Presentation is loading. Please wait.

Presentation is loading. Please wait.

4-1 Dr. Wolf’s CHM 101 Chapter 4 The Major Classes of Chemical Reactions.

Similar presentations


Presentation on theme: "4-1 Dr. Wolf’s CHM 101 Chapter 4 The Major Classes of Chemical Reactions."— Presentation transcript:

1 4-1 Dr. Wolf’s CHM 101 Chapter 4 The Major Classes of Chemical Reactions

2 4-2 Dr. Wolf’s CHM 101 The Major Classes of Chemical Reactions 4.6 Elemental Substances in Redox 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 4.7 Reversible Reactions: An Introduction to Chemical Equilibrium

3 4-3 Dr. Wolf’s CHM 101 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 4-4 Dr. Wolf’s CHM 101 The electrical conductivity of ionic solutions

5 4-5 Dr. Wolf’s CHM 101 Sample Problem 4.1Determining 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) 7.42x10 22 formula units of copper(II) nitrate dissolved in water (d) 35mL of 0.84M zinc chloride PLAN: SOLUTION: We have to relate the information given and the number of moles of ions present when the substance dissolves in water. (a) (NH 4 ) 2 SO 4 ( s ) 2NH 4 + ( aq ) + SO 4 2- ( aq ) 5.0mol (NH 4 ) 2 SO 4 2mol NH 4 + 1mol (NH 4 ) 2 SO 4 = 10.mol NH mol SO 4 2-

6 4-6 Dr. Wolf’s CHM 101 Sample Problem 4.1Determining Moles of Ions in Aqueous Ionic Solutions continued mol CsBr 212.8g CsBr = 0.369mol CsBr = 0.369mol Cs + = 0.369mol Br 1- (b) CsBr( s ) Cs + ( aq ) + Br - ( aq ) 7.42x10 22 formula units Cu(NO 3 ) 2 mol Cu(NO 3 ) x10 23 formula units = 0.123mol Cu(NO 3 ) 2 = 0.123mol Cu 2+ = 0.246mol NO 3 1- (c) Cu(NO 3 ) 2 ( s ) Cu 2+ ( aq ) + 2NO 3 - ( aq ) 35mL ZnCl 2 1L 10 3 mL = 2.9x mol ZnCl 2 (d) ZnCl 2 ( aq ) Zn 2+ ( aq ) + 2Cl - ( aq ) 0.84mol ZnCl 2 L = 2.9x mol Zn 2+ = 5.8x mol Cl g CsBr

7 4-7 Dr. Wolf’s CHM 101 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 4-8 Dr. Wolf’s CHM 101 The Polarity of Water Electron distribution in molecules of H 2 and H 2 O

9 4-9 Dr. Wolf’s CHM 101 The dissolution of an ionic compound

10 4-10 Dr. Wolf’s CHM 101 Acids in water dissociate to give H +. The H + bonds with a water molecule and becomes a hydrated proton, H 3 O +, a hydronium ion.

11 4-11 Dr. Wolf’s CHM 101 Sample Problem 4.2Determining 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: SOLUTION: Use the formula to find the molarity of H +. Nitrate is NO 3 -. HNO 3 ( l ) H + ( aq ) + NO 3 - ( aq ) 1.4M HNO 3 ( aq ) should have 1.4M H + ( aq ).

12 4-12 Dr. Wolf’s CHM 101 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 4-13 Dr. Wolf’s CHM 101 Writing equations for aqueous ionic reactions. A precipitation reaction and its equation. ©

14 4-14 Dr. Wolf’s CHM 101 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 (NH 4 + ) are soluble. 2. All common nitrates (NO 3 - ), acetates (CH 3 COO - or C 2 H 3 O 2 - ) and most perchlorates (ClO 4 - ) are soluble. 3. All common chlorides (Cl - ), bromides (Br - ) and iodides (I - ) are soluble, except those of Ag +, Pb 2+, Cu +, and Hg All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca 2+ ). 2. All common carbonates (CO 3 2- ) and phosphates (PO 4 3- ) are insoluble, except those of Group 1A(1) and NH All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH 4 +. Soluble Ionic Compounds Insoluble Ionic Compounds 4. All common sulfates (SO 4 2- ) are soluble, except those of Ca 2+, Sr 2+, Ba 2+, and Pb 2+.

15 4-15 Dr. Wolf’s CHM 101 Sample Problem 4.3Predicting 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 ) write ions combine anions & cations check for insolubility Table 4.1 eliminate spectator ions for net ionic equation PLAN: SOLUTION: (a) Na 2 SO 4 ( aq ) + Sr(NO 3 ) 2 ( aq ) 2NaNO 3 ( aq ) + SrSO 4 ( s ) 2Na + ( aq ) +SO 4 2- ( aq )+ Sr 2+ ( aq )+2NO 3 - ( aq ) 2Na + ( aq ) +2NO 3 - ( aq )+ SrSO 4 ( s ) SO 4 2- ( aq )+ Sr 2+ ( aq ) SrSO 4 ( s ) (b) NH 4 ClO 4 ( aq ) + NaBr ( aq ) NH 4 Br ( aq ) + NaClO 4 ( aq ) All reactants and products are soluble so no reaction occurs.

16 4-16 Dr. Wolf’s CHM 101 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. Acids Strong hydrochloric acid, HCl hydrobromic acid, HBr hydroiodic acid, HI nitric acid, HNO 3 sulfuric acid, H 2 SO 4 perchloric acid, HClO 4 Weak hydrofluoric acid, HF phosphoric acid, H 3 PO 4 acetic acid, CH 3 COOH (or HC 2 H 3 O 2 ) Bases Strong Weak sodium hydroxide, NaOH calcium hydroxide, Ca(OH) 2 potassium hydroxide, KOH strontium hydroxide, Sr(OH) 2 barium hydroxide, Ba(OH) 2 ammonia, NH 3

17 4-17 Dr. Wolf’s CHM 101 2OH - ( aq )+ 2H + ( aq ) 2H 2 O( l ) Sr 2+ ( aq ) + 2OH - ( aq )+ 2H + ( aq )+ 2ClO 4 - ( aq ) 2H 2 O( l )+Sr 2+ ( aq )+2ClO 4 - ( aq ) Sr 2+ ( aq ) + 2OH - ( aq )+ 2H + ( aq )+ 2ClO 4 - ( aq ) 2H 2 O( l )+Sr 2+ ( aq )+2ClO 4 - ( aq ) Ba 2+ ( aq ) + 2OH - ( aq )+ 2H + ( aq )+ SO 4 2- ( aq ) 2H 2 O( l )+Ba 2+ ( aq )+SO 4 2- ( aq ) 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 ) reactants are strong acids and bases and therefore completely ionized in water PLAN: products are (a) Sr(OH) 2 ( aq )+2HClO 4 ( aq ) 2H 2 O( l )+Sr(ClO 4 ) 2 ( aq ) SOLUTION: (b) Ba(OH) 2 ( aq ) + H 2 SO 4 ( aq ) 2H 2 O( l ) + BaSO 4 ( aq ) 2OH - ( aq )+ 2H + ( aq ) 2H 2 O( l ) water spectator ions 2H + ( aq )+

18 4-18 Dr. Wolf’s CHM 101 An aqueous strong acid-strong base reaction on the atomic scale.

19 4-19 Dr. Wolf’s CHM 101 An acid-base titration

20 4-20 Dr. Wolf’s CHM 101 Sample Problem 4.5Finding 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 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 base mol of base mol of acid M of acid multiply by M of base molar ratio divide by L of acid NaOH( aq ) + HCl( aq ) NaCl( aq ) + H 2 O( l ) ( )mL x 1L 10 3 mL = L LX M= 5.078x10 -3 mol NaOH Molar ratio is 1: x10 -3 mol HCl 0.050L = M HCl

21 4-21 Dr. Wolf’s CHM 101 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 4-22 Dr. Wolf’s CHM 101 The redox process in compound formation

23 4-23 Dr. Wolf’s CHM 101 Table 4.3 Rules for Assigning an Oxidation Number (O.N.) 1. For an atom in its elemental form (Na, O 2, Cl 2, 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. 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 6. For Group 7A(17):O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group 5. For oxygen:O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F) To keep track of the transfer of electrons, atoms are assigned oxidation numbers.

24 4-24 Dr. Wolf’s CHM 101 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: SOLUTION: 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. (a) ZnCl 2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO 3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO 3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.

25 4-25 Dr. Wolf’s CHM 101 Highest and lowest oxidation numbers of reactive main-group elements

26 4-26 Dr. Wolf’s CHM 101 A summary of terminology for oxidation- reduction (redox) reactions X Y e-e- transfer or shift of electrons X loses electron(s)Y gains electron(s) X is oxidizedY is reduced X is the reducing agentY is the oxidizing agent X increases its oxidation number Y decreases its oxidation number

27 4-27 Dr. Wolf’s CHM 101 Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents PROBLEM: Identify the oxidizing agent and reducing agent in each of the following: (a) 2Al( s ) + 3H 2 SO 4 ( aq ) Al 2 (SO 4 ) 3 ( aq ) + 3H 2 ( g ) (b) PbO( s ) + CO( g ) Pb( s ) + CO 2 ( g ) (c) 2H 2 ( g ) + O 2 ( g ) 2H 2 O( 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). (a) 2Al( s ) + 3H 2 SO 4 ( aq ) Al 2 (SO 4 ) 3 ( aq ) + 3H 2 ( g ) 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 4-28 Dr. Wolf’s CHM 101 Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents continued (c) 2H 2 ( g ) + O 2 ( g ) 2H 2 O( g ) (b) PbO( s ) + CO( g ) Pb( s ) + CO 2 ( 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. 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 4-29 Dr. Wolf’s CHM 101 Sample Problem 4.8Balancing Redox Equations by the Oxidation Number Method PROBLEM: Use the oxidation number method to balance the following equations: (a) Cu( s ) + HNO 3 ( aq ) Cu(NO 3 ) 2 ( aq ) + NO 2 ( g ) + H 2 O( l ) (b) PbS( s ) + O 2 ( g ) PbO( s ) + SO 2 ( g ) SOLUTION: (a) Cu( s ) + HNO 3 ( aq ) Cu(NO 3 ) 2 ( aq ) + NO 2 ( g ) + H 2 O( 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. Cu( s ) + HNO 3 ( aq ) Cu(NO 3 ) 2 ( aq ) + NO 2 ( g ) + H 2 O( l ) loses 2e - gains 1e - x2 to balance e - balance unchanged polyatomic ions balance other ions

30 4-30 Dr. Wolf’s CHM 101 Cu( s ) + HNO 3 ( aq ) Cu(NO 3 ) 2 ( aq ) + NO 2 ( g ) + H 2 O( l ) Sample Problem 4.8Balancing Redox Equations by the Oxidation Number Method continued (b) PbS( s ) + O 2 ( g ) PbO( s ) + SO 2 ( g ) PbS( s ) + O 2 ( g ) PbO( s ) + SO 2 ( g ) loses 6e - gains 2e - per O; need 3/2 O 2 to make 3O 2- 3/2 Multiply by 2 to have whole number coefficients. 2PbS( s ) + 3O 2 ( g ) 2PbO( s ) + 2SO 2 ( g )

31 4-31 Dr. Wolf’s CHM 101 A redox titration permanganate ion / oxalate ion

32 4-32 Dr. Wolf’s CHM 101 Sample Problem 4.9Finding an Unknown Concentration by a Redox Titration PROBLEM: Calcium ion (Ca 2+ ) is required for blood to clot and for many other cell processes. An abnormal Ca 2+ concentration is indicative of disease. To measure the Ca 2+ concentration, 1.00mL of human blood was treated with Na 2 C 2 O 4 solution. The resulting CaC 2 O 4 precipitate was filtered and dissolved in dilute H 2 SO 4. This solution required 2.05mL of 4.88x10 -4 M KMnO 4 to reach the end point. The unbalanced equation is KMnO 4 ( aq ) + CaC 2 O 4 ( s ) + H 2 SO 4 ( aq ) MnSO 4 ( aq ) + K 2 SO 4 ( aq ) + CaSO 4 ( s ) + CO 2 ( g ) + H 2 O( l ) (a) Calculate the amount (mol) of Ca 2+. (b) Calculate the amount (mol) of Ca 2+ ion concentration expressed in units of mg Ca 2+ /100mL blood. mol of KMnO 4 volume of KMnO 4 soln mol of CaC 2 O 4 mol of Ca 2+ multiply by M molar ratio ratio of elements in formula PLAN: (a)

33 4-33 Dr. Wolf’s CHM 101 Sample Problem 4.9Finding an Unknown Concentration by a Redox Titration continued 2.05mL soln= 1.00x10 -6 mol KMnO 4 L 10 3 mL 4.88x10 -4 mol KMnO 4 L 1.00x10 -6 mol KMnO 4 5mol CaC 2 O 4 2mol KMnO 4 = 2.50x10 -6 mol CaC 2 O x10 -6 mol CaC 2 O 4 1mol Ca 2+ 1mol CaC 2 O 4 = 2.50x10 -6 mol Ca 2+ PLAN: (b) SOLUTION: mol Ca 2+ /1mL blood mol Ca 2+ /100mL blood g Ca 2+ /100mL blood mg Ca 2+ /100mL blood multiply by 100 multiply by M g = 1mg SOLUTION: x x10 -6 mol Ca 2+ 1mL blood =2.50x10 -4 mol Ca mL blood 2.50x10 -4 mol Ca mL blood 40.08g Ca 2+ mol Ca 2+ mg g =10.0mg Ca 2+ /100mL blood

34 4-34 Dr. Wolf’s CHM 101 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 4-35 Dr. Wolf’s CHM 101 Combining elements to form an ionic compound

36 4-36 Dr. Wolf’s CHM 101 Decomposing a compound to its elements

37 4-37 Dr. Wolf’s CHM 101 Displacing one metal with another

38 4-38 Dr. Wolf’s CHM 101 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: (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 4-39 Dr. Wolf’s CHM 101 Sample Problem 4.10 Identifying the Type of Redox Reaction continued (a) Combination Mg is the reducing agent; N 2 is the oxidizing agent. (b) Decomposition H 2 O 2 ( l ) H 2 O( l ) + O 2 ( g ) /2or Mg( s ) + N 2 ( g ) Mg 3 N 2 ( aq ) 2 H 2 O 2 ( l ) 2 H 2 O( l ) + O 2 ( g ) O is the oxidizing and reducing agent. (c) Displacement Al( s ) + Pb(NO 3 ) 2( aq ) Al(NO 3 ) 3 ( aq ) + Pb( s ) Al( s ) + 3Pb(NO 3 ) 2( aq ) 2Al(NO 3 ) 3 ( aq ) + 3Pb( s ) Pb is the oxidizing and Al is the reducing agent.

40 4-40 Dr. Wolf’s CHM 101 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 3 CaO + CO 2 and CaO + CO 2 CaCO 3 But both reactions are taking place simultaneously and an equilibrium is reached. The equation is written: CaCO 3 CaO + CO 2

41 4-41 Dr. Wolf’s CHM 101 End of Chapter 4


Download ppt "4-1 Dr. Wolf’s CHM 101 Chapter 4 The Major Classes of Chemical Reactions."

Similar presentations


Ads by Google