Chapter 5 Chemical Reactions Chemistry 140 Chapter 5 Chemical Reactions

CHAPTER OUTLINE 5.1 Chemical Equations 5.2 Types of Reactions 5.3 Redox Reactions 5.4 Decomposition Reactions 5.5 Combination Reactions 5.6 Replacement Reactions 5.7 Ionic Equations 5.8 Energy and Reactions 5.9 The Mole and Chemical Equations 5.10 The Limiting Reactant 5.11 Reaction Yields

LEARNING OBJECTIVES/ASSESSMENT When you have completed your study of this chapter, you should be able to: 1. Identify the reactants and products in written reaction equations, and balance the equations by inspection. (Section 5.1; Exercises 5.2 and 5.6) 2. Assign oxidation numbers to elements in chemical formulas, and identify the oxidizing and reducing agents in redox reactions. (Section 5.3; Exercises 5.10 and 5.15) 3. Classify reactions into the categories of redox or nonredox, then into the categories of decomposition, combination, single replacement, or double replacement. (Sections 5.4, 5.5, and 5.6; Exercise 5.20) 4. Write molecular equations in total ionic and net ionic forms. (Section 5.7; Exercise 5.30 a, b, & c) 5. Classify reactions as exothermic or endothermic. (Section 5.8; Exercise 5.34) 6. Use the mole concept to do calculations based on chemical reaction equations. (Section 5.9; Exercise 5.42) 7. Use the mole concept to do calculations based on the limiting‐reactant principle. (Section 5.10; Exercise 5.52) 8. Use the mole concept to do percentage‐yield calculations. (Section 5.11; Exercise 5.56)

Describing Chemical Reactions chemical reaction - a change in matter in which a substance with new identities is formed Indications of a chemical reaction production of heat production of light change in color production of a gas

Facts about chemical reactions. Chemical reactions release or absorb energy. release energy - exothermic absorb energy - endothermic Atoms are rearranged in a chemical change. atoms cannot be created or destroyed coefficients are adjusted to satisfy the law of conservation of matter Particles must collide for a chemical reaction to occur. reactant particles must come in contact with each other

How are reactions written? 1. Equations represent facts - write a word equation. 2. Substitute symbols and formulas for elements and compounds. 3. Use charges of ions to write formulas for compounds. 4. Balance the equation by adjusting coefficients.

Sample Equation 1 1. hydrogen + oxygen water 2. H2 + O2  H2O 3. formulas are correct 4. 2H2 + O2  2H2O

Sample Equation 2 1. calcium carbonate  calcium oxide + carbon dioxide 2. 3. 4

Practice Problems 1. zinc + sulfur  zinc sulfide Zn + S  ZnS 2. sodium chloride + silver nitrate  silver chloride + sodium nitrate NaCl +AgNO3  AgCl + NaNO3 3. potassium + water  potassium hydroxide + hydrogen 2K + 2H2O  2KOH + H2 4. sodium hydroxide + hydrochloric acid (HCl)  sodium chloride + water NaOH + HCl  NaCl + HOH

More Practice Problems 5. magnesium bromide + chlorine  magnesium chloride + bromine MgBr2 + Cl2  MgCl2 + Br2 6. sodium chloride + sulfuric acid (H2SO4)  sodium sulfate + hydrogen chloride 7. aluminum + iron(III) oxide  aluminum oxide + iron 8. butane + oxygen  carbon dioxide + water

What information is in an equation? qualitative - tells what is present names, formulas, etc. (s) solid, (l) liquid, (g) gas (aq) aqueous - in water solution quantitative - tells how much is present coefficients represent quantity of moles energy may be given left side - endothermic right side - exothermic

Describe the following equation in words. C6H12O6(aq) + 6O2(g) 6CO2(g) + 6H2O(l) + 2870kJ One mole of aqueous glucose reacts with 6 moles of oxygen gas to produce 6 moles of carbon dioxide gas, 6 moles of water and 2870kJ of energy is given for each mole of glucose reacting.

Types of Chemical Reactions Combustion - oxygen combines with hydrocarbons C2H6(g) + O2(g) CO2(g) + H2O(l) Any substance that burns in air reacts with oxygen.

Synthesis Synthesis - two or more substances combine to form more complex substances. A + B  AB Example: P4(s) + 6Cl2(g)  4PCl3(l)

Decomposition Reactions Decomposition - complex substances are broken down into simpler substances. AB  A + B Example: 2HgO  2Hg + O2

Single Replacement Reactions Single Replacement - a substance in one compound is replaced by another substance. A + BC  AC + B Example: Mg(s) + 2HCl(aq)  MgCl2(s) + H2(g)

Double Replacement Reactions Double Replacement - two substances exchange components. AB + CD  AD + CB Example: NaCl + AgNO3  AgCl + NaNO3

Oxidation Reduction Chemisty: Redox Chemistry Oxidation and Reduction reactions always take place simultaneously. Loss of electrons – oxidation (Increase in Oxidation Number) Ex:Na ------> Na+1 + e-1 Gain of electrons - reduction ( Decrease in Oxidation Number) Cl2 + 2 e-1 ------> 2 Cl-1

Oxidation occurs when a molecule does any of the following:    Loses electrons Loses hydrogen Gains oxygen If a molecule undergoes oxidation, it has been oxidized and it is the reducing agent (aka reductant).

Reduction occurs when a molecule does any of the following: Gains electrons Gains hydrogen Loses oxygen If a molecule undergoes reduction, it has been reduced and it is the oxidizing agent (aka oxidant).

zinc is being oxidized while the copper is being reduced. Why?

Redox reactions involve electron transfer: Lose e - =Oxidation Cu (s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) Gain e - =Reduction

Oxidation Numbers Rules for Assigning Oxidation States The oxidation state of an atom in an uncombined element is 0. The oxidation state of a monatomic ion is the same as its charge. Oxygen is assigned an oxidation state of –2 in most of its covalent compounds. Important exception: peroxides (compounds containing the O2 2- group), in which each oxygen is assigned an oxidation state of –1) In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1 For a compound, sum total of ON s is zero. For an ionic species (like a polyatomic ion), the sum of the oxidation states must equal the overall charge on that ion.

Redox: Reduction occurs when an atom gains one or more electrons. Ex:   Oxidation occurs when an atom or ion loses one or more electrons. LEO goes GER Copper metal reacts with silver nitrate to form silver metal and copper nitrate: Cu + 2 Ag(NO3)  2 Ag + Cu(NO3)2.

Identifying OX, RD, SI Species Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation = loss of electrons. The species becomes more positive in charge. For example, Ca0  Ca+2, so Ca0 is the species that is oxidized. Reduction = gain of electrons. The species becomes more negative in charge. For example, H+1  H0, so the H+1 is the species that is reduced. Spectator Ion = no change in charge. The species does not gain or lose any electrons. For example, Cl-1  Cl-1, so the Cl-1 is the spectator ion.

Oxidizing Agent and Reducing Agent: Oxidizing agent gets reduced itself and reducing agent gets oxidized itself, so a strong oxidizing agent should have a great tendency to accept e and a strong reducing agent should be willing to lose e easily. What are strong oxidizing agents- metals or non metals? Why? Which is the strongest oxidizing agent and which is the strongest reducing agent?

Agents Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Since Ca0 is being oxidized and H+1 is being reduced, the electrons must be going from the Ca0 to the H+1. Since Ca0 would not lose electrons (be oxidized) if H+1 weren’t there to gain them, H+1 is the cause, or agent, of Ca0’s oxidation. H+1 is the oxidizing agent. Since H+1 would not gain electrons (be reduced) if Ca0 weren’t there to lose them, Ca0 is the cause, or agent, of H+1’s reduction. Ca0 is the reducing agent.

Steps for Balancing a Redox Reaction: Half Reaction Method In half reaction method, oxidation and reduction half- reactions are written and balanced separately before combining them into a balanced redox reaction. It is a good method for balancing redox reactions because this method can be used both for reactions carried out in acidic and basic medium .

Steps for Balancing Redox Reaction Using Half Reaction Method IN ACIDIC MEDIUM: Step 1: Write unbalanced equation in ionic form. Step 2: Write separate half reactions for the oxidation and reduction processes. (Use Oxidation Numbers for identifying oxidation and reduction reactions) Step 3: Balance atoms in the half reactions First, balance all atoms except H and O Balance O by adding H2O Balance H by adding H+ Step 4: Balance Charges on each half reaction, by adding electrons. Step 5: Multiply each half reaction by an appropriate number to make the number of electrons equal in both half reactions. Step 6: Add two half reactions and simplify where possible by canceling species appearing in both sides. Step 7: Check equation for same number of atoms and charges on both sides.

Writing Half-Reactions Ca0 + 2 H+1Cl-1  Ca+2Cl-12 + H20 Oxidation: Ca0  Ca+2 + 2e- Reduction: 2H+1 + 2e-  H20 The two electrons lost by Ca0 are gained by the two H+1 (each H+1 picks up an electron). PRACTICE SOME!

Practice Half-Reactions Don’t forget to determine the charge of each species first! 4 Li + O2  2 Li2O Oxidation Half-Reaction: Reduction Half-Reaction: Zn + Na2SO4  ZnSO4 + 2 Na

Steps for Balancing Redox Reaction Using Half Reaction Method IN BASIC MEDIUM: For balancing redox reactions in basic solutions, all the steps are the same as acidic medium balancing, except you add one more step to it. The H+ ions can then be “neutralized” by adding an equal number of OH- ions to both sides of the equation. Ex.

Standard Cell Potential Just as the water tends to flow from a higher level to a lower level, electrons also move from a higher “potential” to a lower potential. This potential difference is called the electromotive force (EMF) of cell and is written as Ecell. The standard for measuring the cell potentials is called a SHE (Standard Hydrogen Electrode). Description of SHE (Standard Hydrogen Electrode) Reaction 2H+(aq, 1M)+ 2e - H2(g, 101kPa) E0= 0.00 V

Standard Reduction Potentials Many different half cells can be paired with the SHE and the standard reduction potentials for each half cell is obtained. Check the table for values of reduction potential for various substances: Would substances with high reduction potential be strong oxidizing agents or strong reducing agents? Why?

Activity Series For metals, the higher up the chart the element is, the more likely it is to be oxidized. This is because metals like to lose electrons, and the more active a metallic element is, the more easily it can lose them. For nonmetals, the higher up the chart the element is, the more likely it is to be reduced. This is because nonmetals like to gain electrons, and the more active a nonmetallic element is, the more easily it can gain them.

Metal Activity 3 K0 + Fe+3Cl-13 REACTION Metallic elements start out with a charge of ZERO, so they can only be oxidized to form (+) ions. The higher of two metals MUST undergo oxidation in the reaction, or no reaction will happen. The reaction 3 K + FeCl3  3 KCl + Fe WILL happen, because K is being oxidized, and that is what Table J says should happen. The reaction Fe + 3 KCl  FeCl3 + 3 K will NOT happen. Fe0 + 3 K+1Cl-1 NO REACTION

Voltaic Cells (Galvanic Cells) A voltaic cell converts chemical energy from a spontaneous redox reaction into electrical energy. Ex: Cu and Zn voltaic cell (More positive reduction potential is the cathode) Key Words: Cathode Anode Salt Bridge How a Voltaic Cell Works: An Ox, Red Cat A reaction is spontaneous if the metal with higher reduction potential is made cathode.

Voltaic Cells Produce electrical current using a spontaneous redox reaction Used to make batteries! Materials needed: two beakers, piece of the metals (anode, - electrode and cathode + electrode), solution of each metal, porous material (salt bridge), solution of a salt that does not contain either metal in the reaction, wire and a load to make use of the generated current! Use Reference Table J to determine the metals to use Higher = (-) anode (lower reduction potential) Lower = (+) cathode (higher reduction potential)

Making Voltaic Cells

Electrolytic Cells Use electricity to force a nonspontaneous redox reaction to take place. Uses for Electrolytic Cells: Decomposition of Alkali Metal Compounds Decomposition of Water into Hydrogen and Oxygen Electroplating Differences between Voltaic and Electrolytic Cells: ANODE: Voltaic (-) Electrolytic (+) CATHODE: Voltaic (+) Electrolytic (-) Voltaic: 2 half-cells, a salt bridge and a load Electrolytic: 1 cell, no salt bridge, IS the load

Decomposing Alkali Metal Compounds 2 NaCl  2 Na + Cl2 The Na+1 is reduced at the (-) cathode, picking up an e- from the battery The Cl-1 is oxidized at the (+) anode, the e- being pulled off by the battery (DC)

Decomposing Water 2 H2O  2 H2 + O2 The H+ is reduced at the (-) cathode, yielding H2 (g), which is trapped in the tube. The O-2 is oxidized at the (+) anode, yielding O2 (g), which is trapped in the tube.

Electroplating The Ag0 is oxidized to Ag+1 when the (+) end of the battery strips its electrons off. The Ag+1 migrates through the solution towards the (-) charged cathode (ring), where it picks up an electron from the battery and forms Ag0, which coats on to the ring.

Predicting Products of Decomposition Reactions 1. Decomposition of a metallic carbonate yields a metallic oxide and carbon dioxide. 2. Decomposition of a metallic chlorate yields a metallic chloride and oxygen. 3. Decomposition of a metallic hydroxide yields a metallic oxide and water. 4. Decomposition of a metallic nitrate yields a metallic nitrite and oxygen.

How much can a reaction produce? To this point we are able to predict reactants and products of chemical reactions to a relatively high degree of accuracy. These are referred to as qualitative values. We will now concentrate on predicting amounts of substances involved in chemical reactions. These are referred to as quantitative values. stoichiometry- the branch of chemistry dealing with quantitative relations of reactants and products in chemical reactions.

The coefficients of reactants and products in a chemical equation represent the number of moles of substances. In the equation below, determine the number of moles and the mass of each substance represented. 2HCl  H2 + Cl2 moles hydrochloric acid = _____moles, mass = _____g moles of hydrogen gas = _____moles, mass = _____g moles of chlorine gas = _____moles, mass = _____g

Applications of Stoichiometry composition stoichiometry- the mass relationships of elements in chemical compounds. reaction stoichiometry- the mass relationships among reactants and products in chemical reactions.

Reaction-Stoichiometry Problems There are four types of reaction-stoichiometry problems. 1. mole-mole problems given: moles of a substance calculate: moles of a substance 2. mole-mass problems calculate: mass of a substance

Types of reaction-stoichiometry problems(continued) 3. mass-mole problems given: mass of a substance calculate: moles of a substance 4. mass-mass problems calculate: mass of a substance

Mole-Mole Calculations Ammonia (NH3) is widely used as a fertilizer and in many household cleansers. How many moles of ammonia are produced when 6 moles of hydrogen gas react with an excess of nitrogen gas? 3H2 + N2  2NH3 6mol H2 (2mol NH3/3mol H3) = 4mol NH3

Sample Problem 2KClO3  2KCl + 3O2 The decomposition of potassium chlorate (KClO3) is used as a source of oxygen in the laboratory. How many moles of potassium chlorate are needed to produce 15 mol of oxygen? 2KClO3  2KCl + 3O2

Mole-Mass Calculations What mass of MgO is produced when 2.00 mol of Mg burns in air? 2Mg + O2  2MgO 2mol Mg x 2mol MgO/2mol Mg x 40 g/mol = 80g MgO

Sample Problem What mass of oxygen combines with 2.00 mol of magnesium in the reaction in sample problem 1 ? 2MgO  2Mg + O2 2mol Mg x (1mol O2/2mol Mg) x 32g/mol = 32g O2

Mass-Mole Calculations How many moles of mercury(II) oxide (HgO) are needed to decompose and produce 125g of oxygen? 2HgO  2Hg + O2 (125g O2/32g/mol) x (2mol HgO/1mol O2) = 7.8mol HgO

Sample Problem How many moles of mercury are produced in sample problem 1? 2HgO  2Hg + O2 (125g O2/32g/mol) x (2mol Hg/1mol O2) = 7.8mol Hg Notice: the same coefficients mean the same number of moles!

Mass-Mass Calculations How many grams of NH4NO3 are required to produce 33.0 g of N2O by decomposition? NH4NO3  N2O + 2H2O (33g N2O/44g/mol) x (1/1) x 80g/mol = 60g NH4NO3 How many grams of water are produced in the above problem?

Additional Sample Problems 1. Determine the mass of carbon dioxide produced by the decomposition of 50.0 grams of calcium carbonate. 2. Determine the mass of calcium needed to burn in air to produce 14.0 grams of calcium oxide. 3. 60 grams of magnesium ribbon burn in air. a. How many moles of magnesium burns? b. How many moles of magnesium oxide are produced? c. What mass of magnesium oxide is produced?

More Sample Problems 4. 20 grams of sodium hydroxide reacts with an excess of hydrochloric acid (HCl). Find the mass of sodium chloride precipitated from this reaction. 5. 56 grams of nitrogen gas reacts with hydrogen to produce ammonia (NH3). How many grams of ammonia are produced?

Problem Find the mass of calcium oxide produced from the decomposition of 750g of calcium carbonate. Find the number of moles of carbon dioxide produced also. CaCO3 → CaO + CO2

How much does a reaction really produce How much does a reaction really produce? Limiting Reactants and Percent Yield limiting reactant - the reactant that controls the amount of product formed in a chemical reaction excess reactant - the substance that is not used up completely in a reaction

Sample Problem Some rocket engines use a mixture of hydrazine (N2H4) and hydrogen peroxide (H2O2) as the propellant system. The reaction equation is N2H4(l) + 2 H2O2 (l)  N2 (g) + 4H2O (g) a. Which is the limiting reactant in the equation when 0.750 mol of N2H4 reacts with 0.500 mol of H2O2? b. How much of the excess reactant, in moles, remains unchanged? c. How much of each product, in moles, is formed?

Percent Yield theoretical yield - the maximum amount of product that can be produced from a given amount of reactant actual yield - the measured amount of product obtained from a reaction percent yield - the ratio of the actual yield to the theoretical yield, multiplied by 100%

Sample Problem Methanol can be produced through the reaction of CO and H2 in the presence of a catalyst catalyst CO(g) + 2H2(g) ------------> CH3OH(l) If 75.0g of CO reacts to produce 68.4g of CH3OH, what is the percent yield of CH3OH?

Stoichiometry Formulas Mole-Mole coeff unk mol unk = mol kn x ------------------- coeff kn Mole-Mass mass unk = mol unk x ------------------ x mm unk

Stoichiometry Formulas Mass-Mole mass kn coeff unk mol unk = --------------- x --------------------- mm kn coeff kn Mass – Mass mass unk = --------------- x --------------------- x mm unk

Problem Quicklime, CaO, is used by farmers to increase the pH of soil. It reacts with water according to the reaction equation CaO + H2O → Ca(OH)2 Suppose 100Kg of CaO is spread onto a field. A. How many moles of CaO are in 100Kg? B. How many moles of Ca(OH)2 are produced? C. What mass of Ca(OH)2 is produced? Support your answers with detailed calculations.