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Chapter 4: The Major Classes of Chemical Reactions

C142B Lecture Notes Campbell / Callis. Chapter 4: The Major Classes of Chemical Reactions. 4.1 The Role of Water as a Solvent 4.2 Precipitation Reactions and Acid-Base Reactions 4.3 Oxidation - Reduction (Redox) Reactions 4.4 Counting Reactants and Products in Precipitation,

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Chapter 4: The Major Classes of Chemical Reactions

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  1. C142B Lecture Notes Campbell / Callis Chapter 4: The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Precipitation Reactions and Acid-Base Reactions 4.3 Oxidation - Reduction (Redox) Reactions 4.4 Counting Reactants and Products in Precipitation, Acid-Base, and Redox Processes 4.5 Reversible Reactions: An Introduction to Chemical Equilibrium

  2. The Role of Water as a Solvent: The solubility of Ionic Compounds Electrical conductivity - The flow of electrical current in a solution is a measure of the solubility of ionic compounds or a measurement of the presence of ions in solution. Electrolyte - A substance that conducts a current when dissolved in water. Soluble ionic compound dissociate completely and may conduct a large current, and are called Strong Electrolytes. NaCl(s)+ H2O(l) Na+(aq) + Cl -(aq) When sodium chloride dissolves into water the ions become solvated, and are surrounded by water molecules. These ions are called “aqueous” and are free to move through out the solution, and are conducting electricity, or helping electrons to move through out the solution

  3. Fig 4.1 (P135) Electrical Conductivity of Ionic Solutions

  4. Fig. 4.2

  5. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - I Problem: How many moles of each ion are in each of the following: a) 4.0 moles of sodium carbonate dissolved in water b) 46.5 g of rubidium fluoride dissolved in water c) 5.14 x 1021 formula units of iron (III) chloride dissolved in water d) 75.0 mL of 0.56M scandium bromide dissolved in water e) 7.8 moles of ammonium sulfate dissolved in water a) Na2CO3(s) 2 Na+(aq) + CO3-2(aq) moles of Na+ = 4.0 moles Na2CO3 x = 8.0 moles Na+ and 4.0 moles of CO3-2 are present H2O 2 mol Na+ 1 mol Na2CO3

  6. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H2O b) RbF(s) Rb+ (aq)+ F - (aq) moles of RbF = H2O c) FeCl3(s) Fe+3 (aq) + 3 Cl - (aq) moles of FeCl3 = 9.32 x 1021 formula units x moles of Cl - = moles Fe+3 =

  7. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - II H2O b) RbF(s) Rb+ (aq)+ F - (aq) 1 mol RbF 104.47 g RbF moles of RbF = 46.5 g RbF x = 0.445 moles RbF thus, 0.445 mol Rb+and 0.445 mol F -are present H2O c) FeCl3(s) Fe+3 (aq) + 3 Cl - (aq) moles of FeCl3 = 9.32 x 1021 formula units 1 mol FeCl3 6.022 x 1023 formula units FeCl3 x = 0.0155 mol FeCl3 3 mol Cl - 1 mol FeCl3 moles of Cl - = 0.0155 mol FeCl3 x = 0.0465 mol Cl - and 0.0155 mol Fe+3 are also present.

  8. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III H2O d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq) Converting from volume to moles: 1 L 103 mL Moles of ScBr3 = 75.0 mL x x Moles of Br - = H2O e) (NH4)2SO4 (s) 2 NH4+ (aq) + SO4- 2 (aq) 2 mol NH4+ 1 mol(NH4)2SO4 Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4+ and 7.8 mol SO4- 2are also present.

  9. Determining Moles of Ions in Aqueous Solutions of Ionic Compounds - III H2O d) ScBr3 (s) Sc+3(aq) + 3 Br -(aq) Converting from volume to moles: 1 L 103 mL 0.56 mol ScBr3 1 L Moles of ScBr3 = 75.0 mL x x = 0.042 mol ScBr3 3 mol Br - 1 mol ScBr3 Moles of Br - = 0.042 mol ScBr3 x = 0.126 mol Br - 0.042 mol Sc+3 are also present H2O e) (NH4)2SO4 (s) 2 NH4+ (aq) + SO4- 2 (aq) 2 mol NH4+ 1 mol(NH4)2SO4 Moles of NH4+ = 7.8 moles (NH4)2SO4 x = 15.6 mol NH4+ and 7.8 mol SO4- 2are also present.

  10. Fig. 4.3

  11. The Solubility of Ionic Compounds in Water The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20oC = 365 g/L Solubility of MgCl2 in water at 20oC = 542.5 g/L Solubility of AlCl3 in water at 20oC = 699 g/L Solubility of PbCl2 in water at 20oC = 9.9 g/L Solubility of AgCl in water at 20oC = 0.009 g/L Solubility of CuCl in water at 20oC = 0.0062 g/L

  12. The Solubility of Covalent Compounds in Water The covalent compounds that are very soluble in water are the ones with -OH group in them and are called “Polar” and can have strong polar (electrostatic)interactions with water. Examples are compound such as table sugar, sucrose (C12H22O11); beverage alcohol, ethanol (C2H5-OH); and ethylene glycol (C2H6O2) in antifreeze. H Methanol = Methyl Alcohol H C O H H Other covalent compounds that do not contain a polar center, or the -OH group are considered “Non-Polar” , and have little or no interactions with water molecules. Examples are the hydrocarbons in Gasoline and Oil. This leads to the obvious problems in Oil spills, where the oil will not mix with the water and forms a layer on the surface! Octane = C8H18 and / or Benzene = C6H6

  13. Acids - A group of Covalent molecules which loose Hydrogen ions to water molecules in solution When gaseous hydrogen Iodide dissolves in water, the attraction of the oxygen atom of the water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom, and it is lost to the water molecule to form an Hydronium ion and an Iodide ion in solution. We can write the Hydrogen atom in solution as either H+(aq) or as H3O+(aq) they mean the same thing in solution. The presence of a Hydrogen atom that is easily lost in solution is an “Acid” and is called an “acidic” solution. The water (H2O) could also be written above the arrow indicating that the solvent was water in which the HI was dissolved. HI(g) + H2O(l) H+ (aq) + I - (aq) HI(g) + H2O(l) H3O+(aq) + I -(aq) H2O(l) HI(g) H+(aq) + I -(aq)

  14. Fig. 4.4

  15. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of pure H2SO4 into sufficient water to produce 2.30 Liters of sulfuric acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) 1 mole H2SO4 98.09 g H2SO4 Moles H2SO4 = = 1.58 moles H2SO4 155 g H2SO4 x Molarity of SO4- 2 = Molarity of H3O+ (or simply H+) =

  16. Strong Acids and the Molarity of H+ Ions in Aqueous Solutions of Acids Problem: In aqueous solutions, each molecule of sulfuric acid will loose two protons to yield two Hydronium ions, and one sulfate ion. What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of pure H2SO4 into sufficient water to produce 2.30 Liters of sulfuric acid solution? Plan: Determine the number of moles of sulfuric acid, divide the moles by the volume to get the molarity of the acid and the sulfate ion. The hydronium ions concentration will be twice the acid molarity. Solution: Two moles of H+ are released for every mole of acid: H2SO4 (l) + 2 H2O(l) 2 H3O+(aq) + SO4- 2(aq) 1 mole H2SO4 98.09 g H2SO4 Moles H2SO4 = = 1.58 moles H2SO4 155 g H2SO4 x 1.58 mol SO4-2 2.30 L solution Molarity of SO4- 2 = = 0.687 Molar in SO4- 2 Molarity of H+ = 2 x 0.687 mol H+ / 2.30 liters = 0.597 Molar in H+

  17. Fig. 4.5

  18. Precipitation Reactions: A solid product is formed When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together. Pb(NO3)(aq) + NaI(aq) Pb+2(aq) + 2 NO3- (aq) + Na+ (aq) + I- (aq) When we add these two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have Lead Nitrate and Sodium Iodide, or Lead Iodide and Sodium Nitrate formed, to determine which will happen we must look at the Solubility Rules (P 141) to determine what could form. Rule 3a in Table 4.1 indicates that Lead Iodide will be insoluble, so a precipitate will form! The remainder (sodium sulfate) is soluble (Rule 4a). Pb(NO3)2 (aq) + 2 NaI(aq) PbI2 (s) + 2 NaNO3 (aq)

  19. Precipitation Reactions: Will a Precipitate form? If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate? KCl(aq) + NH4NO3(aq) = K+ (aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) By exchanging cations and anions we see that we could have Potassium Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility rules, Table 4.1: If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate, will we get a precipitate? Table 4.1 shows that: Therefore we get: Na2SO4(aq) + Ba(NO3)2(aq) BaSO4( ) + 2 NaNO3( )

  20. Precipitation Reactions: Will a Precipitate form? If we add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate, will we get a precipitate? KCl(aq) + NH4NO3(aq) = K+ (aq) + Cl-(aq) + NH4+(aq) + NO3-(aq) By exchanging cations and anions we see that we could have Potassium Chloride and Ammonium Nitrate, or Potassium Nitrate and Ammonium Chloride. In looking at the solubility rules, Table 4.1 shows all possible products as soluble, so there is no net reaction! KCl(aq) + NH4NO3(aq) = No. Reaction! If we mix a solution of Sodium Sulfate with a solution of Barium Nitrate, will we get a precipitate? Table 4.1 shows that Barium Sulfate is insoluble, therefore we will get a precipitate! Na2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2 NaNO3(aq)

  21. Fig. 4.6

  22. Predicting Whether a Precipitation Reaction Occurs (see Table 4.1); Writing Equations: a) Calcium Nitrate and Sodium Sulfate solutions are added together. Molecular Equation Ca(NO3)2(aq) + Na2SO4(aq) CaSO4(s) + NaNO3(aq) Total Ionic Equation Ca2+(aq)+ 2 NO3-(aq) +2 Na+(aq)+ SO4-2(aq) CaSO4(s) + 2Na+(aq) +2 NO3-(aq) Net Ionic Equation Ca2+(aq)+ SO4-2(aq) CaSO4 (s) Spectator Ions are Na+ and NO3- b) Ammonium Sulfate and Magnesium Chloride are added together. In exchanging ions, no precipitates will be formed, accd. To Table 4.1, so there will be no reactions occurring! All ions are spectator ions!

  23. Acid - Base Reactions : Neutralization Rxns. An Acid is a substance that produces H+ (H3O+) ions when dissolved in water. A Base is a substance that produces OH - ions when dissolved in water. Acids and Bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions, resp.. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate only to a tiny extent (<<100%) and are weak electrolytes. The generalized reaction between an Acid and a Base is: HX(aq) + MOH(aq) MX(aq) + H2O(l) Acid + Base = Salt(aq) + Water

  24. Fig. 4.7

  25. Table 4.2 (P 143) Selected Acids and Bases Acids Bases Strong (100% to H+) Strong (100% to OH-) Hydrochloric, HCl Sodium hydroxide, NaOH Hydrobromic, HBr Potassium hydroxide, KOH Hydroiodoic, HI Calcium hydroxide, Ca(OH)2 Nitric acid, HNO3 Strontium hydroxide, Sr(OH)2 Sulfuric acid, H2SO4 Barium hydroxide, Ba(OH)2 Perchloric acid, HClO4 Weak (tiny % to H+) Weak (tiny % yield of OH-) Hydrofluoric, HF Ammonia, NH3 Phosphoric acid, H3PO4 Acidic acid, CH3COOH (or HC2H3O2)

  26. Writing Balanced Equations for Neutralization Reactions - I Problem:Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) Calcium Hydroxide(aq) and HydroIodic acid(aq) b) Lithium Hydroxide(aq) and Nitric acid(aq) c) Barium Hydroxide(aq) and Sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH)2(aq) + 2HI(aq) CaI2(aq) + 2H2O(l) Ca2+(aq) + 2 OH -(aq) + 2 H+(aq) + 2 I -(aq) Ca2+(aq) + 2 I -(aq) + 2 H2O(l) 2 OH -(aq) + 2 H+(aq) 2 H2O(l)

  27. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3(aq) c) Ba(OH)2(aq) + H2SO4(aq)

  28. Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO3(aq) LiNO3(aq) + H2O(l) Li+(aq) + OH -(aq) + H+(aq) + NO3-(aq) Li+ (aq) + NO3-(aq) + H2O(l) OH -(aq) + H+(aq)H2O(l) c) Ba(OH)2(aq) + H2SO4(aq) BaSO4(s) + 2H2O(l) Ba2+(aq) + 2OH -(aq) + 2 H+(aq) + SO42 -(aq) BaSO4(s) + 2 H2O(l) Ba2 +(aq) + 2 OH -(aq) + 2 H+(aq) + SO42 -(aq) BaSO4(s) + 2 H2O(l)

  29. Fig. 4.8

  30. Finding the Concentration of Acid from an Acid - Base Titration Volume (L) of base (difference in buret readings) needed to titrate M (mol/L) of base Moles of base needed to titrate molar ratio Moles of acid which were titrated volume (L) of acid M (mol/L) of original acid soln.

  31. O C O K+ O H C O Potassium Hydrogenphthalate KHC8H4O4 O C K+ O O C H+ O = “KHP” = a common acid used to titrate bases (M = 204.2 g/mol)

  32. Finding the Concentration of Base from an Acid - Base Titration - I Problem: A titration is performed between Sodium Hydroxide and KHP (204.2 g/mol) to standardize the base solution, by placing 50.00 mg of solid Potassium Hydrogenphthalate in a flask with a few drops of an indicator. A buret is filled with the base, and the initial buret reading is 0.55 mL; at the end of the titration the buret reading is 33.87 mL. What is the concentration of the base? Plan: Use the molar mass of KHP to calculate the number of moles of the acid, from the balanced chemical equation, the reaction is equal molar, so we know the moles of base, and from the difference in the buret readings, we can calculate the molarity of the base. Solution: The rxn is:KHP(aq) + OH-(aq) --> KP-(aq) + H2O or HKC8H4O4(aq) + OH -(aq) KC8H4O4-(aq) + H2O(aq)

  33. Finding the Concentration of Base from an Acid - Base Titration - II 50.00 mg KHP moles KHP = Volume of base = Final buret reading - Initial buret reading = one mole of H+ : one mole of OH-; therefore ____________ moles of KHP will titrate _____________ moles of NaOH. moles L molarity of NaOH = = molarity of base =

  34. Finding the Concentration of Base from an Acid - Base Titration - II 50.00 mg KHP 204.2 g KHP 1 mol KHP 1.00 g 1000 mg moles KHP = x = 0.00024486 mol KHP Volume of base = Final buret reading - Initial buret reading = 33.87 mL - 0.55 mL = 33.32 mL of base one mole of H+ : one mole of OH-; therefore 0.00024486 moles of KHP will titrate 0.00024486 moles of NaOH. 0.00024486 moles 0.03332 L molarity of NaOH = = 0.07348679 moles per liter molarity of base = 0.07349 M

  35. Fig. 4.9

  36. Fig. 4.10

  37. Fig. 4.11

  38. Highest and Lowest oxidation numbers of Chemically reactive main-group Elements +1 +1 -1 -4 1 H Fig. 4.12 1A 2A 3A 4A 5A 6A 7A Period +4 +4 +5 +6 +7 +1 +2 +3 -3 -2 -1 2 Li Be B C N O F 3 Na Mg Al Si P S Cl 4 K Ca Ga Ge As Se Br non-metals 5 Rb Sr In Sn Sb Te I metalloids 6 Cs Ba Tl Pb Bi Po At metals 7 Fr Ra

  39. Fig. 4.13

  40. Period IA VIIIA Common Ox #s: Main Group Elements H He 1 +1 -1 IIA IIIA IVA VA VIA VIIA Li Be B C N O F Ne 2 +4,+2 -1,-4 all from +1 +2 +3 -1,-2 -1 +5 -3 Na Mg Al Si P S -1 Cl Ar 3 +5,+3 -3 +6,+4 +2,-2 +7,+5 +3,+1 +1 +2 +3 +4,-4 Kr K Ca Ga Ge As Se -1 Br 4 +4,+2 -4 +5,+3 -3 +6,+4 -2 +7,+5 +3,+1 +2 +1 +2 +3, +2 Xe Rb Sr In Sn Sb Te -1 I 5 +3,+2 +1 +4,+2, -4 +5,+3 -3 +6,+4 -2 +7,+5 +3,+1 +6,+4 +2 +1 +2 Rn Cs Ba Tl Pb Bi Po -1 At 6 +6,+4 +2,-2 +7,+5 +3,+1 +1 +4,+2 +2 +2 +3,+1 +3

  41. Transition Metals Possible Oxidation States VIIIB IIIB IVB VB VIB VIIB IB IIB Sc Ti V Cr +2 Mn Fe Co Ni Cu Zn +4,+3 +2 +5,+4 +3+2 +6,+3 +2 +7,+6 +4,+3 +3 +3,+2 +3,+2 +2 +2,+1 +2 Y Zr Nb Mo Tc Ru Rh Pd Ag Cd +5,+4 +2 +6,+5 +4,+3 +7,+5 +4 +8,+5 +4,+3 +3 +4,+3 +4,+3 +4,+2 +1 +2 La Hf Ta W Re +2 Os Ir Pt Au Hg +5,+4 +3 +6,+5 +4 +7,+5 +4 +8,+6 +4,+3 +4,+3 +1 +3 +4,+3 +4,+2 +3,+1 +2,+1

  42. Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a)Iron III Chloride b) Nitrogen Dioxide c)Sulfuric acid Plan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: a) FeCl3 This compound is composed of monoatomic ions. The Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe must be +3. b) NO2 c) H2SO4

  43. Determining the Oxidation Number of an Element in a Compound Problem: Determine the oxidation number (Ox. No.) of each element in the following compounds. a)Iron III Chloride b) Nitrogen Dioxide c)Sulfuric acid Plan: We apply the rules in Table 4.3, always making sure that the Ox. No. values in a compound add up to zero, and in a polyatomic ion, to the ion’s charge. Solution: a) FeCl3 This compound is composed of monoatomic ions. The Ox. No. of Cl- is -1, for a total of -3. Therefore the Fe must be +3. b) NO2 The Ox. No. of oxygen is -2 for a total of -4. Since the Ox. No. in a compound must add up to zero, the Ox. No. of N is +4. c) H2SO4 The Ox. No. of H is +1, so the SO42- group must sum to -2. The Ox. No. of each O is -2 for a total of -8. So the Sulfur atom is +6.

  44. Recognizing Oxidizing and Reducing Agents - I Problem: Identify the oxidizing and reducing agent in each of the Rx: a) Zn(s) + 2 HCl(aq) ZnCl2 (aq) + H2 (g) b) S8 (s) + 12 O2 (g) 8 SO3 (g) c) NiO(s) + CO(g) Ni(s) + CO2 (g) Plan: First we assign an oxidation number (O.N.) to each atom (or ion) based on the rules in Table 4.3. A reactant is the reducing agent if it contains an atom that is oxidized (O.N. increased in the reaction). A reactant is the oxidizing agent if it contains an atom that is reduced ( O.N. decreased). Solution: a) Assigning oxidation numbers: -1 -1 +1 0 0 +2 Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2 (g) HCl is the oxidizingagent, andZnis the reducing agent!

  45. Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: S8 (s) + 12 O2 (g) 8 SO3 (g) ____ is the reducing agent and ____ is theoxidizingagent. c) Assigning oxidation numbers: NiO(s) + CO(g) Ni(s) + CO2 (g) ___ is the reducing agent and ____ is the oxidizing agent.

  46. Recognizing Oxidizing and Reducing Agents - II b) Assigning oxidation numbers: S [0] S[+6] -2 S is Oxidized +6 0 0 O[0] O[-2] S8 (s) + 12 O2 (g) 8 SO3 (g) O is Reduced S8 is the reducing agent andO2is theoxidizingagent c) Assigning oxidation numbers: C[+2] C[+4] C is oxidized Ni[+2] Ni[0] Ni is Reduced -2 -2 -2 0 +2 +4 +2 NiO(s) + CO(g) Ni(s) + CO2 (g) CO is the reducing agent and NiO is the oxidizing agent

  47. Balancing REDOX Equations: The oxidation number method Step 1) Assign oxidation numbers to all elements in the equation. Step 2) From the changes in oxidation numbers, identify the oxidized and reduced species. Step 3) Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes. Draw tie-lines between these atoms to show electron changes. Step 4) Multiply one or both of these numbers by appropriate factors to make the electrons lost equal the electrons gained, and use the factors as balancing coefficients. Step 5) Complete the balancing by inspection, adding states of matter.

  48. REDOX Balancing using Ox. No. Method - I +2 e- 0 -2 ___ H2 (g) +___ O2 (g) ___ H2O(g) 2 2 - 1 e- +1 0 electrons lost must = electrons gained therefore multiply Hydrogen reaction by 2! and we are balanced!

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