2=4.5 e−12Kp=(2.8 e−9)¿Note: R=0.082L∗atmmol K,T =300 K , RT=24.8616.4 Manipulating Equilibrium Constant ExpressionsReview Hess’s Law, pg. 230When we write A ↔ B, it is short-hand for 2 separate reactions:Forward: A → BReverse: B → AFor the forward reaction: Kf =[B][ A ]For the reverse reaction: Kr=[ A ][B]So, Kr=1Kfreciprocal relationshipFor N2 + 3H2 ↔ 3NH330°CNHNH[¿¿2][¿¿2]3=2.8 e−9¿[¿¿3 ]2¿Kf =¿HNH[¿¿3 ]2=12.8 e−9=3.6 e8[N2][¿¿2]3¿Kr=¿What if the equilibrium is multiplied by some factor?A ↔ BK 1=[B][ A ]2A ↔ 2BK 2=[B]2[ A ]2=K 12Ex: 2HI (g) ↔ H2 (g) + I2 (g), KP1 compared to:6HI (g) ↔ 3H2 (g) + 3I2 (g), Kp2Kp2 = Kp13Another ex: 2 NO + O2 ↔ 2NO2NO[¿¿2]2[NO ]2[O2]K 1=¿but we want value in terms of 1 mol NO2NO + ½ O2 ↔ NO2Kc=[NO2][NO][O2]12note: [O2]1/2 = √[O2]What if we combine reactions?A ↔ B + C K1B ↔ D K2A ↔ C + D K3K 1=[B][C][ A ]; K 2=[ D][B ]; K 3=[C][ D][ A]K 1+K 2=[B][C ][ A ]∗[D][B]=[C] [D][A]=K 3Given the following:HF (aq) ↔ H+ (aq) + F- (aq) KC1 = 6.8 e-4H2C2O4 (aq) ↔ 2H+ (aq) + C2O42- (aq) KC2 = 3.8 e-6What is the value of Kc for the following?2HF (aq) + C2O4 (aq) ↔ 2F- (aq) + H2C2O4First, work out the chemical equations: 2 * reaction1, reverse reaction 22 HF ↔ 2 H+ + 2 F-2 H+ + C2O42- ↔ H2C2O42 HF + C2O42- ↔ 2 F- + H2C2O4Since reaction 1 was multiplied by 2: KC12 = 13.8 e−6 = 2.6 e5Then multiply the values of Kc: (4.6 e-7)(2.6 e5) = 0.12 for the overall KcThe Rules, pg. 7801. The value of KC for a reaction running in reverse is its reciprocal Kr=1Kf2. If a reaction is multiplied by a factor n, the value of K for that reaction is Kn3. If an overall reaction is the sum of 2 or more reactions, Koverall is the product of the K’s for the individual reactions.16.5 Equilibrium Constant vs. Reaction QuotientH2O + CO ↔ CO2 + H2We used data in Table 16.1 to calculate value of KCThat value is only valid at equilibrium, when the concentrations aren’t changing.Reaction Quotient (Q): numerical value for mass action expression for any concentrations, whether the reaction is at equilibrium or not.At equilibrium, Q = KQ is written just like K:H[¿¿2O][CO]Q=[H2][CO2]¿ but not at equilibriumFig. 16.5 pg 781If Q = K, it is at equilibriumIf Q ≠ K, it is not at equilibriumIf Q < K, need more products, less reactants, shifts to the rightUse data in Table 16.1 to calculate value of Q16.6 Heterogeneous EquilibriaNow, it is very important to consider the phases of reactants and productsHomogenous Equilibria – reactants and products in the same placeHeterogeneous Equilibria – Reactants and products are in different phases.Ex: CaCO3 (s) ↔ CaO (s) + CO2 (g) T = 900 KΔH° = +1781. kJ/molKc=[CaO(s)][C O2(g)][CaC O3(s)]How to handle concentration of a solid?Concentration of pure solid doesn’t change. It’s mass per unit volume doesn’t change. So the concentration of a solid = 1, and it can be omitted from the Kc expressionKc = [CO2 (g)]If pure liquid is involved in the reaction, you can treat it the same way (omit