Chapter 4 Types of Chemical Reactions
Chapter 4
Types of Chemical Reactions
Classifying Chemical Reactions by What Atoms Do
Classification of Reactions
+
Decomposition reaction
Single displacement reaction
+ +
Double displacement reaction
+ +
Synthesis reaction +
Classification of Reactions
4 Al (s) + 3 O2 (g) 2 Al2O3 (s)
2 H2 (g) + O2 (g) ---------> 2 H2O (g)
C2H4 (g) + H2O2 (aq) C2H6O2 (l)
Synthesis reaction 2 HgO (s) ---------> 2 Hg (l) + O2 (g)
CaCO3 (s) ---------> CaO (s) + CO2 (g)
2 NaCl (s) ---------> Cl2 (g) + 2 Na (l)
Decomposition reaction Cu (s) + 2 AgNO3 (aq) ---------> 2 Ag (s) + Cu(NO3)2 (aq)
2 Al (s) + Fe2O3 (s) ---------> Al2O3 (s) + 2 Fe (l)
Mg (s) + 2 HCl (aq) ---------> H2 (g) + MgCl2 (aq)
Single displacement reaction
Ba(NO3)2 (aq) + Na2SO4 (aq) ---------> BaSO4 (s) + 2 NaNO3 (aq)
PCl3 (l) + 3 AgF (s) ---------> PF3 (g) + 3 AgCl (s)
HCl (aq) + NaOH (aq) ---------> H2O(l) + NaCl (aq)
Double displacement reaction
Chemical Reactions Classified by
Reaction Type
Precipitation Reactions
Precipitation Reactions
Precipitation reactions are reactions in which a solid forms when we mix two solutions.
1) reactions between aqueous solutions of ionic compounds 2) produce an ionic compound that is insoluble in water 3) The insoluble product is called a precipitate.
Precipitation Reactions
2 KI(aq) + Pb(NO3)2(aq) ➜ PbI2(s) + 2 KNO3(aq)
No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq) ➜ KCl(aq) + NaI(aq)
KI(aq)
NaCl(aq)
KCl(aq) + NaI(aq)
No precipitate forms, therefore, no reaction.
Process for Predicting the Products ofa Precipitation Reaction
1. Determine which ions are present in each aqueous reactant.
2. Determine formulas of possible products.
3. Determine solubility of each potential product in water.
4. If neither product will precipitate, write no reaction after the arrow.
5. If any of the possible products are insoluble, write their formulas as the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate aqueous.
6. Balance the equation. Remember to only change coefficients, not subscripts
Solubility RulesCompounds Containing theFollowing Ions Are Mostly Soluble Exceptions
Li+, Na+, K+, NH4+ None
NO3-, C2H3O2
-, ClO4- None
Cl- , Br- , I- Ag+, Hg22+, Cu+, Pb2+
SO4 2- Ca2+, Sr2+, Ba2+,Pb2+
Compounds Containing theFollowing Ions Are Mostly Insoluble Exceptions
OH- Group I cations, Ca2+, Sr2+, Ba2+
S2- Group I cations, Ca2+, Sr2+, Ba2+, NH4+
CO32-, PO4
3- Group I cations, NH4+
Practice – Predict the products and balance the equation
K2CO3(aq) + NiCl2(aq) ➜
K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (?) + NiCO3(?)
K2CO3(aq) + NiCl2(aq) ➜ 2 KCl (aq) + NiCO3(s)
K2CO3(aq) + NiCl2(aq) ➜ KCl (?) + NiCO3(?)
Practice – Predict the products and balance the equation
KCl(aq) + AgNO3(aq) ➜ KNO3(?) + AgCl(?)
KCl(aq) + AgNO3(aq) ➜ KNO3(aq) + AgCl(s)
KCl(aq) + AgNO3(aq) ➜
Practice – Predict the products and balance the equation
Na2S(aq) + CaCl2(aq) ➜
Na2S(aq) + CaCl2(aq) ➜ NaCl(?) + CaS(?)
Na2S(aq) + CaCl2(aq) ➜ 2 NaCl(?) + CaS(?)
Na2S(aq) + CaCl2(aq) ➜ 2 NaCl(aq) + CaS(aq)
No Reaction !!!!!
Practice – Predict the products and balance the equation
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ NH4C2H3O2(?) + PbSO4(?)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ 2 NH4C2H3O2(?) + PbSO4(?)
(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ➜ 2 NH4C2H3O2(aq) + PbSO4(s)
Ionic Equations
Equations that describe the material’s structure when dissolved are called complete ionic equations.
Aqueous strong electrolytes are written as ions.
Insoluble substances, weak electrolytes, and nonelectrolytes are written
as molecules.
Equations that describe the chemicals put into the water and the product molecules are called molecular equations.
2 KOH(aq) + Mg(NO3)2(aq) ➜ 2 KNO3(aq) + Mg(OH)2(s)
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ➜ 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)
Ionic Equations
Equations that describe the material’s structure when dissolved are called complete ionic equations.
Aqueous strong electrolytes are written as ions.
Insoluble substances, weak electrolytes, and nonelectrolytes are written
as molecules.
Equations that describe the chemicals put into the water and the product molecules are called molecular equations.
2 KOH(aq) + Mg(NO3)2(aq) ➜ 2 KNO3(aq) + Mg(OH)2(s)
2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ➜ 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s)
Ionic Equations
Ions that are both reactants and products are called spectator ions.
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) ➜ 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
An ionic equation in which the spectator ions are removed is called a net ionic equation.
2 OH−(aq) + Mg2+(aq) ➜ Mg(OH)2(s)
Write the ionic and net ionic equation
K2SO4(aq) + 2 AgNO3(aq) ➜ 2 KNO3(aq) + Ag2SO4(s)
2 Ag+(aq) + SO42−(aq) ➜ Ag2SO4(s)
2K+ (aq) + SO42-(aq) + 2Ag+ (aq) + 2NO3-(aq) ➜ 2K+ (aq) + 2NO3-(aq) + Ag2SO4(s)
Na2CO3(aq) + 2 HCl(aq) ➜ 2 NaCl(aq) + CO2(g) + H2O(l)
CO32−(aq) + 2 H+(aq) ➜ CO2(g) + H2O(l)
2Na+ (aq) + CO32-(aq) + 2H+ (aq) + 2Cl-(aq) ➜ 2Na+ (aq) + 2Cl-(aq) + CO2(g) + H2O(l)
Write the ionic and net ionic equation
Chapter 4 Acid/Base Reactions
Acids and Bases in Solution
Acids ionize in water to form H+ ions. (More precisely, the H+ from the acid molecule is donated
to a water molecule to form hydronium ion, H3O+)
Bases dissociate in water to form OH- ions. (Bases, such as NH3, that do not contain OH- ions, produce
OH- by pulling H+ off water molecules.)
In the reaction of an acid with a base, the H+ from the acid combines with the OH- from the base to make water.
The cation from the base combines with the anion from the acid to make a salt.
acid + base ➜ salt + water
Molecular Models of Selected Acids
Acid-Base ReactionsAlso called neutralization reactions because the acid and base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + 2 H2O(l)
Note that the cation from the base combines with the anion from the acid to make the water soluble salt.
H+(aq) + OH-(aq) ➜ H2O(l)
(as long as the salt that forms is soluble in water)
The net ionic equation for an acid-base reaction is
Common AcidsName Formula Uses Strength
Perchloric HClO4 explosives, catalysts Strong
Nitric HNO3 explosives, fertilizers, dyes, glues Strong
Sulfuric H2SO4 explosives, fertilizers, dyes, glue, batteries Strong
Hydrochloric HCl metal cleaning, food prep, ore refining, stomach acid Strong
Phosphoric H3PO4 fertilizers, plastics, food preservation Moderate
Chloric HClO3 explosives Moderate
Acetic HC2H3O2 plastics, food preservation, vinegar Weak
Hydrofluoric HF metal cleaning, glass etching Weak
Carbonic H2CO3 soda water, blood buffer Weak
Hypochlorous HClO sanitizer Weak
Boric H3BO3 eye wash Weak
Name Formula Common Name Uses Strength
Sodium Hydroxide NaOH Lye, Caustic Soda soap, plastic production, petroleum refining
Strong
Potassium Hydroxide
KOH Caustic Potash soap, cotton processing, electroplating
Strong
Calcium Hydroxide
Ca(OH)2 Slaked Lime cement Strong
Sodium Bicarbonate
NaHCO3 Baking Soda food preparation, antacids Weak
Magnesium Hydroxide
Mg(OH)2 Milk of Magnesia antacids Weak
Ammonium Hydroxide
NH4OH Ammonia Water fertilizers, detergents, explosives Weak
Common Bases
HCl(aq) + NaOH(aq) ➜ NaCl(aq) + H2O(l)
HCl(aq) NaOH(aq)
NaCl(aq) + H2O(l)
Write the molecular, ionic, and net-ionic equation for the acid-base reaction
HNO3(aq) + Ca(OH)2(aq) ➜
2H+ (aq) + 2NO3-(aq) + Ca2+ (aq) + 2OH-(aq) ➜ Ca2+ (aq) + 2NO3-(aq) + 2H2O(l)
HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + H2O(l)
2HNO3(aq) + Ca(OH)2(aq) ➜ Ca(NO3)2(aq) + 2H2O(l)
2H+(aq) + 2OH-(aq) ➜ 2H2O(l)
HCl(aq) + Ba(OH)2(aq) ➜
Write the molecular, ionic, and net-ionic equation for the acid-base reaction
HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + H2O(l)
2H+(aq) + 2OH-(aq) ➜ 2H2O(l)
2H+ (aq) + 2Cl-(aq) + Ba2+ (aq) + 2OH-(aq) ➜ Ba2+ (aq) + 2Cl-(aq) + 2H2O(l)
2HCl(aq) + Ba(OH)2(aq) ➜ BaCl2(aq) + 2H2O(l)
H2SO4(aq) + Sr(OH)2(aq) ➜
2H+ (aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)
2H+(aq) + SO42-(aq) + Sr2+ (aq) + 2OH-(aq) ➜ SrSO4 (s) + 2H2O(l)
H2SO4(aq) + Sr(OH)2(aq) ➜ SrSO4(s) + 2 H2O(l)
Write the molecular, ionic, and net-ionic equation for the acid-base reaction
Chapter 4
Acid/Base Titrations
Titration
A solution’s concentration is determined by reacting it with another solution and using stoichiometry –
this process is called titration.
In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint,
the reactants are in their stoichiometric ratio.
The unknown solution is added slowly from an instrument called a burette.
Acid-Base Titrations
The difficulty is determining when there has been just enough of one solution (the titrant) added to complete the reaction.
In acid-base titrations, because both the reactant and product solutions are colorless, a chemical (indicator) is added that changes color when the solution undergoes large changes in
acidity/alkalinity.
At the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH- . This is the
equivalence point.
TitrationThe titrant is the base solution in the burette.
As the titrant is added to the flask, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change.
At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color.
Titration
The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point.
What is the concentration of the unknown HCl solution?
HCl(aq) + NaOH(aq) ➜ NaCl(aq) + H2O(l)
L
mL
L of NaOH soln
mol of NaOH
L
mol
mol of HCl
mol
mol
mL of NaOH soln
mL of HCl soln
L of HCl soln Molarity =
mol of HCl
L of HCl soln
1.00 mol HCl1.00 mol NaOH
12.54 mL NaOH solution x x
x = mol HCl in the sample
0.100 mol NaOH1.000 L NaOH soln
0.001 L NaOH soln1.000 mL NaOH soln
The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point.
What is the concentration of the unknown HCl solution?
HCl(aq) + NaOH(aq) ➜ NaCl(aq) + H2O(l)
1.25 x 10-3
10.00 mL HCl solution x = L HCl soln0.001 L HCl soln1.000 mL HCl soln
0.0100
Molarity of HCl solution = = 1.25 x 10-3 mol HCl0.0100 L HCl soln 0.125 M
What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of 0.1015 M H2SO4 ?
L
m
L of H2SO4 soln
mol of H2SO4
L
mol
mol of NaOH
mol
mol
mL of H2SO4 soln
mL of NaOH soln
H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)
L of NaOH soln
Molarity =mol of NaOH
L of NaOH soln
27.50 mL NaOH soln x = L NaOH soln
2.00 mol NaOH1.00 mol H2SO4
50.00 mL H2SO4 soln x x
x = mol NaOH in the sample
0.1015 mol H2SO41.000 L H2SO4 soln
0.001 L H2SO4 soln1.000 mL H2SO4 soln
0.1015
0.001 L NaOH soln1.000 mL NaOH soln
0.02750
Molarity of NaOH soln = = 0.1015 mol NaOH0.02750 L NaOH soln
0.3691 M
What is the concentration of NaOH solution that requires 27.50 mL to titrate 50.00 mL of 0.1015 M H2SO4 ?
H2SO4 (aq) + 2 NaOH (aq) ➜ Na2SO4 (aq) + 2 H2O (l)
Chapter 4
Gas-Evolving Reactions
Gas-Evolving Reactions
Some reactions form a gas directly from the ion exchange:
K2S(aq) + H2SO4(aq) ➜ K2SO4(aq) + H2S(g)
Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water.
K2SO3(aq) + H2SO4(aq) ➜ K2SO4(aq) + H2SO3(aq)
H2SO3 ➜ H2O(l) + SO2(g)
NaHCO3(aq) + HCl(aq) ➜ NaCl(aq) + CO2(g) + H2O(l)
NaHCO3(aq)
HCl(aq)
NaCl(aq) + CO2(g) + H2O(l)
Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water.
NaHCO3(aq) + HCl(aq) ➜ NaCl(aq) + H2CO3(aq)
H2CO3 ➜ H2O(l) + CO2(g)
Reactant ReactantExchange Product Gas Formed Example
Metal sulfide Metal hydrogensulfide Acid H2S H2S K2S (g) + HCl (aq)→
H2S (g) + KCl (aq)
Metal carbonate Metal hydrogencarbonate Acid H2CO3 CO2
K2CO3 (aq) + HCl (aq) → CO2 (g) + H2O (l) + KCl (aq)
Metal sulfite Metal hydrogensulfite Acid H2SO3 SO2
K2SO3 (aq) + HCl (aq) → SO2 (g) + H2O (l) + KCl (aq)
Ammonium salt Base NH4OH NH3KOH (aq) + NH4Cl (aq) → NH3 (g) + H2O (l) + KCl (aq)
Compounds that UndergoGas-Evolving Reactions
Practice – Predict the products and balance the equations
Na2CO3(aq) + 2 HNO3(aq) ➜
2 HCl(aq) + Na2SO3(aq) ➜
H2SO4(aq) + CaS(aq) ➜
2 NaNO3(aq) + H2O (l) + CO2(g)
2 NaCl (aq) + H2O (l) + SO2 (g)
CaSO4(aq) + H2S(aq)
“2 NaNO3(aq) + H2CO3”
“2 NaCl(aq) + H2SO3”
Chapter 4
Redox Reactions
Oxidation/Reduction Basic Definitionse-
X loses electrons Y gains electron
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
Oxidation and Reduction - Symbolic Representation
Oxidation and Reduction at the Atomic Level
Oxidation/reduction reactions involve transferring electrons from one atom to another.
Also known as redox reactions
Many involve the reaction of a substance with O2(g).
4 Fe(s) + 3 O2(g) ➜ 2 Fe2O3(s)
Redox Reactions
Atoms in Elements-------> Ions in Compound
Combustion as Redox
2 H2(g) + O2(g) ➜ 2 H2O(g)
Redox without Combustion
2 Na(s) + Cl2(g) ➜ 2 NaCl(s)
Reactions of Metals with Nonmetals
Consider the following reactions:
4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s)
The reactions involve a metal reacting with a nonmetal.
In addition, both reactions involve the conversion of free elements into ions.
Na2O = 2 Na+ + O2-
NaCl = Na+ + Cl-
Oxidation and Reduction
To convert a free element into an ion, the atoms must gain or lose electrons (of course, if one atom
loses electrons, another must accept them).
Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced.
2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction
Chapter 4
Redox Reactions
Electron Bookkeeping
Electron BookkeepingFor reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the
electrons are transferred.
Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine
the electron flow in the reaction.
Even though they look like them, oxidation states are not ion charges!
Oxidation states are imaginary charges assigned based on a set of rules.
Ion charges are real, measurable charges.
Rules for Assigning Oxidation States (in order of priority)
1. Free elements have an oxidation state = 0.
In Na (s), Na = 0 ; In Cl2 (g), Cl2 = 0
2. Monatomic ions have an oxidation state equal to their charge.
In NaCl, Na = +1 and Cl = −1
3. (a) The sum of the oxidation states of all the atoms in a compound is 0.
Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0
Rules for Assigning Oxidation States (in order of priority)
3. (b) The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion.
In NO3–, N = +5 and O = −2 [3 x (2-) + 1 x (5+) = -1]
4. (a) Group I metals have an oxidation state of +1 in all their compounds.
(b) Group II metals have an oxidation state of +2 in all their compounds.
Rules for Assigning Oxidation States (in order of priority)
5. In their compounds, nonmetals have oxidation states according to the table below
Assign an oxidation state to each element in the following
Br2
K+
LiF
CO2
SO42−
Na2O2
Br = 0, (Rule 1)
K = +1, (Rule 2)
Li = +1, (Rule 4a) & F = −1, (Rule 5)
O = −2, (Rule 5) & C = +4, (Rule 3a)
O = −2, (Rule 5) & S = +6, (Rule 3b)
Na = +1, (Rule 4a) & O = −1 , (Rule 3a)
Determine the oxidation states of all the atoms in a propanoate polyatomic anion, C3H5O2–
There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b
(C3) + (H5) + (O2) = −1
Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order: H = +1 O = −2
(C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔
reduction
Oxidation and Reduction Another Definition
Oxidation occurs when an atom’s oxidation state increases during a reaction.
Reduction occurs when an atom’s oxidation state decreases during a reaction.
CH4 + 2 O2 → CO2 + 2 H2O-4 +4 0 -2
oxidation
Oxidation–Reduction
2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized Cl is reduced
Na is the reducing agent Cl2 is the oxidizing agent
Oxidation and reduction must occur simultaneously.
If an atom “loses” electrons another atom must “take” them.
The reactant that reduces an element in another reactant is called the reducing agent.
The reactant that oxidizes an element in another reactant is called the oxidizing agent.
Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing
agent in the following reactions:
Sn4+ + Ca → Sn2+ + Ca2+
2 F2 + S → SF4
Sn4+ is being reduced; Sn4+ is the oxidizing agent. Ca is being oxidized; Ca is the reducing agent.
F is being reduced from F0 to F-;F2 is the oxidizing agent.
S is being oxidized from S0 to S+4;S is the reducing agent.
0
0 0 S +4
F -
reduction
Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and
reducing agent in the following reactions:
0 +3+7 +4
Fe is the reducing agent. MnO4− is the oxidizing agent.
oxidation