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Chemical Equilibrium

Chemical Equilibrium

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Chemical Equilibrium

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  1. Chemical Equilibrium • Concept of Equilibrium • Equilibrium constant • Law of mass action • Writing equilibrium constant • Equilibrium constant and balanced equation • Equilibrium concentrations • Reaction quotient • Factors affecting equilibrium • Le Châtelier’s Principle

  2. NaCl(aq) Na+(aq) + Cl–(aq) • HC2H3O2(aq) H+(aq) + C2H3O2–(aq) H2O (l) H2O (g) N2O4(g) 2NO2(g) I. Concept of Equilibrium • Equilibrium: a state in which no change is observed Ex: Consider following reactions: NaCl in water HC2H3O2 in water • Chemical equilibrium is reached when • The forward rates = rates of reverse reactions • No change in concentrations of the reactants and products • Physicalequilibrium

  3. equilibrium equilibrium equilibrium Example: N2O4(g) 2NO2(g) Start with NO2 Start with N2O4 Start with NO2 & N2O4 N2O4 NO2

  4. N2O4(g) 2NO2(g) constant

  5. N2O4(g) 2NO2(g) K = aA + bB cC + dD = 4.63 x 10-3 [NO2]2 [C]c[D]d [N2O4] K = [A]a[B]b Ex. For the reaction H2(g)+ I2(g) 2 HI (g), K=794 at 298 K and K=54 at 700 K. The formation of HI is more favored at ____ K. II. Equilibrium constant A. Law of mass action • Describes the relationship between concentrations of reactants and products Lie to the right Favor products K >> 1 Lie to the left Favor reactants K << 1 298

  6. aA + bB cC + dD [C]c[D]d K = [A]a[B]b B. Writing Equilibrium Constant Expressions • Start with a balanced chemical equation • Reactants in the denominator, products in the numerator • Conc. is raised to the power as the stoichiometric coeff. • Conc. is expressed in M in condensed phase, in M or atm in gas phase • Independent of reaction mechanism • Omit concentrations of pure solids, pure liquids, and solvents in equilibrium constant expression • The equilibrium constant has no unit • The value of the equilibrium constant depends onthe balanced equationand thetemperature.

  7. N2O4(g) 2NO2(g) Kc = Kp = In most cases Kc Kp P2 [NO2]2 NO2 [N2O4] aA (g) + bB (g)cC (g) + dD (g) P N2O4 (1) Homogeneous Equilibrium • All reacting species are in the same phase Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

  8. CO (g) + Cl2(g) COCl2(g) = 0.14 0.012 x 0.054 [COCl2] [CO][Cl2] The equilibrium concentrations for the reaction between carbon monoxide and chlorine to form COCl2 (g) at 74oC are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. Q: = 220 Kc= Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = 0.0821 T = 273 + 74 = 347 K Kp= 220 x (0.0821 x 347)-1 = 7.7

  9. CH3COOH (aq) + H2O (l) CH3COO-(aq) + H3O+ (aq) ' ' Kc= Kc [H2O] *No units for the equilibrium constant, K. [CH3COO-][H3O+] [CH3COO-][H3O+] = [CH3COOH][H2O] [CH3COOH] (1). Homogeneous Equilibrium [H2O] = constant Kc =

  10. 2 PNO PO = Kp 2 PNO 2 Kp = 2NO2 (g) 2NO (g) + O2 (g) PO PO = 158 x (0.240)2/(0.136)2 2 2 2 2 PNO PNO 2 2 The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO2 = 0.240 atm and PNO = 0.136 atm? Q: = 492 atm

  11. CaCO3(s) CaO (s) + CO2(g) ‘ ‘ Kc = Kc x [CaO][CO2] [CaCO3] [CaCO3] [CaO] Kp = PCO 2 (2). Heterogeneous Equilibrium • Reactants and products are in different phases [CaCO3] = constant [CaO] = constant Kc= [CO2] = The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

  12. PCO PCO 2 2 does not depend on the amount of CaCO3 or CaO CaCO3(s) CaO (s) + CO2(g) = Kp

  13. Kp = P NH3 NH4HS (s) NH3(g) + H2S (g) P H2S Consider the following equilibrium at 295 K: The partial pressure of each gas is 0.265 atm. Calculate Kp and Kcfor the reaction? Q: = 0.265 x 0.265 = 0.0702 Kp = Kc(RT)Dn Kc = Kp(RT)-Dn Dn = 2 – 0 = 2 T = 295 K Kc = 0.0702 x (0.0821 x 295)-2 = 1.20 x 10-4

  14. N2O4(g) 2NO2(g) K = K = = 4.63 x 10-3 CO(g)+ Cl2(g) COCl2(g) COCl2(g) CO(g)+ Cl2(g) Kc= =220 Kc= = =0.0045 [NO2] [NO2]2 [N2O4]1/2 [N2O4] 1/2 N2O4(g) NO2(g) [COCl2] [CO][Cl2] 1 [COCl2] 220 [CO][Cl2] C. Equilibrium constant and balanced equation • Kc or Kp depends on the balanced equation • Kof the reverse reaction is the reciprocal of the forward reaction • If a equation is multiplied by a factor of n, then the K is raised to a power of n => Kn = (4.63 x 10-3)1/2 = 0.0680

  15. A + B C + D C + D E + F A + B E + F [C][D] [E][F] [E][F] [A][B] [A][B] [C][D] Kc = ‘ ‘ ‘ Kc Kc Kc = ‘ ‘ Kc = 2 PNOPBr x Kc = 2 Kp1= Ex. If Kp=0.42 for 2 NOBr (g) 2 NO (g) + Br2 (g) Kp=7.20 for Br2(g) + Cl2(g) 2 BrCl (g) What is Kp for 2 NOBr (g)+ Cl2 (g) 2 NO(g) + 2 BrCl(g) ‘ ‘ ‘ ‘ Kc Kc 2 2 2 PNOPBrCl PBrCl Kp3= Kp2= 2 PNOBr 2 PNOBrPCl PBr PCl 2 2 2 C. Equilibrium constant and balanced equation • When a reaction can be expressed as the sum of several reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. Kc = kp1x kp2= 3.0

  16. If Kc= 0.420 for N2(g) + 3 H2(g) 2 NH3 (g), what is the Kc for NH3(g) 1/2 N2 (g) + 3/2 H2(g)? (1) 0.420 (2) 0.650 (3) 1.54 (4) 2.38 Examples: K expression • What is the equilibrium constant, Kc for the following reaction? H2(g) + I2 (g) 2 HI(g)

  17. III. Equilibrium concentration A. Reaction quotient, Qc • Predict direction of a reaction • Calculated by substituting the initial conc. of the reactants and products into the equilibrium constant (Kc) expression. • Qc > Kcsystem proceeds from right to left to reach equilibrium • Qc = Kcthe system is at equilibrium • Qc < Kcsystem proceeds from left to right to reach equilibrium

  18. (12.6)2(4.65) = = 299 (1.57)2 2 PNO PO 2 Qp = Qp > Kp 2 PNO 2 Q: At 1000K, 2NO2 (g) 2NO (g) + O2 (g), Kp=158 If a reaction mixture contains PNO2 = 1.57 atm, PNO = 12.6 atm, and PO2= 4.65 atm, will the reaction go forward? No, reaction shifts to the reactant side

  19. B. Calculating equilibrium concentrations • Start with balanced equation • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. • Determine the equilibrium concentrations of all species. • Check answer

  20. Q: [Br]2 [Br2] Kc = Br2 (g) 2Br (g) Br2 (g) 2Br (g) = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x At 12800C the equilibrium constant (Kc) for the reaction is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Let x be the change in concentration of Br2 Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x Solve for x

  21. -b ± b2 – 4ac x = 2a Br2 (g) 2Br (g) Initial (M) 0.063 0.012 Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0 ax2 + bx + c =0 x = -0.0105 x = -0.00178 At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.063 – x = 0.0648 M

  22. Q: N2O4 (g) 2NO2(g) Kp = N2O4 (g) 2NO2 (g) P2 NO2 = 0.113 = (2.00- x)2 x P N2O4 At 25oC the equilibrium constant (Kc) for the reaction is Kc=0.00462. If the total pressure is 2.00 atm at equilibrium, calculate the partial pressure of these species at equilibrium. First figure out Kp Kp= Kc(RT)Dn =0.00462x (0.0821 x 298)1 = 0.113 Let x be the partial pressure of N2O4 Initial (P) Change (P) Equilibrium (P) x 2.00 - x Solve for x

  23. -b ± b2 – 4ac x = 2a N2O4(g) 2NO2 (g) Initial (P) Change (P) x 2.00 - x Equilibrium (M) = 0.113 Kc = (2.00 - x)2 x x2 - 4.00 x + 4.00 = 0.113x x2 - 4.113x + 4.00 = 0 ax2 + bx + c =0 x = 2.54 x = 1.58 At equilibrium, PN2O4 = 2.54 atm or 1.58 atm At equilibrium, PNO2 = 2.00 – x = -0.54 atm or 0.42 atm

  24. N2(g) + 3H2(g) 2NH3(g) Equilibrium shifts left to offset stress Add NH3 IV. Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. A. Changes in Concentration

  25. Remove Remove Add Add aA + bB cC + dD A. Changes in Concentration Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left *Removing or adding pure solid/liquid does NOT shift equilibrium

  26. A (g) + B (g) C (g) Example: H2(g) + I2 (g) 2 HI (g) Equilibrium will ______________ if volume is reduced. B. Changes in Volume and Pressure Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas *Change pressure or volume does NOT change K not be affected

  27. *In general, temperature affect the value of K Example: For the reaction C(s) + H2O (g) CO(g) + H2(g) Kp=1.7 x 10-21 at 25oC and Kp= 14.1 at 800oC. This reaction is ____________ (endothermic and exothermic). C. Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases (toward reactant) K increases (toward product) Decrease temperature K increases (toward product) K decreases (toward reactant) endothermic

  28. uncatalyzed catalyzed D. Effects of Catalyst • – does not change K • – does not shift the position of an equilibrium system • – system will reach equilibrium sooner Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

  29. Change Equilibrium Constant Summary of Le Châtelier’s Principle Change Shift Equilibrium Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no

  30. CaCO3(s) CaO (s) + CO2(g) DHo=178.32 kJ Example: Predict the effect of the following changes Change Shifts the Equilibrium Increase pressure by adding inert gas no effect Increase temperature (keeping pressure constant) Decrease volume Add CaCO3 no effect (pure solid) Remove CO2 Add catalyst no effect