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Chemical Equilibrium: Types, Calculations, and Le Chatelier's Principle

Learn about chemical equilibrium, including the law of mass action, equilibrium constants, calculations using the ICE table, and how changes in concentration, volume, pressure, and temperature affect equilibrium. Understand the role of catalysts in equilibrium reactions.

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Chemical Equilibrium: Types, Calculations, and Le Chatelier's Principle

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  1. Chapter 15 Chemical Equilibrium

  2. 15.1

  3. 15.1

  4. Cold Temp Hot Temp 15.1

  5. 15.1

  6. 15.1

  7. 15.2: Law of Mass Action • Derived from rate laws by Guldberg andWaage (1864) • For a balanced chemical reactionin equilibrium: a A + b B ↔ c C + d D • Equilibrium constant expression (Keq): Cato Guldberg Peter Waage (1836-1902) (1833-1900) or • Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism). • Units: Keq is considered dimensionless (no units)

  8. **Relates Kc to Kp 15.2

  9. For Example Write the equilibrium expression Kc for the following reactions: 15.2

  10. Example #2 In the synthesis of ammonia from nitrogen and hydrogen, Kc = 9.60 at 300°C: N2(g) + 3H2(g) 2NH3(g) Calculate Kp for this reaction at this temperature Hint..you will need to use.. 15.2

  11. What does the Equilibrium constant tell us?? 15.2

  12. Looking at reversible reactions • So far we have only written the expression forwards, but it can also be written equally backwards! • The constants are the recipricals of each other Kc = 0.212 Kc = 4.72 15.2

  13. 15.3: Types of Equilibria Homogeneous: all components in same phase (usually g or aq) N2 (g) + H2 (g) ↔ NH3 (g) 1 3 2 Fritz Haber(1868 – 1934) 15.3

  14. Heterogeneous:different phases CaCO3 (s) ↔ CaO (s) + CO2 (g) Definition: What we use: • Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq • Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium. 15.3

  15. For Example • Write equilibrium-constant expressions for Kc and Kp for each of the following reactions: Click for answers 15.3

  16. 15.4: Calculating Equilibrium Constants Steps to use “ICE” table: • “I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression • “C” = Determine the concentration change for the species where initial and equilibrium are known • Use stoichiometry to calculate concentration changes for all other species involved in equilibrium • “E” = Calculate the equilibrium concentrations

  17. Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is 0.0124 M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction: NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

  18. NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq) X 0.0124 M 0 M 0 M X - x + x + x X 0.0119 M 4.64 x 10-4 M 4.64 x 10-4 M x = 4.64 x 10-4 M

  19. Ex: A 5.000-L flask is filled with 5.000 x 10-3 mol of H2 and 1.000 x 10-2 mol of I2 at 448ºC. The value of Keq is 1.33. What are the concentrations of each substance at equilibrium? H2 (g) + I2 (g) ↔ 2 HI (g)

  20. H2 (g) + I2 (g) ↔ 2 HI (g) 1.000x10-3 M 2.000x10-3 M 0 M - x M - x M + 2x M (1.000x10-3 – x) M (2.000x10-3 – x) M 2x M 4x2 = 1.33[x2 + (-3.000x10-3)x + 2.000x10-6] 0 = -2.67x2 – 3.99x10-3x + 2.66x10-6 Using quadratic eq’n: x = 5.00x10-4or –1.99x10-3; x = 5.00x10-4 Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M

  21. 15.6 Le Chatelier’s Principle

  22. Changes that do not affect Keq: • Concentration • Upon addition of a reactant or product, equilibrium shifts to re-establish equilibrium by consuming part of the added substance. • Addition of solids/liquids do not appreciably shift the system and can be ignored. But they are still made/consumed! • Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance. • Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l) • Add HCl, temporarily inc forward rate

  23. Volume, with a gas present (T is constant) • Upon a decrease in V (thereby increasing P),equilibrium shifts to reduce the number of moles of gas. • Upon an increase in V (thereby decreasing P),equilibrium shifts to produce more moles of gas. Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g) • If V of container is decreased, equilibrium shifts right. • XN2 and XH2dec • XNH3 inc Since PT also inc, KP remains constant.

  24. 3. Pressure, but not Volume • Usually addition of a noble gas, p. 560 • Avogadro’s law: adding more non-reacting particles “fills in” the empty space between particles. • In the mixture of red and blue gas particles, below, adding green particles does not stress the system, so there is no Le Châtelier shift.

  25. Catalysts • Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse rxn rates. • Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq) Ea, uncatalyzed Ea, catalyzed Energy Rxn coordinate

  26. Change that does affect Keq: • Temperature:consider “heat” as a part of the reaction • Upon an increase in T, endothermic reaction is favored (equilibrium shifts to “consume the extra heat”) • Upon a decrease in T, equilibrium shifts to produce more heat. Effect on Keq • Exothermic equilibria: Reactants ↔ Products + heat • Inc T increases reverse reaction rate which decreases Keq • Endothermic equilibria: Reactants + heat ↔ Products • Inc T increases forward reaction rate increases Keq • Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l); DH=+? • Inc T temporarily inc forward rate • Dec T temporarily inc reverse rate

  27. Effect of Various Changes on Equilibrium

  28. Effect of Pressure on Equilb. Disturbance Direction of Reaction Effect of K

  29. Effect of Temperature on Equilb Disturbance Direction of Reaction Effect of K

  30. For Example For the following reaction 5 CO(g) + I2O5(s) I2(g) + 5 CO2(g) + 1175 kJ • for each change listed, predict the equilibrium shift and the effect on the indicated quantity.

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