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SCH 4U REVIEW
CHAPTER 1: ORGANIC COMPOUNDS
organic compound – a compound that contains carbon and usually hydrogen
catenation – the property of carbon to form a covalent bond with another carbon atom, forming
long chains or rings
functional group – a group of atoms in an organic molecule that impart particular physical and
chemical characteristics to that molecule
– there are three main components:
 multiple bonds between C atoms (e.g. – C ≡ C – )
 C bonded to a more electronegative atom (i.e., N, O, OH, or a halogen)
 C double bonded to O
BASIC NAMING
NUMBER
1
2
3
4
5
6
7
8
9
10
PREFIX FOR
NAMING
methethpropbutpenthexheptoctnondec-
ALKYL GROUP
NAME
methylethylpropylbutylpentylhexylheptyloctylnonyldecyl-
FUNCTIONAL
GROUP
-F
- Cl
- Br
-I
- OH
- NO2
- NH2
ISOMERS OF ALKYL GROUPS
butyl (or n-butyl)
CH3 – CH2 – CH2 – CH2
CH3 – CH – CH3
isobutyl (or i-butyl)
s-butyl (secondary butyl)
CH2
CH3 – CH – CH2 – CH3
CH3
t-butyl (tertiary butyl)
CH3 – C – CH3
SYMBOLS
R, R΄, R˝  alkyl group
X  halogen atom
(O)  oxidizing agent (like KMnO4 or Cr2O7 2- in H2SO4)
PREFIX FOR
NAMING
fluoro
chloro
bromo
iodo
hydroxy
nitro
amino
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PRIORITY FOR NAMING (FROM HIGHEST TO LOWEST)
OH
hydroxyl
NH2
amino
F, Cl, Br, I
fluoro, chloro, bromo, iodo
CH2CH2CH3
propyl
CH2CH3
ethyl
CH3
methyl
ALKANES
DEFINITION
NAMING
a hydrocarbon with only single bonds between carbon atoms; general formula,
CnH2n+2
prefix referring to number of carbons in the longest continuous chain + “-ane”
 alkyl groups are named in alphabetical order, numbered from the end that
will give the lowest combination of numbers
 the presence of two or more of the same alkyl groups requires a “di” or “tri”
prefix before the alkyl group name
 cyclic hydrocarbons have the carbon ring become the parent chain, and the
prefix “cyclo” is used before the parent name
GENERAL
FORMULA
–C–C–
propane
CH3 – CH2 – CH3
cyclohexane
EXAMPLE(S)
CH2
CH2
|
CH2
CH2
|
CH2
CH2
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
non-polar
dispersion forces
 relatively low
 increases with chain length
 straight chains have higher b.p.s than branched chains
immiscible in water and other polar solvents
combustion
C3H8 + 5 O2  3 CO2 + 4 H2O
REACTIONS
substitution (halogenation)
1) CH4 + Cl2  CH3Cl + HCl
2) CH3Cl + Cl2  CH2Cl2 + HCl
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ALKENES
DEFINITION
NAMING
a hydrocarbon that contains at least one carbon-carbon double bond; general
formula, CnH2n
prefix referring to number of carbons in the longest continuous chain that
contains the double bond + “-ene”
 alkyl groups are named in alphabetical order, numbered from the end
that is closest to the double bond
|
GENERAL
FORMULA
EXAMPLE(S)
POLARITY
FORCES
BOILING POINT
SOLUBILITY
|
–C=C–
CH2 = CH – CH3
propene
non-polar
dispersion forces
relatively low
immiscible in water and other polar solvents
halogenation (with Br2, Cl2, etc.)
ethene
room temperature
+ bromine
CH2 = CH2 +
1,2 – dibromoethane
Br
Br

Br2
CH2 – CH2
hydrogenation (with H2)
ethyne
+ hydrogen
CH2 = CH2 +
REACTIONS
catalyst, heat, pressure

2 H2
ethane
H
H
CH2 – CH2
H
H
hydrohalogenation (with hydrogen halides)
propene
room temperature
+ hydrogen bromine
H – C = CH – CH3
|
H
+ HBr
PREPARATION
hydration (with H2O)
REACTION FOR
ALCOHOLS
1-butene
+
CH3 – CH2 – CH = CH2 +

2-bromopropane
H Br
|
|
H – C – CH – CH3
|
H
H2SO4 catalyst
water
HOH
2-butanol

CH3 – CH2 – CH – CH2
OH
H
“The rich get richer” … when a hydrogen halide or water is added to an
MARKOVNIKOV’S
alkene or alkyne, the hydrogen atom bonds to the carbon atom within the
RULE
double bond that already has more hydrogen bonds.
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
a cis isomer has both alkyl groups on the same side of the molecular
strucutre
CH3
CH3
C=C
cis-2-butene
H
GEOMETRIC
ISOMERS

H
a trans isomer has alkyl groups on the opposite side of the molecular
strucutre
CH3
H
C=C
H
trans-2-butene
CH3
ALKYNES
DEFINITION
NAMING
a hydrocarbon that contains at least one carbon-carbon triple bond; general
formula, CnH2n-2
prefix referring to number of carbons in the longest continuous chain that
contains the triple bond + “-yne”
 alkyl groups are named in alphabetical order, numbered from the end that is
closest to the triple bond
GENERAL
FORMULA
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
REACTIONS
–C≡C–
propyne
non-polar
dispersion forces
CH ≡ C – CH3
relatively low
immiscible in water and other polar solvents
Same as Alkenes (resulting in double bonds rather than single)
AROMATIC HYDROCARBONS
DEFINITION
NAMING
GENERAL
FORMULA
a compound with a structure based on benzene (a ring of six carbons)
consider the benzene ring to be the parent molecule
 alkyl groups are named to give the lowest combination of numbers, with no
particular starting carbon (as it is a ring)
 when it is easier to consider the benzene ring as an alkyl group, we use the
name “phenyl” to refer to it
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methylbenzene
CH3
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
REACTIONS
non-polar
dispersion forces
relatively low
immiscible in water and other polar solvents
Same as Alkanes (performs substitution reactions)
ORGANIC HALIDES
DEFINITION
NAMING
a compound of carbon and hydrogen in which one or more hydrogen atoms have
been replaced by halogen atoms
halogen atoms are considered to be attachments to the parent chain and are
numbered and named with a prefix as such
GENERAL
FORMULA
R–X
H H
| |
H–C–C–H
| |
Cl Cl
1,2-dichloroethane
EXAMPLE(S)
Cl
1,2-dichlorobenzene
Cl
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
polar (due to halogen atom)
dispersion forces (increased strength b/c of carbon-halogen bonds)
higher than the corresponding hydrocarbons
more soluble in polar solvents
addition
+ hydrogen idodie 
ethylene
H–C≡C–H
REACTIONS
+
H–I

1-iodoethene
H I
| |
H–C=C–H
substitution
ethane
CH3 – CH3
+ chlorine 
1-chloroethane
H Cl
+ hydrogen chloride
Cl – Cl 
H–C–C–H
+
+
H H
H – Cl
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elimination
2-bromopropane + hydroxyl group 
H Br
H
H H H
| | |
H–C–C=C–H +
H – C – C – C – H + OH- 
H H
propene + bromine ion + water
H
Br-
+
H2O
H
ALCOHOLS
an organic compound characterized by the presence of a hydroxyl functional
group (OH-)
-
S
O
DEFINITION
O
< 109.5°

NAMING


H
S+
the “e” ending of the parent hydrocarbon is changed to “ol” to indicate the
presence of the OH- group
the chain is numbered to give the OH- group the smallest possible number
when there is more than one OH- group, the endings “diol” and “triol” are
used, and each is indicated with a numerical prefix, however the “e” ending
remain then (e.g., 1,3-propanediol)
GENERAL
FORMULA
R – OH
propanol
CH3 – CH2 – CH2 – OH
OH
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
phenol
polar (due to hydroxyl group)
hydrogen bonding and dispersion forces
high (due to capacity for hydrogen bonding)
very soluble in polar solvents and nonpolar solvents (due to OH- group)
hydration (preparation)
H2SO4 catalyst
1-butene
+
CH3 – CH2 – CH = CH2 +
water
HOH
2-butanol

CH3 – CH2 – CH – CH2
OH
H
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dehydration (elimination)
H2SO4 catalyst
propanol

CH3 – CH – CH2
OH
propene
CH3 – CH = CH2
+
water
+
HOH
H
Primary Alcohol (1°)
CH3 – CH2 – OH
 1 other carbon group attached to carbon with OH
OH
1°, 2°, AND 3°
ALCOHOLS
Secondary Alcohol (2°)
CH3 – CH – CH3
 2 other carbon groups attached to carbon with OH
OH
Tertiary Alcohol (3°)
 3 other carbon groups attached to carbon with OH
CH3 – C – CH3
CH3
ETHERS
an organic compound with two alkyl groups (the same or different) attached to
an oxygen atom
-
C
DEFINITION
O
116°



NAMING
C
S
O
C S+
C
the longer of the two alkyl groups is considered the parent chain
the other alkyl group with the oxygen is considered to be the substituent
group (with prefix of carbons and “oxy”)
numbering of C atoms starts at the O
propane
ethoxy
CH3CH2CH2 – O – CH2CH3
3
2
1
1
2
ethoxypropane
GENERAL
FORMULA
R – O – R΄
CH3 – O – CH2 – CH3
EXAMPLE(S)
methoxyethane
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
polar (due to the V-shape and C - O bonds)
dispersion forces
medium (higher than hydrocarbons, lower than alcohols of similar length)
soluble in polar solvents and nonpolar solvents
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condensation (preparation)
* addition of two alcohols (same or different) *
H2SO4 catalyst
methanol +
methanol
CH3 – OH +
CH3 – OH
methoxymethane + water

CH3 – O – CH3
+ HOH
ALDEHYDES
DEFINITION
NAMING
an organic compound that contains the carbonyl group (–C=O) on the end
carbon of a chain
the “e” ending of the parent hydrocarbon is changed to “al” to indicate the
presence of the R-C=O
O
||
R[H] – C – H
GENERAL
FORMULA
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
O
||
CH3 – CH2 – C – H
propanal
polar (due to carbonyl group)
dispersion forces
medium (higher than ethers, lower than alcohols of similar length)
similar solubility to alcohols
oxidation (preparation)
1° alcohol + mild oxidizing agent  aldehyde + water
+ (O) 
ethanol
OH
|
CH3 – C – H
REACTIONS
+ (O) 
ethanal
+
water
O
||
CH3 – C – H +
HOH
|
H
reduction (hydrogenation)
aldehyde + hydrogen  1° alcohol
+ hydrogen 
proponal
O
CH3 – CH2 – C – H
1-propanol
OH
+
H2

CH3 – CH2 – C – H
KETONES
DEFINITION
an organic compound that contains the carbonyl group (–C=O) on a carbon
other than those on the end of a carbon chain (in the middle)
SCH 4U REVIEW
NAMING
GENERAL
FORMULA
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
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the “e” ending of the parent hydrocarbon is changed to “one”
O
||
R – C – R΄
O
||
CH3 – C – CH3
propanone (acetone)
polar (due to carbonyl group)
dispersion forces
medium (higher than ethers, lower than alcohols of similar length)
similar solubility to alcohols
oxidation (preparation)
2° alcohol + oxidizing agent 
2-propanol
+ (O) 
propanone
+
water
+ (O) 
O
||
CH3 – C – CH3
+
HOH
OH
CH3 – C – CH3
ketone + water
H
REACTIONS
reduction (hydrogenation)
ketone + hydrogen  2° alcohol
+ hydrogen 
butanone
O
2-butanal
OH
CH3 – CH2 – C – CH3
+
H2

CH3 – CH2 – C – CH3
NOTE: Tertiary alcohols do not undergo oxidation reactions; no H atom is
available on the central C atom.
CARBOXYLIC ACIDS
DEFINITION
one of a family of organic compounds that is characterized by the presence of a
carboxyl group:
O
–C
NAMING
GENERAL
FORMULA


O–H
the “e” ending of the parent alkane is changed to “oic acid”
numbering starts with the C of the carboxyl group
O
||
R[H] – C – OH
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CH2CH3
CH3 – CH2 – CH – CH – CH2 – COOH
EXAMPLE(S)
4-ethyl-3-methylhexanoic acid
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
CH3
polar (due to carboxyl group)
dispersion forces and hydrogen bonding
very high (higher than hydrocarbons of similar length due to –CO and –OH
groups)
similar solubility to alcohols
dissolution in water (proof of acidity)
ethanoic acid + water 
carboxylate ion
+
hydronium ion
CH3COOH + H2O 
CH3COO-
+
H3O+
O–
CH3 – C
* shares the double bond
O
oxidation (with weak oxidizer)
1) 1° alcohol + weak oxidizing agent  aldehyde + water
2) aldehyde + weak oxidizing agent  carboxylic acid + water
O
REACTIONS
1) CH3 – CH2 – OH + Cr2O72- + H+

CH3 – C – H + H2O + Cr3+
O
2) CH3 – C – H
+ Cr2O72- + 2 H+  3 CH3 – COOH + 4 H2O
oxidation (with strong oxidizer)
1° alcohol + strong oxidizing agent  carboxylic acid + water
ethanol + oxidizing agent + hydrogen  ethanoic acid + water + manganese dioxide
O
3 CH3 – CH2 – OH +
NOTE:
4 MnO4- + 4 H+
 3 CH3 – C – OH + 5 H2O + 4 MnO2
Ketones are not readily oxidizing, as in the oxidation of aldehydes.
ESTERS
DEFINITION
NAMING
GENERAL
FORMULA
an organic compound characterized by the presence of a carbonyl group bonded
to an oxygen atom
 the group that is attached to the double-bonded O becomes the parent chain
with the “e” ending changed to “oate”
 the other group is named as a substituent group
O
||
R[H] – C – O – R΄
SCH 4U REVIEW
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
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2-methylbutyl propanoate
O
CH3
CH3 – CH2 – C – O – CH2 – CH – CH2 – CH3
less polar than carboxylic acids (loss of OH- group)
dispersion forces
medium (lower than carboxylic acids, higher than aldehydes / ketones of similar
length due to extra O)
less soluble than acids
condensation (formation)
carboxylic acid + alcohol  ester + water
ethanoic acid
+ methanol

O
||
CH3 – C – OH + CH3 – OH 
methyl ethanoate
O
||
CH3 – C – O – CH3
+
water
+
H2O
+
R΄ – OH
REACTIONS
hydrolysis (saponification)
ester + NaOH  sodium salt of acid
O
+ alcohol
O
R – C – O – R΄ + Na+ + OH-  R – C – O-
NOTE:
+
Na+
Esters generally have nice odours and are used to create artificial flavours.
AMINES
DEFINITION
NAMING
an ammonia molecule in which one or more H atoms are substituted by alkyl or
aromatic groups
1) nitrogen group is named as a substituent group using “amino-”
2) alkyl group is named as a substituent group from “-amine”
R΄ [H]
|
R – N – R΄΄ [H]
GENERAL
FORMULA
1) 1-aminopropane
2) propylamine
NH2
CH2 – CH2 – CH3
EXAMPLE(S)
N-ethyl-N-methyl-1-aminobutane
CH2CH3
|
CH3 – N – CH2 – CH2 – CH2 – CH3
POLARITY
polar (not as polar as alcohols, because N is less polar than O)
SCH 4U REVIEW
FORCES
BOILING
POINT
SOLUBILITY
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dispersion forces and N-H bonds
medium (lower than alcohols of similar length, higher than hydrocarbons)
soluble in water
formation
alkyl halide + ammonia  amine + ammonium halide
REACTIONS
iodoethane
+ ammonia 

CH3 – CH2 – I + 2 NH3
1°, 2°, AND 3°
AMINES
aminoethane
+
ammonium iodide
CH3 – CH2 – NH2 +
NH4I
Primary Amine (1°)
 1 alkyl group attached to N
CH3 – N – H
|
H
methylamine
Secondary Amine (2°)
 2 alkyl groups attached to N
CH3 – N – CH3
|
H dimethylamine
Tertiary Amine (3°)
 3 alkyl groups attached to N
CH3 – N – CH3
|
CH3 trimethylamine
AMIDES
DEFINITION
NAMING
an organic compound characterized by the presence of a carbonyl functional
group (C=O) bonded to a nitrogen atom
alkyl group attached to double-bonded O is considered to be the substituent
group, attached to the parent “-amide”
O R΄΄ [H]
|| |
R[H] – C – N – R΄ [H]
GENERAL
FORMULA
EXAMPLE(S)
POLARITY
FORCES
BOILING
POINT
SOLUBILITY
REACTIONS
propanamide
O
||
CH3 – CH2 – C – NH2
slightly more polar than amine of similar length (extra O)
dispersion forces and N-H bonds
same as amines
soluble in water
formation
H2SO4 catalyst
carboxylic acid + amine
ethanoic acid
+ ammonia
amide + water

ethanamide
+
water
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O
H
||
|
CH3 – C – OH + H – N – H 
O
||
CH3 – C – NH2
+
HOH
EXAMPLE OF A STEPPED SYNTHESIS REACTION
Write a series of equations for a method of synthesis for N-ethylethanamide from an alkane and ammonia.
1.
ethane + chlorine  chloroethane + hydrogen chloride
CH3 – CH3 + Cl – Cl  CH3 – CH2 +
H – Cl
|
Cl
2.
chloroethane + water 
ethanol
CH3 – CH2
HOH  CH3 – CH2
|
|
Cl
OH
3.
+
+
hydrogen chloride
H – Cl
ethanol
+ strong oxidizer  ethanoic acid
CH3 – CH2 +
MnO4 CH3 – C – OH
|
||
OH
O
+ water + manganese dioxide
+ HOH +
MnO2
ammonia  aminoethane
+ ammonium chlorine
2 H – N – H  CH3 – CH2 – NH2 +
NH4Cl
|
H
4.
chloroethane +
CH3 – CH2 +
|
Cl
5.
ethanoic acid
+
aminoethane
CH3 – C – OH
||
O
+
CH3 – CH2 – NH2

N-ethylethanamide
O H
|| |
 CH3 – C – N – CH2 – CH3
+ water
+ HOH
COMMON NAMES
COMMON NAME IUPAC NAME
ethylene
ethene
propylene
propene
acetylene
ethyne
formaldehyde
methanal
acetaldehyde
ethanal
acetone
propanone
COMMON NAME
IUPAC NAME
formic acid
carboxyl group (COOH)
acetic acid
ethanoic acid
toluene / phenyl methane
methyl benzene
acetate
ethanoate
acetamide
ethanamide
FLOW CHART OF ORGANIC REACTIONS
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CHAPTER 2: POLYMERS
polymer – a molecule of large molar mass that consists of many repeating subunits called
monomers; two types: addition and condensation
monomer – a molecule or compound usually containing carbon and of relatively low molecular
weight and simple structure which is capable of conversion to polymers by
combination with itself or other similar molecules or compounds
dimer – a molecule made up of two monomers
ADDITION POLYMERS
addition polymer – a polymer formed when monomer units are linked through addition
reactions; all atoms present in the monomer are retained in the polymer
monomer


alkene
+
alkene

polymer
| |
C=C
| |
+
| |
C=C
|
|

| | | |
–C–C–C–C–
|
| | |
or
|
|
– C– C –
|
| n
less reactive than their monomers, because the unsaturated alkene monomers have been
transformed into saturated carbon skeletons of alkanes
forces of attraction are largely van der Waals attractions, which are individually weak,
allowing the polymer chains to slide along each other, rendering them flexible and
stretchable
CONDENSATION POLYMERS
condensation polymer – a polymer formed when monomer units are linked through
condensation reactions; a small molecule is formed as a byproduct
polyester – a polymer formed by condensation reactions resulting in ester linkages between
monomers
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dialcohol
+

dicarboxylic acid
O O
|| ||
C–C
| |
HO OH
HO – CH2 – CH2 – OH +

polyester (dimer)
+ water
O O
|| ||
– O – CH2 CH2 – O – C – C –
+ 2 H2O
n
polyamide – a polymer formed by condensation reactions resulting in amide linkages between
monomers; also known as a nylon
diamine
+
H
H
|
|
N – CH2 – N
|
|
H
H

dicarboxylic acid
O O
|| ||
C–C
| |
HO OH
+

polyamide (dimer)
+ water
H
H O O
|
| || ||
– N – CH2 – N – C – C –
+ 2 H2O
n
COMPOUNDS OF LIFE
protein – a large complex molecule made up of one or more chains of amino acids (an amino
group and carboxyl group attached to the same carbon atom)
– perform a wide variety of activities in the cell, including muscular growth, cellular
repair, and serve as building blocks for all body tissue
O
||
– NH – CH – C –
|
n
R
carbohydrate – a compound of carbon, hydrogen, and oxygen, with a general formula Cx(H2O)y
– a major source of food energy, including sugars, starches, and cellulose
– produced through photosynthesis in plants
– provides an equivalent amount of energy as an equal mass of fatty acids
fat – known chemically as a triglyceride (an ester of three fatty acids which are long-chain
carboxylic acids and one glycerol molecule)
– serves as a storage system, reserve supply of energy, and insulation
glycerol
H
|
H – C – OH
|
H – C – OH
|
H – C – OH
+
fatty acids
3
O
||
HO – C – R

fat (triglyceride)
O
||
H–C–O–C–R
|
H O
||
H – C – O – C – R’
|
H O
||
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H – C – O – C – R”
|
H
nucleic acid – hereditary information stored in all living cells from which the information can
be transferred; the chief types being DNA and RNA
– DNA is created from four different nucleotides (monomer consisting of a ribose
sugar, a phosphate group, and one of four possible nitrogenous bases)
– carries the genetic information that encodes proteins and enables cells to
reproduce and perform their functions
nitrogenous
base
phosphate
sugar
CHAPTER 3: ATOMIC THEORY
THE BOHR ATOMIC THEORY




electrons travel in the atom in circular orbits with quantized energy – energy is restricted to
only certain discrete quantities
there is a maximum number of electrons allowed in each orbit
electrons “jump” to a higher level when a photon (a quantum of light energy) is absorbed,
resulting in absorption spectrum (series of dark lines)
electrons “drop” to a lower level when a photon is emitted, resulting in bright-line spectrum
(series of bright lines)
ORBITS VS. ORBITALS
orbit – 2-D path; fixed distance from nucleus; circular or elliptical path; 2n2 electrons per orbit
orbital – 3-D region in space; variable distance from nucleus; no path and varied shape of
region; 2 electrons per orbital; predicted by Schrodinger’s equation
QUANTUM NUMBERS TO DESCRIBE ORBITALS
n – principal quantum number or energy level
l – secondary quantum number or subshell (s, p, d, or f)
m1 – magnetic quantum number (direction of the electron orbit)
m2 – spin quantum number (can only be +½ or –½ to describe spin of electron)
VALUE OF l
SUBLEVEL
SYMBOL
s
p
d
f
0
1
2
3
6p
32
e-
6s
5p
5d
NUMBER OF
ORBITALS
1
3
5
7
MAX. NUMBER
OF ELECTRONS
2
6
10
14
4f
PRESENT AT
n=
1-7
2-7
3-7
4-7
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18 e-
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4d
5s
4p
18 e-
3d
4s
3p
8 e-
3s
2p
8 e-
2s
2 e1s
CREATING ENERGY-LEVEL DIAGRAMS
energy-level diagram – interpretation of which orbital energy levels are occupied by electrons
for a particular atom or ion; also called an orbital diagram
m2 (spin) m1 (one orbital is distinguished from another at the same sublevel)
e.g., 9F
2p
2s
1s
l (secondary quantum number or sublevel)
n (primary quantum number or energy level)
RULES FOR ENERGY-LEVEL DIAGRAMS


Start adding electrons to the lowest energy level (1s) and build up from the bottom until the
limit on the number of electrons for the particle is reached – the aufbau principle
To obtain the correct order of orbitals for any atom, start at the hydrogen and move from left
to right across the periodic table, filling the orbitals; see below:
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



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For anions, add extra electrons to the number for the atom. For cations, do the neutral atom
first, and then subtract the required number of electrons from the orbitals with the highest
principle quantum number, n (i.e., you would remove the electrons from 4s, rather than 3d.)
Each orbital will hold a maximum of two electrons that spin in opposite directions – the
Pauli exclusion principle
Electrons must be distributed among orbitals of equal energy so that as many electrons
remain unpaired as possible – Hund’s rule
Half-filled and filled subshells are more stable than unfilled subshells as the overall energy
state of the atom is lower after the electron is promoted to a lower energy level:
Predicted
Actual
Cr:
[Ar]
Cr: [Ar]
Cu:
[Ar]
Cu: [Ar]
4s
3d
ELECTRON CONFIGURATION
4s
3d
electron configuration – a method for communicating the location and number of electrons in
electron energy levels
principal quantum number
e.g., O: 1s2 2s2 2p4
3p5
number of electrons
in orbital(s)
S2-: 1s2 2s2 2p6 3s2 3p6
orbital
shorthand electron configuration – when the electron configuration is written with the
preceding noble gas placed before the subshell information
(e.g., Cl: 1s2 2s2 2p6 3s2 3p5 becomes Cl: [Ne] 3s2 3p5)
isoelectronic – when two atoms/ions have the same electron configuration (e.g., Ne, F-, Na+)
ferromagnetism – exhibited by the metals iron, cobalt, nickel and a number of alloys that
become magnetized in a magnetic field and retain their magnetism when the
field is removed
paramagnetism – exhibited by materials like aluminum or platinum that become magnetized in
a magnetic field but it disappears when the field is removed (caused by
unpaired electrons)
QUANTUM MECHANIC THEORY
quantum mechanics – the current theory of atomic structure based on wave properties of
electrons; also known as wave mechanics
Heisenberg uncertainty princple – it is impossible to simultaneously know exact position and
speed of a particle
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electron probablity density – a mathematical or graphical representation of the chance of
finding an electron in a given space; see below for example of a
2s orbital:
CHAPTER 4: CHEMICAL BONDING
chemical bond – formed as the result of the simultaneous
attraction of two or more nuclei
TYPES OF BONDS
1. Ionic Bonding – when one atom has low ionization energy and low En, while the other has
high ionization energy and high En (i.e., metal and nonmetal)
– Δ En > 1.7
– the electrostatic attraction between positive and negative ions
2. Covalent Bonding – when both atoms have high ionization energy and high En (i.e., two
nonmetals)
– Δ En ≤ 1.7
– the sharing of valence electrons between atomic nuclei
3. Metallic Bonding – when both atoms have low ionization energy and low En (i.e., two
metals)
LEWIS THEORY OF BONDING





Atoms and ions are stable if they have a noble gas-like electron structure (i.e., a stable octet
of electrons).
Electrons are most stable when they are paired.
Atoms form chemical bonds to achieve a stable octet of electrons.
A stable octet may be achieved by an exchange of electrons between metal and nonmetal
atoms.
A stable octet may be achieved by the sharing of electrons between nonmetal atoms,
resulting in a covalent bond.
LEWIS STRUCTURES
Lewis structure – a symbolic depiction of the distribution of valence electrons in a molecule
1.
2.
Arrange atoms symmetrically around the central atom (usually listed first in the formula,
not usually oxygen and never hydrogen).
Count the number of valence electrons of all atoms. For polyatomic ions, add electrons
corresponding to the negative charge, and subtract electrons corresponding to the positive
charge on the ion.
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3.
4.
5.
6.
7.
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Determine the number of valence electrons all the atoms “want”, and subtract the number
of valence electrons it has (result from step 2). Divide this number by 2 to determine the
number of bonds the molecule will have.
Place a bonding pair of electrons between the central atom and each of the surrounding
atoms.
Complete the octets of the surrounding atoms using lone pairs of electrons. Remember
hydrogen (2 e-), beryllium (4 e-), and boron (6 e-) are the exceptions. Any remaining
electrons go on the central atom.
If the central atom does not have an octet, move lone pairs from the surrounding atoms to
form double or triple bonds until the central atom has a complete octet. Confirm the
number of bonds is correct by comparing it to the result in step 3.
Draw the Lewis structure and enclose polyatomic ions within square brackets showing the
ion chare.
e.g., SO3
has
wants
difference
O
S
6
8
O3
3 (6)
3 (8)
6 + 18
8 + 24
= 24 e= 32 e= 8 e= 4 bonds
S
O
O
VALENCE BOND THEORY







Covalent bonds form when atomic or hybrid orbitals with one electron overlap to share e-.
Bonding occurs with the highest energy (valence shell) electrons.
Normally, the s orbitals (sphere shape) and p orbitals (dumb-bell shape) overlap with each
other to form bonds between atoms.
sp3, sp2, and sp hybrid orbitals are formed from one s orbital and three, two, and one p
orbital, respectively, with orientations of tetrahedral (109.5°), trigonal planar (120°), and
linear (180°), respectively.
End-to-end overlap of orbitals or single covalent bonds is called a sigma (σ) bond.
Side-by-side overlap of unhybridized p orbitals is called a pi (Π) bond (weaker than σ bond).
Double bonds have one pi bond, while triple bonds have two pi bonds.
VSEPR THEORY
Valence-Shell Electron-Pair Repulsion Theory – pairs of electrons in the valence shell of an
atom stay as far apart as possible to minimize repulsion of their negative charges
1.
2.
3.
Draw the Lewis structure for the molecule, including the e- pairs around the central atom.
Count the total number of bonding pairs and lone pairs of electrons around the central atom.
Use the table below to predict the shape of the molecule.
# OF e- PAIRS
AROUND
CENTRAL
ATOM
2
3
ORIENTATION
OF
ELECTRON
PAIRS
linear
triangular
planar
NUMBER OF
BONDING
AND LONE
PAIRS
2 BP, 0 LP
3 BP, 0 LP
BOND
ANGLES
180°
120°
SHAPE
EXAMPLE
linear
trigonal
planar
BeF2
BF3
MOLECULAR
GEOMETRY
X–A–X
X
|
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A
X
tetrahedral
4 BP, 0 LP
109.5°
tetrahedral
X
X
|
A
CH4
X
X
X
4
tetrahedral
3 BP, 1 LP
< 109.5°
trigonal
pyramidal
NH3
PCl3
A
X
X
X
tetrahedral
2 BP, 2 LP
< 109.5°
v-shaped
or bent
H2O
OF2
tetrahedral
1 BP, 3 LP
180°
linear
HCl
5
trigonal
bipyramidal
5 BP, 0 LP
120° &
90°
trigonal
bipyramidal
PCl5
6
octahedral
6 BP, 0 LP
90°
octahedral
SF6
A
X
e.g., HOF(l)
.. ..
H : .O. : .F. :
X
A–X
X
X
|
X
A
| X
X
X
X
|
X
A
X | X
X
2 BP and 2 LP, therefore HOF(l) is v-shaped:
O
H
F
Note: – ions are treated in the same way, but square brackets are placed around the diagram
with the charge placed in the upper right hand corner
– double and triple bonds are treated as one group of electrons when using VSEPR
theory, and most of those molecules take the same shape as their Lewis structure
POLAR MOLECULES
bond dipole – the electronegativity difference of two bonded atoms represented by an arrow
pointing from the lower (∂+) to the higher (∂-) electronegativity
nonpolar molecule – a molecule that has either nonpolar bonds (≤0.5 Δ En) or polar bonds
whose bond dipoles cancel to zero (i.e., VSEPR diagram is symmetrical)
polar molecule – a molecule that has polar bonds with dipoles that do not cancel to zero
. . ∂-
e.g., NH3
N 3.0
∂+
INTERMOLECULAR FORCES
2.1 H
H 2.1
H 2.1 ∂+
therefore, it is a polar molecule
∂+
intermolecular force – the force of attraction and repulsion between molecules; much weaker
than covalent bonds
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dipole-dipole force – a force of attraction due to the simultaneous attraction of one dipole by its
surrounding dipoles
– the more polar the molecules, the
stronger the force
– the shape of the molecule also affect
the dipole-dipole strength (i.e., closer together = stronger force)
London dispersion force – the simultaneous attraction of an electron by the positive nuclei in
the surrounding molecules
– the greater the number of
electrons per molecule, the
stronger the force

As dipole-dipole force or London dispersion force
increases, the boiling point increases, allowing you to
predict relative boiling point, if one of the forces is the
same between the two substances, or if both forces are
moving in the same direction.
hydrogen bonding – the attraction of a hydrogen atom to a
lone pair of electrons in N, O, or F atoms in adjacent molecules (possible
because H has no valence e- to “shield” nucleus)
..
..
H – .F. : ------------ H – .F. :
STRUCTURE AND PROPERTIES OF CRYSTALLINE SOLIDS
CRYSTAL
PARTICLES
FORCE/
BOND
PROPERTIES
SOLUBILITY
MELTING CONDUCIN LIQUID/
POINT
TIVITY
SOLUTION
EXAMPLES
Ionic
medium
ions
(+ and -)
ionic
(higher as
ionic
charge
increases,
size of ion
decreases)
yes
yes
NaCl,
Na3PO4,
CaF2, MgO,
CuSO4•5H2O
yes
no
Pb, Fe,
Cu, Al, Ag,
Au
metallic
Metallic
cations
(fixed
nuclei with
mobile
delocalized
electrons,
allowing for
conduction
of heat and
electricity)
high
(higher as
number of
valence
electrons
increases)
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Polar Molecular
low
molecules
London,
dipoledipole,
hydrogen
Nonpolar Molecular
molecules
or single
atoms
(higher as
number of
electrons,
polarity of
molecules,
and
presence
of
hydrogen
bonds
increases)
yes
PF3, ICl,
CHCl3, H2O,
NH3, SO3
no
inert gases,
diatomic
elements,
CO2, CCl4,
SF6, BF3
no
diamond, SiC,
AlN, BeO,
SiO2, CuCl2,
Mg2S
no
covalent
Covalent Network
Crystal
atoms
(a 3-D
arrangement
of strong
covalent
bonds
between
atoms;
resulting in
hardness and
high melting
point)
very high
(melting
point
increases
as strength
of
covalent
bonds
increase )
no
NETWORK SOLIDS (AKA “SUPERMOLECULES”)
ALLOTROPES OF CARBON:
– a 3-D network solid
– each C atom is in the centre of a tetrahedron whose vertices are occupied by
other C atoms
– each C atom shares its valence e- with 4 other C atoms
– bonding e- are tightly bound and highly localized
a)
diamond
a)
graphite – a 2-D network solid
– each C atom is surrounded by a 3 others in a plane
– the double bond consists of delocalized electrons, therefore a good conductor
– separate layers are held together by dispersion forces and are easily separated
SILICA AND THE SILICATES:


silicon combines with oxygen to form silicon dioxide, SiO2 (silica)
it further reacts with metal compounds to produce metal silicates
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a)
b)
c)
24 of 39
quartz – a 3-D network solid
mica
– a 2-D network solid
asbestos – a 1-D network solid
CHAPTER 5: THERMOCHEMISTRY
thermochemistry – the study of the energy changes that accompany physical or chemical
changes of matter
thermal energy – energy available from a substance as a result of the motion of its molecules
chemical system – a set of reactants and products under study, usually represented by a chemical
equation
surroundings – all matter around the system that is capable of absorbing or releasing thermal
energy
heat – amount of energy transferred between substances
exothermic – releasing thermal energy as heat flows out of the system; negative molar enthalpy
endothermic – absorbing thermal energy as heat flows into the system; positive molar enthalpy
temperature – average kinetic energy of the particles in a sample of matter
open system – one in which both matter and energy can move in or out (e.g., burning
marshmallow)
isolated system – an ideal system in which neither matter nor energy can move in or out (e.g.,
ideal calorimeter)
closed system – one in which energy can move in or out, but not matter (e.g., realistic
calorimeter)
calorimetry – the process of measuring energy changes in a chemical system
enthalpy change (ΔH) – the difference in enthalpies of reactants and products during a change
q = quantity of heat (J) = m c ΔT
m
n M
MOLAR ENTHALPY
molar enthalpy (ΔHx) – the enthalpy change associated with a physical, chemical, or nuclear
change involving one mole of a substance; examples are below:
TYPE OF MOLAR ENTHALPY
solution (ΔHsol)
combustion (ΔHcomb)
vaporization (ΔHvap)
freezing (ΔHfr)
neutralization (ΔHneut)
formation (ΔHf)
ΔH = n ΔHvap or sol = mcΔT
EXAMPLE OF CHANGE
NaBr(s)  Na (aq) + Br-(aq)
CH4(g) + 2 O2(g)  CO2(g) + H2O(l)
CH3OH(l)  CH3OH(g)
H2O(l)  H2O(s)
2 NaOH(aq) + H2SO4(aq)  2 Na2SO4(aq) + 2 H2O(l)
C(s) + 2 H2(g) + ½ O2(g)  CH3OH(l)
+
ΔH = q
n
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ASSUMPTIONS USED IN CALORIMETRY:
1. No heat is transferred between the calorimeter and the outside environment.
2. Any heat absorbed or released by the calorimeter materials is usually negligible.
3. A dilute aqueous solution is assumed to have a density and specific heat capacity equal to
that of pure water (1.00 g/mL and 4.18 J/g•°C).
LAB: COMBUSTION OF ALCOHOLS





Mass of Fuel Burned:
Mass of Water Heated:
Mass of Water Vapourized:
Mass of Pop Can:
Temperature Change of Water:
1.41 g
97.00 g
0.18 g
16.07 g
22.2 °C

Heat Absorbed by Water:
q = m c Δt
= (97.00 g) (4.18 J / g · °C) (22.2 °C)
= 9.00 x 103 J

Heat Absorbed by Can:
q = m c Δt
= (16.07 g) (0.900 J / g · °C) (22.2 °C)
= 321 J

Heat Used to Vapourize Water: q = m · LV
= (0.18 g) (2268 J / g)
= 4.0 x 102 J

Total Heat Evolved by Fuel:
qtotal = (9.00 x 103 J) + (321 J) + (4.0 x 102 J)
= 9.72 x 103 J
= 9.72 kJ

Number of Moles of Fuel:
n=m
M
= 1.41 g
60.11 g / mol
= 0.0235 mol

Molar Heat of Combustion of Fuel:

% error: 100% –
ΔH = qtotal
n
= 9.72 kJ
0.0235 mol
= 414 kJ/mol
experimental enthalpy x 100%
actual enthalpy
= 100% –
414 kJ x 100%
490 kJ
= 15.5%, therefore unacceptable (>10%)
REPRESENTING ENTHALPY CHANGES
METHOD 1: THERMOCHEMICAL EQUATIONS WITH ENERGY TERMS
e.g., C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l) + 2802.7 kJ
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METHOD 2: THERMOCHEMICAL EQUATIONS WITH ΔH VALUES
e.g., C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
ΔH = -2802.7 kJ
METHOD 3: MOLAR ENTHALPIES OF REACTION
e.g., ΔHrespiration = -2802.7 kJ/mol glucose
METHOD 4: POTENTIAL ENERGY DIAGRAM
e.g.,
Ep (kJ)
C6H12O6(s) + 6 O2(g)
ΔH = -2802.7 kJ
6 CO2(g) + 6 H2O(l)
Reaction Progress
STANDARD ENTHALPY OF FORMATION
standard enthalpy of formation (ΔH°f) – the quantity of energy associated with the formation
of one mole of a substance from its elements in
standard state; zero for elements in standard state
HESS’S LAW (ADDITIVITY OF REACTION ENTHALPIES)
Hess’s law – the value of the ΔH for any reaction that can be written in steps equals the sum of
the values of ΔH for each of the individual steps (i.e., ΔHtarget = Σ ΔHknown)
e.g.
Determine the enthalpy change involved in the formation of two moles nitrogen monoxide from
its elements.
N2(g) + O2(g)  2 NO(g)
(1) ½ N2(g) + O2(g)  NO2(g)
(2) NO(g) + ½ O2(g)  NO2(g)
ΔH°1 = +34 kJ
ΔH°2 = -56 kJ
2 x (1): N2(g) + 2 O2(g)  2 NO2(g)
-2 x (2): 2 NO2(g)  2 NO(g) + O2(g)
ΔH°1 = 2(+34) kJ
ΔH°2 = -2(-56) kJ
N2(g) + O2(g)  2 NO(g)
ΔH° = +68 kJ + 112 kJ = +180 kJ
USING STANDARD ENTHALPIES OF FORMATION TO DETERMINE ΔH
ΔH =
Σ nΔH°f (products)
–
Σ nΔH°f (reactants)
e.g., MULTI-STEP CALCULATION
If 3.20 g of propane burns, what temperature change will be observed if all of the heat from combustion
transfers into 4.0 kg of water?
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
ΔH°f (CO2) = -393.5 kJ/mol
ΔH°f (H20) = -285.8 kJ/mol
ΔH°f (C3H8) = -104.7 kJ/mol
ΔH°f (O2) = 0.0 kJ/mol
mpropane = 3.20 g
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mH2O = 4.0 kg
cH2O = 4.18 J/(g•°C)
ΔH = Σ nΔH°f (products) – Σ nΔH°f (reactants)
= 3 mol x -393.5 kJ
1 mol
= -2219 kJ
+ 4 mol x -285.8 kJ
1 mol
– 1 mol x -104.7 kJ
1 mol
+ 5 mol x 0.0 kJ
1 mol
ΔHc (propane) = qwater
n ΔHc
= mcΔT
ΔT
= n ΔHc
mc
=
mpropane ΔHc
Mpropane mwater c
=
(3.20 g) (2219 kJ)
(44.11 g) (4.0 kg) (4.18 J/g•°C)
= 9.6°C
CHAPTER 6: CHEMICAL KINETICS
chemical kinetics – the area of chemistry that deals with rates of reaction
rate of reaction – the speed at which a chemical change occurs, generally expressed in
concentration per unit time, such as mol/(L•s)
rate = Δc
Δt
average rate of reaction – the speed at which a reaction proceeds over a period of time;
determined using slope of a secant (line between two points)
instantaneous rate of reaction – the speed at which a reaction is proceeding at a particular point
in time; determined using a tangent
METHODS TO MEASURE RATE: pH change; conductivity; volume of gas produced; change in
mass of products; change in colour
RATE LAW EQUATION
r = k [X]m [Y]n
k = rate constant; valid only for a specific temperature
[X] and [Y] = concentrations of reactants
m and n = order of reaction (describes the initial concentration dependence
of a particular reactant)
overall order of reaction – the sum of the exponents in the rate law equation
e.g., r = k[BrO3(aq)-]1 [HSO3(aq)-]2, therefore overall order is 3 (1 + 2)
zeroth-order reaction – the rate does not depend on [A]
e.g., if the initial concentration of A is doubled,
the rate will multiply by 1 (20), and so will be
unchanged
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first-order reaction – the rate is directly proportional to [A]
e.g., if the initial concentration of A is doubled,
the rate will multiply by 2 (21)
second-order reaction – the rate is proportional to the
square of [A]
e.g., if the initial concentration
of A is doubled, the rate will
multiply by 4 (22)
e.g.,
NO(g) + H2(g)  HNO2(g)
EXPERIMENT
NO (MOL/L)
H2 (MOL/L)
1
2
3
4
5
6
0.001
0.002
0.003
0.004
0.004
0.004
0.004
0.004
0.004
0.001
0.002
0.003
a) Write the rate law for the reaction.
r = k [NO(g)]2 [H2(g)]1
b) Write the overall order of the reaction.
3rd order
c) Calculate k for the reaction (use for experiment’s values).
r = 0.02 mol/L•s
r = k [NO(g)]2 [H2(g)]
[NO] = 0.001 mol/L
(0.02) = k (0.001)2 (0.004)
[H2] = 0.004 mol/L
k = 5 x 106 L2/(mol2•s)
INITIAL RATE OF
REACTION (MOL/(L•S)
0.002
0.008
0.018
0.008
0.016
0.024
The units for rate constant, k, are
related to overall order of reaction:
 first order overall, the units are 1/s
or s-1
 second order overall, the units are
L/(mol•s) or L/(mol-1•s-1)
 third order overall, the units are
L2/(mol2•s) or L/(mol-2•s-1)
COLLISION THEORY
CONCEPTS OF THE COLLISION THEORY:
 A chemical system consists of particles (atoms, ions, or molecules) that are in constant
random motion at various speeds. The average kinetic energy of the particles is proportional
to the temperature of the sample.
 A chemical reaction must involve collisions of particles with each other or the walls of the
container.
 An effective collision is one that has sufficient energy and correct orientation of the colliding
particles so that bonds can be broken and new bonds formed.
 Ineffective collisions involve particles that rebound from the collision unchanged.
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
29 of 39
The rate of a given reaction depends on the frequency of collision and the fraction of those
collisions that are effective.
activation energy – the minimum
increase in
potential energy
of a system
Ep
required for
molecules to react
activated complex
activation
energy
products
net potential energy
reactants
change, ΔH
activated complex – an unstable
chemical species
containing
Reaction Progress
partially broken
and partially
formed bonds representing the maximum potential energy point in the
change; also called the transition state
reaction mechanism – a series of elementary steps that makes up an overall reaction
elementary step – a step in a reaction mechanism that only involves one-, two-, or three-particle
collisions
rate-determining step – the slowest step in a reaction mechanism
reaction intermediates – molecules formed as short-lived products in reaction mechanisms
e.g.,
elementary step
rate-determining step
HBr(g) + O2(g)
 HOOBr(g)
HOOBr(g) + HBr(g)
 2 HOBr(g)
(fast)
2 HOBr(g) + HBr(g)
 H2O(g) + Br2(g)
(fast)
4 HBr(g) + O2(g)
 2 H2O(g) + 2 Br2(g)
(slow)
reaction
mechanism
reaction intermediate
FACTORS AFFECTING RATE OF REACTION
1. NATURE OF THE REACTANTS:
 each reactant contains a different number of bonds, each with differing bond strengths, that
must be broken for the reaction to proceed
 each reactant has a different threshold energy (minimum kinetic energy required to convert
kinetic energy to activation energy)
 each reactant requires a different collision geometry that can be simple or complex
2. TEMPERATURE :
 an increase in temperature increases the rate of reaction
 as temperature rises, the reactant particles gain kinetic energy, moving faster, colliding more
frequently, and thus reacting more quickly
 with a higher temperature, a larger fraction of the molecules will have the required kinetic
energy to have effective collisions
3. CONCENTRATION:
 an increase in reactant concentration increases the rate of reaction
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
30 of 39
the greater the concentration, the greater the number of particles per unit volume, which are
more likely to collide as they move randomly within a fixed space
4. SURFACE AREA:
 an increase in reactant surface area increases the rate of reaction
 reactants can collide only at the surface where the substances are in contact, and by
increasing the surface area, you are increasing the number of particles in an area, thereby
increasing the probability of an effective collision
5. CATALYST:
 a catalyst is a substance that increases the rate of a chemical reaction without itself being
permanently changed
 a catalyst provides an alternate “pathway”, with lower activation energy, to the same product
formation, meaning a much larger fraction of collisions are effective
 the catalyst can help break the bonds in the reactant particles, provide a surface for the
necessary collisions, and allow the reactants’ atoms to recombine in new ways
 catalysts are involved in the reaction mechanism at some point, but are regenerated before
the reaction is complete
CHAPTER 7: CHEMICAL SYSTEMS IN EQUILIBRIUM
equilibrium – the balanced state of a reversible reaction or process where there is no net
observable change; the rate of the forward reaction equals that of the reverse
reaction (A ↔ B)
– can be approached from either side of the reaction equation
– the concentration of the reactants and products do not change (are constant)
solubility equilibrium – an equilibrium between a solute and a solvent in a saturated solution
phase equilibrium – an equilibrium between different physical states of a pure substance (e.g.,
ice over a lake)
chemical reaction equilibrium – an equilibrium between reactants and products of a chemical
reaction
EQUILIBRIUM LAW EQUATION
For the general chemical reaction:
K = [C]c [D]d
[A]a [B]b
aA + bB ↔ cC + dD
where: • A, B, C, D are chemical entities in gas or aqueous phases (liquids and
solids are omitted from the equation)
• a, b, c, and d are the coefficients in the balanced chemical equation
• K is the equilibrium constant (temperature and pressure specific)
- if K is large, reaction’s concentration of products greater than reactants
- if K is small, reaction’s concentration of reactants greater than products
- K is inversely proportional to the K value of the reverse reaction
LE CHÂTELIER’S PRINCIPLE
Le Châtelier’s Principle – when a chemical system at equilibrium is disturbed by a change in a
property, the system adjusts in a way that opposes the change
equilibrium shift – movement of a system at equilibrium resulting in a change in the
concentrations of reactants and products
VARIABLES AFFECTING CHEMICAL EQUILIBRIA:
VARIABLE
TYPE OF CHANGE
RESPONSE OF SYSTEM
SCH 4U REVIEW
31 of 39
concentration
increase
decrease
increase
temperature
shifts to consume added reactant
shifts to replace removed reactant
shifts to consume added thermal energy
(away from heat term)
shifts to replace removed thermal energy
(towards heat term)
shifts towards side with larger total number
of gaseous entities
shifts towards side with smaller total
number of gaseous entities
shifts away from common ion to consume
the added reactant
decrease
volume
increase (decrease in
pressure)
decrease (increase in
pressure)
common ion effect dissolving a compound into
solution that adds a
common ion
VARIABLES THAT DO NOT AFFECT CHEMICAL EQUILIBRIA
catalysts
no effect
adding inert gases
no effect
SOLVING EQUILIBRIUM PROBLEMS
1.
2.
Write a balanced equation for the reaction and list the known values.
If the direction the system must go to attain equilibrium is not obvious (i.e., one entity is not
present initially), calculate Q with the initial concentrations and compare it to the value of K
to determine which direction the system will proceed to attain equilibrium.
3. Construct an ICE (Initial concentration, Change in concentration, Equilibrium
concentration) table and input the initial concentrations.
4. Let x represent the changes in concentration, multiplying it by the coefficient in the
balanced equation. The reactants should all change in the same way and all the products
should proceed in the opposite way.
5. Rewrite the E row using the x values.
6. Substitute equilibrium concentrations into the equilibrium constant equation.
7. Apply appropriate simplifying assumptions, if possible (e.g., 4x3 ÷ (0.4 – 2x)2 can be
simplified to 4x3 ÷ (0.4)2 because x value is so small in comparison).
8. Solve for x.
9. Justify any assumptions you have made (i.e., the x value you get should be plugged into the
original equation and the difference between the two must be less than 5%).
10. Calculate the equilibrium concentrations by substituting x into equilibrium concentration
expressions from the E row.
e.g., 4.00 mol of hydrogen and 2.00 mol of iodine are placed in a 2.00-L reaction vessel at 440°C and
react to form hydrogen iodide. At this temperature, the K is 49.7. Determine the concentrations of
all entities.
[H2(g)] = 4.00 mol
2.00 L
= 2.00 mol/L
H2(g) + I2(g) ↔ 2 HI(g)
K = 49.7
[I2(g)] = 2.00 mol
2.00 L
= 1.00 mol/L
[HI(g)] = 0.00 mol/L
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
H2(g)
2.00
–x
2.00 – x
+
I2(g)
1.00
–x
1.00 – x
↔
2 HI(g)
0.00
+2 x
2x
SCH 4U REVIEW
K =
49.7 =
2
[HI]
[H2] [I2]
(2x)2
(2.00 – x) (1.00 – x)
32 of 39
[H2(g)] = 2.00 mol/L - x
= 2.00 mol/L - (0.93 mol/L)
= 1.07 mol/L
0.92x2 – 3.00x + 2.00 = 0
[I2(g)] = 1.00 mol/L - x
= 1.00 mol/L - (0.93 mol/L)
= 0.07 mol/L
x = –b ± √ b2 – 4ac
2a
[HI(g)] = 2x
= 2(0.93 mol/L)
= 1.87 mol/L
4x = 49.7 (2.00 – x) (1.00 – x)
2
= 3.00 ± √ 9.00 – 7.36
1.84
= 2.33 or 0.93
2.33 is rejected, because concentrations cannot have a negative value (i.e., 2.00 – 2.33 = - 0.33)
SOLUBILITY PRODUCT CONSTANT
solubility – the concentration of a saturated solution of a solute in a particular solvent at a
particular temperature; specific maximum concentration
solubility product constant (Ksp) – the value obtained from the equilibrium law applied to
saturated solution (remember solids are not included in the
equation); omit units as with all K values
– can only be determined for ionic compounds that are
classified as insoluble or slightly soluble
1.
2.
3.
4.
Write a balanced equation and list the known values.
Use the solid product to write an equilibrium equation for it dissolving into ions.
Find the Ksp value for the solid product and write it next to the equilibrium equation.
Determine the number of moles of both ions, by using the mole ratios and initial
concentrations of reactants.
5. Determine the concentration upon mixing, by dividing the number of moles by the new
volume.
6. Plug these new concentrations into the Ksp equation to determine the Q (experimental value).
7. Compare the Q value to the Ksp to predict whether a precipitate will form (see below).
USING Q TO PREDICT SOLUBILITY
Ion product, Q > Ksp
precipitate will form (supersaturated solution)
Ion product, Q = Ksp
precipitate will not form (saturated solution)
Ion product, Q < Ksp
precipitate will not form (unsaturated solution)
e.g., 20.0 mL of 0.20 mol/L ammonium sulfate solution is added to 130 mL of 0.50 mol/L barium nitrate
solution. What are the concentrations of the ions and will a precipitate form?
(NH4)2SO4(aq) + Ba(NO3)2(aq)  BaSO4(s) + 2 NH4NO3(aq)
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
Ksp = 1.08 x 10-10
Choose the solid to write the
equilibrium equation (remember any
NO3- molecule is aqueous)
nBa2NO3 = c v
= (0.50 mol/L) (0.130 L)
= 0.065 mol
= nBa
nNH4NO3 = c v
= (0.20 mol/L) (0.020 L)
= 0.0040 mol
= nSO4
[Ba2+(aq)] = n
v
[SO42-(aq)] = n
v
Divide by the
“new” volume (i.e.,
20 mL and 130 mL
equals 150 mL)
SCH 4U REVIEW
33 of 39
= 0.065 mol
0.150 L
= 0.43 mol/L
Ksp = [Ba2+] [SO42-]
= (0.43) (0.027)
= 0.011
= 0.004 mol
0.150 L
= 0.027 mol/L
0.011 > 1.08 x 10-10, therefore it will precipitate
ENTROPY
spontaneous reaction – one that, given the necessary activation energy, proceeds without
continuous outside assistance
entropy, S – a measure of the randomness or disorder of a system or the surroundings
– equals 0 when the temperature is at absolute zero (0 K)
FACTORS THAT INCREASE ENTROPY (S)
 the volume of a gaseous system increases (i.e., pressure decreases)
 the temperature of a system increases
 the physical state of a system changes from solid to liquid to gas, or liquid to gas (i.e., Sgas >
Sliquid > Ssolid)
 fewer moles of reactant molecules form a greater number of moles of product molecules
 complex molecules are broken down into simpler subunits (e.g., combustion of organic fuels
into carbon dioxide and water)
CLASSIFICATION OF SPONTANEOUS AND NONSPONTANEOUS REACTIONS
Entropy increases (ΔS > 0)
Endothermic (ΔH > 0)
spontaneous at high temps.
nonspontaneous at low temps.
Exothermic (ΔH < 0)
spontaneous
C(s) + O2(g)  CO2(g)
H2O(s)  H2O(l)
Entropy decreases (ΔS < 0)
spontaneous at low temps.
nonspontaneous at high temps.
nonspontaneous
3 O2(g)  2 O3(g)
2 SO2(g) + O2(g)  2 SO3(g)
Gibb’s free energy, G – energy that is available to do useful work; ΔG° = ΔH° – (T ΔS°)
CHAPTER 8: ACID-BASE EQUILIBRIUM
BRØNSTED-LOWRY THEORY
Brønsted-Lowry acid – proton donor
Brønsted-Lowry base – proton acceptor
amphoteric (amphiprotic) – a substance capable of acting as an acid or a base in different
chemical reactions;conjugate
a substance
pair that may accept or donate a proton
e.g.,
acid
HC2H3O2(aq) + H2O(l) ↔ C2H3O2-(aq) + H3O+(aq)
base
strong acid – an acid with a very weak attraction for protons and easily donates it to a base
strong base – a base with a very strong attraction for protons

The stronger an acid, the weaker its conjugate base, and vice versa.
AUTOIONIZATION OF WATER
SCH 4U REVIEW
34 of 39
autoionization of water – the reaction between two water molecules producing a hydronium ion
and a hydroxide ion (H2O(l) ↔ H+(aq) + OH-(aq)
Kw = [H+(aq)] [OH-(aq)]
= (1.0 x 10-7 mol/L) (1.0 x 10-7 mol/L)
= 1.0 x 10-14
[H+(aq)] =
Kw
[OH-(aq)]
In neutral solutions
In acidic solutions
In basic solutions
[OH-(aq)] =
Kw
[H+(aq)]
[H+(aq)] = [OH-(aq)]
[H+(aq)] > [OH-(aq)]
[H+(aq)] < [OH-(aq)]
PH
pH – the negative of the logarithm to the base ten of the concentration of hydrogen (hydronium)
ions in a solution
pOH – the negative of the logarithm to the base ten of the concentration of hydroxide ions in a
solution
pH = –log [H+(aq)]
[H+(aq)] = 10–pH
pOH = –log [OH-(aq)]
[OH-(aq)] = 10–pOH
pH + pOH = 14.00
e.g., Calculate the pH of a solution prepared by dissolving 4.3 g of Ba(OH)2(s) in water to form 1.5 L of
solution.
Ba(OH)2(aq)
100%
Ba2+(aq) + 2 OH–(aq)
nBa(OH)2 = 4.3 g x 1 mol
171.3 g
= 2.5 x 10-2 mol
[Ba(OH)2(aq)] = 2.5 x 10-2 mol
1.5 L
= 1.7 x 10-2 mol/L
[OH-(aq)] = 2 (1.7 x 10-2 mol/L)
= 3.3 x 10-2 mol/L
pOH = –log [OH-(aq)]
= –log (3.3 x 10-2)
= 1.47
pH = 14.00 – pOH
= 14.00 – 1.47
= 12.53
STRONG VS. WEAK ACIDS AND BASES
strong acid – an acid that is assumed to ionize quantitatively (completely) in aqueous solution
(i.e., percent ionization is > 99%); HCl(aq), HNO3(aq), and H2SO4(aq) are the only
strong acids we will work with
strong base – an ionic substance that dissociates completely in water to release hydroxide ions;
all of the metal hydroxides are strong bases
weak acid – an acid that partially ionizes in solution but exists primarily in the form of
molecules
weak base – a base that has a weak attraction for protons
percent ionization (p) = concentration of acid ionized x 100%
concentration of acid solute
[H+(aq)] = p x [HA(aq)]
100
SCH 4U REVIEW
35 of 39
acid ionization constant (Ka) – equilibrium constant for the ionization of an acid
e.g., Calculate the Ka and pH of hydrofluoric acid if a 0.100 mol/L solution at equilibrium at SATP has a
percent ionization of 7.8%.
HF(aq)
7.8% H+
(aq) + F (aq)
Ka = [H+(aq)] [F-(aq)]
[HF(aq)]
[H+(aq)] = (p/100) [HA(aq)]
= (7.8 / 100) (0.100 mol/L)
= 0.0078 mol/L
Initial concentration (mol/L)
Change in concentration (mol/L)
Equilibrium concentration (mol/L)
Ka = [H+(aq)] [F-(aq)]
[HF(aq)]
= 0.00782
0.0922
= 6.6 x 10-4
HF(aq) ↔
0.100
–x
0.100 – 0.0078
= 0.0922
H+(aq)
0.000
+x
0.0078
+
F-(aq)
0.000
+x
0.0078
pH = –log [H+(aq)]
= –log (0.0078)
= 2.1
CHAPTER 9: ELECTROCHEMISTRY (ELECTRIC CELLS)
OXIDATION NUMBER
oxidation number – an integer that is assigned to each atom in a compound when considering
redox reactions
– a positive or negative number corresponding to the apparent charge that an
atom in a molecule or ion would have if the electron pairs in covalent bonds
belonged entirely to the more electronegative atom
RULES FOR ASSIGNING OXIDATION NUMBERS
1. The oxidation number of an atom in an uncombined element is always 0 (e.g., H2 is 0).
2. The oxidation number of a simple ion is the charge of ion (e.g., Ca2+ is +2).
3. The oxidation number of hydrogen is +1, except in metal hydrides when it is -1 (e.g., the H
in NaH is -1).
4. The oxidation number of oxygen is -2, except in peroxides when it is -1 (e.g., the O in H2O2
is -1).
5. The oxidation number of Group 1 element ions is +1. The oxidation number of Group 2
element ions is +2.
6. The sum of oxidation numbers in a compound must equal 0.
7. The sum of oxidation numbers in a polyatomic ion must equal the charge on the ion (e.g.,
OH- is -1).
OXIDATION-REDUCTION
Oxidation can be defined as a reaction in which:
1) an element is chemically united with oxygen (e.g., C + O2  CO2; carbon is oxidized)
2) a metal is changed from an uncombined to a combined state (e.g., Zn + Cl2  ZnCl2)
3) an element loses electrons, and therefore has an increase in oxidation number
(e.g., Ca  Ca2+ + 2e-)
SCH 4U REVIEW
36 of 39
Loss of Electrons is Oxidation
“LEO”
Reduction can be defined as a reaction in which:
1) an element loses oxygen (e.g., Fe2O3  2 FeO + ½ O2)
2) a metal is changed from a combined to a uncombined state (e.g., FeO  Fe + ½ O2)
3) an element gains electrons, and therefore has a decrease in oxidation number
(e.g., Cl2 + 2e-  2 Cl-)
Gain of Electrons is Reduction “GER”
redox reaction – a chemical reaction in which electrons are transferred between particles; two or
more atoms undergo a change in oxidation number; also known as oxidationreduction reactions
– all single displacement reactions are redox, while some combination and
decomposition reactions are; double displacement reactions are never redox
e.g.,
oxidation
+1 -2
H2S(g)
0
+
O2(g)
+4 -2

SO2(g)
+1 -2
+
H2O(g)
reduction
BALANCING REDOX EQUATIONS USING OXIDATION NUMBERS

this method is most appropriate when dealing with covalent compounds
1. Assign oxidation numbers to all the atoms in the equation.
2. Identify which atoms undergo a change in oxidation number.
3. Determine the ratio in which these atoms must react so that the total increase in oxidation
number equals the decrease (i.e., the total number of electrons lost and gained is equal).
4. Balance the redox participants in the equation using this ratio.
5. Balance the other atoms by the inspection method.
6. Add H+(aq) or OH-(aq) to balance the charge, depending on if it is an acidic or a basic solution
(the total charge on each side must be the same).
7. Add H2O(l) to balance the O atoms.
e.g.,
Ag(s) + Cr2O72-(aq)  Ag+(aq) + Cr3+(aq)
oxidation: lost 1 e- (x 6)
0
6 Ag(s)
+
+6 -2
Cr2O72-(aq)
+ 14 H+(aq) 
+1
6 Ag+(aq)
+
+3
2 Cr3+(aq) + H2O(l)
reduction: gained 2(3 e-) = 6 e- (x 1)
BALANCING REDOX EQUATIONS USING HALF-REACTIONS

this method is most appropriate for ionic reactions in solution and for relating to electrical
processes
1. Separate the skeleton equation into the start of two half-reaction equations (one for the
oxidation reaction and one for the reduction reaction).
2. Balance all species, other than O and H.
3. Balance the oxygen, by adding H2O(l) for acidic solutions or OH-(aq) for basic solutions.
4. Balance the hydrogen, by adding H+(aq) for acidic solutions or H2O(l) for basic solutions.
SCH 4U REVIEW
37 of 39
5. Balance the charge on each side by adding electrons and canceling anything that is in equal
amounts on both sides.
6. Multiply each half-reaction equation by simple whole numbers to balance the electrons lost
and gained.
7. Add the two half-reaction equations, canceling the electrons and anything else that appears
in equal amounts on both sides.
8. Check to ensure all entities and the overall charge on both sides balance.
MnO4– + N2H4  MnO2 + N2
e.g.,
4 [MnO4– + 2 H2O + 3 e-  MnO2
3 [N2H4 + 4 OH–  N2 + 4 H2O
+ 4 OH–]
+ 4 e-]
4 MnO4– + 8 H2O + 12 e-  4 MnO2 + 16 OH–]
3 N2H4 + 12 OH–  3 N2
+ 12 H2O
+ 12 e-]
4 MnO4– + 3 N2H4  4 MnO2 + 3 N2 + 4 H2O + 4 OH–
TECHNOLOGY OF CELLS AND BATTERIES
electric cell – a device that continuously converts chemical energy into electrical energy;
contains two electrodes (solid conductor) and one electrolyte (aqueous conductor);
each electrode has a cathode (+) and anode (–); electrons flow from the anode to
the cathode
battery – a group of two or more electric cells connected in series
voltage – the potential energy difference per unit charge; measured in volts (V), 1 J/C
electric current – the rate of flow of charge past a point; measured in amperes (A), 1 C/s
TYPE
primary cell
– electric cell
that cannot
be recharged
secondary
cell – electric
cell that can
be recharged
fuel cell –
electric cell
that produces
NAME OF
CELL
dry cell
(1.5 V)
2 MnO2 + 2 NH4+ + 2 e-  Mn2O3 + 2 NH3 + H2O
Zn  Zn2+ + 2 e-
alkaline dry
cell (1.5 V)
2 MnO2 + H2O + 2 e-  Mn2O3 + 2 OHZn + 2 OH-  ZnO + H2O + 2 e-
mercury cell
(1.35 V)
HgO + H2O + 2 e-  Hg + 2 OHZn + 2 OH-  ZnO + H2O + 2 e-
Ni-Cad cell
(1.25 V)
2 NiO(OH) + 2 H2O + 2 e-  2 Ni(OH)2 + 2 OHCd + 2 OH-  Cd(OH)2 + 2 e-
lead-acid cell
(2.0 V)
PbO2 + 4 H+ + SO42- + 2 e-  PbSO4 + 2 H2O
Pb + SO42-  PbSO4 + 2 e-
aluminum-air
cell (2 V)
3 O2 + 6 H2O + 12 e-  12 OH4 Al  4 Al3+ + 12 e-
HALF-REACTIONS
CHARACTERISTICS
AND USES
 inexpensive, portable
 flashlights, radios
 longer shelf life,
higher currents for
longer periods
 same uses as dry cell
 small cell, constant
voltage during life
 hearing aids,
watches
 completely sealed,
lightweight
 power tools, shavers,
portable computers
 large currents,
reliable for recharges
 all vehicles
 high energy density,
readily available
aluminum alloys
 electric cars
SCH 4U REVIEW
electricity by
a continually
supplied fuel
38 of 39
hydrogenoxygen cell
(1.2 V)
O2 + 2 H2O + 4 e-  4 OH2 H2 + 4 OH-  4 H2O + 4 e-
 lightweight, high
efficiency, can be
adapted to use
hydrogen-rich fuels
 vehicles and space
shuttle
GALVANIC CELLS
galvanic cell – consists of two half-cells
separated by a porous
boundary with solid
electrodes connected by
an external circuit;
standard cells occur at
SATP and with
concentrations of 1.0
mol/L
cathode – positive electrode; reduction
(gaining electrons because
electronegativity is higher) of
the strongest oxidizing agent
occurs here; the half-cell that
is higher on the “Relative
Strengths of Oxidizing and
Reducing Agents” table; “red
cat on the roof”  reduction
at the cathode, higher half-cell
anode – negative electrode; oxidation (losing electrons because electronegativity is lower) of the
strongest reducing agent occurs here; the half-cell that is lower on the table
inert electrode – a solid conductor that will not react with any substance present in a cell
(usually carbon or platinum)


Electrons travel in the external circuit from the anode to the cathode
Internally, anions from the salt bridge move toward the anode and cations from the salt
bridge move toward the cathode as the cell operates, keeping the solution electrically neutral
CELL NOTATION
electrons
cathode (+) | electrolyte || electrolyte | anode (-)
(reduction)
(oxidation)
CELL POTENTIALS
standard cell potential (ΔE°) – the maximum electric potential difference (voltage) of a cell
operating under standard conditions
reference half-cell – a half-cell assigned an electrode potential of exactly 0.00 volts; the
standard hydrogen half-cell, Pt(s) | H2(g), H+(aq)
standard reduction potential (ΔEr°) – represents the tendency of a standard half-cell to attract
electrons in a reduction half-reaction, compared to the
reference half-cell
SCH 4U REVIEW
39 of 39
standard oxidation potential
ΔE° =
cell

(ΔEo°)
ΔEr°
– ΔEr°
cathode
anode
– represents the tendency of a standard half-cell to lose
electrons in an oxidation half-reaction; the value of the
reverse reaction with an opposite sign
A positive standard cell potential (ΔE° > 0) indicates that the overall cell reaction is
spontaneous.
e.g.,
SOA
+
Ag(s) | Ag
(aq)
OA
|| Cu2+(aq) | Cu(s)
RA
SRA
2 [Ag+(aq) + e-  Ag(s) ]
Cu(s)  Cu2+(aq) + 2 eCu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
ΔE° =
cell
ΔE =
=
=
°
ΔEr° –
ΔEr°
cathode
anode
ΔE Ag+ – ΔE°Cu2+
0.80
–
0.34
0.46 V
°