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2. Resources Chapter Presentation Bellringer Transparencies Sample Problems Visual Concepts Standardized Test Prep

3. Chemical Equilibrium Chapter 14 Table of Contents Section 1Reversible Reactions and Equilibrium Section 2Systems at Equilibrium Section 3Equilibrium Systems and Stress

4. Section1 Reversible Reactions and Equilibrium Chapter 14 Bellringer • Describe what reversiblemeans. • Find a synonym for reversible.

5. Section1 Reversible Reactions and Equilibrium Chapter 14 Objectives • Contrast reactions that go to completion with reversible ones. • Describe chemical equilibrium. • Give examples of chemical equilibria that involve complex ions.

6. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions • If enough oxygen gas is provided for the following reaction, almost all of the sulfur will react: • S8(s) + 8O2(g) → 8SO2(g) • Reactions such as this one, in which almost all of the reactants react, are called completion reactions. • In other reactions, called reversible reactions, the products can re-form reactants.

7. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium • One reversible reaction occurs when you mix solutions of calcium chloride and sodium sulfate. • CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq) • The net ionic equation best describes what happens.

8. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium, continued • Solid calcium sulfate, the product, can break down to make calcium ions and sulfate ions in a reaction that is the reverse of the previous one. • Use arrows that point in opposite directions when writing a chemical equation for a reversible reaction.

9. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium, continued • The reactions occur at the same rate after the initial mixing of CaCl2 and Na2SO4. • The amounts of the products and reactants do not change. • Chemical equilibrium is a state of balance in which the rate of a forward reaction equals the rate of the reverse reaction and the concentrations of products and reactants remain unchanged.

10. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium • The reaction of hydrogen, H2, and iodine, I2, to form hydrogen iodide, HI, reaches chemical equilibrium. • Only a very small fraction of the collisions between H2 and I2 result in the formation of HI. • H2(g) + I2(g) → 2HI(g)

11. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued • After some time, the concentration of HI goes up. • As a result, fewer collisions occur between H2 and I2 molecules, and the rate of the forward reaction drops. • Similarly, in the beginning, few HI molecules exist in the system, so they rarely collide with each other.

12. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued • As more HI molecules are made, they collide more often and form H2 and I2 by the reverse reaction. • 2HI(g) → H2(g) + I2(g) • The greater the number of HI molecules that form, the more often the reverse reaction occurs.

13. Rate Comparison for H2(g) + I2(g)  2HI(g) Section1 Reversible Reactions and Equilibrium Chapter 14

14. Section1 Reversible Reactions and Equilibrium Chapter 14 Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued • When the forward rate and the reverse rate are equal, the system is at chemical equilibrium. • If you repeated this experiment at the same temperature, starting with a similar amount of pure HI instead of the H2 and I2, the reaction would reach chemical equilibrium again and produce the same concentrations of each substance.

15. Visual Concepts Chapter 14 Chemical Equilibrium

16. Section1 Reversible Reactions and Equilibrium Chapter 14 Chemical Equilibria Are Dynamic • If you drop a ball into a bowl, it will bounce. • When the ball comes to a stop it has reached static equilibrium, a state in which nothing changes. • Chemical equilibrium is different from static equilibrium because it is dynamic. • In a dynamic equilibrium, there is no net change in the system. • Two opposite changes occur at the same time.

17. Section1 Reversible Reactions and Equilibrium Chapter 14 Chemical Equilibria Are Dynamic, continued • In equilibrium, an atom may change from being part of the products to part of the reactants many times. • But the overall concentrations of products and reactants stay the same. • For chemical equilibrium to be maintained, the rates of the forward and reverse reactions must be equal. • Arrows of equal length also show equilibrium.

18. Section1 Reversible Reactions and Equilibrium Chapter 14 Chemical Equilibria Are Dynamic, continued • In some cases, the equilibrium has a higher concentration of products than reactants. • This type of equilibrium is also shown by using two arrows. • The forward reaction has a longer arrow to show that the products are favored.

19. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria • Even when systems are not in equilibrium, they are continuously changing in order to reach equilibrium. • For example, combustion produces carbon dioxide, CO2, and poisonous carbon monoxide, CO. After combustion, a reversible reaction produces soot. • This reaction of gases and a solid will reach chemical equilibrium. • Equilibria can involve any state of matter, including aqueous solutions.

20. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions • Complex ion, or coordination compound, is the name given to any metal atom or ion that is bonded to more than one atom or molecule. • Some ions have a metal ion surrounded by ligands, molecules or anions that readily bond to metal ions. • Complex ions may be positively charged cations or negatively charged anions.

21. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • In this complex ion, [Cu(NH3)4]2+, ammonia molecules bond to the central copper(II) ion.

22. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued Complex ions formed from transition metals are often deeply colored.

23. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • The charge on a complex ion is a sum of the charges on the species from which the complex ion forms. • For example, when the cobalt ion, Co2+, bonds with four Cl− ligands, the total charge is (+2) + 4(−1) = −2. • Metal ions and ligands can form complexes that have no charge. These are not complex ions. • Complex ions often form in systems that reach equilibrium.

24. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • Consider zinc nitrate dissolving in water: • In the absence of other ligands, water molecules bond with zinc ions. So, this reaction can be written: • If another ligand, such as CN−, is added, the new system will again reach chemical equilibrium.

25. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • Both water molecules and cyanide ions “compete” to bond with zinc ions, as shown in the equation below. • All of these ions are colorless, so you cannot see which complex ion has the greater concentration.

26. Section1 Reversible Reactions and Equilibrium Chapter 14 More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • In the chemical equilibrium of nickel ions, ammonia, and water, the complex ions have different colors. • You can see which ion has the greater concentration. • greenblue-violet • The starting concentration of NH3 will determine which one will have the greater concentration.

27. Section2 Systems at Equilibrium Chapter 14 Bellringer • Make a list of numbers that are “constants” under constant conditions. • An example is the speed of light. • What do these constants have in common? • Answer:Each constant is always the same number for a certain and constant set of conditions.

28. Section2 Systems at Equilibrium Chapter 14 Objectives • WriteKeq expressions for reactions in equilibrium, and perform calculations with them. • WriteKsp expressions for the solubility of slightly soluble salts, and perform calculations with them.

29. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq • Limestone caverns form as rainwater, slightly acidified by H3O+, dissolves calcium carbonate. • The reverse reaction also takes place, depositing calcium carbonate and forming stalactites and stalagmites. • When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium.

30. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued • There is a mathematical relationship between product and reactant concentrations at equilibrium. • For limestone reacting with acidified water at 25°C: • Keqis the equilibrium constant of the reaction. • Keqfor a reaction is unitless, applies only to systems in equilibrium, anddepends on temperature and must be found experimentally or from tables.

31. Visual Concepts Chapter 14 Equilibrium Constant

32. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued Determining Keqfor Reactions at Chemical Equilibrium • Write a balanced chemical equation. • Make sure that the reaction is at equilibrium before you write a chemical equation. • Write an equilibrium expression. • To write the expression, place the product concentrations in the numerator and the reactant concentrations in the denominator.

33. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued Determining Keqfor Reactions at Chemical Equilibrium, continued • The concentration of any solid or a pure liquid that takes part in the reaction is left out. • For a reaction occurring in aqueous solution, water is omitted. • Complete the equilibrium expression. • Finally, raise each substance’sconcentration to the power equal to the substance’s coefficient in the balanced chemical equation.

34. Visual Concepts Chapter 14 Equilibrium Constant

35. Section2 Systems at Equilibrium Chapter 14 Calculating Keq from Concentrations of Reactants and Products Sample Problem A An aqueous solution of carbonic acid reacts to reach equilibrium as described below. The solution contains the following solution concentrations: carbonic acid, 3.3 × 10−2 mol/L; bicarbonate ion, 1.19 × 10−4 mol/L; and hydronium ion, 1.19 × 10−4 mol/L. Determine the Keq.

36. Section2 Systems at Equilibrium Chapter 14 Calculating Keq from Concentrations of Reactants and Products Sample Problem A Solution For this reaction, the equilibrium constant expression is Substitute the concentrations into the expression.

37. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued KeqShows If the Reaction Is Favorable • When Keq is large, the numerator of the equilibrium constant expression is larger than the denominator. • Thus, the concentrations of the products will usually be greater than those of the reactants. • In other words, when a reaction that has a large Keq reaches equilibrium, there will be mostly products. • Reactions in which more products form than reactants form are said to be “favorable.”

38. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued • The synthesis of ammonia is very favorable at 25°C and has a large Keq value.

39. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued • However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C.

40. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued • When Keqis small, the denominator of the equilibrium constant expression is larger than the numerator. • The larger denominator shows that the concentrations of reactants at chemical equilibrium may be greater than those of products. • A reaction that has larger concentrations of reactants than concentrations of products is an “unfavorable” reaction.

41. Section2 Systems at Equilibrium Chapter 14 The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued These pie charts show the relative amounts of reactants and products for three Keq values of a reaction.

42. Section2 Systems at Equilibrium Chapter 14 Calculating Concentrations of Products from Keq and Concentrations of Reactants Sample Problem B Keqfor the equilibrium below is 1.8 × 10−5 at a temperature of 25°C. Calculate when [NH3] = 6.82 × 10−3.

43. Section2 Systems at Equilibrium Chapter 14 Calculating Concentrations of Products from Keq and Concentrations of Reactants, continued Sample Problem B Solution The equilibrium expression is and OH− ions are produced in equal numbers, so So, the numerator can be written as x2.

44. Section2 Systems at Equilibrium Chapter 14 Calculating Concentrations of Products from Keq and Concentrations of Reactants, continued Sample Problem B Solution, continued Keq and [NH3] are known and can be put into the expression. x2= (1.8  10−5)  (6.82  10−3) = 1.2 × 10−7

45. Section2 Systems at Equilibrium Chapter 14 Calculating Concentrations of Products from Keq and Concentrations of Reactants, continued Sample Problem B Solution, continued Take the square root of x2.

46. Section2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, Ksp • The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water. • Solubilities can be expressed in moles of solute per liter of solution (mol/L or M). • For example, the solubility of calcium fluoride in water is 3.4 × 10−4 mol/L. • So, 0.00034 mol of CaF2 will dissolve in 1 L of water to give a saturated solution. • If you try to dissolve 0.00100 mol of CaF2 in 1 L of water, 0.00066 mol of CaF2 will remain undissolved.

47. Visual Concepts Chapter 14 Solution Equilibrium

48. Section2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, Ksp, continued • Calcium fluoride is one of a large class of salts that are said to be slightly soluble in water. • The ions in solution and any solid salt are at equilibrium. • Solids are not a part of equilibrium constant expressions, so Keq for this reaction is the product of [Ca2+] and [F−]2, which is equal to a constant.

49. Section2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, Ksp, continued • Equilibrium constants for the dissolution of slightly soluble salts are called solubility product constants,Ksp, and have no units. • The Ksp for calcium fluoride at 25°C is 1.6  10−10. • Ksp= [Ca2+][F−]2 = 1.6  10−10 • This relationship is true whenever calcium ions and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves.

50. Section2 Systems at Equilibrium Chapter 14 The Solubility Product Constant, Ksp, continued • For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates. • The net ionic equation for this precipitation is the reverse of the dissolution. • This equation is the same equilibrium. So, the Kspfor the dissolution of CaF2 in this system is the same and is 1.6 × 10−10.