Principles of Chemistry II © Vanden BoutFigure Copyright Houghton Mifflin Company. All rights reservedLast TimeEach equilibrium has different concentrations, but the same value for KcPrinciples of Chemistry II © Vanden BoutRelating ∆RG° to K∆RG° = -RTlnKK = exp(-∆RG°/RT)∆RG°< 0 then K>1 favors products∆RG°> 0 then K<1 favors reactantsK can be very large or very smallPrinciples of Chemistry II © Vanden Bout! A.! ! K = 1 ! B.! ! K = 0.25! C.! ! K = 1.55 x 10-83! D.! K=! 6.83 x 10822H2(g) + O2(g) 2H2O(l)for the following reaction ∆RG° = -474 kJ mol-1at 300K the equilibrium constant isthis is the only one >1-474 kJ mol-1 is “big”K = e+190Principles of Chemistry II © Vanden Bout! A.! ! K = 1 ! B.! ! K = 85,432! C.! ! K = 1.66 x 10-34! D.! K=! 7.23 x 10332HgO(s) O2(g) + 2Hg(l)for the following reaction ∆RG° = +194 kJ mol-1at 300K the equilibrium constant isthis is the only one <1194 kJ mol-1 is “big”K = e-78Principles of Chemistry II © Vanden Bout! A.! ! ∆RG° = 0! B.! ! ∆RG° = -10.4 kJ mol-1! C.! ! ∆RG° = -312 kJ mol-1! D.! ∆RG° = +3.28 kJ mol-1CH3COOH(aq) CH3COO-(aq) + H+(aq)for the following reaction at 300K K = 1.78 x 10-5∆RG° for this reaction isthis is the only one >0equilibrium constant nothugely small or hugely largePrinciples of Chemistry II © Vanden BoutEquilibria and Perturbations (Stress)What happens to a system at equilibrium if I change something likeThe concentration of one of the chemicalsThe PressureThe TemperaturePrinciples of Chemistry II © Vanden BoutQualitatively Understanding "stress"Le Chatlier's PrincipleIf a chemical system at equilibrium experiences a change, then the equilibrium shifts to partially counter-act the imposed change.Principles of Chemistry II © Vanden BoutN2(g) + 3H2(g) 2NH3(g)You find the system at equilibrium,then you decide to add more H2 to the mixtureWhat happens as the reaction goes to a new equilibrium?! A.! ! the concentration of N2 decreases ! B.! ! the concentration of NH3 decreases! C.! ! nothing when equilibrium is reached all the concentrationswill the same as beforeThe system will compensate by moving to "reduce" the stress. You added H2The reaction will try to reduce the amount of H2Principles of Chemistry II © Vanden BoutStressing the concentrationsAdd ReactantsReaction Shifts towards Product Add ProductsReaction Shifts towards Reactants What if I increase the pressure?Principles of Chemistry II © Vanden BoutN2(g) + 3H2(g) 2NH3(g)You find the system at equilibrium at 1 atm,then you decide to increase the pressure to 2 atm.What happens as the reaction goes to a new equilibrium?! A.! ! moves towards the products as they have fewer molecules ! B.! ! moves towards the reactants as they have more molecules! C.! ! moves towards the products as they have a lower free energyYou increased the pressureThe reaction will try to reduce the reduce the pressurethe only way to do this is to reduce the number of molecules (move toward products)Principles of Chemistry II © Vanden BoutDealing with Stress from a Quantitative PerspectiveN2(g) + 3H2(g) 2NH3(g)Equilibrium[N2] = 0.921 M[H2] = 0.763 M[NH3] = 0.157 MKc =[N2][H2]3[NH3]2[.921][.763]3[0.157]2Kc = = 0.06If I increase [N2] to 3 M the system will no longer be at equilibrium. Which way will it shift to get back to equilibrium?Principles of Chemistry II © Vanden BoutN2(g) + 3H2(g) 2NH3(g)Not at equilibrium[N2] = 3 M[H2] = 0.763 M[NH3] = 0.157 MQ =[N2][H2]3[NH3]2[3][.763]3[0.157]2Q = =.0185Q = 0.0185K = 0.06Q < K therefore reaction needs to increase products to get to equilibriumnot at equilibriumPrinciples of Chemistry II © Vanden BoutK is constantK =ProductsReactantsConstant!So if products goes upthe reaction will shift to get back to the same constant ratioThis can happen if Product goes down slightly and Reactant goes up slightlyPrinciples of Chemistry II © Vanden BoutTwo equilibrium constantsN2(g) + 3H2(g) 2NH3(g)ConcentrationsKc =[NH3]2[N2][H2]3solutionsPartial PressuresKp =PNH32PN2PH23gasPrinciples of Chemistry II © Vanden BoutIncreasing PressureKp =PN2O4PNO22=XN2O4 PXNO22 P22NO2(g) N2O4(g)If you increase PThen the mole fraction of NO2 must go down since K is constantXN2O4 XNO22 P=Principles of Chemistry II © Vanden BoutRelating Kp and Kc2NO2(g) N2O4(g)Kc =[NO2]2[N2O4]Kp =PN2O4PNO22PN2O4 = nN204RTV= [N2O4]RTconcentrationPrinciples of Chemistry II © Vanden BoutRelating Kp and Kc2NO2(g) N2O4(g)Kc =[NO2]2[N2O4]Kp =PN2O4PNO22=[N2O4]RT[NO2]2(RT)2= Kc1RTIn general KP = Kc(RT)ΔnΔn is the change in the number of moles of gasPrinciples of Chemistry II © Vanden BoutTemperature ChangeN2(g) + 3H2(g) 2NH3(g) + heatthis reaction is exothermicIf you increase T then to "partially compensate" the reactionsshifts to the reactants (consuming heat)Principles of Chemistry II © Vanden BoutHow to change the pressure (constant T)Increase P (decrease V) Shifts to side with fewer gas moleculesDecrease P (increase V) Shifts to side with more gas moleculesAdd an inert gas (one that doesn't react. Like He) Constant PThis is like diluting the systemincrease in Vlike lowering Pshift to side with more gas moleculesConstant VThis is like essentially doing nothingThe partial pressures of all themolecules that matter are unchanged(the number of collisions
View Full Document
Unlocking...