4-1 dr. wolf’s chm 101 chapter 4 the major classes of chemical reactions

4-1 Dr. Wolf’s CHM 101

Chapter 4The Major Classes of Chemical Reactions


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


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.


The electrical conductivity of ionic solutions


Sample Problem 4.1 Determining 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) 7.42x1022 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) (NH4)2SO4(s) 2NH4+(aq) + SO4

2-(aq)

5.0mol (NH4)2SO4

2mol NH4+

1mol (NH4)2SO4

= 10.mol NH4+

5.0mol SO42-


Sample Problem 4.1 Determining Moles of Ions in Aqueous Ionic Solutions

continued

mol CsBr 212.8g CsBr

= 0.369mol CsBr

= 0.369mol Cs+

= 0.369mol Br1-

(b) CsBr(s) Cs+(aq) + Br-(aq)

7.42x1022 formula units Cu(NO3)2

mol Cu(NO3)2

6.022x1023 formula units

= 0.123mol Cu(NO3)2

= 0.123mol Cu2+

= 0.246mol NO3 1-

(c) Cu(NO3)2(s) Cu2+(aq) + 2NO3-(aq)

35mL ZnCl2 1L

103mL

= 2.9x110-2mol ZnCl2

(d) ZnCl2(aq) Zn2+(aq) + 2Cl-(aq)

0.84mol ZnCl2L

= 2.9x110-2mol Zn2+ = 5.8x110-2mol Cl-

78.5g CsBr


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.


The Polarity of WaterElectron distribution in molecules of H2 and H2O


The dissolution of an ionic compound


Acids in water dissociate to give H+. The H+ bonds with a water molecule and becomes a hydrated proton, H3O+ , a hydronium ion.


Sample Problem 4.2 Determining 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 NO3-.

HNO3(l) H+(aq) + NO3

-(aq)

1.4M HNO3(aq) should have 1.4M H+(aq).


Writing Equations for Aqueous Ionic ReactionsMolecular 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)


Writing equations for aqueous ionic reactions. A precipitation reaction and its equation.

©


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 (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 Hg2

2+.

1. 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 (PO4

3-) 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

+.

Soluble Ionic Compounds

Insoluble Ionic Compounds

4. All common sulfates (SO42-) are soluble,

except those of Ca2+ , Sr2+ , Ba2+ , and Pb2+ .


Sample Problem 4.3 Predicting 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 insolubilityTable 4.1

eliminate spectator ions for net ionic equation

PLAN:SOLUTION:

(a) Na2SO4(aq) + Sr(NO3)2 (aq) 2NaNO3(aq) + SrSO4(s)

2Na+(aq) +SO42-(aq)+ Sr2+(aq)+2NO3

-(aq)

2Na+(aq) +2NO3-(aq)+ SrSO4(s)

SO42-(aq)+ Sr2+(aq) SrSO4(s)

(b) NH4ClO4(aq) + NaBr (aq) NH4Br (aq) + NaClO4(aq)

All reactants and products are soluble so no reaction

occurs.


Strong / Weak Acids and BasesStrong 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, HNO3

sulfuric acid, H2SO4

perchloric acid, HClO4

Weak

hydrofluoric acid, HF

phosphoric acid, H3PO4

acetic acid, CH3COOH (or HC2H3O2)

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, NH3


2OH-(aq)+ 2H+(aq) 2H2O(l)

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)

Ba2+(aq) + 2OH-(aq)+ 2H+(aq)+ SO42-(aq)

2H2O(l)+Ba2+(aq)+SO42-(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)+2HClO4(aq) 2H2O(l)+Sr(ClO4)2(aq)

SOLUTION:

(b) Ba(OH)2(aq) + H2SO4(aq) 2H2O(l) + BaSO4(aq)

2OH-(aq)+ 2H+(aq) 2H2O(l)

water

spectator ions

2H+(aq)+


An aqueous strong acid-strong base reaction on the atomic scale.


An acid-base titration


Sample Problem 4.5 Finding 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 0.1524M 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) + H2O(l)

(33.87-0.55)mL x 1L

103mL= 0.03332L

0.03332L X 0.1524M = 5.078x10-3molNaOH

Molar ratio is 1:1

5.078x10-3mol HCl

0.050L= 0.1016M HCl


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.


The redox process in compound formation


Table 4.3 Rules for Assigning an Oxidation Number (O.N.)

1. 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.

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 peroxidesO.N. = -2 in all other compounds(except with F)

To keep track of the transfer of electrons, atoms are assigned oxidation numbers.


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) 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.


Highest and lowest oxidation numbers of reactive main-group elements


A summary of terminology for oxidation-reduction (redox) reactions

X Y

e-

transfer or shift of electrons

X loses electron(s) Y gains electron(s)

X is oxidized Y is reduced

X is the reducing agent Y is the oxidizing agent

X increases its oxidation number

Y decreases its oxidation number


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) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

(b) PbO(s) + CO(g) Pb(s) + CO2(g)

(c) 2H2(g) + O2(g) 2H2O(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) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2(g)

0 0+6+1 -2 +3 +6 -2SOLUTION:

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.


Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents

continued

(c) 2H2(g) + O2(g) 2H2O(g)

(b) PbO(s) + CO(g) Pb(s) + CO2(g)

+2 -2 +2 -2 0 +4 -2

0 0 +1 -2

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.


Sample Problem 4.8 Balancing Redox Equations by the Oxidation Number Method

PROBLEM: 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:

(a) Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l)

0 +1 -2+5 +2 +5 -2 +4 -2 +1 -2

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) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l)

loses 2e-

gains 1e- x2 to balance e-

balance unchanged polyatomic ions

balance other ions


Cu(s) + HNO3(aq) Cu(NO3)2(aq) + NO2(g) + H2O(l)

Sample Problem 4.8 Balancing Redox Equations by the Oxidation Number Method

continued

2 2

2

24

(b) PbS(s) + O2(g) PbO(s) + SO2(g)

+2 -2 0 +2 -2 +4 -2

PbS(s) + O2(g) PbO(s) + SO2(g)

loses 6e-

gains 2e- per O; need 3/2 O2 to make 3O2-

3/2

Multiply by 2 to have whole number coefficients.

2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)


A redox titrationpermanganate ion / oxalate ion


Sample Problem 4.9 Finding an Unknown Concentration by a Redox Titration

PROBLEM: 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 is

KMnO4(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.

mol of KMnO4

volume of KMnO4 soln

mol of CaC2O4

mol of Ca2+

multiply by M

molar ratio

ratio of elements in formula

PLAN:

(a)


Sample Problem 4.9 Finding an Unknown Concentration by a Redox Titration

continued

2.05mL soln = 1.00x10-6mol KMnO4L

103 mL

4.88x10-4mol KMnO4

L

1.00x10-6mol KMnO4 5mol CaC2O4

2mol KMnO4

= 2.50x10-6 mol CaC2O4

2.50x10-6 mol CaC2O4 1mol Ca2+

1mol CaC2O4

= 2.50x10-6 mol Ca2+

PLAN:

(b)

SOLUTION:

mol Ca2+/1mL blood

mol Ca2+/100mL blood

g Ca2+/100mL blood

mg Ca2+/100mL blood

multiply by 100

multiply by M

10-3g = 1mg

SOLUTION:x1002.50x10-6 mol Ca2+

1mL blood=2.50x10-4 mol Ca2+

100mL blood

2.50x10-4 mol Ca2+

100mL blood40.08g Ca2+

mol Ca2+

mg

10-3g

=10.0mg Ca2+/100mL blood


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.


Combining elements to form an ionic compound


Decomposing a compound to its elements


Displacing one metal with another


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.


Sample Problem 4.10 Identifying the Type of Redox Reaction

continued (a) Combination

3

0 0 +2 -3

Mg is the reducing agent; N2 is the oxidizing agent.

(b) Decomposition

H2O2(l) H2O(l) + O2(g)

+1 -1 +1 -2 0

1/2 or

Mg(s) + N2(g) Mg3N2 (aq)

2 H2O2(l) 2 H2O(l) + O2(g)

O is the oxidizing and reducing agent.

(c) Displacement

Al(s) + Pb(NO3)2(aq) Al(NO3)3(aq) + Pb(s)

0 +2 +5 -2 +3 +5 -2 0

2Al(s) + 3Pb(NO3)2(aq) 2Al(NO3)3(aq) + 3Pb(s)

Pb is the oxidizing and Al is the reducing agent.


Reversible ReactionsChemical Equilibrium

For some reactions, which proceed from reactants to products, can also undergo the reverse reaction of “products” to “reactants”.

For example:CaCO3 CaO + CO2 and

CaO + CO2 CaCO3 But both reactions are taking place simultaneously and an equilibrium is reached. The equation is written:

CaCO3 CaO + CO2


End of Chapter 4

4-1 dr. wolf’s chm 101 chapter 4 the major classes of chemical reactions

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