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Chapter 4 The Major Classes of Chemical Reactions

Chapter 4 The Major Classes of Chemical Reactions. The Major Classes of Chemical Reactions. 4.1 The Role of Water as a Solvent. 4.2 Writing Equations for Aqueous Ionic Reactions. 4.3 Precipitation Reactions. 4.4 Acid-Base Reactions. 4.5 Oxidation-Reduction (Redox) Reactions.

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Chapter 4 The Major Classes of Chemical Reactions

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  1. Chapter 4 The Major Classes of Chemical Reactions

  2. The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions 4.6 Elemental Substances in Redox Reactions 4.7 Reversible Reactions: An Introduction to Chemical Equilibrium

  3. Water As a Solvent Water will dissolve ionic compounds. The charged ions become separated from each other and become surrounded by water molecules. One demonstration of this is the conduction of electricity through an ionic solution.

  4. The electrical conductivity of ionic solutions

  5. (a) (NH4)2SO4(s) 2NH4+(aq) + SO42-(aq) 2mol NH4+ 1mol (NH4)2SO4 Sample Problem 4.1 Determining Moles of Ions in Aqueous Ionic Solutions PROBLEM: How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water (b) 78.5g of cesium bromide dissolved in water (c) 7.42x1022 formula units of copper(II) nitrate dissolved in water (d) 35mL of 0.84M zinc chloride PLAN: We have to relate the information given and the number of moles of ions present when the substance dissolves in water. SOLUTION: = 10.mol NH4+ 5.0mol (NH4)2SO4 5.0mol SO42-

  6. (b) CsBr(s) Cs+(aq) + Br-(aq) mol CsBr 212.8g CsBr (c) Cu(NO3)2(s) Cu2+(aq) + 2NO3-(aq) mol Cu(NO3)2 6.022x1023 formula units (d) ZnCl2(aq) Zn2+(aq) + 2Cl-(aq) 1L 0.84mol ZnCl2 103mL L Sample Problem 4.1 Determining Moles of Ions in Aqueous Ionic Solutions continued = 0.369mol Cs+ 78.5g CsBr = 0.369mol CsBr = 0.369mol Br1- = 0.123mol Cu2+ 7.42x1022 formula units Cu(NO3)2 = 0.123mol Cu(NO3)2 = 0.246mol NO31- 35mL ZnCl2 = 2.9x110-2mol ZnCl2 = 2.9x110-2mol Zn2+ = 5.8x110-2mol Cl-

  7. Water is a Polar Solvent The electron distribution in a water molecule is such that an excess of electron density is on the oxygen atom and an electron deficiency is on the hydrogen atoms. The polarity of water is what allows it to separate charged particles in ionic solutions. It’s also what allows covalent compounds that are polar to dissolve in water, e.g. sugar. Although these polar compounds dissolve in water they are non-electrolytes. They do not conduct electricity since they are not charged. Water will also dissolve compounds with a polar bond to a H atom. These compounds, called acids, dissociate to give protons, H+, and anions.

  8. The Polarity of WaterElectron distribution in molecules of H2 and H2O

  9. The dissolution of an ionic compound

  10. Acids in water dissociate to give H+. The H+ bonds with a water molecule and becomes a hydrated proton, H3O+ , a hydronium ion.

  11. HNO3(l) H+(aq) + NO3-(aq) Sample Problem 4.2 Determining the Molarity of H+ Ions in Aqueous Solutions of Acids PROBLEM: Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H+ ion. What is the molarity of H+(aq) in 1.4M nitric acid? PLAN: Use the formula to find the molarity of H+. SOLUTION: Nitrate is NO3-. 1.4M HNO3(aq) should have 1.4M H+(aq).

  12. Writing Equations for Aqueous Ionic Reactions Molecular Equation - shows all reactants and products as if they were undissociated compounds. Total Ionic Equation - shows any compounds that are soluble ionic compounds dissociated into ions. Net Ionic Equation - eliminates spectator ions showing only actual chemical reaction taking place. Precipitation Reactions - Reactant ions form an insoluble product which precipitates from solution. Acid-Base Reactions - (neutralization) - A reaction between H+ ions from an acid dissolved in water and OH- ions from a base dissolved in water. The product is a molecule of water. Redox Reactions - (later in chapter)

  13. Writing equations for aqueous ionic reactions. A precipitation reaction and its equation. ©

  14. Soluble Ionic Compounds Insoluble Ionic Compounds Table 4.1 Solubility Rules For Ionic Compounds in Water 1. All common compounds of Group 1A(1) ions (Li+, Na+, K+, etc.) and ammonium ion (NH4+) are soluble. 2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2-) and most perchlorates (ClO4-) are soluble. 3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble, except those of Ag+, Pb2+, Cu+, and Hg22+. 4. All common sulfates (SO42-) are soluble,except those of Ca2+ , Sr2+ , Ba2+ , and Pb2+ . 1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca2+). 2. All common carbonates (CO32-) and phosphates (PO43-) are insoluble, except those of Group 1A(1) and NH4+. 3. All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH4+.

  15. Sample Problem 4.3 Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations PROBLEM: Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) sodium sulfate(aq) + strontium nitrate(aq) (b) ammonium perchlorate(aq) + sodium bromide(aq) SOLUTION: PLAN: (a) Na2SO4(aq) + Sr(NO3)2 (aq) 2NaNO3(aq) + SrSO4(s) write ions 2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq) 2Na+(aq) +2NO3-(aq)+ SrSO4(s) combine anions & cations SO42-(aq)+ Sr2+(aq) SrSO4(s) check for insolubility Table 4.1 (b) NH4ClO4(aq) + NaBr(aq) NH4Br(aq) + NaClO4(aq) eliminate spectator ions for net ionic equation All reactants and products are soluble so no reaction occurs.

  16. Strong / Weak Acids and BasesStrong acids and bases dissociate completely into ions in water.Weak acids and bases only slightly dissociate when dissolved in water. Acids Bases Strong Strong hydrochloric acid, HCl sodium hydroxide, NaOH hydrobromic acid, HBr potassium hydroxide, KOH hydroiodic acid, 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 Weak hydrofluoric acid, HF ammonia, NH3 phosphoric acid, H3PO4 acetic acid, CH3COOH (or HC2H3O2)

  17. Sample Problem 4.4 Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions. (a) strontium hydroxide(aq) + perchloric acid(aq) (b) barium hydroxide(aq) + sulfuric acid(aq) PLAN: SOLUTION: reactants are strong acids and bases and therefore completely ionized in water (a) Sr(OH)2(aq)+2HClO4(aq) 2H2O(l)+Sr(ClO4)2(aq) Sr2+(aq) + 2OH-(aq)+ 2H+(aq)+ 2ClO4-(aq) 2H2O(l)+Sr2+(aq)+2ClO4-(aq) Sr2+(aq) + 2OH-(aq)+ 2H+(aq)+ 2ClO4-(aq) 2H2O(l)+Sr2+(aq)+2ClO4-(aq) 2H+(aq)+ products are 2OH-(aq)+ 2H+(aq) 2H2O(l) water (b) Ba(OH)2(aq) + H2SO4(aq) 2H2O(l) + BaSO4(aq) spectator ions Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq) 2H2O(l)+Ba2+(aq)+SO42-(aq) 2OH-(aq)+ 2H+(aq) 2H2O(l)

  18. An aqueous strong acid-strong base reaction on the atomic scale.

  19. An acid-base titration

  20. 1L 103mL 5.078x10-3mol HCl 0.050L Sample Problem 4.5 Finding the Concentration of Acid from an Acid-Base Titration PROBLEM: You perform an acid-base titration to standardize an HCl solution by placing 50.00mL of HCl in a flask with a few drops of indicator solution. You put 0.1524M NaOH into the buret, and the initial reading is 0.55mL. At the end point, the buret reading is 33.87mL. What is the concentration of the HCl solution? PLAN: SOLUTION: volume(L) of base NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) multiply by M of base (33.87-0.55)mL x = 0.03332L mol of base molar ratio 0.03332L X 0.1524M = 5.078x10-3mol NaOH mol of acid Molar ratio is 1:1 divide by L of acid = 0.1016M HCl M of acid

  21. Writing Equations for Aqueous Ionic Reactions Molecular Equation - shows all reactants and products as if they were undissociated compounds. Total Ionic Equation - shows any compounds that are soluble ionic compounds dissociated into ions. Net Ionic Equation - eliminates spectator ions showing only actual chemical reaction taking place. Oxidation-Reduction (Redox) Reactions - Reactions where there is movement of electrons from one reactant to another. It can be in the form of a transfer of electrons from one atom to another or in the case of many covalent compounds, an uneven distribution of electron density in the covalent bond.

  22. The redox process in compound formation

  23. General rules Rules for specific atoms or periodic table groups 1. For Group 1A(1): O.N. = +1 in all compounds 2. For Group 2A(2): O.N. = +2 in all compounds 3. For hydrogen: O.N. = +1 in combination with nonmetals 4. For fluorine: O.N. = -1 in combination with metals and boron 5. For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F) 6. For Group 7A(17): O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group To keep track of the transfer of electrons, atoms are assigned oxidation numbers. Table 4.3 Rules for Assigning an Oxidation Number (O.N.) 1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge.

  24. Sample Problem 4.6 Determining the Oxidation Number of an Element PROBLEM: Determine the oxidation number (O.N.) of each element in these compounds: (a) zinc chloride (b) sulfur trioxide (c) nitric acid PLAN: The O.N.s of the ions in a polyatomic ion add up to the charge of the ion and the O.N.s of the ions in the compound add up to zero. SOLUTION: (a) ZnCl2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.

  25. Highest and lowest oxidation numbers of reactive main-group elements

  26. e- Y X A summary of terminology for oxidation-reduction (redox) reactions transfer or shift of electrons X loses electron(s) Y gains electron(s) X is oxidized Y is reduced X is the reducing agent Y is the oxidizing agent X increases its oxidation number Y decreases its oxidation number

  27. (a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) (a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g) (b) PbO(s) + CO(g) Pb(s) + CO2(g) (c) 2H2(g) + O2(g) 2H2O(g) 0 +1 +6 -2 +3 +6 -2 0 Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents PROBLEM: Identify the oxidizing agent and reducing agent in each of the following: PLAN: Assign an O.N. for each atom and see which gained and which lost electrons in going from reactants to products. An increase in O.N. means the species was oxidized (and is the reducing agent) and a decrease in O.N. means the species was reduced (is the oxidizing agent). SOLUTION: The O.N. of Al increases; it is oxidized; it is the reducing agent. The O.N. of H decreases; it is reduced; it is the oxidizing agent.

  28. +2 -2 +2 -2 0 +4 -2 (b) PbO(s) + CO(g) Pb(s) + CO2(g) (c) 2H2(g) + O2(g) 2H2O(g) 0 0 +1 -2 Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents continued The O.N. of C increases; it is oxidized; it is the reducing agent. The O.N. of Pb decreases; it is reduced; it is the oxidizing agent. The O.N. of H increases; it is oxidized; it is the reducing agent. The O.N. of O decreases; it is reduced; it is the oxidizing agent.

  29. Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l) (a) Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l) (a) Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l) (b) PbS(s) + O2(g) PbO(s) + SO2(g) 0 +1 +5 -2 +2 +5 -2 +4 -2 +1 -2 Sample Problem 4.8 Balancing Redox Equations by the Oxidation Number Method PROBLEM: Use the oxidation number method to balance the following equations: SOLUTION: O.N. of Cu increases because it loses 2e-; it is oxidized and is the reducing agent. O.N. of N decreases because it gains 1e-; it is reduced and is the oxidizing agent. loses 2e- balance other ions gains 1e- x2 to balance e- balance unchanged polyatomic ions

  30. Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l) (b) PbS(s) + O2(g) PbO(s) + SO2(g) +2 -2 0 +2 -2 +4 -2 PbS(s) + O2(g) PbO(s) + SO2(g) 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) Sample Problem 4.8 Balancing Redox Equations by the Oxidation Number Method continued 4 2 2 2 2 loses 6e- 3/2 gains 2e- per O; need 3/2 O2 to make 3O2- Multiply by 2 to have whole number coefficients.

  31. A redox titrationpermanganate ion / oxalate ion

  32. KMnO4(aq) + CaC2O4(s) + H2SO4(aq) MnSO4(aq) + K2SO4(aq) + CaSO4(s) + CO2(g) + H2O(l) PLAN: (a) Sample Problem 4.9 Finding an Unknown Concentration by a Redox Titration PROBLEM: Calcium ion (Ca2+) is required for blood to clot and for many other cell processes. An abnormal Ca2+ concentration is indicative of disease. To measure the Ca2+ concentration, 1.00mL of human blood was treated with Na2C2O4 solution. The resulting CaC2O4 precipitate was filtered and dissolved in dilute H2SO4. This solution required 2.05mL of 4.88x10-4M KMnO4 to reach the end point. The unbalanced equation is (a) Calculate the amount (mol) of Ca2+. (b) Calculate the amount (mol) of Ca2+ ion concentration expressed in units of mg Ca2+/100mL blood. volume of KMnO4 soln mol of Ca2+ multiply by M ratio of elements in formula mol of KMnO4 mol of CaC2O4 molar ratio

  33. L 4.88x10-4mol KMnO4 103 mL L 1mol Ca2+ 5mol CaC2O4 2mol KMnO4 1mol CaC2O4 PLAN: 2.50x10-6 mol Ca2+ 2.50x10-4 mol Ca2+ =2.50x10-4 mol Ca2+ (b) 1mL blood 100mL blood 100mL blood 40.08g Ca2+ mg mol Ca2+ 10-3g Sample Problem 4.9 Finding an Unknown Concentration by a Redox Titration continued SOLUTION: 2.05mL soln = 1.00x10-6mol KMnO4 1.00x10-6mol KMnO4 = 2.50x10-6 mol CaC2O4 2.50x10-6 mol CaC2O4 = 2.50x10-6 mol Ca2+ mol Ca2+/1mL blood SOLUTION: multiply by 100 x100 mol Ca2+/100mL blood multiply by M g Ca2+/100mL blood 10-3g = 1mg =10.0mg Ca2+/100mL blood mg Ca2+/100mL blood

  34. Older Approach to Classify of Chemical Reactions Combination Reactions - Where two or more reactants form one product. Decomposition Reactions - Where one reactant formed two or more products. Displacement Reactions - Where atoms or ions of reactants exchange places in products. Also - Combustion Reactions - Special redox reactions with oxygen a reactant which serves as the oxidant.

  35. Combining elements to form an ionic compound

  36. Decomposing a compound to its elements

  37. Displacing one metal with another

  38. (a) magnesium(s) + nitrogen(g) magnesium nitride (aq) (b) hydrogen peroxide(l) water(l) + oxygen gas (c) aluminum(s) + lead(II) nitrate(aq) aluminum nitrate(aq) + lead(s) Sample Problem 4.10 Identifying the Type of Redox Reaction PROBLEM: Classify each of the following redox reactions as a combination, decomposition, or displacement reaction, write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: PLAN: Combination reactions produce fewer products than reactants. Decomposition reactions produce more products than reactants. Displacement reactions have the same number of products and reactants.

  39. 0 0 +2 -3 Mg(s) + N2(g) Mg3N2 (aq) +1 -1 +1 -2 0 H2O2(l) H2O(l) + O2(g) 2 H2O2(l) 2 H2O(l) + O2(g) 0 +2 +5 -2 +3 +5 -2 0 Al(s) + Pb(NO3)2(aq) Al(NO3)3(aq) + Pb(s) 2Al(s) + 3Pb(NO3)2(aq) 2Al(NO3)3(aq) + 3Pb(s) Sample Problem 4.10 Identifying the Type of Redox Reaction (a) Combination continued 3 Mg is the reducing agent; N2 is the oxidizing agent. (b) Decomposition 1/2 or O is the oxidizing and reducing agent. (c) Displacement Pb is the oxidizing and Al is the reducing agent.

  40. Reversible ReactionsChemical Equilibrium For some reactions, which proceed from reactants to products, can also undergo the reverse reaction of “products” to “reactants”. For example:CaCO3 CaO + CO2 and CaO + CO2 CaCO3 But both reactions are taking place simultaneously and an equilibrium is reached. The equation is written: CaCO3 CaO + CO2

  41. End of Chapter 4

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