Chemical EquilibriumEquilibrium vs. KineticsSlide 3Law of Mass ActionSlide 5Slide 6Direction of Reaction and QN2O4(g) 2 NO2(g)Slide 9Slide 10Slide 11Slide 12PracticeWrite KSlide 15Slide 16K vs. QSlide 18K for multistep reactionsSlide 20Slide 21Multiples of KSlide 23MultiplesHeterogeneous EquilibriumKc vs. KpKp and KcSlide 28Slide 29Slide 30ProblemsSlide 32Le Châtelier’s PrincipleSlide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Chemical EquilibriumChapter 17Chapter 17Equilibrium vs. KineticsKinetics:Kinetics:speed of a reaction or processspeed of a reaction or processhow fast?how fast?Equilibrium:Equilibrium:extent of reaction or processextent of reaction or processhow much?how much?Chemical EquilibriumReactant and product concentrations remain constantReactant and product concentrations remain constantMolecular level:Molecular level:rapid activity (dynamic)rapid activity (dynamic)Macroscopic level:Macroscopic level:unchangingunchangingAt equilibrium:At equilibrium:raterateforwardforward= rate= ratereversereverseDoes not limit timeDoes not limit timeLaw of Mass ActionIf:If:Then:Then:And:And:reverseforwardrate rate nm]k[products ]reactants[k K]reactants[k]k[products rateratemnreverseforwardK = equilibrium constantm, n = coefficients in balanced equationLaw of Mass ActionFor:For:Equilibrium Equilibrium expression:expression:mD lC kB jA kjmlc[B][A][D][C] K Value of K Favors K < 0.01 ReactantsK > 100 Products0.01 < K < 100 NeitherDirection of Reaction and QFor:For:Equilibrium Equilibrium expression:expression:Reaction Reaction Quotient:Quotient:mD lC kB jA kjmlc[B][A][D][C] K  tat time, [B][A][D][C] Qkjmlc]O[N][NO K4222cN2O4(g) 2 NO2(g)Practice1.1.Writing expression for KWriting expression for K2. 2. Q vs. K and reaction directionQ vs. K and reaction direction3.3.K for a multistep processK for a multistep process4.4.K for reaction “multiples”K for reaction “multiples”Write KThe decomposition of dinitrogen pentoxide:N2O5(g) NO2(g) + O2(g)The combustion of propane gas:C3H8(g) + O2(g) CO2(g) + H2O(g)Write KThe decomposition of dinitrogen pentoxide:2 N2O5(g) 4 NO2(g) + O2(g)The combustion of propane gas:C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)252242c]O[N][O][NO K 52834232c]][OH[CO][H][CO K Write K4 NH3(g) + O2(g) 4 NO(g) + 6 H2O(g)2 NH3(g) + 5/2 O2(g) 2 NO(g) + 3 H2OK vs. QFor the reaction: N2O4(g) 2NO2(g)Kc = 0.21 at 1000C. At a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M. (a) Find Q. Is the reaction at equilibrium?(b) If not, in which direction is it progressing?K vs. QN2O4(g) 2NO2(g) Kc = 0.21 at 1000C. At a point, [N2O4] = 0.12M and [NO2] = 0.55M. (a) Find Q. Is the reaction at equilibrium?(b) If not, in which direction is it progressing?5.2(.12)(0.55) ]O[N][NO Q24222cleft Right toso K QccK for multistep reactionsNitrogen dioxide, a toxic pollutant that contributes to photochemical smog, can develop in combustion engines from N2 and O2.(1) N2 + O2 2NO Kc1 = 4.3 x 10-25(2) 2NO + O2 2NO2 Kc2 = 6.4 x 109(a) Show that Qc for the overall reaction is the same as the product of Qcs of the individual reactions.(b) Calculate Kc for the overall reaction.K for multistep reactions(1) N2 + O2 2NO Kc1 = 4.3 x 10-25(2) 2NO + O2 2NO2 Kc2 = 6.4 x 109 N2 + 2 O22 NO2(a) Qc(b) Kc,overall222222222222c2c1c]][O[N][NO][O[NO]][NO ]][O[N[NO]QQ Q    15925c2c1c108.2106.4 103.4K KKK for multistep reactionsFor the following(1) Br2 2 Br(2) Br + H2 HBr + H (3) H + Br HBr(a) Write the overall balanced reaction.(b) Write out the individual expressions for Qc and show that their product is equivalent to the overall Qc.Multiples of KFor the ammonia reaction:N2(g) + 3H2(g) 2NH3(g)Kc is 2.4x10-3 at 1000K.Find K for the following:(a) 1/3 N2+ H22/3 NH3(b) NH3 1/2 N2 + 3/2 H2Multiples of KN2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3(a) 1/3 N2+ H22/3 NH3(b) NH3 1/2 N2 + 3/2 H2 13.0104.2 KK31331cac, .20104.2 KK21321cbc,MultiplesN2(g) + O2(g) 2 NO(g)Kc = 1 x 10-30Write the expression for Q and determine its value for ½ N2(g) + ½ O2(g) 2 NO(g)H2(g) + Cl2(g) 2 HCl(g)Kc = 7.6 x 108Write the expression for Q and determine its value for 2/3 HCl(g) 1/3 H2(g) + 1/3 Cl2(g)Heterogeneous EquilibriumPURE solids and liquids do not appear in expression for K (or Q). 2CO K Kc vs. KpFor:For:mD lC kB jA np K(RT)KmolKatmL 0.08206 Rgas moles # in change )nn( nreactantsproductsKp and KcFor the ammonia reaction, N2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3Find Kp at 1000 K.Kp and KcN2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3Find Kp at 1000 K.2)4()2( n)nn( nreactantsproducts7p23p2cΔncp103.56 K1000) )(0.08206104.2( K(RT)K(RT) K KKp and KcFor the following reaction, PCl3(g) + Cl2(g) PCl5(g),Kc = 1.67 at 500 KFind Kp at 500 K.K vs. QFor the reaction: CH4(g) + Cl2 CH3Cl(g) + HClKp = 1.6x104 at 1500 K. At a point during the reaction, PCH4 = 0.13 atm, PCl2= 0.035 atm, PCH3Cl = 0.24 atm, and PHCl = 0.47 atm.(a) Find Q. Is the reaction at equilibrium?(b) If not, in which direction is it progressing?Problems1.1.Given equilibrium concentrations or pressures, Given equilibrium concentrations or pressures, find K or Q.find K or Q.2. Given K and initial conditions (conc’s or P’s), 2. Given K and initial conditions (conc’s or P’s), find equilibrium quantities (conc’s or P’s).find equilibrium quantities (conc’s or P’s).Le Châtelier’s Principle. . . if a change is imposed on a . . . if a change is imposed on a system at equilibrium, the system at equilibrium, the position of the equilibrium will position of the equilibrium will shift in a direction that tends to shift in a direction that tends to reduce that change.reduce that change.Le Châtelier’s