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Chapter 15

Chapter 15. Chemical Equilibrium. Reversible Reactions. Most natural processes are reversible reactions reactants  products Example: sodium carbonate + calcium chloride  table salt + limestone. Reversible Reaction. Reactants & products are together as a mixture

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Chapter 15

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  1. Chapter 15 Chemical Equilibrium

  2. Reversible Reactions • Most natural processes are reversible reactions reactants  products Example: sodium carbonate + calcium chloride table salt + limestone

  3. Reversible Reaction • Reactants & products are together as a mixture • Ratio of reactants & products depends on the speed of forward & reverse reactions • Changing rate of forward or reverse reaction changes the ratio of products and reactants in the reaction mixture

  4. Chemical Equilibrium • Forward and reverse reactions happen at the same time (simultaneous) & the same rate (balanced) • Dynamic • Involves constant change, but no net change

  5. Chemical Equilibrium

  6. Equilibrium Concentrations • Will the concentration of reactants and products be the same at equilibrium? • Story of Miss Boykov’s “Magic Salt” (lots dissolves!) and Miss Boykov’s brother’s not-so-magic salt (little dissolved) • Miss Boykov’s salt was product-favored • Her brother’s salt was reactant-favored

  7. Product favored reaction • Concentration of products higher than concentration of reactants at equilibrium • Equilibrium lies to the right • Reactant favored reaction • Concentration of reactants is higher than concentration of products at equilibrium • Equilibrium lies to the left

  8. Product favored reaction: H2 + I2  2HI A • What does a reactant favored chart look like? What happens to concentration of reactants in A&B? Products? B

  9. Homework: finish reading 15A

  10. Write out 3 questions you had after reading the rest of 15A *(questions should be specific enough to convince your teacher you did the reading) • Hand in your 1/4 sheet • Read SRQ15A 2, 3, & 5

  11. Equilibrium Constant Reaction: I2 (g) + H2 (g)  2HI (g) Equilibrium Constant: Reaction: 3H2 (g) + N2 (g)  2NH3 (g) Equilibrium Constant:

  12. Equilibrium Constants • Equilibrium constant (Keq) describes a reaction’s equilibrium • Gives product to reactant ratio • More reactants if Keq is less than 1 • Reactant favored, equilibrium lies to left • More products if Keq is more than 1 • Product favored, equilibrium lies to the right, reaction almost goes to completion

  13. Law of Chemical Equilibrium • At a fixed temperature, a chemical system may reach a point at which the concentration of products to reactants is constant • Square brackets show moles per liter (molarity) • Homogeneous equilibrium • Reactants and products are in the same state • Heterogeneous equilibrium • Not all substances are in the same state

  14. Tips for writing Keq of heterogeneous reactions • Solids and liquid solvents to not affect the equilibrium constant • Ex: AgCl (s) Ag+ (aq) + Cl- (aq) Keq = [Ag+][Cl-] • Ex: I2 (s)  I2 (g) Keq = [I2]

  15. Solving for Keq • 2SO2 (g) + O2 2SO3 (g) • At equilibrium, the concentration of SO2 is 4 mol, the concentration of O2is 3.5 mol, and the concentration of SO3 is 1 mol • What is the equilibrium constant?

  16. Le Chatelier’s Principle • When a reversible process is disturbed, it will proceed in the direction that relieves the stress • Causes of stress: concentration, pressure, temperature, action of a catalyst

  17. Concentration 2SO2 (g) + O2 2SO3 (g) If more is added… If some is taken away…

  18. Pressure 2SO2 (g) + O2 2SO3 (g) • 1 mol takes up 22.4 L at STP • Must consider total moles of reactants & total moles of products • Apply pressure  shifts toward fewer-molecule sides • Reduce pressure  shifts toward more-molecules side

  19. Temperature 2SO2 (g) + O2 2SO3 (g) + heat • Shifts equilibrium away from the “heat” side

  20. Catalyst 2SO2 (g) + O2 2SO3 (g) • Causes equilibrium to be reached more quickly • No effect on equilibrium constant

  21. 15B Applications of Equilibrium Chemistry alcohol + organic acid  ester + water • Esters • Give fruits their odors/flavors • Made for artificial flavoring of things like candy through a reversible reaction • Reaction does not go through to completion • Manufacturers want to force it to make more esters • What should they do? They can add reactants or remove products

  22. Ammonia N2 (g) + 3H2 (g) 2NH3(g) + heat • Ammonia (NH3) is used in fertilizers, cleaning products, & explosives • Reaction has large equilibrium constant (product favored) but is very slow • Fritz Haber invented the Haber process to speed it up

  23. Haber Process for Ammonia N2 (g) + 3H2 (g) 2NH3(g) + heat • Uses high pressure • Uses high temperature • Actually low temperature would help the forward reaction & high temperature hinders it • Uses a catalyst • Fe3O4 • Removes ammonia (NH3) as it forms

  24. Haber Process

  25. Homework: • Read the facet on p. 384-385 and answer the following questions: • 1) Who was Haber? • 2) What is the Haber process? • 3) List 2 good things and 2 bad things that happened because Haber invented the Haber process. Analyze the good/bad outcomes and explain if he should have kept the Haber process a secret.

  26. 15B-2 Solubility & Equilibria • Some salts form many ions • Lots of reactant dissolves, forming product • High Ksp • Some salts form few ions • Little reactant dissolves, forms little product • Low Ksp Ksp is related to number of ions formed Number of ions can be described by solubility

  27. Finding Ksp when given solubility aka “ion concentration” • BaSO4 has a solubility of 1.1x10^-5 M (1.1x10^-5 M BaSO4 will dissolve, forming ions) How many ions will form? BaSO4 Ba & SO4 1.1x10^-5 BaSO4 formula units  1.1x10^-5 Ba ions & 1.1x10^-5 SO4 ions 2- 2- 2+ 2+ 2- 2+

  28. Finding Ksp when given solubility aka “ion concentration” 2+ 2+ 2- 2- BaSO4 (s)  Ba (aq) + SO4 (aq) • Ksp = [Ba ][SO4 ] Remember: 1.1x10^-5 BaSO4 formula units  1.1x10^-5 Ba ions & 1.1x10^-5 SO4 ions • Ksp = [1.1x10^-5] [1.1x10^-5] • Ksp = 1.2x10^-10 How to find Ksp: • Write out the Ksp equation from the balanced formula • Plug in the number of ions • Multiply & solve 2- 2+

  29. Now you try it… • Solve CRQ 20 • Answer: 1.19x10^-11

  30. What if the problem didn’t give you the concentrations of each ion? Finding Ksp when given solubility aka “ion concentration” 3- 2+ Pb3(PO4)2(s)3Pb (aq) + 2PO4(aq) Lead (II) phosphate has a solubility of 7.7x10^-10. What is the equilibrium product constant of lead (II) phosphate? • Ksp = [Pb ] [PO4 ] • Ksp = [(3x7.7)x10^-10] [(2x7.7)x10^-10] Ksp = [23.1x10^-10] [15.4x10^-10] • Ksp = 3.0x10^-44 How to find Ksp: • Write out the Ksp equation from the balanced formula • Plug in the number of ions • Multiply & solve 3 2 3- 2+ 2 3 2 3

  31. Finding solubility when given Ksp aka “ion concentration” 3- 2+ Pb3(PO4)2(s)3Pb (aq) + 2PO4(aq) The solubility product constant of lead (II) phosphate is 3.0x10^-44 mol/L. What is the solubility? • 3.0x10^-44 = [Pb ] [PO4 ] • 3.0x10^-44 = [(3S) ] [(2S) ] • 3.0x10^-44 =[3 S ][2 S ] 3.0x10^-44 = [27S ][4S ] 3.0x10^-44 = 108S How to find Solubility: • Write out the Ksp equation from the balanced formula • Plug in the Ksp & replace each ion name with S • Solve for S 3 2 2+ 3- (Step 3 continued…) 3.0x10^-44 / 108 = S 2.78x10^-46 = S √ 2.78x10^-46 = √S Answer: 7.7x10^-10 = S 2 5 3 5 3 3 2 2 5 5 3 2 5

  32. Finding solubility when given Ksp aka “ion concentration” AgCl (s) Ag+ (aq) + Cl- (aq) Given that the Ksp of silver chloride is 1.8x10^-10, determine its solubility. • 1.8x10^-10 = [Ag+][Cl-] • 1.8x10^-10 = [S][S] • 1.8x10^-10 = S √ 1.8x10^-10 = √S 1.3x10^-5 = S How to find Solubility: • Write out the Ksp equation from the balanced formula • Plug in the Ksp & replace each ion name with S • Solve for S 2 2

  33. What can Ksp tell you? Ksp = how much it can hold • Saturated solution • Product of ion concentration = Ksp • (solution has as many ions as it can hold) • Unsaturated solution • Product of ion concentration is less than Ksp • (solution has fewer ions than it can hold) • Supersaturated solution • Product of ion concentration is greater than Ksp • (solution has more ions than it can hold)

  34. How to find Ksp: • Write out the Ksp equation from the balanced formula • Plug in the number of ions • Multiply & solve Calculate the Ksp of a 2.5x10^-6 M zinc hydroxide solution. Determine whether the solution is unsaturated, saturated, or supersaturated. 1) Determine the method: 2) Write out the formula Zn(OH)2(s)Zn (aq) + 2OH- (aq) 3) Follow the steps of the method Ksp = [2.5x10^-6][(2)(2.5x10^-6)] Ksp = 6.3x10^-17 how many ions solution is holding 4) Compare amount of ions the solution is holding (aka product of ion concentration) to how many ions it can hold (the Ksp in given in the table on p. 388) p.388 says Ksp=3.0x10^-17 2

  35. What the solution is holding: Ksp = 6.3x10^-17 What the solution can hold: Ksp=3.0x10^-17 How would you describe the solution? It is supersaturated.

  36. Now you try it… • Solve CRQ 26a • Answer: 7.5x10^-3 M

  37. Common Ion Effect AgCl (s)  Ag+ (aq) + Cl- (aq) • What would happen if more Cl- was added to a solution containing this reversible reaction? • Answer: Adding more product (Cl-) forces the reverse reaction to occur, which causes AgCl to precipitate out of the solution. • How can more Cl- be added? • Answer: Dissolve another salt that produces Cl- ions in the same solution.

  38. Common Ion Effect AgCl (s)  Ag+ (aq) + Cl- (aq) • This solution is saturated at 1.8x10^-10 M • What would happen if NaCl was dissolved in the same container? NaCl (s) Na+ (aq) +Cl- (aq) • The two solutions have a common ion (Cl-) • The common ion may cause the less soluble salt to precipitate by raising that number of ions above what the solution can hold

  39. Precipitation Reactions • Mixing two solutions may cause a precipitate to form • HCl – very soluble, forms H+ & Cl- ions • AgNO3 – very soluble, forms Ag+ & NO3 ions • If there are enough Ag+ & Cl- ions, AgCl forms • AgCl is not very soluble • If [Ag+][Cl-] (the ion concentration) is greater than Ksp of AgCl, a precipitate will form

  40. How to solve precipitate problems: • Write out the ions • Use p.388 table to find least soluble ion combination • Find the ion concentration • Compare Ksp & concentration Will a precipitate form when equal volumes of a 5.0x10^-5 M barium nitrate solution and a 1.0x10^-3 M sodium carbonate solution are mixed? 2+ 1) Ba(NO3)2 Ba & 2NO3 - Na2CO3 2Na+ & CO3 2) Least soluble combination: BaCO3 3) BaCO3 (s)  Ba (aq) + CO3 (aq) Ksp = [Ba][CO3] Ksp = [2.3x10^-5][5.0x10^-4] *molarities divided in half since solution has twice the volume when combined Ksp = 1.2x10^-8 4) Ksp of BaCO3 in the table is 5.1x10^-9 Actual Ksp is 1.2x10^-8 2- 2+ 2- 1.2x10^-8 > 5.1x10^-9 therefore a precipitate will form

  41. How to solve precipitate problems: • Write out the ions • Use p.388 table to find least soluble ion combination • Find the total ion concentration • Compare Ksp & concentration 2+ 1) MgCO3 Mg + CO3 - Mg(OH)2 Mg + 2(OH-) 2) Least soluble combination: Mg(OH)2 3) Mg(OH)2 Mg + 2(OH-) Ksp = [Mg][OH] Ksp = [(7.5x10^-4+1.2x10^-4)/2][(1.2x10^-4+1.2x10^-4)/2] ] *molarities added for each time the ion appears, then divided in half since solution has twice the volume when combined Ksp = 6.26x10^-12 4) Ksp of Mg(OH)2 in the table is 5.6x10^-12 Actual Ksp is greater, therefore Mg(OH)2 will precipitate Will a precipitate form when equal volumes of 7.5x10^-4 M MgCO3 & 1.2x10^-4 M Mg(OH)2 are mixed together? 2+ 2+ 2 2 6.3x10^-12 > 5.6x10^-12 therefore a precipitate will form

  42. Now you try it… • Solve CRQ 25c • Answer: Ksp = 1.0x10^-9, therefore no precipitate will form

  43. Homework: CRQ 21-26 • Helpful hints: 21) A. solubility / molar mass = molar solubility (these two from table) (use this to find answer) B. Once you know molar solubility, write out reactants & products equation. (Watch the coefficients/exponents!) C. Plug molar solubility value for each product & solve 22 & 24) Hint: think “common ion” 23) We will do as a class 25) Write the equation for AgCl, plug in values to compare Ksp & concentration 26) Write the Ksp equation, plug in Ksp, solve for S

  44. Homework at-a-glance • SRQ 15A1-9 • Explain: Who was Haber? What is the Haber Process? • SRQ15B • CRQ CRQ12-15 (a&b only) • CRQ 16-19 • CRQ 21-26 • Finish Analysis/Discussion of Lab 14 • Complete prelab Lab 15

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