Chapters 11 and12

Chemical Reaction One or more substance(s) change into one or more new substances Reactants Products Exothermic- energy is product (on right) products are more stable Endothermic- energy is reactant (on left) Products are less stable

Balancing Chemical Equations Includes kinds and parts of substances involved **LAW OF CONSERVATION OF MASS Mass (matter), charge and energy are always conserved in a chemical rxn

In a balanced equation, Each side of the equation has the same number and type of atoms Remember Dalton’s Theory: Bonds broken and formed; atoms rearranged

To write a balanced chemical equation: First write the skeleton equation using the correct formulas for elements and compounds (remember diatomics and criss-cross method) H 2 + O 2 --> H 2 O Then use coefficients to balance the equation so that it obeys The Law of Conservation of Mass 2H 2 + O 2 --> 2H 2 O

The Rules 1. Determine the correct formulas for all the reactants and products Molecular (covalent)- use prefixes Ionic- balance charges Diatomics!! BrOFINClH

2. Write the skeleton equation by placing the formulas for the reactants on the left and the formulas for the products on the right with a “yields” sign in between. If two or more reactants or products are involved, separate their formulas with (+) signs

3. Determine the number of atoms of each element in the reactants and products. TIP: Count a polyatomic ion as a single unit if it appears unchanged on both sides of the equation

4. Balance the elements one at a time by using coefficients. Begin by balancing elements that appear only once on each side of the equation. Unwritten coefficients are assumed to be 1 Once you are certain you have the correct chemical formula for the substances involved, NEVER change the subscripts in a chemical formula

5. Check each atom or polyatomic ion to be sure they are equal on both sides of the equation. *** Add phase symbols to substances if necessary (s, l, g, aq)

6. Make sure all of the coefficients are in the lowest possible ratio

Example #1 The reaction of zinc with aqueous hydrochloric acid produces a solution of zinc chloride and hydrogen gas

Example #2 Sodium reacts with sulfur to produce sodium sulfide

Example #3 Aqueous nitric acid (HNO 3 ) reacts with aqueous magnesium hydroxide to produce aqueous magnesium nitrate and water

Classifying Reactions The five general types of reactions are: Combination/Synthesis Decomposition Single-replacement (displacement) Double-replacement (displacement) Combustion

Synthesis Reactions A chemical change in which two or more substances react to form a single new substance Zn (s) + I 2(g)  ZnI 2(s)

Combination (synthesis) Reactions Group A metal + nonmetal  metal cation and nonmetal anion (ionic compound) 2 K (s) + Cl 2(g)  2 KCl (s)

Combination (synthesis) Reactions 2 nonmetals  more than one product is often produced S (s) + O 2(g) --> SO 2(g) (sulfur dioxide) 2S (s) + 3 O 2(g) --> 2SO 3(g) (sulfur trioxide)

Combination (synthesis) Reactions Transition metal + nonmetal  could produce more than one product Fe (s) + S (s)  FeS (s) (iron (II) sulfide) 2Fe (s) + 3S (s)  Fe 2 S 3(s) (iron (III) sulfide)

Decomposition Reactions A chemical change in which a single compound breaks down into two or more simpler products

Decomposition Reactions One reactant and two or more products Difficult to predict products Most require energy in the form of heat, light or electricity (endothermic) 2HgO (s) --> 2Hg (l) + O 2(g)

Combustion Reactions A chemical change in which an element or a compound reacts with oxygen, often producing energy in the form of heat and light

Combustion oxygen + hydrocarbon  water + carbon dioxide + energy

Combustion Magnesium and sulfur will also burn in the presence of oxygen. 2Mg (s) + O 2(g) --> 2MgO (s) For our purposes, we’ll call these reactions synthesis.

Single Replacement Reactions One element replaces a second element in a compound NOTE: both the reactants AND the products consist of an element and a compound K + AgCl  Ag + KCl

Single Replacement Reactions Zn (s) + Cu(NO 3 ) 2(aq) --> Cu (s) + Zn(NO 3 ) 2(aq) Whether one metal will displace another metal from a compound depends on the relative reactivities of the two metals **TABLE J**

The ACTIVITY SERIES Lists elements in order of decreasing reactivity A halogen can also replace a halogen- reactivity decreases as you go down the group Metals- want to lose electrons Nonmetals- want to gain electrons

Double Replacement Reactions An exchange of positive ions between two compounds Generally takes place in aqueous solutions and often produce a precipitate, a gas, or a molecular compound such as water.

Double Replacement Reactions One of the products is insoluble and precipitates from solution **TABLE F** Na 2 S (aq) + Cd(NO 3 ) 2(aq)  CdS (s) + 2NaNO 3(aq)

Double Replacement Reactions One of the products is a gas 2NaCN (aq) + H 2 SO 4(aq)  2HCN (g) + Na 2 SO 4(aq)

Double Replacement Reactions One product is a molecular compound such as water (acid-base rxn: neutralization) Ca(OH) 2(aq) + 2HCl (aq)  CaCl 2(aq) + 2H 2 O (l)

Examples CaBr 2(aq) + AgNO 3(aq)  FeS (s) + HCl (aq) 

Reactions in Aqueous Solutions AgNO 3(aq) + NaCl (aq)  AgCl (s) + NaNO 3(aq) Most ionic compounds dissociate into cations and anions when dissolved in water We can write a complete ionic equation to show the dissolved ionic compounds as dissociated free ions

Complete Ionic Equation Ag + (aq) + NO 3 - (aq) + Na + (aq) + Cl - (aq)  AgCl (s) + Na + (aq) + NO 3 - (aq) Na + (aq) and NO 3 - (aq) appear unchanged on either side of the equations; they did not technically participate in the reaction.

Spectator Ion An ion that appears on both sides of an equation and is not directly involved in the reaction

Net Ionic Equation An equation for a reaction in solution that shows only those particles directly involved in the chemical change Ag + (aq) + Cl - (aq) --> AgCl (s) **You must make sure that the charges on either side are balanced

Try this one Pb (s) + AgNO 3(aq)  Ag (s) + Pb(NO 3 ) 2(aq)

Predicting the Formation of a Precipitate Table F - Solubility Guidelines

2Na + (aq) + CO 3 2- (aq) + Ba 2+ (aq) + 2NO 3 - (aq)  Sodium is an alkali metal- soluble Carbonates are generally insoluble BaCO 3 will precipitate Net: Ba 2+ (aq) + CO 3 2- (aq) --> BaCO 3(s)

Types of Reactions The number of elements and/or the compounds reacting is a good indicator of possible reaction types and thus possible products

Mole-Mole Relationships We can learn to use a balanced equation to determine relationships between moles of reactants and moles of products stoichiometry

Use equations to determine the number of moles that can be produced from certain numbers of moles of reactants 2H 2 O  2H 2 + O 2 2 moles of water produces 2 moles of H mole of O 2

Example What number of moles of products will be produced by the decomposition of 5.8 moles of water? 5.8 moles H 2 O  5.8 H O 2

Example Calculate the number of moles of O 2 required to react exactly with 4.30 moles of propane, C 3 H 8, in the reaction described by the following unbalanced equation: __ C 3 H 8 + __ O 2  __ CO 2 + __ H 2 O

For each of the following unbalanced equations, indicate how many moles of the second reactant would be required to react exactly with 0.25 moles of the first reactant: 1. CO + O 2  CO 2 2. CH 4 + Cl 2  CCl 4 + HCl

Limiting and Excess Reagents In a chemical rxn, not enough of any of the reactants will limit the amount of product that forms Excess means that substance is NOT the limiting reagent

N 2(g) + 3H 2(g)  2NH 3(g) When 1 mol of N 2 reacts with 3 moles of H 2, 2 mols of NH 3 are produced What would happen if two moles of N 2 reacted with 3 moles of H 2 ?

In this reaction Only the H is completely used up Limiting Reagent: reagent that determines the amount of product that can be formed by a reaction Excess reagent: reactant that is not completely used up

The first step is always to Convert the quantity of each reactant to number of moles so that the limiting reagent can be identified The amount of product formed in a reaction can be determined from the given amount of limiting reagent.

Example 2Cu (s) + S (s)  Cu 2 S (s) What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? ***The reactant that is present in the smaller amount by mass or volume is NOT necessarily the l. r.

Percent Yield Theoretical yield: the maximum amount of product that could be formed from the given amounts of reactants Actual yield: the amount of product that actually forms when the rxn is carried out in the lab

Percent Yield The ratio of the actual yield to the theoretical yield expressed as a percent Percent Yield= actual yield x 100% theoretical yield

Should normally be near 100% But there are many reasons why it could be less: Accuracy of measurement Reaction does not go to completion Sloppy lab procedures Side reactions may occur

Example What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO 3 is heated? CaCO 3  CaO + CO 2