Chapter 6

Chemical Reactions

Chemical Reactions

 In a chemical reaction, one or more reactants is converted to one or more products

Reactant(s) Product(s)

 In this chapter we discuss three aspects of chemical reactions (a) mass relationships (stoichiometry) (b) types of reactions (c) heat gain and loss accompanying reactions

Chemical Equations

 The following chemical equation tells us that propane gas and oxygen gas react to form carbon dioxide gas and water vapor

C 3 H 8 ( g) Propane + O 2 (g) Oxygen CO 2 (g) Carbon dioxid e + H 2 O( g) Water

 But while it tells us what the reactants and products are and the physical state of each, it is incomplete because it is not balanced

Balancing Equations

 To balance a chemical equation – begin with atoms that appear only in one compound on the left and one on the right; in this case, begin with carbon (C) which occurs in C 3 H 8 and CO 2

C 3 H 8 (g) + O 2 (g) 3 CO 2 ( g) + H 2 O(g)

– now balance hydrogens, which occur in C 3 H 8 and H 2 O

C 3 H 8 (g) + O 2 (g) 3CO 2 ( g) + 4 H 2 O(g)

– if an atom occurs as a free element, as for example Mg or O 2 , balance this element last; in this case O 2

C 3 H 8 (g) + 5 O 2 ( g) 3CO 2 (g) + 4H 2 O( g)

Balancing Equations

 Practice problems: balance these equations

Ca( OH) 2 ( s) Calcium hydroxide + HCl( g) CaCl 2 (s) Calcium chlorid e + H 2 O( l) CO 2 ( g) + H 2 O(l) ph otosynthes is C 6 H 1 2 O 6 (aq) Glucose + O 2 (g) C 4 H 1 0 ( g) + O 2 (g) Butane CO 2 (g) + H 2 O(g)

Balancing Equations

 Solutions to practice problems

6 CO 2 Ca( OH) 2 ( s) Calcium hydroxide (g) + + 6 H 2 O(l) 2 HCl(g) ph otosynthes is CaCl 2 (s) Calcium chlorid e + H 2 O( l) C 6 H 1 2 O 6 (aq) + Glucose C 4 H 1 0 ( l) Bu tane + 13 2 O 2 (g) 4 CO 2 (g) + 5 H 2 O( g) 6 O 2 (g)

– it is common practice to use only whole numbers; therefore, multiply all coefficients by 2, which gives

2 C 4 H 1 0 ( l) Butane + 13 O 2 (g) 8 CO 2 ( g) + 10 H 2 O(g)

Formula Weight

 Formula weight: the sum of the atomic weights in atomic mass units (amu) of all atoms in a compound’s formula

Ionic Comp ou nds Sod ium chlorid e (N aCl) 23.0 amu N a + 35.5 amu Cl = 58.5 amu Nickel(II) ch loride h yd rate (N iCl 2 • 6H 2 O) 58.7 amu N i + 2(35.5 amu Cl) + 12(1.0) amu H) + 6(12.0 amu O) = 237.7 amu Molecu lar Comp ou nds Water (H 2 O) Aspirin (C 9 H 8 O 4 ) 2(1.0 amu H) + 16.0 amu O = 18.0 amu 9(12.0 amu C) + 8(1.0 amu H) + 4(16.0 amu O) = 180.0 = amu

Formula Weight

 formula weight can be used for both ionic and molecular compounds; it tells nothing about whether a compound is ionic or molecular  molecular weight should be used only for molecular compounds  in this text, we use formula weight for ionic compounds and molecular weight for molecular compounds

The Mole

 Mole (mol) – a mole of the amount of substance that contains as many atoms, molecules, or ions as are in exactly 12 g of carbon-12 – a mole, whether it is a mole of iron atoms, a mole of methane molecules, or a mole of sodium ions, always contains the same number of formula units – – – the number of formula units in a mole is known as Avogadro’s number Avogadro’s number has been measured experimentally its value is 6.02214199 x 10 23 formula units per mole

Molar Mass

 Molar mass: the formula weight of a substance expressed in grams  Glucose, C 6 H 12 O 6 – molecular weight: 180 amu – molar mass: 180 g/mol – one mole of glucose has a mass of 180 g  Urea, (NH 2 ) 2 CO – molecular weight 60.0 amu – molar mass: 60.0 g/mol – one mole of urea has a mass of 60.0 g

Molar Mass

 We can use molar mass to convert from grams to moles, and from moles to grams

You are given one of these and asked to find the other Grams of A Moles of A Use molar mass (g/mol) as the conversion factor

– calculate the number of moles of water in 36.0 g water

36.0 g H 2 O x 1 mol H 2 O 18.0 g H 2 O = 2.00 mol H 2 O

Grams to Moles

 Calculate the number of moles of sodium ions, Na + , in 5.63 g of sodium sulfate, Na 2 SO 4 – first we find the how many moles of sodium sulfate – the formula weight of Na 2 SO 4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 amu – therefore, 1 mol of Na 2 SO 4 = 142.1 g Na 2 SO 4 –

5.63 g Na 2 SO 4 x 1 mol Na 2 SO 4 142.1 g Na 2 SO 4

the formula Na 2 SO 4 tells us there are two moles of Na ions per mole of Na 2 SO 4

= 0.0396 mol Na 2 SO 4

+

2 mol N a + 0.0396 mol N a 2 SO 4 x = 0.0792 mol N a + 1 mol N a 2 SO 4

Grams to Molecules

 A tablet of aspirin, C 9 H 8 O 4 , contains 0.360 g of aspirin. How many aspirin molecules is this?

– first we find how many mol of aspirin are in 0.360 g

0.360 g aspirin x 1 mol asp irin 180.0 g asp irin = 0.00200 mol as pirin

– – each mole of aspirin contains 6.02 x 10 23 molecules the number of molecules of aspirin in the tablet is

0.00200 mole x 6.02 x 10 23 molecules mole = 1.20 x 10 21 molecules

Stoichiometry

 Stoichiometry: the study of mass relationships in chemical reactions – following is an overview of the the types of calculations we study

You are given one of these And asked to find one of these Grams of A Moles of A Moles of B Grams of B From grams to moles, use molar mass (g/mol) as a conversion factor From moles to moles, use the coefficients in the balanced equation as a conversion factor From moles to grams, use molar mass (g/mol) as a conversion factor

Stoichiometry

 Problem: how many grams of nitrogen, N 2 , are required to produce 7.50 g of ammonia, NH 3

N 2 (g) + 3H 2 (g) 2NH 3 ( g)

– first find how many moles of NH 3

7.50 g N H 3 x 1 mol N H 3 17.0 g N H 3

are in 7.50 g of NH

= mol NH 3

3 – next find how many moles of N 2 produce this many moles of NH 3 are required to

7.50 g N H 3 x 1 mol N H 3 17.0 g N H 3 x 1 mol N 2 2 mol N H 3 = mol N 2

Stoichiometry

 Practice problem (cont’d) – finally convert moles of N 2 now do the math to grams of N 2 and

7.50 g N H 3 x 1 mol N H 3 17.0 g N H 3 x 1 mol N 2 2 mol N H 3 x 28.0 g N 2 1 mol N 2 = 6.18 g N 2

Stoichiometry

 Practice problems: – what mass of aluminum oxide is required to prepare 27 g of aluminum?

Al 2 O 3 ( s) electrolysis Al( s)

+

O 2 ( g)

– how many grams each of CO 2 and NH 3 produced from 0.83 mol of urea?

ureas e ( NH 2 ) 2 CO(aq) + H 2 O 2NH 3 (aq) +

are

CO 2 (g) Urea

Limiting Reagent

 Limiting reagent : the reagent that is used up first in a chemical reaction – consider this reaction of N 2

N 2 (g)

and O 2

+ O 2 (g) 2 NO( g) before reaction (moles) 5.0

1.0

0 after reaction (moles) 4.0

0 2.0

– – – in this experiment, there is only enough O 2 1.0 mole of N 2 O 2 is used up first; it the limiting reagent 4.0 moles of N 2 remain unreacted to react with

Limiting Reagent

 Practice Problem – suppose 12 g of carbon is mixed with 64 g of oxygen and the following reaction takes place

C(s) + O 2 ( g) CO 2 ( g)

– complete the following table. Which is the limiting reagent?

C + O 2 CO 2 before reaction (g) before reaction (mol) 12 g 64 g 0 after reaction (mol) after reaction (g)

Percent Yield

 Actual yield: the mass of product formed in a chemical reaction  Theoretical yield: the mass of product that should be formed according to the stoichiometry of the balanced chemical equation  Percent yield: actual yield divided by theoretical yield times 100

Percent yield = Actu al yield Theoretical yield x 100

Percent Yield

 Practice problem: – suppose we react 32.0 g of methanol with excess carbon monoxide and get 58.7 g of acetic acid – complete this table

CH 3 OH + CO CH 3 COOH before reaction (g) 32.0

before reaction (mol) th eoretical yield (mol) th eoretical yield (g) actual yield (g) percent yield (%) excess 0 58.7



Reactions Between Ions

Ionic compounds, also called salts, consist of both positive and negative ions  When an ionic compound dissolves in water, it dissociates to aqueous ions

H 2 O NaCl(s) Na + (aq) + Cl (aq)

 What happens when we mix aqueous solutions of two different ionic compounds?

– if two of the ions combine to form a water-insoluble compound, a precipitate will form – otherwise no physical change will be observed

Reactions Between Ions

 Example: – suppose we prepare these two aqueous solutions

Solu tion 1 AgNO 3 (s) H 2 O Ag + ( aq) + NO 3 (aq) Solu tion 2 NaCl(s) H 2 O Na + ( aq) + Cl ( aq)

– if we then mix the two solutions, we have four ions present; of these, Ag + and Cl react to form AgCl(s)

Ag + (aq)

which precipitates

+ NO 3 ( aq) + Na + (aq) + Cl ( aq) AgCl(s) + Na + (aq) + NO 3 ( aq)

–

Reactions Between Ions

we can simplify the equation for the formation of AgCl by omitting all ions that do not participate in the reaction

Net ionic equation: Ag + ( aq) + Cl (aq) AgCl( s)

– the simplified equation is called a net ionic equation ; it shows only the ions that react – ions that do not participate in a reaction are called spectator ions

Reactions Between Ions

 In general, ions in solution react with each other when one of the following can happen – two of them form a compound that is insoluble in water – two of them react to form a gas that escapes from the reaction mixture as bubbles, as for example when we mix aqueous solutions of sodium bicarbonate and hydrochloric acid

HCO 3 ( aq) + H 3 O + (aq) CO 2 ( g) + 2 H 2 O( l) Bicarbonate ion

–

Carbon d ioxide

an acid neutralizes a base (Chapter 8) – one of the ions can oxidize another (Section 4.7)

Reactions Between Ions

 Following are some generalizations about which ionic solids are soluble in water and which are insoluble – all compounds containing Na + , K + , and NH 4 + soluble in water are – all nitrates (NO 3 ) and acetates (CH 3 COO ) are soluble in water – – most chlorides (Cl ) and sulfates (SO 4 2 ) are soluble; exceptions are AgCl, BaSO 4 , and PbSO 4 most carbonates (CO 3 2 ), phosphates (PO 4 3 ), sulfides (S 2 ), and hydroxides (OH ) are insoluble in water; exceptions are LiOH, NaOH, KOH, and NH 4 OH which are soluble in water

Oxidation-Reduction

 Oxidation: the loss of electrons  Reduction: the gain of electrons  Oxidation-reduction (redox) reaction: any reaction in which electrons are transferred from one species to another

Oxidation-Reduction

 Example: if we put a piece of zinc metal in a beaker containing a solution of copper(II) sulfate – some of the zinc metal dissolves – – some of the copper ions deposit on the zinc metal the blue color of Cu 2+ ions gradually disappears  In this oxidation-reduction reaction – zinc metal loses electrons to copper ions

Zn(s) Zn 2 + (aq) + 2 e Zn is oxidized

– copper ions gain electrons from the zinc

Cu 2 + ( aq) + 2 e Cu( s) Cu 2 + is red uced

Oxidation-Reduction

– we summarize these oxidation-reduction relationships in this way

electrons flow from Zn to Cu 2 + Zn 2 + ( aq) + Cu( s) Zn(s) + loses electrons ; is oxidized gives electrons to Cu 2+ ; is th e red ucing agent Cu 2 + (aq) gains electrons ; is red uced tak es electrons from Zn; is th e oxidizin g agent

Oxidation-Reduction

 Although the definitions of oxidation (loss of electrons) and reduction (gain of electrons) are easy to apply to many redox reactions, they are not easy to apply to others – for example, the combustion of methane

CH 4 (g) Methan e + O 2 ( g) CO 2 (g) + H 2 O( g)

 An alternative definition of oxidation-reduction is – oxidation: the gain of oxygen or loss of hydrogen – reduction: the loss of oxygen or gain of hydrogen

Oxidation-Reduction

– using these alternative definitions for the combustion of methane

electrons are trans ferred from carb on to oxygen CH 4 (g) gain s O and los es H; is oxidized + is the reducin g agent O 2 (g) gain s H; is reduced is th e oxid izing agent CO 2 (g) + H 2 O(g)

Oxidation-Reduction

 Five important types of redox reactions – – – combustion: burning in air. The products of complete combustion of carbon compounds are CO 2 and H 2 O.

respiration: the process by which living organisms use O 2 CO to oxidize carbon-containing compounds to produce 2 and H 2 O. The importance of these reaction is not the CO 2 produced, but the energy released.

rusting: the oxidation of iron to a mixture of iron oxides –

4Fe(s)

bleaching:

+ 3O 2 ( g) 2Fe 2 O 3 ( s)

the oxidation of colored compounds to products which are colorless – batteries: in most cases, the reaction taking place in a battery is a redox-reaction



Heat of Reaction

In almost all chemical reactions, heat is either given off or absorbed – example: the combustion (oxidation) of carbon liberates 94.0 kcal per mole of carbon oxidized

C( s) + O 2 (g) CO 2 (g) + 94.0 kcal/mole C

 Heat of reaction: the heat given off or absorbed in a chemical reaction – – exothermic reaction: one that gives off heat endothermic reaction: one that absorbs heat – heat of combustion: the heat given off in a combustion reaction; all combustion reactions are exothermic

Chemical Reactions

End Chapter 6