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Lecture 16.1

Lecture 16.1. Chapter 16: Chemical Equilibria. All chemical reactions are reversible, at least in principle.

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Lecture 16.1

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  1. Lecture 16.1 • Chapter 16: Chemical Equilibria

  2. All chemical reactions are reversible, at least in principle. Equilibrium is a dynamic situation. It appears that no change is occurring. However, reactants are combining together to from products and the products are combining to reform the reactions. The two reactions, the forward one and the reverse one, occur at the same rate so there is no NET change in the concentrations. H2O(g) + CO(g) CO2(g) + H 2(g)

  3. Dynamite exploding-- equilibrium? The equilibrium position of a reaction– left right, or somewhere in between is determined by many factors: the initial concentrations, the relative energies of the products and reactants, and the relative degree of “organization” of the reactants and products. Energy and organization come into play because nature tries to achieve minimum energy and maximum disorder. (Chapter 20)

  4. The Equilibrium ConstantThe Law of Mass Action Two Norwegian chemist, Cato Maximilian Guildberg (1836-1902) and Peter Waage (1833-1900) proposed in 1864 a general description of the equilibrium condition (called the Law of Mass Action). They proposed for the rxn of the type a A + b B  c C + d D The equilibrium concentrations of rxn and products can be represented by the equilibrium constant expression Equilibrium Constant = K = [C]c[D]d PRODUCTS [A]a[B]b REACTANTS [ ] represent concentration at equilibrium A, B, C, and D represent chemical species and a,b, c, and d are their coefficients in the balanced equation.

  5. Write the equilibrium expression for the rxn:2NH3(g) N2(g) + 3H2(g) Keq= [N2][H2]3 PRODUCTS [NH3]2 REACTANTS Keq at a given T can be calculated if equilibrium concentrations of reaction components are known. NOTE:Keqconstants are without units. The reason is beyond this course, but involves corrections for nonideal behavior of substances taking part in the reaction. When corrections are made, the units cancel out and the corrected K has no units. So, we will NOT use units for K.

  6. Writing Equilibrium Constant ExpressionsRule for Rxn Involving SolidsR.O.T. Keq is in same phase In a heterogeneous rxn., equilibrium does NOT depend on amounts of pure solids or liquids present. The concentration of pure liquids and solids does NOT change. (C = constant) Keq =CCaO[CO2(g)] /CCaCO3 = [CO2(g)]

  7. Rules for Rxns Involving WaterR.O.T. Keq is in same phase 2H2O(l) 2 H2(g) + O2(g) Keq = [H2(g)]2[O2(g)] However, if the rxn were carried out under conditions where the water is a gas rather than a liquid, that is then it is included in Keq b/c the concentration OR pressure of water vapor can change. If pure solids or pure liquids are involved in a chem. rxn, their concentrations are not included in the equilibrium expression. This simplification occurs ONLY with pure solids or liquids, not with solutions or gases, since the last two cases the concentrations vary. 2H2O(g)2 H2(g)+ O2(g) Keq =[H2(g)]2[O2(g)]/[H2O(g)]

  8. Equilibrium Expressions Involving Pressures The relationship b/t P and concentration PV= nRT or P = (n/V)RT or P = CRT Write the equilibrium expression in terms of partial pressures and the equilibrium expression in terms of concentration for the rxn:2NH3(g) N2(g) + 3H2(g) Keq= [N2][H2]3 [NH3]2 Kp= PN2PH23 PNH32

  9. Write the expressions for K and Kp for the following processes: PCl5(s) PCl3(l) + Cl2(g) CuSO4 5H2O(s)  CuSO4(s) + 5 H2O(g) Keq= [Cl2] Kp= PCl2 Keq= [H2O]5 Kp= PH2O5

  10. Calculating Kp values The reaction for the formation of nitrosyl chloride 2NO(g) + Cl2(g) 2 NOCl(g) was studied at 25oC. The pressures at equilibrium were found to be PNOCl = 1.2 atm PNO= 0.050 PCl2 = 0.30 atm. Calculate the value of Kp for the reaction. Kp = PNOCl2 /(PNO)2(PCl2) = (1.2)2/(0.05)2(0.3) = 1.9 x 10 3

  11. The relationship between K and KpKp = K(RT)Dn For the derivation read A Closer Look p. 753 in text. Remember P = CRT or C = P/RT and must be substituted in equilibrium expression. The relationship is Kp = K(RT)Dnwhere Dn is the difference in the sums of the coefficients for the gaseous products and reactants. NOTE: Kp = Kc when Dn = 0

  12. Example: The value of Kp was 1.9 x 103 in the previous example. Calculate the value of K at 25oC for the reaction 2NO (g) + Cl2(g)  2NOCl(g) 1.9 x 103 = K((0.0821 Latm/mol K)(273 + 25))(2-3) 1.9 x 103 = K (24 –1) K = 45600 = 4.6 x 10 4 NOTE: If Dn = 0, then Kp = Kc Kp = K(RT)Dn

  13. Manipulating Equilibrium Expressions2NO (g+ Cl2(g)  2NOCl(g)K = 4.6 x 10 4 Forward and reverse equilibrium expressions Kforward = [NOCl]2/[NO]2[Cl2] = 4.6 x 10 4 Kreverse = [NO]2[Cl2]/[NOCl]2= 1/4.6 x 10 4 If stoichiometric coefficients of balanced equations are multiplied by some factor, n, the equilibrium constant for new equation is raised to the power of the new factor K new = (Kold)n Kforward = [NOCl]6/[NO]6[Cl2]3 = (4.6 x 10 4)3 3 (2 NO (g) + Cl2(g)  2 NOCl(g) )

  14. Write the equilibrium expression for the reverse direction for the following reactionBr2(g)  2 Br(g) K = 2.2 x 10 –15 Keq = [Br2]/[Br]2 = 1/ 2.2 x 10 –15 Write the equilibrium expression for the reverse direction for the following reaction½ (Br2(g)  2 Br(g)) K = 2.2 x 10 –15 K new = (Kold)n note: n is multiplication factor Keq = [Br2] 1/2/[Br]= 1/(2.2 x 10 –15)1/2

  15. In general, when two or more equations are added to produce a net equation, K for the net equation is the product of the K for added equations. AgCl(s) is dissolved in water (to a small extent). Ammonia is then added to form the complex Ag(NH3)2+ AgCl(s) Ag+(aq) + Cl-(aq) K1 = [Ag+][Cl-] = 1.8 x 10 –10 Ag+(aq)+ 2 NH3(aq)Ag(NH3)2+ (aq) K2 =[Ag(NH3)2+(aq)]= 1.6 x 107 [Ag+][NH3]2 NET AgCl (s) + 2 NH3(aq)  Ag(NH3)2+ (aq) + Cl -(aq) Knet=[Ag(NH3)2+][Cl-] =K1K2=(1.6 x 107)(1.8 x 10 –10 )=2.9x 10 -3[NH3 ]2

  16. Exercise 16.3 p. 756 Manipulating Equilibrium Constant ExpressionsThe following equilibrium constants are given at 500 K. H2(g) + Br2(g) 2 HBr(g) K = 7.9 x 10 11H2(g)  2 H(g) K = 4.8 x 10 –41Br2(g)  2 Br(g) K = 2.2 x 10 –15Calculate K for the reaction of H and Br atoms to give HBr H(g) + Br(g)  HBr (g) K = ? H2(g) + Br2(g) 2 HBr(g) K1 = 7.9 x 10 11 2 H(g) H2(g) K2 = 1/4.8 x 10 –41 2 Br(g)  Br2(g) K3 = 1/2.2 x 10 –15 2 H(g)+2 Br(g)  2 HBr Knet = [HBr]2/[H]2[Br]2 WantKnet = [HBr]/[H][Br]so… Knet=(K1K2K3 )½=((7.9 x10 11)(1/4.8 x10 –41)(1/2.2 x10 –15)) ½ = (7.5 x 10 66 )½ = 2.7 x 1033

  17. Work on HW • ↔

  18. a. b. c. d. H2O + CO  CO2 + H2 (a) H2O and CO are mixed in equal numbers and begin to react (b) to from CO2 and H2. After time has passed, equilibrium is reached (c) and the number of reactant and product molecules then remain constant over time (d)

  19. Demonstration: Limewater and CO2 Saturated solution of limewater Ca(OH)2(s) Ca(OH)2(aq) Add CO2(g) CO2 + Ca(OH)2(aq) CaCO3(s) + H2O Add MORE CO2(g) to the CaCO3 (s) CO2 + H2O + CaCO3(s) Ca(HCO3)2(aq)

  20. Equilibria in CO2/Ca2+/H2O System CaCl2+ 2NaHCO3 2NaCl+ Ca(HCO3)2  CaCO3(s) + 2 H2O + 2 CO2 Net: Ca2++ HCO3- CaCO3(s) + H2O +CO2 Add CO2(s) Ca2++ HCO3–  CaCO3(s) + H2O +CO2

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