CHEM 106 1nd Edition Lecture 21Outline of Last Lecture II. Relationship between Kc and KpIII. Manipulating Equilibrium Constant ExpressionsIV. Combining ReactionsV. Reaction Quotient (Q)VI. heterogeneous EquilibriaOutline of Current Lecture VII. Using K to Calculate [Equilibrium]VIII. Le Chatelier’s PrincipleIX. Calculations based on Ka. ICE table with Quadratic FormulaCurrent LectureChapters 16: Chemical Equilibria (continued)Using K to calculate [equilibrium]:Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) Kp = 1.45 e-5 @ 500°C[Equilibrium] PH2 = 0.928 atmPN2 = 0.432 atmCalculate [Equilibrium] of NH3:N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)0.928 ¿3¿(0.432)¿Kp=PN H32PN 2PH23=X2¿0.928 ¿3=5.01 e−6X2=(1.45 e−5)(0.431)¿X =2.24 e−3atm=PNH 3You can check you answer!2.24 eo .928 ¿3¿(0.432 )¿(¿¿−3)2¿Kp=¿16.7 Le Chatelier’s Principle If a system at equilibrium is perturbed, it will respond in a way that returns it to equilibrium.Qualitative; similar to comparing Q & KFig. 16.5 pg. 781Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) initially @ equilibriumWhat happens if we add more H2?Using Le Chatelier (qualitative)Excess of reactants, must consume, shift to the rightComparing Q vs K (quantitative)N H3¿2¿H2¿3[N2]¿¿Kp=¿increases denominator!Q < K, shifts RIGHT Fig. 16.5Can also shift equilibrium by changing T or VLet’s do volume first:Following the reaction at equilibrium:2 NO2 (g) ↔ N2O4 (g)“Squeeze it”, e.g. decrease volume by pushing in piston, pressure increasesHow does the system respond?Le Chatelier (Qualitative)System will want to minimize # of molecules, so, shifts rightCompare Q & K (Quantitative)At equilibrium, let [NO2] = X, [N2O4] = Y K = YX2When V is reduced, [species] doubles so [NO2] = 2X, [N2O2] = 2YQ = 2 X ¿2¿¿2 Y¿Q < K, reaction shifts RIGHTNote: Not every perturbation results in a shift towards equilibrium.If # molecules for reactants and products are equal, changing the volume does not disturb the equilibrium. NO shift left or right.Ex: A ↔ B K = [B][ A ]=XYDecrease volume by 1/3:Q = 3 X3 Y=XY=Kno shift in equilibriumWhat is the effect of changing T?Think of it as adding or removing heat. Treat heat as a reactant or product:Exothermic Reaction: heat as product2 H2 + O2 ↔ 2 H2O + heatPerturb reaction by lowering T, remove heatLe Chatelier (qualitative)System responds to produce more heat, reaction shifts to the RIGHTCompare Q & K (quantitative)It’s possible to show mathematically but we need more theory to explain: we will come back to this later.Another Ex: Cl2 (g) ↔ 2 Cl (g) ΔH > 0, endothermicSince endothermic, heat is reactantSuppose we raise T, what happens?Le Chatelier (qualitative)Heat + Cl2 ↔ 2ClAdding more heat shifts reaction RIGHTSo, what if a catalyst is present?Catalyst: 1. Speed up reaction2. Change mechanism3. Neither created nor destroyed4. Does not appear in KCatalysis has no effect on K!If catalyst speeds forward reaction, it also speeds the reverse reaction.16.8 Calculations based on KAlready did example where you calculate [equilibrium], given value for K & some of the [equilibrium], e.g. [reactant] given, calculate [products].Now let’s calculate [equilibrium] that are produced from [initial].ICE Table: Initial, Change, EquilibriumAlways start with a balanced chemical reaction!H2 (g) + I2 (g) ↔ 2 HI (g) Kc = 50.5[H2]I = 1.0M, [I2]I = 2.0 M @ T = 448°CWhat are the [equilibrium] of H2, I2, HI, in moles per liter?H2 (g) + I2 (g) ↔ 2 HI (g)I 1.0 M 2.0 M OC -X -X + 2XE 1.0 - X 2.0 - X 2X2 x¿¿¿2¿Kc=[ HI ]2[H2][ I2]= ¿(2 x)2(1.0−x )(2.0−x ) = 50.54 x2=50.5(x2−3.0 x +2.0)46.5 x2−151.5 x+101.0=0Quadratic Equation:On calculator (App-3)x=−b±√b2−4 ac2 aQuadratic gives 2 solutions for XX = 2.323 or x = 0.935Which value do you choose?[I2] = 2.0 – x = 2.0 – 2.32 = Negative #Meaningless to have a negative concentration!So, instead we use x = 0.935:[H2] = 1.000 – X = 0.065 M[I2] = 2.000 – X = 1.065 M[HI] = 2X = 1.87 MCheck your answer:Kc=[ HI ]2[H2][ I2]=(1.87)2(0.065)(1.065)=51Another example: N2 (g) + O2 (g) ↔ 2NO (g)Kp = 1 e-5, initial PN2 = 0.79 atmPO2 = 0.21 atmWhat are the [equilibrium] of N2, O2, NO?N2 (g) + O2 (g) ↔ 2 NO (g)I 0.79 M 0.21 M OC -X -X + 2XE 0.79 - X 0.21 - X +2XKp=(PNO)2PN2PO2=1 e−5(2 x)2(0.79− x)(0.21−x )=1 e−54 x2=(1e−5) (0.166−1.00 x +x2) 4 x2=1.66 e−6−1e−5x +1 e−5x2When you simplify, no need to carry out math beyond 2 sig figs!4 x2+1 e−5x−1.66 e−6=0Use quadratic equationX = 6.43 e-4 atmSo, PO2 = 0.21 – X = 0.21 atmPN2 = 0.79 – X = 0.79 atmPNO = 2X = 1.3 * 10-3