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CHEMICAL EQUILIBRIUM Unit 11, Part II

CHEMICAL EQUILIBRIUM Unit 11, Part II. Pb 2+ (aq) + 2 Cl – (aq)  PbCl 2 (s). What is equilibrium?. Definition (dictionary.com): a state of rest or balance due to the equal action of opposing forces.

cynthia-pittman
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CHEMICAL EQUILIBRIUM Unit 11, Part II

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  1. CHEMICAL EQUILIBRIUMUnit 11, Part II Pb2+(aq) + 2 Cl–(aq) PbCl2(s)

  2. What is equilibrium? Definition (dictionary.com): a state of rest or balance due to the equal action of opposing forces Chemical Equilibrium: A process where a forward and reverse reaction occur at equal rates Not all chemical reactions are reversible!!!

  3. General Characteristics of Equilibrium DYNAMIC (in constant motion) REVERSIBLE can be approached from either direction (reaction can run in the forward direction or the reverse direction) Cobalt Chloride Complex Equilibrium Pink to blue Co(H2O)6Cl2 Co(H2O)4Cl2 + 2 H2O Blue to pink Co(H2O)4Cl2 + 2 H2O Co(H2O)6Cl2

  4. Characteristics of Dynamic Equilibrium After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue afterequilibrium is attained. + Fe3+ + SCN- FeSCN2+

  5. Examples of Chemical Equilibria Phase changes such as H2O(s) H2O(liq)

  6. Examples of Chemical Equilibria Formation of stalactites and stalagmites CaCO3(s) + H2O(liq) + CO2(g)Ca2+(aq) + 2 HCO3-(aq)

  7. Graphing Dynamic Equilibrium Equilibrium achieved Product conc. increases and then becomes constant at equilibrium Equilibrium achieved when product and reactant concentrations remain constant!! Reactant conc. declines and then becomes constant at equilibrium

  8. The Equilibrium Expression, Keq Keq = equilibrium constant (for a given T) Brackets "[ ]" = concentration (molarity) "a, b, c, and d" = coefficients from balanced equation The "c" in Kc = concentration (Kc = a special Keq based on concentration)

  9. The Equilibrium Constant for the Synthesis of HI Equilibrium achieved Note that Equilibrium Constants have NO units!! In the equilibrium region

  10. There are two cases when a species is not shown in the equilibrium expression: #1: SOLIDS – (s) after the formula #2: pure LIQUIDS – (l) after the formula

  11. Examples of Equilibrium Expressions: The oxidation-reduction reaction occurring between iron(III) chloride and tin(II) chloride: 2 FeCl3 (aq) + SnCl2 (aq) 2 FeCl2 (aq) + SnCl4 (aq) The replacement of silver ions by copper: Cu (s) + 2 Ag+ (aq)  Cu2+ (aq) + 2 Ag (s)

  12. Equilibrium Constants K is independent of: • Original amounts of reactants or products • Size of container • Pressure K is dependent on: • Temperature • States of reactants and products

  13. Can tell if a reaction is product-favored or reactant-favored For N2(g) + 3 H2(g)  2 NH3(g) The Meaning of K If K is large, the conc. of products is much greaterthan that of the reactants at equilibrium. The reaction is strongly product-favored (right side of equation is favored)

  14. For AgCl(s)  Ag+(aq) + Cl-(aq) Kc = [Ag+] [Cl-] = 1.8 x 10-5 If K is small, the conc. of the products is much less than that of the reactants at equilibrium. The reaction is strongly reactant-favored(left side of equation is favored). The Meaning of K AgCl(s)  Ag+(aq) + Cl-(aq) is reactant-favored.

  15. Product- or Reactant Favored Product-favored K >> 1 Reactant-favored K << 1

  16. Determining an Equilibrium Constant if all Concentrations are known: Kc = [ NH3]2 [N2] [ H2]3 Can substitute numbers into concentrations; if all concentrations = 2.0M, Then Kc =(2)2 \ (2) x (2)3 = ¼ or .25 (no units on K)

  17. Example: For NH4Cl(s)  NH3(g) + HCl(g) at 500oC, at equilibrium there are 2.0 mol of ammonia, 2.0 mol of hydrochloric acid and 1.0 mol of ammonium chloride present in a 5.0 L container. Calculate the equilibrium constant of the system at this temperature.

  18. Example: For the system 2SO3(g)  2SO2(g) + O2(g) at a given temperature the equilibrium concentrations are [SO2] = [O2] = 0.10M and [SO3] = 0.20M. Calculate the Kc at this temperature.

  19. Example: For N2(g) + O2(g)  2NO(g) Kc = 1.0 x 10-30 at 25°C Calculate the equilibrium concentration of NO(g) if at equilibrium the concentration of N2(g) is 0.04 M and that of O2 is 0.01 M.

  20. LE CHATELIER’S PRINCIPLE “If a system at equilibrium is stressed, the system tends to shift its equilibrium position to counter the effect of the stress.”

  21. How does a “stress” influence equilibrium? The impact of addition of reactants on reaction rate

  22. The Seesaw Analogy

  23. The “stresses” (factors) that can cause changes at equilibrium Changes in amount of species A. Add reactant; system shifts to right (produces more products) B. Add product; system shifts to left (produces more reactants) C. Remove reactant; system shifts to left D. Remove product; system shifts to right

  24. Example: Predict the direction of shift of the following concentration changes on the reaction: CH4(g) + 2S2(g)  CS2(g) + 2H2S(g) (A) Some S2(g) is added. (B) Some CS2(g) is added. (C) Some H2S(g) is removed. (D) Some argon gas (an inert gas) is added.

  25. Changes in Pressure/Volume If P goes down (same as V goes up), system shifts to increased # of moles of gas If P goes up (same as V goes down), system shifts to decreased # of moles of gas

  26. Changes in Temperature Write heat as a product (exothermic) or reactant (endothermic). System shifts to get rid of added heat: will shift LEFT for exo reactions as T goes up RIGHT for endo reactions as T goes up

  27. Changes in Amount of Pure Solid/Liquid or Inert Substance If a substance is NOT in Keq, changes will have NO EFFECT on equilibrium!

  28. Example Predict the effect of increasing pressure (decreasing volume) on each of the following reactions. (a) CH4(g) + 2S2(g)  CS2(g) + 2H2S(g) (b) H2(g) + Br2(g)  2HBr(g) (c) CO2(g) + C(s)  2CO(g) (d) PCl5(g)  PCl3(g) + Cl2(g)

  29. Le Chatelier Example EXAMPLE: CO(g) + 2 H2 (g) <---> CH3OH (g) How can the reaction be shifted to the right?

  30. Example 16.17 What conditions would be most ideal for shifting the following equilibrium as far to the right as possible? N2(g) + 3H2(g)  2NH3(g)∆H = negative

  31. N2(g) + 3 H2(g)  2 NH3(g) + heat K = 3.5 x 108 at 298 K Haber-Bosch Process for NH3

  32. Haber-Bosch Ammonia Synthesis Fritz Haber 1868-1934 Nobel Prize, 1918 Carl Bosch 1874-1940 Nobel Prize, 1931

  33. Addition of a Catalyst (NOT in Keq) Add catalyst ---> no change in K A catalyst only affects the RATE of approach to equilibrium. Catalytic exhaust system

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