Lecture 3

© Professor Geanangel, CHEM 1331 Chapter 4 1

4.1 The Role of Water as a Solvent4.2 Writing Equations for Aqueous Ionic Reactions4.3 Precipitation Reactions4.4 Acid-Base Reactions4.5 Oxidation-Reduction (Redox) Reactions4.6 Elements in Redox Reactions4.7 Reaction Reversibilty and the Equilibrium State

Chapter 4 Three Major Classes of Chemical Reactions

Learn the material in Tables 4.1, 4.2 and 4.3.


The Role of Water as a SolventPure water conducts very little electric current due to the lack of charged particles (ions).



When an ionic compound dissolves in water, the light demonstrates that an electric current flows in the solution.

As water dissolves ionic compounds, it dissociates them into ions.

Movement of + and – ions in an electrolyte solution carries the electric current.

NaCl(s) ! Na+(aq) + Cl –(aq)H2O


A substance, like KBr, that conducts a current when dissolved in water is an electrolyte.In dissolving, each ion in KBr becomes solvated,i.e., surrounded by solvent molecules.

For KBr, one mole of compound dissociates into two moles of ions. Depends on the formula.

Solubility of Ionic Compounds

KBr(s) ! K+(aq) + Br –(aq)H2O

(NH4)2SO4(s) ! 2NH4+(aq) + SO4

2–(aq)H2O

AlCl3(s) ! Al3+(aq) + 3Cl –(aq)H2O

Soluble ionic compounds dissociate fully in solution


Problem: How many moles of each ion are in each solution?

Skill: Using the formula to find the number of moles of ions in solution


35 mL of 0.84 M zinc chloride


Using Molecular Depictions to Understand Solution Reactions

A: KCl, Na2SO4, MgBr2 or AgNO3?

B: NH4NO3, MgSO4, Ba(NO3)2 or KF?

Which compounds are present in each beaker?


Its ionizing power is due to the distribution of its electrons and its shape.

The Polar Nature of WaterWater separates ions in ionic compounds by overcoming their electrostatic attractions.

Polar covalent bonds, create partially charged poles (!) on a polar H2O molecule.

In covalent bonds between different atoms, the sharing of electrons is unequal.

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© Professor Geanangel, CHEM 1331 Chapter 4

Water molecule has a bent shape: O-H bonds at an angle.

Combined effect of its shape and polar covalent bonds makes water polar and able to attract ions.

10

Each bond’s polarity is indicated by a polar arrow, with the arrowhead pointing to the negative pole.

δ–δ+


Electrostatic charges drive the separation of ions from the solid to the liquid.

Water molecules congregate over an ionic crystal’s surface.


Some covalent substances like benzene (C6H6) and octane (C8H18) have nonpolar molecules that do not dissolve appreciably in water.

Covalent Compounds in Water Water dissolves many molecular substancesNH3 sucrose ethanol (C2H6O) antifreeze (ethylene glycol, C2H6O2)

Their solutions do not conduct electric current so the substances are nonelectrolytes.

Acids are covalent compounds that interact so strongly with H2O that they dissociate into ions.

Each has polar molecules that attractH2O’s but do not dissociate into ions.


Workshop What volume (mL) of 0.625 M HCl(aq) can be prepared from 13.5 mL of 6.02 M HCl(aq)?


Problem: In water, strong acids form H+(aq) ion. What is molarity of H+(aq) in 1.4 M sulfuric acid?

Solution: 2 mol of H+(aq) formed per mol of H2SO4

H+ is simply a proton; positive charge in a tiny volumeConcentrated + charge in H+ strongly attracts the !– pole of H2O giving a covalent H-O bond.

Thus, 1.4 M H2SO4 contains 2.8 M H+(aq).

H2SO4(aq) " 2H+(aq) + [SO4]2-(aq)

H+(aq) is written as H3O +, the hydronium ion.

[H7O3]+


Writing Equations for Ionic ReactionsIf solutions of two ionic compounds are mixed, a reaction occurs only if ions leave solution.Mixing aqueous solutions of CaCl2 and KNO3 just causes their ions to disperse within the solution.

No reaction occurs since all possible combinations, CaCl2, KNO3, Ca(NO3)2, and KCl, are water soluble.

Three major types of aqueous ionic reactions:(1) precipitation reactions (2) neutralization reactions (3) reactions that form a gaseous product.

CaCl2(s) ! Ca2+(aq) + 2Cl –(aq)H2O

KNO3(s) ! K+(aq) + NO3–(aq)

H2O


Precipitation ReactionsSolutions of ionic compounds may react to form an insoluble product, a precipitate.Mixing AgNO3 and Na2CrO4 solutionsgives a reddish precipitate of Ag2CrO4.

Molecular equation shows reactants and products as undissociated compounds:2AgNO3(aq)+ Na2CrO4(aq) ! Ag2CrO4(s) + 2NaNO3(aq)

Total ionic equation shows all the soluble ionic substances dissociated into ions.

2Ag+(aq) +2NO3

-(aq) + 2Na+

(aq) + CrO42-

(aq) ! Ag2CrO4(s)+ 2Na+

(aq) + 2NO3-(aq)


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Spectator ions; not involved in the chemical change.

Note: equal numbers of Na+ and NO3

– present on both sides.

+

What Really Happens?

Net ionic equation: spectator ions are eliminated, leaving the actual chemical change that occurs:

2Ag+(aq) +2NO3

-(aq) + 2Na+

(aq) + CrO42-

(aq) ! Ag2CrO4(s)+ 2Na+

(aq) + 2NO3-(aq)


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+. All common fluorides (F-) are soluble except those of Pb2+ and Group 2A(2).

Soluble Ionic Compounds

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 (CO3

2-) and phosphates (PO43-) 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

+.

Insoluble Ionic Compounds

4. All common sulfates (SO22-) are soluble, except those of Ca2+, Sr2+,

Ba2+, Ag+, and Pb2+.

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Problem: Predict whether a reaction occurs when solutions below are mixed. If so, show molecular, total ionic, net ionic equations and spectator ions, if any.

Plan: In each case, write cation-anion combinations and determine if any are insoluble.

Predicting formula of precipitate and writing ionic equations for precipitation and acid-base reactions

(a) sodium sulfate (aq) + lead(II) nitrate(aq) " ?

Na2SO4(aq) + Pb(NO3)2 (aq) " 2 NaNO3 + PbSO4

(b) ammonium perchlorate (aq) + sodium bromide(aq)

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


Na + and NO3

– are spectator ions.

(a) Beside Na2SO4 and Pb(NO3)2, PbSO4 and NaNO3. PbSO4 is insoluble so a precipitation reaction occurs. Molecular equation:Na2SO4(aq) + Pb(NO3)2 (aq) # 2 NaNO3(aq) + PbSO4(s)

Total ionic equation:2 Na+(aq) + SO4

2-(aq) + Pb2+(aq) + 2 NO3-(aq) #

2 Na+(aq) + 2 NO3-(aq) + PbSO4(s)

Net ionic equation:(b) Possible product ion combinations are ammonium bromide and sodium perchlorate.

Ammonium and sodium salts are soluble so no reaction occurs. All are spectator ions!

NH4+(aq) + ClO4

-(aq) + Na+ (aq) + Br-(aq) #

NH4+ (aq) + Br-(aq) + Na+ (aq) + ClO4

- (aq)


Neutralization (Acid-Base) Reactions

Acids produce H+ ions when dissolved in H2O.

Bases produce OH– in water.

Acids and bases are electrolytes - grouped by “strength”- how extensively they dissociate into ions in solution.

Table 4.2 Selected Acids and Bases

Acids

Strong

hydrochloric acid, HCl

hydrobromic acid, HBr

hydriodic acid, HI

nitric acid, HNO3

sulfuric acid, H2SO4

perchloric acid, HClO4

Weak

hydrofluoric acid, HF

phosphoric acid, H3PO4

acetic acid, CH3COOH or HC2H3O2)

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Weak acids and bases dissociate less; most molecules remain intact.

Strong bases contain either the OH– or O 2– ions.

Weak base ammonia does not contain O2– or OH–, but produces some OH– ions in reaction with water

Strong acids and bases dissociate fully into ions when dissolved in water.

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

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Consider the molecular equation for reaction of strong acid HCl with strong base Ba(OH)2 :

The cation of the salt (Ba2+) is contributed by the base and the anion (Cl–) by the acid.

Both HCl and Ba(OH)2 dissociate completely to ions, so the total ionic equation is:

The net ionic equation is:


Acid-Base Titrations: Neutralization reactions done quantitatively

Acid-base titration, a standardized base solution is slowly added to an acid solution of unknown concentration.

Known volume of acid is placed in a flask with some indicator sol’n.

Base solution is added slowly from a buret clamped above the flask.


mol H+ (acid) = mol of OH– (buret)

25

Near the end point, indicator changes color as a drop of base is added.

At equivalence point:

Titration end point occurs when a tiny excess of OH– changes indicator to its basic color permanently.

2HCl(aq) + Ba(OH)2(aq) " BaCl2(aq) + 2H2O(l)


Problem: 50.00 mL HCl(aq) is placed in a flask. A buret holds 0.1524 M NaOH; vol. = 0.55 mL. At end point, vol. = 33.87 mL. What is HCl concentration?

Skill: Calculating an unknown concentration from an acid-base or redox titration

26


Find moles of NaOH added:

Find moles of HCl originally present:

Calculate molarity of HCl:

27


Consider the flashbulb reaction, in which an ionic compound, MgO, forms from its elements:

Each Mg atom loses two electrons and each O atom gains them, forming Mg 2+ and O 2– ions.

Redox Reactions: the importance of the net movement of electrons in the redox process

2+ 2-

!

2 Mg + O2 " 2 MgO


Some Essential Redox Terminology “oil rig”Oxidation is loss of electrons; reduction is gain of e–

In the formation of MgO, Mg undergoes oxidation (e– loss) and O2 undergoes reduction (e– gain).

O2 gains electrons, so O2 is oxidizing agent.Mg loses electrons, so Mg is reducing agent.

Remember:•Oxidizing agent gets reduced, accepting electrons. •Reducing agent gets oxidized as it loses electrons.

Oxidation half rxn: Mg $ Mg2+ + 2e–

Reduction half rxn: 1/2 O2 + 2e- $ O2–

The no. of e– moving is given by oxidation numbers


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. Sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for atoms in a polyatomic ion equals ion’s charge.

General rules

Rules for specific atoms or periodic table groups

1. For Group 1A(1): O.N. = +1 in all compounds2. For Group 2A(2): O.N. = +2 in all compounds

4. For fluorine: O.N. = -1 in all compounds

3. For hydrogen: O.N. = +1 in combination with nonmetalsO.N. = -1 in combination with metals and boron

5. For oxygen: O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F)

6. For Group 7A(17): O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group

30

Σ(ON) = 0

Σ(ON) = ion chg.


(c) HNO3 ON of H= +1; NO3 group sums to -1

Problem: Determine the O.N. of each element in: (a) zinc chloride (b) sulfur trioxide (c) nitric acidSolution: (a) ZnCl2 Contains a Zn2+ and two Cl– ON of Zn2+ is +2. ON of each Cl– is -1.(b) SO3 The ON of each oxygen is -2. The !(ON’s) in a compound = 0

Skill: Determining the oxidation number (O.N.) of any atom in a compound.


Skill: Identifying Redox Reactions (Samp. Prob. 4.8)There is a change in oxidation number in all redox reactions. Which of these are redox reactions?

4.5 Oxidation-Reduction (Redox) Reactions 161

By assigning an oxidation number to each atom, we can see which specieswas oxidized and which reduced and, from that, which is the oxidizing agent andwhich the reducing agent:• If a given atom has a higher (more positive or less negative) oxidation num-

ber in the product than it had in the reactant, the reactant species that containsthe atom was oxidized (lost electrons) and is the reducing agent. Thus, oxida-tion is represented by an increase in oxidation number.

• If an atom has a lower (more negative or less positive) oxidation number inthe product than it had in the reactant, the reactant species that contains theatom was reduced (gained electrons) and is the oxidizing agent. Thus, the gainof electrons is represented by a decrease (a “reduction”) in oxidation number.

SAMPLE PROBLEM 4.8 Identifying Redox ReactionsPROBLEM Use oxidation numbers to decide which of the following are redox reactions:(a) CaO(s) ! CO2(g) CaCO3(s)(b) 4KNO3(s) 2K2O(s) ! 2N2(g) ! 5O2(g)(c) NaHSO4(aq) ! NaOH(aq) Na2SO4(aq) ! H2O(l)PLAN To determine whether a reaction is an oxidation-reduction process, we use Table 4.3to assign each atom an O.N. and see if it changes as the reactants become products.SOLUTION

(a)Because each atom in the product has the same O.N. that it had in the reactants, we con-clude that this is not a redox reaction.

(b)

In this case, the O.N. of N changes from !5 to 0, and the O.N. of O changes from "2to 0, so this is a redox reaction.

(c)

The O.N. values do not change, so this is not a redox reaction.COMMENT The reaction in part (c) is an acid-base reaction in which HSO4

" transfers anH! to OH" to form H2O. In the net ionic equation for a strong acid–strong base reaction,

we see that the O.N. values remain the same on both sides of the equation. Therefore, anacid-base reaction is not a redox reaction.

FOLLOW-UP PROBLEM 4.8 Use oxidation numbers to decide which, if any, of thefollowing equations represents a redox reaction:(a) NCl3(l) ! 3H2O(l) NH3(aq) ! 3HOCl(aq)(b) AgNO3(aq) ! NH4I(aq) AgI(s) ! NH4NO3(aq)(c) 2H2S(g) ! 3O2(g) 2SO2(g) ! 2H2O(g)±£

±£±£

H!(aq) ! OH"(aq) ±£ H2O(l)

!1 !1"2!1

"2

NaHSO4(aq) ! NaOH(aq) ±£ Na2SO4(aq) ! H2O(l)

!1!1!6

"2!1!1

"2 !1!6

"2 "2!1

4KNO3(s) ±£ 2K2O(s) ! 2N2(g) ! 5O2(g)

0 0!1!5

!1 "2"2

#

O.N. decreased: reduction

O.N. increased: oxidation

CaO(s) ! CO2(g) ±£ CaCO3(s)

!2"2

!4"2 !2

!4"2

±£±£¢

±£

As Sample Problem 4.8 shows, a redox reaction can be defined as one in whichthe oxidation numbers of the species change. (In the rest of Section 4.5 andSection 4.6, blue type indicates oxidation, and red type indicates reduction.)

siL48593_ch04_140-185 3:12:07 11:30pm Page 161

4.5 Oxidation-Reduction (Redox) Reactions 161






(b)


(c)





±£±£

H!(aq) ! OH"(aq) ±£ H2O(l)

!1 !1"2!1

"2


!1!1!6

"2!1!1

"2 !1!6

"2 "2!1

4KNO3(s) ±£ 2K2O(s) ! 2N2(g) ! 5O2(g)

0 0!1!5

!1 "2"2

#




!2"2

!4"2 !2

!4"2

±£±£¢

±£


siL48593_ch04_140-185 3:12:07 11:30pm Page 161 4.5 Oxidation-Reduction (Redox) Reactions 161






(b)


(c)





±£±£

H!(aq) ! OH"(aq) ±£ H2O(l)

!1 !1"2!1

"2


!1!1!6

"2!1!1

"2 !1!6

"2 "2!1

4KNO3(s) ±£ 2K2O(s) ! 2N2(g) ! 5O2(g)

0 0!1!5

!1 "2"2

#




!2"2

!4"2 !2

!4"2

±£±£¢

±£


siL48593_ch04_140-185 3:12:07 11:30pm Page 161


33

Oxidation: increase in oxidation number.Reduction, gain of electrons, is represented by a decrease in O.N.

Plan: Assign O.N. to atoms (or ion) based on rules.

reactant = red. agent if it has an atom that is oxidized

reactant = ox. agent if it has an atom that is reduced

Skill: Selecting oxidizing and reducing agents.

Problem: Identify the oxidizing and reducing agents in:

It is essential to realize that the transferred electrons are never free, whichmeans that the reducing agent loses electrons and the oxidizing agent gains themsimultaneously. In other words, a complete reaction cannot be just an “oxidation”or a “reduction”; it must be an “oxidation-reduction.” Figure 4.14 summarizesredox terminology.

162 Chapter 4 Three Major Classes of Chemical Reactions

X loseselectron(s)

X is oxidized

X is thereducing agent

X increasesits oxidationnumber

Y gainselectron(s)

Y is reduced

Y is theoxidizing agent

Y decreasesits oxidationnumber

Transferor shift ofelectrons

e–

X Y

Figure 4.14 A summary of terminol-ogy for oxidation-reduction (redox)reactions.

Balancing Redox EquationsWe balance a redox reaction by making sure that the number of electrons lost bythe reducing agent equals the number of electrons gained by the oxidizing agent.

Two methods used to balance redox equations are the oxidation numbermethod and the half-reaction method. This section describes the oxidation number

SAMPLE PROBLEM 4.9 Recognizing Oxidizing and Reducing AgentsPROBLEM 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 We first assign an oxidation number (O.N.) to each atom (or ion) based on the rulesin Table 4.3. The reactant is the reducing agent if it contains an atom that is oxidized(O.N. increased from left side to right side of the equation). The reactant is the oxidizingagent if it contains an atom that is reduced (O.N. decreased).SOLUTION (a) Assigning oxidation numbers:

The O.N. of Al increased from 0 to !3 (Al lost electrons), so Al was oxidized;Al is the reducing agent.

The O.N. of H decreased from !1 to 0 (H gained electrons), so H! was reduced;H2SO4 is the oxidizing agent.

(b) Assigning oxidation numbers:

Pb decreased its O.N. from !2 to 0, so PbO was reduced; PbO is the oxidizing agent.C increased its O.N. from !2 to !4, so CO was oxidized; CO is the reducing agent.In general, when a substance (such as CO) becomes one with more O atoms (such asCO2), it is oxidized; and when a substance (such as PbO) becomes one with fewer Oatoms (such as Pb), it is reduced.(c) Assigning oxidation numbers:

O2 was reduced (O.N. of O decreased from 0 to "2); O2 is the oxidizing agent.

H2 was oxidized (O.N. of H increased from 0 to !1); H2 is the reducing agent.Oxygen is always the oxidizing agent in a combustion reaction.

FOLLOW-UP PROBLEM 4.9 Identify each oxidizing agent and each reducing agent:(a) 2Fe(s) ! 3Cl2(g) 2FeCl3(s)(b) 2C2H6(g) ! 7O2(g) 4CO2(g) ! 6H2O(g)(c) 5CO(g) ! I2O5(s) I2(s) ! 5CO2(g)±£

±£±£

2H2(g) ! O2(g) ±£ 2H2O(g)

0 "2

PbO(s) ! CO(g) ±£ Pb(s) ! CO2(g)

!2 0"2"2 "2

2Al(s) ! 3H2SO4(aq) ±£ Al2(SO4)3(aq) ! 3H2(g)

0!1 "2!6

"2!6

±£±£±£

siL48593_ch04_140-185 4:12:07 04:26am Page 162


Solution: (a) Assign oxidation numbers:

ON Al 0 to +3, Al is oxidized so Al is reducing agent

ON H +1 to 0, H reduced so H2SO4 is oxidizing agent

Pb is reduced; ON from +2 to 0; PbO = ox. agentC is oxidized, ON from +2 to +4; CO = red. agent

(a) 2Al(s) + 3H2SO4(aq) $ Al2(SO4)3(aq) + 3H2(g)

0 0+6+1 -2 +3 +6-2

(b) Assigning ON’s:

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

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


Balancing Redox Reactions by Oxidation No. MethodNo. of e– lost by the oxidized reactant must equal the number of e– gained by the reduced reactant.

Changes in oxidation numbers give the number of e– transferred; can be used to balance redox reactions.Five steps in the method listed on text page 163.

PbO(s) + NH3(aq) $ Pb(s) + N2(g) + H2O(l)

+2 0 0-2 -3 +1 +1 -2

3PbO(s) + 2NH3(aq) $ 3Pb(s) + N2(g) + H2O(l)

PbO(s) + 2NH3(aq) $ Pb(s) + N2(g) + H2O(l)

2

3


Problem: Use oxidation no. method to balance:

Assign ON’s to all atoms:

Skill: Balancing redox equations

ON of Cu: 0 in Cu to +2 in Cu2+; Cu oxidized ON of N: +5 in HNO3 to +4 in NO2; N reduced

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


loses 2e-

gains 1e- x2 to balance e-

36

2 2


Multiply by factors to make e – lost equal e– gained, and use the factors as coefficients.


Complete the balancing by inspection.


1. Assign ON’s:

2. S ( –2 in PbS) is oxidized (+4 in SO2) O (0 in O2 ) is reduced (–2 in PbO and in SO2 )

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

3. PbS(s) + O2(g) $ PbO(s) + SO2(g)

loses 6e-

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

total of 6e- transferred

3/2

4. Multiply all by 2 to have whole number coefficients.

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


MnO4–(aq) + Fe2+(aq) " Mn2+(aq) + Fe3+(aq) (acidic)

Workshop: Use ON change to balance this equation+7 +2 +2 +3

gains 5e-loses 1e-

MnO4–(aq) + 5Fe2+(aq) " Mn2+(aq) + 5Fe3+(aq)

MnO4–(aq) + 5Fe2+(aq) " Mn2+(aq) + 5Fe3+(aq) + 4H2O


Cr2O72- + I– " Cr3+ + I2 (acidic aqueous soln, H+/H2O)

Cr2O72- + 2I– = 2Cr3+ + I2

Cr2O72- + 6I- = 2Cr3+ + 3I2 (electrons balanced!)

Cr2O72- + 6I- = 2Cr3+ + 3I2 + 7H2O

Cr2O72- + 6I- + 14H+ = 2Cr3+ + 3I2 + 7H2O

Using ON change to balance in acidic solution

+6 +3-1 0-2

22


4.6 Summary of General Reaction TypesCombination (synthesis) X + Y " Z

Decompositon Z " X + Y

Single displacement X + YZ " XZ + Y

Double displacement WX + YZ " WZ + YX(metathesis)

Combustion 2C4H10 + 13O2 " 8CO2 + 10H2O S8 + 8O2 " 8SO2


Combination: Two nonmetals form a binary covalent compound.

Industrial production of ammonia (Haber):

Some nonmetal oxides react with O2 to form higher oxides.

Metal oxide and nonmetal oxide react to form an ionic compound with an oxoanion.

Note: its not a redox reaction.


Thermal decomposition: Z " X + YHeating ionic compounds with oxoanions may form a metal oxide and a non-metal oxide.

Hydroxides, hydrates and some oxoacids release water on heating

Some metal oxides, chlorates, and perchlorates release O2 on heating


Displacement Reactions: Number of Reactants Equals The Number of Products

An atom or ion in a compound is displaced by an atom or ion of another element.

Single-displacement reactions

1. Some metals can displace hydrogen from either water or an acid.

Ni(s)+2HCl(aq) $ NiCl2(aq) + H2(g)


A metal may displace another metal from solution.2AgNO3(aq) + Cu(s) " Cu(NO3)2(aq) + 2Ag(s)


Skill: Identifying combination, decomposition, and displacement reactionsProblem: Classify each of the following reactions and write a balanced equation for each:

ammonium dichromate(s) " nitrogen(g) +chromium(III) oxide(s) +water(g)


Combination: two substances form one.

Decomposition: one substance forms two. H2O2 is sold in brown bottles because it decomposes in light

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

b) magnesium(s) + nitrogen(g) $ magnesium nitride(s)

(c) hydrogen peroxide(l) $ water(l) + oxygen (g)


Section 4.7. Reversible Reactions is omitted

END OF CHAPTER 4

(d) aluminum(s) + lead(II) nitrate(aq) $ aluminum nitrate(aq) + lead(s)

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