Presentation on theme: "The Major Classes of Chemical Reactions"— Presentation transcript:
1 The Major Classes of Chemical Reactions Chapter 4The Major Classes of Chemical Reactions
2 The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent4.2 Writing Equations for Aqueous Ionic Reactions4.3 Precipitation Reactions4.4 Acid-Base Reactions4.5 Oxidation-Reduction (Redox) Reactions4.6 Elemental Substances in Redox Reactions4.7 Reversible Reactions: An Introduction to ChemicalEquilibrium
3 Water As a SolventWater 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.
5 Sample Problem 4.1Determining Moles of Ions in Aqueous Ionic SolutionsPROBLEM: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 chloridePLAN:We have to relate the information given and the number of moles of ions present when the substance dissolves in water.SOLUTION:(a) (NH4)2SO4(s) NH4+(aq) + SO42-(aq)2mol NH4+1mol (NH4)2SO4= 10.mol NH4+5.0mol (NH4)2SO45.0mol SO42-
6 Sample Problem 4.1Determining Moles of Ions in Aqueous Ionic Solutionscontinued(b) CsBr(s) Cs+(aq) + Br-(aq)= 0.369mol Cs+mol CsBr212.8g CsBr78.5g CsBr= 0.369mol CsBr= 0.369mol Br1-(c) Cu(NO3)2(s) Cu2+(aq) + 2NO3-(aq)= 0.123mol Cu2+7.42x1022 formula units Cu(NO3)2mol Cu(NO3)26.022x1023 formula units= 0.123mol Cu(NO3)2= 0.246mol NO3 1-(d) ZnCl2(aq) Zn2+(aq) + 2Cl-(aq)1L103mL0.84mol ZnCl2L35mL 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 Water Electron distribution in molecules of H2 and H2O
10 Acids in water dissociate to give H+ Acids in water dissociate to give H+. The H+ bonds with a water molecule and becomes a hydrated proton, H3O+ , a hydronium ion.
11 Sample Problem 4.2Determining the Molarity of H+ Ions in Aqueous Solutions of AcidsPROBLEM: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-.HNO3(l) H+(aq) + NO3-(aq)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 Writing equations for aqueous ionic reactions. A precipitation reaction and its equation.
14 Table 4.1 Solubility Rules For Ionic Compounds in Water Soluble Ionic Compounds1. 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+ .Insoluble Ionic Compounds1. 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 4.3Predicting Whether a Precipitation Reaction Occurs; Writing Ionic EquationsPROBLEM: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) NaNO3(aq) + SrSO4(s)write ions2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3-(aq)2Na+(aq) +2NO3-(aq)+ SrSO4(s)combine anions & cationsSO42-(aq)+ Sr2+(aq) SrSO4(s)check for insolubilityTable 4.1(b) NH4ClO4(aq) + NaBr (aq) NH4Br (aq) + NaClO4(aq)eliminate spectator ions for net ionic equationAll reactants and products are soluble so no reaction occurs.
16 Strong / Weak Acids and Bases Strong acids and bases dissociate completely into ions in water. Weak acids and bases only slightly dissociate when dissolved in water.AcidsBasesStrongStronghydrochloric acid, HClsodium hydroxide, NaOHhydrobromic acid, HBrpotassium hydroxide, KOHhydroiodic acid, HIcalcium hydroxide, Ca(OH)2nitric acid, HNO3strontium hydroxide, Sr(OH)2sulfuric acid, H2SO4barium hydroxide, Ba(OH)2perchloric acid, HClO4WeakWeakhydrofluoric acid, HFammonia, NH3phosphoric acid, H3PO4acetic acid, CH3COOH (or HC2H3O2)
17 Sample Problem 4.4Writing Ionic Equations for Acid-Base ReactionsPROBLEM: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 are2OH-(aq)+ 2H+(aq) H2O(l)water(b) Ba(OH)2(aq) + H2SO4(aq) H2O(l) + BaSO4(aq)spectator ionsBa2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq)2H2O(l)+Ba2+(aq)+SO42-(aq)2OH-(aq)+ 2H+(aq) H2O(l)
18 An aqueous strong acid-strong base reaction on the atomic scale.
20 Sample Problem 4.5Finding the Concentration of Acid from an Acid-Base TitrationPROBLEM: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 M 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 baseNaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)multiply by M of base( )mL x1L103mL= Lmol of basemolar ratioLX M= 5.078x10-3molNaOHmol of acidMolar ratio is 1:1divide by L of acid5.078x10-3mol HCl0.050L= M HClM 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.
23 To keep track of the transfer of electrons, atoms are assigned oxidation numbers. Table 4.3 Rules for Assigning an Oxidation Number (O.N.)General rules1. For an atom in its elemental form (Na, O2, Cl2, etc.): O.N. = 02. For a monoatomic ion: O.N. = ion charge3. 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.Rules for specific atoms or periodic table groups1. For Group 1A(1):O.N. = +1 in all compounds2. For Group 2A(2):O.N. = +2 in all compounds3. For hydrogen:O.N. = +1 in combination with nonmetals4. For fluorine:O.N. = -1 in combination with metals and boron5. For oxygen:O.N. = -1 in peroxidesO.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
24 Sample Problem 4.6Determining the Oxidation Number of an ElementPROBLEM:Determine the oxidation number (O.N.) of each element in these compounds:(a) zinc chloride(b) sulfur trioxide(c) nitric acidPLAN: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 A summary of terminology for oxidation-reduction (redox) reactions transfer or shift of electronsYX loses electron(s)Y gains electron(s)X is oxidizedY is reducedX is the reducing agentY is the oxidizing agentX increases its oxidation numberY decreases its oxidation number
27 Sample Problem 4.7Recognizing Oxidizing and Reducing AgentsPROBLEM:Identify the oxidizing agent and reducing agent in each of the following:(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) H2O(g)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:+1+6-2+3+6-2(a) 2Al(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)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 Sample Problem 4.7Recognizing Oxidizing and Reducing Agentscontinued+2-2+2-2+4-2(b) PbO(s) + CO(g) Pb(s) + CO2(g)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.+1-2(c) 2H2(g) + O2(g) H2O(g)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 Sample Problem 4.8Balancing Redox Equations by the Oxidation Number MethodPROBLEM:Use the oxidation number method to balance the following equations:(a) Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l)(b) PbS(s) + O2(g) PbO(s) + SO2(g)SOLUTION:+1+5-2+2+5-2+4-2+1-2(a) Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l)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 ionsCu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l)gains 1e-x2 to balance e-balance unchanged polyatomic ions
30 Sample Problem 4.8Balancing Redox Equations by the Oxidation Number MethodcontinuedCu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) H2O(l)42222+2-2+2-2+4-2(b) PbS(s) + O2(g) PbO(s) + SO2(g)loses 6e-PbS(s) O2(g) PbO(s) + SO2(g)3/2gains 2e- per O; need 3/2 O2 to make 3O2-Multiply by 2 to have whole number coefficients.2PbS(s) + 3O2(g) PbO(s) + 2SO2(g)
31 A redox titration permanganate ion / oxalate ion
32 Sample Problem 4.9Finding an Unknown Concentration by a Redox TitrationPROBLEM: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 isKMnO4(aq) + CaC2O4(s) + H2SO4(aq)MnSO4(aq) + K2SO4(aq) + CaSO4(s) + CO2(g) + H2O(l)(a) Calculate the amount (mol) of Ca2+.(b) Calculate the amount (mol) of Ca2+ ion concentration expressed in units of mg Ca2+/100mL blood.PLAN:(a)volume of KMnO4 solnmol of Ca2+multiply by Mratio of elements in formulamol of KMnO4mol of CaC2O4molar ratio
33 Sample Problem 4.9Finding an Unknown Concentration by a Redox TitrationcontinuedSOLUTION:2.05mL solnL103 mL4.88x10-4mol KMnO4L= 1.00x10-6mol KMnO41.00x10-6mol KMnO45mol CaC2O42mol KMnO4= 2.50x10-6 mol CaC2O42.50x10-6 mol CaC2O41mol Ca2+1mol CaC2O4= 2.50x10-6 mol Ca2+PLAN:(b)mol Ca2+/1mL bloodSOLUTION:multiply by 1002.50x10-6 mol Ca2+1mL bloodx100=2.50x10-4 mol Ca2+100mL bloodmol Ca2+/100mL bloodmultiply by M2.50x10-4 mol Ca2+100mL blood40.08g Ca2+mol Ca2+mg10-3gg Ca2+/100mL blood10-3g = 1mg=10.0mg Ca2+/100mL bloodmg 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.
38 Sample Problem 4.10Identifying the Type of Redox ReactionPROBLEM: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:(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)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 Sample Problem 4.10Identifying the Type of Redox Reaction+2-3(a) CombinationcontinuedMg(s) N2(g) Mg3N2 (aq)3Mg is the reducing agent; N2 is the oxidizing agent.+1-1+1-2(b) DecompositionH2O2(l) H2O(l) O2(g)1/2or2 H2O2(l) H2O(l) O2(g)O is the oxidizing and reducing agent.(c) Displacement+2+5-2+3+5-2Al(s) + Pb(NO3)2(aq) Al(NO3)3(aq) + Pb(s)2Al(s) + 3Pb(NO3)2(aq) Al(NO3)3(aq) + 3Pb(s)Pb is the oxidizing and Al is the reducing agent.
40 Reversible Reactions Chemical Equilibrium For some reactions, which proceed from reactants to products, can also undergo the reverse reaction of “products” to “reactants”.For example: CaCO CaO CO andCaO CO CaCO3But both reactions are taking place simultaneously and an equilibrium is reached. The equation is written:CaCO CaO CO2