1 / 54

Chapter 14

Chapter 14. Chemical Equilibrium. 14.1: Equilibrium Constant, K eq. Objective: To write the equilibrium constant expression for a chemical reaction. Reversible Reactions and Equilibrium. Reversible Reaction: A chemical reactions in which products re-form the original reactants.

macon
Télécharger la présentation

Chapter 14

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 14 Chemical Equilibrium

  2. 14.1: Equilibrium Constant, Keq • Objective: • To write the equilibrium constant expression for a chemical reaction.

  3. Reversible Reactions and Equilibrium • Reversible Reaction: A chemical reactions in which products re-form the original reactants. • Arrows that point in opposite directions are used to indicate a reaction is reversible. • Example: H2(g) + I2(g) 2HI(g) • Chemical Equilibrium: A state of balance in which the rate of a forward reaction equals the rate of the reverse reactions and the concentrations of products and reactants remain unchanged.

  4. Equilibrium Constant, Keq • Equilibrium Constant, Keq: a number that relates that concentrations of starting materials and products of a reversible chemical reaction to one another at a given temperature. aA + bB  cC + dD coefficient concentration

  5. Writing an Equilibrium Constant Expression • Step 1: Balance the chemical equation. • Step 2: Set up your Keq expression with the products on the top of a fraction and the reactants on the bottom of a fraction. • Step 3: Raise each substance's concentration to the power equal to the substance’s coefficient in the balanced equation. • Note: Solids (s) and pure liquids (l) are not used in the expression because their concentrations do not change.

  6. Example • Write the equilibrium constant expression for the following reaction: CaCO3(s) + CO2(aq)+ H2O(l) Ca2+(aq) + 2HCO3-(aq)

  7. Practice • Write the equilibrium constant expression for the following chemical reactions at equilibrium (don’t forget to balance the equation): 1.) H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+ (aq) 2.) COCl2 (g) CO(g) + Cl2 (g) 3.) CO(g) C(s) + CO2 (g)

  8. Answers 1.) H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+ (aq) 2.) COCl2 (g) CO(g) + Cl2 (g) 3.) 2CO(g) C(s) + CO2 (g)

  9. 14.1: Equilibrium Constant, Keq • Objective: • To calculate the equilibrium constant.

  10. What does the Keq tell us? • Keq < 1 Favors Reactants • Keq = 1 Same amount of Reactants and Products • Keq > 1 Favors Products • Practice: Determine if the following Keq values favor the reactants, products, or neither. 1.) Keq = 0.02 2.) Keq = 1 3.) Keq = 50

  11. Calculating Keq • Step 1: Write the balanced chemical equation. • Step 2: Set up your Keq expression. • Step 3: Substitute concentrations. • Step 4: Calculate!

  12. Example • An aqueous solution of carbonic acid reacts to reach equilibrium as described below: H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+ (aq) The solution contains the following solute concentrations: H2CO3 = 3.3 x 10-2 M; HCO3- = 1.19 x 10-4 M; H3O+ = 1.19 x 10-4 M. Determine the Keq. Note: Keq does not have units!

  13. Practice 1.a. Calculate the equilibrium constant for the following reaction: COCl2(g) CO(g) + Cl2(g) [CO] = 0.0178 M [Cl2] = 0.0178 M [COCl2] = 0.00740 M b. Are the reactants for products favored?

  14. Practice 2.a. For the system involving dinitrogen tetraoxide and nitrogen dioxide at equilibrium at a temperature of 100⁰C, the product concentration of N2O4 is 4.0 x 10-2 M and the reactant concentration of NO2 is 1.4x 10-1 M. What is the Keq value for this reaction? NO2(g) N2O4(g) b. Are the reactants or products favored?

  15. Practice 3.a. An equilibrium mixture at 852 K is found to contain 3.61 x 10-3 M of SO2, 6.11 x 10-4 M of O2, and 1.01 x 10-2 M of SO3. Calculate the equilibrium constant for the reaction. SO2 (g) + O2 (g) SO3 (g) b. Are the reactants or products favored?

  16. Calculating Concentrations from Keq 4. Keq for the equilibrium below is 1.8 x 10-5 at a temperature of 25⁰C. Calculate [NH3] when [NH4+] and [OH-] are 3.5 x 10-4 M. NH3(aq) + H2O(l) NH4+ (aq) + OH-(aq) [NH3] = 6.8 x10-3 M

  17. Practice 5. a. If the equilibrium constant is 1.65 x 10-3 at 2027⁰C for the reaction below, what is the equilibrium concentration of NO when [N2] = 1.8 x 10-3 M and [O2] = 4.2 x 10-3 M. N2(g) + O2(g) NO(g) b. Are the reactants for products favored?

  18. Practice 6.a. At 600⁰C, the Keq for the reaction below is 4.32 when [SO3] = 0.260 M and [O2] = 0.045 M. Calculate the equilibrium concentration for sulfur dioxide. SO2(g) + O2(g) SO3(g) b. Are the reactants or products favored?

  19. 14.2 Solubility Product Constant, Ksp • Objective: (1) To calculate the solubility product constant, Ksp.

  20. Solubility • The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water.

  21. Solubility Product Constant, Ksp • Solubility Product Constant, Ksp: the equilibrium constant for a solid that is in equilibrium with the solid’s dissolved ions. • How much of a partially soluble salt will dissolve? AaBb (s) aA (aq) + bB (aq)

  22. Calculating Ksp • The lower the value of Ksp, the less soluble the substance. • Practice: Rank the following substances from least soluble to most soluble:

  23. Calculating Ksp • The lower the value of Ksp, the less soluble the substance. • Practice: Rank the following substances from least soluble to most soluble: CuS Least soluble Ca3(PO4)2 Ag2CO3 BaSO4 Most soluble

  24. Calculating Ksp • Step 1: Write and Balance the equation. • Step 2: Determine the concentration of the ions. • Step 3: Write the solubility product expression. • Step 4: Substitute values and calculate.

  25. Example • Most parts of oceans are nearly saturated with calcium fluoride. A saturated solution of CaF2 at 25⁰C has a solubility of 3.4 x 10-4 M. Calculate the solubility product constant for CaF2. CaF2(s) Ca2+(aq) + F-(aq)

  26. Solution • Balance equation: CaF2(s) Ca2+(aq) + 2F-(aq) • Determine Concentrations: CaF2(s) Ca2+(aq) + 2F-(aq) 3.4 x 10-4 3.4 x 10-4 6.8 x 10-4 3. Write solubility product expression: 4. Substitute values and calculate: Note: Ksp does not have units!

  27. Practice 1. Copper(I) bromide is dissolved in water to saturation at 25⁰C. The concentration of Cu+ and Br- ions in solution is 7.9 x 10-5 M. Calculate the Ksp for copper(I) bromide at this temperature.

  28. Practice 2. What is the Ksp value for calcium phosphate at 298 K if the concentrations in a solution at equilibrium with excess solid are 3.42 x 10-7 M for Ca2+ and 2.28 x 10-7 M for PO43- ions?

  29. Practice 3. If a saturated solution of silver chloride contains an AgCl concentration of 1.34 x 10-5 M, what is the solubility product constant?

  30. Practice 4. A saturated solution of magnesium fluoride contains a MgCl2 concentration of 1.19x10-3 M. What is the Ksp for magnesium fluoride?

  31. Calculating Concentration from Ksp 5. What is the concentration of Ca2+ in a saturated solution of CaF2 if the concentration of F- is 2.20 x 10-3M and Ksp = 5.30 x 10-9.

  32. Practice 6. What is the concentration of Al3+ in a saturated solution of Al(OH)3 if the OH- concentration is 7.90 x 10-9 M. Ksp = 1.30 x 10-33.

  33. Practice: Chem 331 7. The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution?

  34. The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution? Step 1: Write and Balance Equation PbI2 (s) Pb2+ (aq) + 2I- (aq)

  35. The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution? Step 1: Write and Balance Equation PbI2 (s) Pb2+ (aq) + 2I- (aq) Step 2: Write the Ksp expression

  36. The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution? Step 1: Write and Balance Equation PbI2 (s) Pb2+ (aq) + 2I- (aq) Step 2: Write the Ksp expression Step 3: Assign x values to concentrations PbI2 (s) Pb2+ (aq) + 2I- (aq) x x 2x

  37. The Ksp for lead(II) iodide is 7.08 x 10-9 at 25⁰C. What is the molar concentration of PbI2 in a saturated solution? Step 1: Write and Balance Equation PbI2 (s) Pb2+ (aq) + 2I- (aq) Step 2: Write the Ksp expression Step 3: Assign x values to concentrations PbI2 (s) Pb2+ (aq) + 2I- (aq) x x 2x Step 4: Solve x = [PbI2] = 1.21 x 10-3 M

  38. Practice 8. The Ksp of calcium sulfate is 9.1 x 10-6. What is the molar concentration of calcium sulfate in a saturated solution?

  39. Practice 9. The Ksp of CdF2 is 6.4 x 10-3. What is the molar concentration of cadmium fluoride in a saturated solution?

  40. 14.3 LeChatelier’s Principle • Objective: (1) To use LeChatelier’s Principle to determine how a system at equilibrium will respond to an external stress.

  41. LeChatelier’s Principle • LeChatelier’s Principle: When a system at equilibrium is disturbed, the system adjusts in a way to reduce the change. • There are 3 possible disturbances: Change in (1) concentration, (2) temperature, or (3) pressure

  42. 1. Change in Concentration • Increase concentration of reactant  Equilibrium shifts toward products • Decrease concentration of reactant  Equilibrium shifts toward reactants • Increase concentration of product  Equilibrium shifts toward reactants • Decrease concentration of product  Equilibrium shifts toward products

  43. Use the following reaction to answer the questions below: H2 (g) + I2 (g) 2HI (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase H2 2.) Decrease I2 3.) Increase HI 4.) Decrease HI

  44. Use the following reaction to answer the questions below: H2 (g) + I2 (g) 2HI (g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase H2RIGHT 2.) Decrease I2LEFT 3.) Increase HI LEFT 4.) Decrease HI RIGHT

  45. 2. Change in Temperature • Think of heat as a reactant or product • Exothermic: heat is a product • Endothermic: heat is a reactant • For an exothermic reaction: • Increasing temperature  equilibrium favors reactants • Decreasing temperature  equilibrium favors products • For an endothermic reaction • Increasing temperature  equilibrium favors products • Decreasing temperature  equilibrium favors reactants

  46. Use the following reaction to answer the questions below: 2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the temperature 2.) Decrease the temperature

  47. Use the following reaction to answer the questions below: 2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the temperature RIGHT 2.) Decrease the temperature LEFT

  48. 3. Change in Pressure • Only affects gases! • Increasing pressure  Equilibrium shifts toward the side with fewer moles of gas • Decreasing pressure  Equilibrium shifts toward the side with more moles of gas

  49. Use the following reaction to answer the questions below: 2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the pressure 2.) Decrease the pressure

  50. Use the following reaction to answer the questions below: 2SO3(g) + CO2 (g) + heat CS2 (g) + 4O2(g) In which direction (left or right) does the equilibrium shift in each of the following situations: 1.) Increase the pressure LEFT (3 moles gas) 2.) Decrease the pressure RIGHT (5 moles gas)

More Related