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Gr.12 chemistry - Exam Review
Chemistry (High School - Canada)
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Chemistry - Reference page
Partial Charges:
• 𝒮+ (partial postive)
• if something is a partial positive it’s electronegativity will be less than the other
elements electronegativity
• if the molecule has more than two elements you look at the individual bonds
• 𝒮- (partial negative)
• if something is a partial negative it’s electronegativity will be higher than the other
elements
Partial Charges:
• Bonds
• to know if a molecule has polar or non-polar bonds you look at the difference in
electronegativity
• if the difference is above 0.3 its polar bonds
• if the difference is below 0.3 its non-polar bonds
• Molecules
• to know if a molecule is polar or non-polar, you can look at its symmetry
• if its symmetrical the molecule is non-polar
• if its asymmetrical the molecule is polar
Forces:
• London
• if the molecule is non-polar you will only have london forces
• Dipole Dipole
• if the molecule is polar you will have dipole dipole forces (only if you don’t have
hydrogen bonds)
Hydrogen
Bonds
•
• if you only have a H with a 𝒮+and a N, O, or F with a 𝒮- you will have a hydrogen
bond
Density:
Example;
d = Mass
———
Volume
dwater = 1.00g/mL
( = 1.00g/mL³)
( = 1mL = 1 g/cm³)
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Chemistry
History of Atomic Theories: (https://www.youtube.com/watch?v=-4Us5PTb4j8)
Scientist Name
Dates
Theory of Model
Diagram of Model
- proposed that all matter was
Dalton
1803
composed of atoms, indivisible and
indestructible building blocks
- he was the first to discover a sub
atomic particle, the electron
Thompson
1897
- proposed the plum pudding model of
the atom in 1904
- in Thompson’s model the atom is
composed of electrons
- gold foil experiment in which he
Rutherford
1911
demonstrated that the atom ha a tiny
and heavy nucleus
- depicts the atom as a small, positively
Bohr
1913
charged nucleus surrounded by
electrons that travel in circular orbits
around the nucleus
Ions:
ion - an atom that loses or gains electrons (to form a full valence shell) and results in a
charge. It is a charged particle.
• Cations
• lose electrons
• become more positive
• Anions
• gains electrons
• becomes more negative
orbitals - 3D space around a nucleus where there is
a probability of finding electrons.
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Bohr Rutherford Diagram:
Line Spectra - Bohr’s Evidence
for Electron Energy Levels
Energy Shells & Sub-Shells:
Principal Energy Level (shell)
Bohr’s First Postulate:
• Main energy levels for electrons
• electrons do not radiate energy as they orbit the
• Corresponds to the period on the periodic
table
nucleus. each orbital corresponds to a state of
constant energy (stationary state)
• example; period 1 is principal energy level 1
Bohr’s Second Postulate:
Electron Energy Sub-Levels (sub-shells)
• electrons can change their energy only by
• orbital - 3D space mourned a nucleus
undergoing a transition from one stationary state
where there is a probability of finding
to another.
electrons
Summary:
Orbital Principal
(SubEnergy
Shell)
Level
• electrons absorb and emit very specific amounts
Shape
# of
orbitals
Max # of
electrons
S
1+
spherical
1
2
P
2+
dumb-bell like
3
5
D
3+
shape can vary
5
10
F
4+
shape can vary
7
14
of energy.
photon - a quantum of light energy
quantum - a small, discrete (specific), indivisible
quantity.
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Energy Level Diagrams & Electron Configurations:
- electrons in atoms have different energies. Each (neutral) type
of atom has a unique number of electrons. However, many
elements have similar properties. scientists has studied atomic
structure (electron configuration or arrangement) in order to
understand properties.
3 Rules for Energy Level Diagrams:
• Aufbau Principal (means “building up”)
• each electron is added to the lowest energy level available.
Hund’s
Rule
•
• for orbitals at the same energy level, one electron occupies each sub orbital before
the electrons pair up.
• Pauli Exclusion Principal
• only two electrons (with opposite spins) can occupy each orbital.
Electron Configuration For Ions:
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Lewis Bonding Theory:
• atoms/ions are stable if they have an octet of electrons
• electrons are most stable in pairs
• atoms form chemical bonds (ionic or covalent) to become stable
Linking Lewis Theory to Quantum Mechanics:
• the “octet” of electrons in Lewis Theory comes from the maximum of 2 electrons in
the S orbital and 6 in the P orbital.
Explaining Ionic Charges:
• Zinc forms a 2+ ion. The electron configuration for this is: [Ar]4s², 3d¹⁰ this shows that
there are 12 valence electrons. If another atom were to take the 4s electrons this would
leave zinc with a filled 3d orbital - a relatively stable state.
Electron Configuration for Ions:
Rules for Drawing Lewis Structures:
STEP 1: arrange the atoms symmetrically around the central atoms
- usually first atom
- usually not oxygen
- never hydrogen
STEP 2: count the numbers of total valence electrons (for polyatomic ions, add or substance as necessary)
STEP 3: place one bonded pair of electrons (straight line) between the central atom and each surrounding
atom
STEP 4: complete the octet of each surrounding atom using lone pairs of electrons (any remaining electron
pairs go on central atom)
STEP 5: if central atom doesn’t have an octet, move lone pairs of electrons from a surrounding atom to form
double/triple bonds
STEP 6: draw the final lewis structure. (polyatomic ion: enclose in square brackets and write ion charges
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Exceptions to the Octet Rule:
BeCl2
(*NOT ionic*)
BCl3
(*some elements have too few electrons to form an octet (H, Be, B)*)
SF6
(*some elements have more than an octet*)
PCl5
(*◣ direction is back into the page*)
(*\\\ direction is out of the page (toward you)*)
(add 3-D concept to 2-D drawing)
XeF6
(*some noble gases can form compounds*)
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The VSEPR Theory:
• VSEPR (Valence Shell Electron Pair Repulsion)
• valence electrons stay as far away as possible to minimize repulsion
• when looking at a molecule we look specifically at the central atom (the one that has the
most bonding electrons) to determine the 3-D geometry
Rules of VSEPR:
•
•
•
•
•
only the valence shell electrons of the central atom(s) are important to molecular shape
valence shell electrons are paired or will be paired in a molecule or polyatomic ion
bonded pairs of electrons and lone pairs of electrons are treated approximately equally
valence shell electron pairs repel each other electrostatically
the molecular shape is determined by the positions of the electron pairs when they are a
maximum distance apart (with the lowest repulsion possible)
Representing 3-D Structures:
• the solid wedge represents atom/pairs of electrons coming TOWARDS the viewer
• the dashed wedge represents atom/pairs of electrons going AWAY from the viewer
How to Use VSEPR:
1. draw lewis structures of the molecule, including the electron pairs around the central
atom
2. count the total number of bonding pairs and lone pairs around the central atom
3. use the chart provided to determine the
geometry (shape name) and draw a shape
diagram
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Polarity:
Polar Bonds:
• Polar bonds are created if there is an equal attraction of an atom for electrons
(electronegativity).
• Use periodic table to look up electronegativity values for each element in a bond,
then determine the difference.
• < 0.45
→ non-polar bond (covalent)
• 0.45-1.7 → polar bond (covalent)
• 1.7+
→ ionic bond (metal and nonmetal - involve electron transfer, not sharing)
Examples;
HBr
BH3
▲ENEG = 3.0 - 2.2 = 0.8
∴ there is polar bonds
the molecule is polar as well
▲ENEG = 2.0 - 2.2 = 0.2
∴ there is non- polar bonds
the molecule is non-polar as well
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Polarity:
Polar Molecules:
• Bond Dipole - the electronegativity difference of two bonded atoms represented by an
arrow pointing from the lower (𝒮-) to the higher electronegativity (𝒮+)
• Polar Molecule - a molecule that has polar bonds who dipoles that do not cancel
to zero
• overall a polar molecule has a pastiche end and a negative end
• if the molecule is asymmetrical (for example has a lone pair of electrons) it
will be polar
Non-Polar Molecules:
- a molecule that has either non-pear bonds, or polar bonds whose bond dipoles cancel
to zero
- non-polar molecules that have polar bonds are completely symmetrical molecules
Theoretical Prediction of Molecular Polarity:
STEP 1: draw a lewis structure for the molecule
STEP 2: use the number of electron pairs and VSEPR rules to determine the shape
around each central atom
STEP 3: use electronegativity differences to determine the polarity of each bond
STEP 4: determine if there is an overall dipole for the molecule (example; is it
symmetrical or asymmetrical?)
STEP 5: classify as polar or non-polar
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Intermolecular Forces:
• intermolecular forces are forces of attraction (and repulsion) between molecules
• they involve an electrostatic force (positive and negative charges)
• overall they are weak compared to intramolecular forces (ionic and covalent bonds). if
we say that a covalent bond has a strength of 100 then intermolecular forces would
range from 0.001 to 15.
London Dispersion Forces:
• Simultaneous attraction of electrons in one molecule for the nuclei in a different
molecule
• all types of molecules exhibit london forces
• they are the weakest of all the intermolecular forces
• they are very weak because electrons are always moving so the attraction comes and
goes as they move around.
Dipole Dipole Forces:
• exist between polar molecules. (a polar molecule is created when there is enough of a
difference in electronegativity between atoms in a molecule).
• occurs when the dipole of one molecule is attracted to the dipole in a different molecule
• are stronger than london forces
• an ion-dipole exists between a dipole in a molecule and an ion
Hydrogen Bonds:
• exists when a strong enough dipole is created in molecules (hydrogen atoms are bonded
to highly electronegative atoms like N, O, or F)
• the hydrogen (+ charge) on one molecule is attracted to a pair of lone electrons
(- charge) on an adjacent molecule
• are the strongest of all the intermolecular forces
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Types of Solids:
• Ionic Solid - a 3-D lattice of cations and anions, held together by ionic
bonds (strong electrostatic forces), with an overall neutral charge.
- particles involved:
- ions (cations & anions)
- forces involved:
- ionic (both within & between)
- properties:
- hard
- strong force (crystal lattice)
- brittle
- once initially broken ions easily repel each other
- high melting point
- strong force, lots of energy needs to break the bonds
- soluble in water
- ions form forces of attraction with water (polar)
- conduct electricity
- can easily transfer electrons
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• Metallic Solid - a lot us of nuclei surrounded by a “sea of electrons”.
Held together by metallic bonds.
- particles involved:
- cations (with a sea of electrons)
- forces involved:
- metallic (both within & between)
- properties:
- soft - very hard
- flexible, but strong bonds (sea of electrons)
- conduct electricity
- solids conduct
{valence electrons can freely move around}
- liquids conducts
- ductile/malleable
- non-directional bonds mean planes of atoms can slide over each other
and remain bonded
- lustrous
- valence electrons absorb and emit energy from light
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• Molecular Solid - separate molecules attracted by week into
molecular forces called van der Waals forces. If molecules are polar,
they are attracted by stronger intermolecular forces, called dipoledipole forces.
- particles involved:
- molecule
- forces involved:
- intramolecular forces (covalent bonds
- intermolecular forces (london, dipole-dipole, & hydrogen bonds)
- properties:
- soft
- weak intermolecular forces
- low melting point
- not much energy required to break these forces
- some soluble in water
- can form adequate electrostatic forces (which is whats needed to dissolve
in water
- non conducting electricity
- individual particles are neutral (no ions)
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Covalent Network Solid - large molecules held together by continuous
covalent bonds.
- particles involved:
- atoms (neautral)
- forces involved:
- covalent (both within & between)
- properties:
- very hard
- strong force between all atoms (crystal structure)
- very high melting point
- strong force between all atoms (crystal structure)
- not soluble in water
- there bonds are too strong to break
- non conduct electricity
- neutral particles
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Organic Chemistry
Organic Chemistry:
organic chemistry - is the study of structure, properties, and reactions of organic
molecules (contains carbon and hydrogens).
• it is important because:
• they make up living things
• medical applications
• huge variety of compounds
• different structures
• different properties
• different uses
• carbon features
• can form 4 covalent bonds
• can form single, double and triple bonds
IUPAC Naming:
• international union of pure and applied chemistry
• specific naming system
Chemical Formula:
• C5H10O2
• CH3(CH2)3COOH
Structural Formula:
or
Organic Family:
organic family - a group of organic compounds with common structural features that
import certain physical and chemical properties
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Functionals Groups:
functional group - a structural arrangement of atoms
hydrocarbons - only hydrogens and carbons
example; methane, ethene, ethyne
Alcohols
Aldehydes
Ketones
Structure DICTATES Function
Carboxylic Acid
Esters
Aromatic
Ethers
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Naming Hydrocarbons:
hydrocarbon - compound that only contains hydrogen and carbon
Alkane - hydrocarbon with only single carbon bonds
Alkene - hydrocarbon with at least one double carbon bond
Alkyne - hydrocarbon with at least one triple carbon bond
Cyclic Hydrocarbon - closed ring structure
Aromatic Compound - structure based on benzene
Properties of Benzene Uses of Benzene
Colorless
manufacturing of:
Flammable
• plastics
toxic
• dyes
melting point: 5.5℃
• rubber
boiling point: 80.1℃
• medications
CH4
C2H6
C4H10
C7H16
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>>
>>
>>
>>
methane (natural gas)
ethane
butane (lighters)
heptane
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Hydrocarbons can be straight chained or branched:
General prefixes used in IUPAC Naming:
meth - 1
hex - 6
eth - 2
hept - 7
prop - 3
oct - 8
but - 4
non - 9
pent - 5
dec - 10
4. use the prefix that corresponds to the number of carbon atoms in the chain
5. use the suffix “ane”
example;
Example - Structural Diagrams of Hexane:
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Rules for Branched Alkanes:
• branches alkanes have alkyl groups
• these groups are named using the organic prefixes, with the suffix “yl”
•
•
•
•
•
CH3
CH2 - CH3
CH2 - CH2 - CH3
C7H15
C9H19
>> methyl
>> ethyl
>> propyl
>> heptyl
>> nonyl
Rules
Examples
1. identify the longest carbon chain. (this may
travel through one or more branches - need to
check all possibilities).
Heptane (chain)
2. number the carbons starting with the end
closest to the branches
3. name each branch (alkyl group) and identify its
location on the parent chain (always wants the
lowest numbering)
• ethyl off carbon 5
• methyl off carbon 3
4. write the name according to the general format:
(number of location) - (branch name)(parent chain)
* if more than one branch is present, list them in
alphabetical order according to ally group
* is the same alkyl group is present more than
once use di, tri, tetra etc. before the alkyl name
to indicate how many there are.
* list each number location on the parent chain at
the start of naming.
5 - ethyl - 3 - methylheptane
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Naming Alkenes & Alkynes:
* the rules are similar to alkanes, but use the endings “ene” for double bonds and “yne” for triple bonds
Rules
Examples
1. find the longest parent chain that includes the
double or triple bond
2. number the carbon atoms starting at the end
closest to the double or triple bond
3. the location of the multiple bond is indicated by
the number of the C atom that starts the bond
• C = C off C1
4. identify any branches and name accordingly
• methyl off carbon 3
5. write the full name following proper conversions
* hyphens are used between numbers and words
3 - methyl - 1 - butene
Aromatic Hydrocarbons:
* benzene ring is the parent molecule
* when a branch is not easily named, benzene can become the branch. it is then called phenyl
Name of Molecule
Structure
methyl benzene
1,2 - diethyl benzene
1,3 - diethyl benzene
1,4 - diethyl benzene
2 - phenylbutane
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Reactions of Hydrocarbons:
• Combustion
fuel hydrocarbon + oxygen gas → carbon dioxide + water
C6H14(l) + 19O2 → 12CO2 + 14H2O
• Substitution
• reaction with a halogen:
→
+
+
Hydrogen Bromide
ethane
Bromoethane
• reaction of an aromatic hydrocarbon with a halogen:
→
+
cyclohexane
chlorine
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+
hydrogen acid
chlorocyclohexane
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• Addition
• halogenation (with Br2 or Cl2)
→
+
bromine
ethene
1,2 - dibromoethane
• hydrogenation (with H2)
→
+
ethyne
hydrogen gas
ethane
• hydrohalogenation (HBr)
→
+
propene
hydrogen bromide
2 - bromopropane
• hydration (with H2O)
+
propene
→
water
2 - hydroxypropane
How do you know which Carbon to put the C and OH on?
Markovnikov’s Rule
“The rich get richer” (with reference to hydrogens atoms)
* when a hydrogen halide or water is added to an alkene or akin the hydrogen atoms bonds to the carbon that already had the
most hydrogen atoms
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Naming Alcohols:
alcohol - OH group
Rules For Naming:
• “ol” is added to the end of the parent chain
example;
Common Alcohols:
• ethanol (wine)
• glycerol
• isopropanol
• 2 - propanol (rubbing alcohol)
• retinol (vitamin A)
• cholesterol
methanol
- meth = 1 carbon
- an = single carbon bonds
- ol = alcohol (OH)
• if more than 2 carbon atoms or more than two —OH groups are present use the
numbering system to indicate location
example;
example;
1 - propanol
- prop = 3 carbons
- an = single carbon bonds
-ol = alcohol (OH)
- brake fluid
2 - propanol
(isopropanol)
- 2 (iso) = OH coming
off the second carbon
- an = single carbon
bonds
- ol = alcohol
- cleaning cuts
Sub-Categories of Alcohols:
Primary
- one other carbon
Secondary
- 2 other carbons
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Tertiary
- 3 other carbons
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Poly-Alcohols:
- more than 1 OH group
- use di, tri, tetra, etc
- use numbering
Cyclic Alcohols:
2 - isopropyl - 5 - methyl - 1 - cyclohexanol
Reactions of Alcohols:
preparing alcohol: alkene → alcohol
Hydration Reaction:
Follows
Markovnikov’s
Rule
* the simplest (methanol) is toxic to humans. even small amounts or swallowed can lead to
blindness and death.
Combustion Reaction:
C3H8O
propanol
+
5O2 → 4H2O
oxygen
water
+
3CO2
+
carbon dioxide
Elimination Reaction: (dehydration (opposite of hydration reaction)
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energy
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Naming Ethers:
ether - R - O - R’
(R & R’ could be different or the same)
Rules for Naming:
Properties of Ethers:
1. add foxy to the end of the smaller group,
then name the larger group
2. name each group alphabetically and add
ether to the end
* if two alkyl groups are the same, the prefix
• volatile (easily evaporates)
• flammable
• used in he olden day as an anesthetics
• used as solvents for fats and oils
“di” is used
example;
example;
1.
methoxyethane
1. ethoxyethane
2. ethyl methyl ether
2. diethyl ether
Reactions of Ethers:
formed when two alcohols react
Condensation Reaction:
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Naming Ketones:
Functional Group -
Ketone - carbonyl group bonded to two other carbon atoms
* insects use ketones to communicate with each other
Rules for Naming:
Example;
1. replace “e” ending of the name of an
alkane with “one”
propanone
Naming Aldehydes:
functional groups -
aldehyde - carbonyl group on a terminal carbon
* aldehydes are detectable over one distances by our sense of smell.
Rules for Naming:
Example:
1. take parent alkane name, drop the final
methanal
“e” and add “al”
ethanal
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Reactions of Aldehydes & Ketones:
oxidation
- historically, any reactions involving oxygen
- broadened to all chemical processes that involves a loss of electrons
- partner molecule that undergoes reduction (gain of electrons)
- oxidation mean gain of oxygen or loss of hydrogen (alcohols gain oxygen)
Making an Aldehyde:
+
alcohol → aldehyde
(need a primary alcohol)
(O)
→
H
H3C - C = O + H2O
Making a Ketone:
alcohol → ketone
(need a secondary alcohol)
+ (O)
→
* tertiary alcohols do not undergo this oxidation reaction. WHY NOT?
* there are no other H to form water
* there aren’t enough bonds to have a double bonded O (O=)
Making an Alcohol From Aldehyde:
H
H3C - C = O + H2
→
Making an Alcohol From Ketone:
+
H2
→
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+
H 2O
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Naming Carboxylic Acids:
Functional Group -
carboxyl group (this is made up of conbonyl and
hydroxyl groups
When wine is left exposed to air, it turns “sour”. In this case the alcohol in the wine has
turns into a carboxylic acid.
-
carboxyl acids are weak acids (only partially ionize)
found in citrus fruits, rhubarb, crab apples
found in sour milk and yogurt (lactic acid)
lactic acid also form in muscles when you workout, and you need to give your body
time to get rid of it to recover
Rules for Naming:
1. take the longest alkane/ alkaline
parent chain and replace “e” ending
with “oic acid”
Example;
methanoic acid
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2 - methylbutanoic acid
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Reactions of Carboxylic Acids:
Example 1:
+
aldehyde
(O)
→
oxidizing agent
- ethanal
carboxylic acid
- ethanoic acid
Example 2:
+
butanal
(O)
→
oxidizing agent
butanoic acid
Series of Equations:
+
methanol
(O)
oxidizing agent
+
methanal
→
(O)
+
methanal
→
oxidizing agent
methanoic acid
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water
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Naming Esters:
Functional Group -
formed from an alcohol and a carboxylic acid
• occurs naturally in plants
• responsible for odours of fruits and flowers
• synthetic esters are added to flavouring and cosmetics/perfumes
Examples:
Odour
Name
Apple
Methyl butanoate
Cherry
Methyl benzoate
Rum
Ethyl methanoate
Pineapple
Ethyl butanoate
Formula
Rules for Naming:
Rules for Drawing:
1. first part of the name is alkyl group
1. when drawing, the acid is drawn first
from alcohol
2. second part of the name is from the
acid
3. ending is “oate”
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Reactions of Esters:
Example 1: Esterification (condensation)
→
+
butanoic acid
+ H2O
ethyl butanoate
ethanol
Example 2:
+ H2O
→
+
benzoic acid
propyl benzoate
propanol
Hydrolysis (reversal of esterification):
+
Ester
NaOH
Base
→
+
alcohol
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R - C - O- Na+
salt of an acid
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Polymerization:
Polymers - are large molecules composed of a repeating sequence of monomers. A
monomer is usually a small molecule or compound. A common monomer is ethene
(ethylene), C2H4. Polymers are typically produced using two types of reactions; addition
polymerization and condensation polymerization reactions.
Polymerization - is a process of reacting monomer molecules together in a chemical
reaction to form polymer chains or three-dimensional networks.
Polyethylene: polymer of ethene
PolyPropene: polymer of propene
** You need to have an
alkene because you need to
have the electron potential
from the double bond.
Polyvinyl Chloride: polymer of vinyl chloride
Polystyrene: polymer of styrene
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Organic Solvents:
A solvent is a substance that dissolves a solute (a chemically distinct liquid, solid, or gas)
resulting in a solution. A solvent is usually a liquid it but can also be a solid or gas. The
quantity of solute that can result in a specific volume of solutes varies with temperature
(the higher the temperature, the more solute can dissolve because the electrons have more
energy and are moving around more which give it the ability to form attractions with
water).
Use
Solvent Used
Dry cleaning
Terechloroethelene
Paint cleaners
Toluene, Turpentine
** they have
low boiling
points
because
they’re nonpolar
Nail polis remover, glue solvents Acetone, Methyl acetate, Ethyl acetate
Spot removers
Hexene
Detergents
Citrus terpenes
Perfumes
Ethanol
Properties:
• Many solvents are volatile (many solvents are liquid with low boiling points, which
means that they will evaporate form vapour in the air when containers are left open.
There vapour can be breathed in and lead to health problems).
• Flammable (flashpoint below 38°C)
flashpoint - lowest temperature
• Combustible (flashpoint about 38°C)
at which vapours will ignite
(given a source)
Health Concerns:
• Exposure to different solvents can cause a host of health issues: toxicity to the nervous
system, reproductive damage, kidney and liver damage, respiratory impairment, cancer,
and dermatitis.
Safety Considerations:
• eye protection - prevents splashes from getting in eyes
• have proper ventilation - prevents vapours from being breathed in
• store properly - prevents volatile solvents from being vaporized, and also prevents spills
• read the safety information - be aware of all safety concerns for each solvent
• wear gloves and proper footwear- prevents from skin contact and getting in cuts
• wash hands after handling solvents - keep from prolonged skin exposure
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Organic
Family
Forces
(intramolecular/
intermolecular)
Hydrocarbons
• intramolecular
• ethane
• covalent (strong) • -89℃
• Intermolecular
H3C - H3
• london force
(weak)
Alcohol
Boiling Point
(based on 2
carbon molecule)
Solubility in Polar Solubility in NonSolvents
Polar Solvents
H2O
Very low solubility
High solubility
• Intermolecular
• hydrogen bonds
(strong)
• dipole-dipole
• ethanol
• 78.5℃
Ethers
• Intermolecular
• dipole-dipole
• london force
• dimethylether
• small chain
• small chain
methoxymethane
high
•
•
• low solubility
-23℃
solubility
long
chain
•
•
long
chain
high
•
•
H3C - O - CH3
solubility
• low solubility
Aldehydes
• small chain
• polar
• dipole-dipole
• long chain
• non-polar
• ethanal
• -21℃
• small chain
• small chain
• high
• low solubility
solubility
• long chain
(1-3)
• high
solubility
• long chain
• low solubility
Ketones
• small chain
• polar
• dipole-dipole
• ethanal
• -21℃
• small chain
• small chain
• high
• low solubility
solubility
• long chain
• long chain
• high
solubility
• low solubility
Carboxylic
Acid
• Intermolecular
• hydrogen bonds
(strong)
• dipole-dipole
• ethanoic acid
• vinegar
• 118℃
• small chain
• small chain
• high
• low solubility
solubility
• long chain
• long chain
• high
solubility
• low solubility
Esters
• Intermolecular
• dipole-dipole
• methyl ethanoate • small chain
• 31.8℃
• some
solubility
H3C - CH3
• small chain
• small chain
• high
• low solubility
solubility
• long chain
• long chain
• high
solubility
• low solubility
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• long chain
• high
solubility
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Isomers:
Isomer - each of two or more compounds with the same formula but a different
arrangement of atoms in the module and different properties
Isomers of C6H14
• hexane
• 2 - methylpentane
• 3 - methylpentane
• 2- dimethylbutane
• 2,3 - methybutane
Isomers of C6H12
• cyclo hexane
• 1 - hexene
• 2 - hexene
• 3- hexene
Molecules
Molecule Name
Structural Diagram
Common Name
IUPAC Name
Natural Gas
Methane
Vinegar (CH3COOH)
Acetic Acid
or
Ethanoic Acid
Acetone
Propanone
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Energy Changes & Rates of Reaction:
Energy Changes in Matter:
Thermochemistry - the study of energy changes that company physical or chemical changes in
matter
Thermal Energy - energy available form 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 a system that is capable of absorbing/ releasing thermal energy
Temperature - the average kinetic energy of the particles in a sample of matter
Heat - the amount of energy transferred between substances
Exothermic - releasing thermal energy as heat flows out of a system
Endothermic - Absorbing thermal energy as heat flows into a system
Open System - matter and energy can move in and out of a system
Closed System - energy can move in and out of a system, but not matter
Isolated System - an ideal system; neither energy nor matter can move in or out of a system
Enthalpy - an amount of heat energy (usually measured in kilojoules)
Physical Change - a change in the form of a substance in which no chemical bonds are broken
Chemical Change - a change in the chemical bonds between atoms, resulting in the rearrangement
in an atom, resulting in the formation of new atoms
Nuclear Change - a change in the protons or neutrons in an atom, resulting in the formation of new
atoms
Activated Complex - intermediate sate that is formed during the conversions of reactants to products
(high energy)
Activation Energy - the minimum energy required for a reaction to occur
Catalyst - substance that alters the rate of a chemical reaction without being consumed or
permanently changed
Hess’s Law - the value of the ∆H for any reaction that can be written in steps, equals the sum of the
values of ∆H for each individual step
Specific Heat Capacity - the heat required to raise the temperature of the unit mass of a given
substance by a given amount
Reaction Mechanism - a series of elementary steps that makes up an overall reaction
Reaction Intermediates - molecules formed as short-lived products in section mechanisms
Rate-Determining Step - the slowest step in a reaction mechanism
Kinetic Energy - (motion) moving electrons, vibrating atoms and molecules, rotation of molecules
Potential Energy - nuclear potential (protons/neutrons), electric potential of atoms in bonds, stored
chemical energy, energy is stored and released as the position of particles change
Effective Collisions - one that has sufficient energy and correct orientation of the colliding particles
so that bonds can be broken and new bonds can be formed
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Types of Enthalpy Changes:
Physical
Chemical
Nuclear
How is Energy Used?
• energy is used to
overcome or allow
intermolecular forces
to act
fundamental
particles
•
remain unchanged at
the molecular level
• energy changes
overcome the
electronic structure
and chemical bonds
within the particles
(atoms or ions)
• energy changes
overcome the forces
between protons and
neutrons in the nuclei
Range of Typical
Enthalpy Changes (kJ/
mol)
∆H = 100 - 102 kJ/mol
∆H = 102 - 104 kJ/mol
∆H = 1010 - 1012 kJ/mol
Measurement
Number of Sig Figs
32.07 m
4
0.0041 g
2
5 x 105 kg
1
6400 s
2
204.0 cm
4
10.0 kJ
3
100 people
infinite
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Calorimetry:
- the technological process of measuring energy changes in a chemical system
ideal system
what we will see
Measuring Energy Changes:
3 Factors:
1. temperature change (how much energy (heat) was put in/out)
2. mass (surroundings) (given the same amount of heat a small mass will
undergo a larger temperature change than a bigger mass )
3. types of substance (different substances can absorb different amounts of
energy)
relationships between amount of heat transferred. mass, type of substance, temperature change
q = heat transferred (to/from surroundings) → Joules
m = mass → grams
q = mc∆T
c = specific heat capacity → J/g·℃
∆T = temperature change (T2 - T1) → ℃
Density of water = 1g/mL
(1mL = 1cm2)
Specific Heat Capacities of Substances (table 1 on page 301)
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Enthalpy Change (∆H):
∆H - the difference in enthalpies of reactants and products during a chemical change
→ broad category of types of energies:
• Kinetic energy:
- moving electrons
- vibrating atoms and bonds
- rotation of molecules
• Potential energy
- nuclear potential (protons & neutrons)
- electric potential of atoms in bonds
Exothermic Reactions:
the release of energy the the surroundings ∆H ⊖ (system)
∆H ⊕ (surroundings)
Endothermic Reactions:
the absorption of energy from the surroundings ∆H ⊕ (system)
∆H ⊖ (surroundings)
Law of Conservation of Energy - the total energy of the system and surroundings remains the same
Reverse processes have an equal but opposite change in enthalpy
example;
H2O(l) → H2(g) ½O2(g)
∆Hdecomposition
H2(g) + ½O2(g) → H2O(g)
∆Hcombustion = -285.8kJ
= +285.8kJ
* to controls for variables, energy changes in chemical systems are measured at STP or SATP
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Molar Enthalpies (∆Hx):
enthalpy change is associated with a physical, chemical, or nuclear change involving one mole
of a substance
example; combustion of hydrogen
H2(g) + ½O2(g) → H2O(g) + 285.8kJ
1 mol ½ mol 1 mol release of energy
Exothermic Reaction
(table 1 - page 306: molar enthalpies reaction)
example; vaporization & condensation of water
H2O(l) + 40.kJ → H2O(g)
∆Hvapoization = +40.8kJ
H2O(g) → H2O(l) + 40.kJ
∆Hcondensation = -40.8kJ
* your breaking up the intermolecular forces NOT the covalent bonds
*molar enthalpies are obtained EMPIRICALLY and listed in reference tables
Generally:
To BREAK bonds, energy has to be put into a system which is ENDOTHERMIC
To FORM bonds, energy is released out of a system which is EXOTHERMIC
Using Molar Enthalpies in Heat Calculations:
a common refrigerant (Freon-12, molar mass 120.91g/mol) is alternately vaporized in tubes
inside a refrigerator, absorbing heat. and condensed in tubes outside the refrigerator, releasing
heat. this results in energy being trans formed from the inside to the outside of the refrigerator.
the molar enthalpy for the refrigerant is 34.99kJ/mol. if 500.0g of the refrigerant is vaporized
what is the expected enthalpy change ∆H?
∆Hvaporization = 34.99kJ/mol
m = 500.0g
Molar mass = 120.91g/mol
∆H = ?
nFreon12 = 500.0g ÷ 120.91g/mol
= 4.135mols
∆H = 4.135mols x 24.99kJ
= 144.7kJ
∆H = n∆Hx
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Representing Enthalpy Changes:
Method 1
• include an energy value as a term in the thermochemical equation
CH4O(l) + 3/2O2(g) → CO2 + 2H2O(g) + 726kJ
Method 2
• write a chemical equation and state the enthalpy change
CH4O(l) + 3/2O2(g) → CO2 + 2H2O(g)
∆H = -726kJ
Method 3
• state the solar enthalpy of a specific reaction
∆Hcombustion = -726kJ/mol CH4O(l)
Method 4
• potential energy diagram
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Calorimetry of Physical and Chemical Changes:
- The values for molar enthalpies are obtained by carefully designed experiments and precise
measurements/
- when we investigate energy changes we base out analysis on the law of conservation of
energy: the total energy change of the chemical system is equal to the total energy change
of the surroundings.
∆Hsystem = qsurroudings
H → enthalpy of system
q → surroundings
* one quantity is ⊖, gives off heat
* one quantity is ⊕, absorbs heat
Assumptions used in calorimetry:
1. no heat is transferred between calorimeter and the outside environment
2. any heat absorbed or released by the calorimeter materials, such as the container, is
negligible
3. a dilute aqueous solution is assumed to have a density and specific heat capacity equal
to that of water (1.00g/mL and 4.18J/g℃)
Example;
in a calorimetry experiment, 7.46g of potassium chloride is dissolved in 100.0mL (100.0g) of
water at an initial temperature of 24.1℃. The final temperature of the solution is 20.0℃. What
is the molar enthalpy of the solution of potassium chloride?
system: KCl(s) → K+(aq) + Cl-(aq)
c = 4.18J/g℃
mass = 100.0g
∆T = -4.1℃
∆Hsolution = ?
1. q=mc∆T
q = (100.0g)(4.18J/g℃)(-4.1℃)
q = -1713.8 J
q = -1.7kJ
2. n = m÷M = 7.46÷74.55 = 0.1000670
3. ∆Hsolution = 1.7kJ÷0.100067mols = 17kJ
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Potential Energy Diagrams:
- a graphical representation of the energy transferred during a physical or chemical
change
Potential Energy - stored chemical energy, energy is stored or released as the portion of
particles change.
- bonds breaking requires energy
- bonds forming released energy
A - exothermic reaction
combustion reaction
B - exothermic reaction
because the products have less EP than the reactants (∆H = ⊖)
C - endothermic
because the products have more EP than the reactants (∆H = ⊕)
Activation energy (Ea): the minimum energy required for a reaction to occur
Activated Complex: intermediate state that is formed during the conversion of reactants to products (high energy)
∆H: enthalpy change of the system
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Hess’s Law:
- Not all chemical reactions can be analyzed directly from calorimetry experiments.
Some reactions are very slow or other produce too many products to clearly
distinguish between in a calorimeter.
Hess’s Law: The value of the ∆H for any reaction that can be written in steps, equals the sum of the
values of ∆H for each individual step.
For example; consider the following Potential Energy Diagrams:
If two or more equations with known enthalpy changes can be added together to from a
new “target” equation, their enthalpy chnages may be added together to yield the overall
enthalpy change to the “target” equation.
∆HTARGET = ∑∆HKNOWN
Rules for Manipulating Equations:
1. if a chemical equation is reversed,
the sign of ∆H changes.
2. if coefficients are altered by
multiplying or dividing, ∆H is
altered in the same way.
example;
What is the enthalpy change (∆H) for the formula of two moles of nitrogen monoxide
from its elements?
Referenece Equations:
(1) ½N2(g) + O2(g) → NO2(g)
(2) NO(g) + ½O2(g) → NO2(g))
∆H = +34kJ
∆H = -56kJ
Target Equation: N2(g) + O2(g) → 2NO(g) ∆H = ?
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Multistep Energy Calculations:
- in practise, energy calculations rarely involve only a single step calculation of heat or
enthalpy change. Several energy calculations may be required, involving a
combination of energy change definitions such as:
- Heat flows:
q = mC∆T
- Enthalpy Change: ∆H = n∆Hx
- Hess’s Law:
∆Htarget = ∑∆Hknwon
- In these types of problems, ∆H is often found by using standard molar enthalpies or
Hess’s Law and then equated to the transfer of heat, q. As well, conversions between
moles and mass may be required to find the china in enthalpy of a specific amount
of a substance.
example;
in the productions of sodium carbonate one step in the endothermic decomposition of
sodium hydrogen carbonate:
2 NaHCO3 + 129.2kJ → Na2CO3(g) + CO2(g) + H2O(g)
what quantity of chemical energy ∆H, is required to decompose 100.0kg of sodium
hydrogen carbonate?
100kg = 100,000gNaHCO3
MNaHCO3 = 84.01g/mol
n = m÷n
= 100,000 ÷ 84.01
= 1190.33 mols
129.2kJ ÷ 2 = 64.6kJ
∆H = n • ∆Hx
= (1190.33mols) (64.6kJ)
= 76,895.3 kJ
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Enthalpies of Formation:
- Standard Enthalpy of Formation: the quantity of energy associated with the
formation of one mole of a substance from its element in their standard states.
Writing Formation Equations:
1. write one mole of product in the state that has been specified.
2. write the reactant elements in their standard states (solid, liquid, gas)
3. choose equations coefficients for the reactants to give a balanced equations yielding
one mole of product (fractions may be used)
example; write the formation equation of liquid ethanol
2C(s) + 3H2(g) + ½O2(g) → C2H6O(l)
*∆Hf of an element already in its standard state is 0kJ
Use pg. 799-800 for enthalpies
∆H = ∑n∆H°f(products) - ∑n∆H°f(reactants)
example;
the main component in natural gas used in home heating or laboratory burners is
methane. what is the molar enthalpy of combustion of methane fuel?
C2H4 + 2O2 → 2CO2 + 2H2O
∆H = [∑n∆H°f(products)] - [∑n∆H°f(reactants)]
∆H = [1mol(-393.5kJ/mol) + 2mol(-285.8kJ/mol)] - [1mol(-74.4kJ/mol) + 2 (0kJ/mol)]
∆H = -890.7kJ
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Collision Theory & Factors Affecting Reaction Rate:
Collision Theory:
• a chemical system consists of particles (atoms, ions, or molecules) that are in
constant random motion at various speeds
• a chemical reaction must involve collisions of particles with each other
• 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 can be formed
• ineffective collisions involve particles that rebound from the collision,
essentially unchanged in nature
The rate of a reaction depends on:
1. frequency of collisions
2. fraction (proportion) of those collisions that are effective
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Factors that Affect the Rate of a Reaction:
• Temperature
- if you increase the temperature, you’re also increasing the potential
energy, which causes more collisions, then that increases the probability
of a higher frequency of effective collisions
• Surface Area of Reactants
- by increasing the surface area your getting more particles that can
participate in collisions (at a faster rate)
- many small pieces have a greater surface area than just one large piece
• Catalyst
- a catalyst is a substance that alters the rate of a chemical reaction
without being consumed or permanently changed
- catalysts work by orienting the reactions molecules properly, by orienting
the reactants in the proper way, less energy is needed
- enzymes are proteins that speed up chemical reactions in the body
• Concentration of Reactants
- if you increase the concentration, then you have more reactants, which
means more collisions, which increases the chances of getting more
effective collisions
• Nature of Reactants
- different elements naturally react at different rates
- some react very slow (gold/silver)
- some are very reactive (sodium)
Analogy of Video
Factor Affecting Reaction Rate
Hire a match maker
catalyst
Increase population of school
concentration
Shrink size of hallways
concentration
Half the time between classes
temperature
Break up students in packs
surface area
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Reaction Mechanism:
• If you have a collision involving one particle this means you are decomposing or
breaking it down
• The single particle reaction mechanisms involve a step in which the single
molecule hits container walls or other particles
• This is converting kinetic energy to potential energy
• The collision of these particles simultaneously is much less frequent then he two
particle collision
• Most chemical reactions occur as a sequence of elementary steps
Elementary Steps - a step in a reaction mechanism that only involves one-, two-, or
three-particle collisions
• This overall sequence is called the reaction mechanism
Reaction Mechanism - a series of elementary steps that makes up an overall
reaction
• Increasing the concentration of the fast steps will have no effect on the chemical
reaction
• Increasing concentration concentration at the slow steps should increase the rate
of chemical reaction
Rate Determining Step - the slowest step in the reaction mechanism
Reaction Intermediates - molecules formed as a short-lived products in reaction
mechanisms
example;
HBr(g) + O2(g) → HOOBr(g)
HOOBr(g) +HBr(g) → 2HOBr(g)
2{HOBr(g) + HBr(g) → H2O(g) + Br2(g)}
4HBr(g) + O2(g) → 2H2O(g) + 2Br2(g)
slow
fast
fast
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Chemical Systems & Equilibrium:
Dynamic Equilibrium - a balance between forward and reverse processes occurring at the
same time
Chemical Equilibrium - a chemical equilibrium between reactants and products of the
chemical reaction in a closed system
Phase Equilibrium - a dynamic equilibrium between different physical states of a pure
substance in a closed system
Homogenous Equilibrium - one in which all of the reactants and products are present in
a single solution
Heterogenous Equilibrium - a system whose reactants, products, or both are in more than
one phase
Ionization - is the process by which an atom or a molecule acquires a negative or positive
charge by gaining or losing electrons to form ions
Dissociation - the splitting of a molecule into smaller molecules, atoms, or ions, especially
by a reversible process
Concentrated - high molarity
Dilute - low molarity
Strong - 100% ionization/dissociation (NO equilibrium)
Weak - less than 50% ionization/disassociation (equilibrium form)
Equilibrium Constant - a number that expresses the relationship between the amounts of
products and reactants present at equilibrium
Acid - proton donor
Base - proton acceptor
Conjugate Acid - contains one more + charge than the base that formed it
Conjugate Base - contains one more - charge than the acid that formed it
Auto-ionization - the reaction between two water molecules producing a hydronium ion
and a hydroxide ion
Hydronium Ion - the ion H3O+
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Chemical Equilibrium:
Reference Video: Crash Course Chemistry #28 (up to 4:25)
Throughout this unit when we speak of equilibrium reactions, we are usually referring to a
closed system: the system can exchange energy, but not matter (no atoms can escape).
Equilibrium Overview:
Up to this point, you have generally been told that reactions occur in one direction only:
reactants turn into product or products turn into reactants. This is actually not the case. In many
natural chemical reactions, forward and reverse processes are both occurring simultaneously.
example;
3H2(g) + N2(g) ⇌ 2NH3(g)
Forward reaction: left-to-right reaction →
Reverse reaction: right-to-left reaction ←
The rates of these forward and reverses reactions are dependant on several factors, but
mostly depend on the concentration of the substances. Typically when a reaction begins, the
forward and reverse reaction rates are not equal. (If you start with only reactants, then the rate of
forming the products will be fast due to the high amount of reactants particles available to collide
and react. As the concentration of these reactant particles decreases, and the concentration of
the product particles increases, the rate of the forward reactions flows, and threat of the reverse
reaction speeds up.)
Eventually a balance is formed between the forward and reverse reactions and the rate of
the forward reaction matches the rate of the reverse reaction. When this happens, we say that the
system is at equilibrium. Qualitatively, at equilibrium the system will look stable. We will se no
observable changes (ie. colour change, gas bubbles forming etc). This is because the number of
particles changing into products will equal the number o particles changing into reactants. The
systems as a whole will look constant.
It is important to note that equilibrium does not mean there are equal concentrations of
reactants and products (we very rarely see this).
Dynamic Equilibrium:
Solubility Equilibrium:
example;
NaCl(s) ⇌ Na+(aq) + Cl-(aq)
dissolving and crystallizing taking place at
the same rate
A dynamic equilibrium between different physical states of
a pure substance in a closed system
Phase Equilibrium:
example;
A balance between forward and reverse processes
occurring at the same rate
A dynamic equilibrium between a solute and a solvent
in a saturated solution in a closed system
H2O(l) ⇌ H2O(g)
evaporation and condensation at the same rates
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Equilibrium Stoichiometry:
ICE tables (math strategy)
I → initial concentrations of reactants and products [2.0mol/L] (square brackets imply concentration)
C → change in concentrations
E → equilibrium concentrations of reactants and products
example 1;
H2(g) + F2(g) ⇌ 2 HF(g)
**balance**
the reaction begins with 1.00mol/L of H2(g) and F2(g), calculate the concentrations of the H2(g) and HF(g)
at equilibrium if the concentration of F2(g) at equilibrium is 0.24mol/L
mol/L
H2(g)
F2(g)
2HI(g)
I
1.00
1.00
0
C
-x
-x
+2x
E
1.00-x 1.00-x
F2(g)
0.24 = 1.00 - x
x = 0.76 mol/L
2x
H2(g)
1.00 - x = [H2(g)]
1.00 - 0.76 = [H2(g)]
0.24 mol/L = [H2(g)]
example 2;
2NH3(g) ⇌ 2 3H2(g) + N2(g)
2HF(g)
2 (0.76) = [HF(g))]
1.52 mol/L = [HF(g))]
**balance**
when 4.0mol of NH3(g) is introduced into a 2.0L container and heated to a particular temperature, the
amount of ammonia changes to 2.0mol. Determine the equilibrium concentrations of the two entities
mol/L
2NH3(g)
3H2(g)
N2(g)
I
2.0
0
0
C
-2x
+3x
+x
E
2.0-2x
3x
x
[NH3(g) initial] = 4.0mol ÷ 2.0L
= 2.0mol/L
[NH3(g) equil] = 2.0mol ÷ 2.0L
= 1.0mol/L
NH3(g)
1.0 = 2.0 - 2x
x = 2.0 - 1.0
x = 0.5mol/L
H2(g)
[H2(g)] = 3x
[H2(g)] = 3 (0.5)
[H2(g)] = 1.5mol/L
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N2(g)
[N2(g)] = x
[N2(g)] = 0.5mol/L
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Equilibrium Constant (K):
When chemical reactions take place in a closed system, forward and reverse
reactions occur continuously and the system always contain products and reactants. In
previous studies, you used stoichiometry to calculate amounts of reactants and products.
In equilibrium systems, equilibrium law is used:
For the general chemical reaction
aA + bB ⇌ cC + dD
K = [C]c[D]d
————
[A]a[B]b
A,B,C,D are: chemical entities in gas/aqueous phases (concentration)
a,b,c,d are: coefficients in the balanced equations
(products)
(reactants)
K is constant called the equilibrium constant. Each chemical system has a unique K value
that is dependant on the temperature.
H2(g) + I2(g) ⇌ 2HI(g)
Exp.
#
Initial
concentrations
Equilibrium
Concentrations
[H2(g)]
[I2(g)]
[HI(g)]
[H2(g)]
[I2(g)]
[HI(g)]
1
2.000
2.000
0
0.442
0.442
3.116
2
0
0
2.000
0.221
0.221
1.560
3
0
0.010
0.350
0.035
0.045
0.280
Experiment Ratio of Equilibrium Concentrations
K =
1
K =
2
K =
3
Value of K
[HI(g)]2
—————
[H2(g)][HI(g)]
(3.116)2
————— = 49.7
(0.442)(0.442)
49.7
[HI(g)]2
—————
[H2(g)][HI(g)]
(1.560)2
————— = 49.8
(0.221)(0.221)
49.8
[HI(g)]2
—————
[H2(g)][HI(g)]
(0.280)2
————— = 49.7
(0.035)(0.045)
49.8
Note the following characteristics of equilibrium law:
- the molar concentrations of the products are always multiplied by one another and written in the
-
numerator, and the molar concentrations of the reactants are always multiplied by each other and written
in the denominator
the coefficients in the balanced chemical equation are equal to the exponents of the equilibrium expression
the concentrations in the equilibrium law expression are the molar concentrations of the entities at
equilibrium (mol/L - for gases, this means the moles of gas per litre of space occupied)
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Example 1:
write the equilibrium law expression for the reaction of nitrogen gas with hydrogen
gas to produce ammonia gas in a closed system.
N2(g) + 3H2(g) ⇌ 2NH2(g)
K =
[NH3(g)]2
—————
[N2(g)][H2(g)]3
Example 2:
calculate the value (K) of the equilibrium constant for the reaction of nitrogen gas
with hydrogen gas to produce ammonia gas if the equilibrium concentrations are 1.50 x
10-5 mol/L, 3.45 x 10-1 mol/L, and 2.00 x 10-4 mol/L respectively.
N2(g) + 3H2(g) ⇌ 2NH2(g)
K =
[NH3(g)]2
—————
[N2(g)][H2(g)]3
** favours
(2.00 x 10-4)2
———————————
(1.50 x 10-5)(3.45 x 10-1)3
K = 0.065 or 6.49 x 10-2
reactants
Forward & Reverse Reactions:
- K represents the equilibrium constant for the forward reaction
- Kʹ represents the equilibrium constant for the reverse reaction
Forward Reaction Example;
N2(g) + 3H2(g) ⇌ 2NH3(g)
[NH3(g)]2
K = —————
[N2(g)][H2(g)]3
= 6.49 x
⇌ 2NH3(g) ⇌N2(g) + 3H2(g)
10-2
[N2(g)][H2(g)]3
Kʹ = —————
[NH3(g)]2
= 15.4
K=1
1
— = — = 6.49 x 10-2
Kʹ
15.4
The equilibrium constant of a forward reaction and the equilibrium constant of
the reverse reaction are reciprocal quantities
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Equilibrium Law in Chemical Reactions:
Limitations of Equilibrium Constants and Precent Reaction Values:
- the position of the equilibrium is a measure of the extent to which reactants
become products in a closed system
- the value of the equilibrium constant, K, depends on the temperature
example:
percent reaction values are dependant on temperature and
concentration
N2(g) + 3H2(g) ⇌ 2NH3(g)
K = 4.26 x 108 at 25℃
K = 1.02 x 10-5 at 300℃
K = 8.00 x 10-7 at 400℃
Homogenous Equilibria:
- the equilibria in which all entities are in the same phase (all gases or all aqueous
solution)
- however in some systems reactants and produce are in different phases, this is
called heterogenous equilibria
-
Heterogeneous Equilibria:
- equilibrium systems can involve (s), (l), (g), (aq), most will be homogenous solutions
- equilibria in which reactants and products are in more than one phase
example;
2H2O(l) + 2H2(g) ⇌ O2(g)
[H2(g)]2[O2(g)]
K = —————
[H2O(l)]2
= 6.49 x 10-2
- liquid water written in the denominator is a problem because the concentration of
-
liquid cannot change; it is fixed and equal to the substances density
1L of liquid water has a mass of 1.00kg = 55.5mols ∴ the [H2O(l)] = 55.5mol/L
adding.removing water doesn’t change the concentration
adding/ removing H2(g) or O2(g) does change the concentration
the [H2O] is a constant , its value is incorporated into the K value shown
example;
K[H2O(g)] = [H2(g)]2[O2(g)]
K 55.52 = [H2(g)]2[O2(g)]
K = [H2(g)]2[O2(g)]
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- the concentrations of entities in a condenses state (s) or (l) are not included as
variables in the equilibrium law expression, but rather incorporated into the value
of the equilibrium constant
- water vapour is a gas just like H2 or O2, its concentration varies
- equilibrium law expressions are always written from the net ionic form, balanced
with simplest numbers
Magnitude of K:
- the magnitude of the equilibrium constant provides a measure of the extent to
which the reaction has gone to completion when equilibrium is reached
Equilibrium Constant:
Limitations of Equilibrium Constant
- the position of equilibrium is a measure of the extent to which reactants become
products in a closed system
- it is important to note that the value of the equilibrium constant, K, depends on
temperature
N2(g) + 3H2(g) ⇌ 2NH3(g)
K = 4.26 x 108 at 25℃
Generally, any calculation involving an
equilibrium law equation must specify a
temperature.
K = 1.02 x 10-5 at 300℃
K = 8.00 x 108 at 400℃
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Heterogeneous Equilibria
equilibrium systems can involve solids, liquids, gases, and aqueous solutions
Homogenous Equilibria - all entities are in the same phase
Heterogeneous Equilibria - reactants and/or products are in more than one phase
most of the systems we will study will be homogenous equilibria. however, in some
systems, the reactants and products are in different phases. the concentration of solids,
and pure liquids (ie. water) cannot change, their concentrations are fixed and equal to
their densities. because these substances have fixed concentrations, their values become
part of the value K, and not written in the equilibrium law expression. only substances in
the gas or aqueous phases are included, as their concentrations can change.
example;
Zn(s) + CuCl2(aq) ⇌ Cu(s) + ZnCl2(g)
Zn(s) + Cu2+(aq) + 2Cl-(aq) ⇌ Cu(s) + Zn2+(aq) + 2Cl-(g)
Zn(s) + Cu2+(aq) ⇌ Cu(s) + Zn2+(aq)
K = [Zn2+(aq)]
————
[Cu2+(aq)]
Magnitude
Explanation
∴ product favoured
K >> 1
[product] much greater than [reactants]
K=1
[product] and [reactants] are equal
K << 1
[reactants] much greater than [products] ∴ reactant favoured
Calculations involving equilibrium systems
in the equilibrium law expression, the variables are K, and the concentrations of
reactants and products in the gas or aqueous phases. as long as you know all values except
one, you can mathematically calculate the one unknown
example;
sulfur trioxide gas is produced when sulfur dioxide gas and oxygen gas react.
calculate the equilibrium concentration of oxygen gas if 1.50mol/L sulfur dioxide and
3.50 mol/L sulfur trioxide are found in an equilibrium mixture at 600℃. the equilibrium
constant for the reaction at 600℃ is 4.30
2SO2(g) + O2(g)⇌ 2SO3(g)
[SO2(g)]equil = 1.50 mol/L
[SO3(g)]equil = 3.50 mol/L
K = 4.30 @ 600℃
K=
[SO3(g)]2
[O2(g)] = 3.50 ÷ 1.50
——————
= 1.26 mol/L
[SO2(g)]2[O2(g)]
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Le Chatelier’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
concentration of reactants and products.
generally, Le Chatelier’s Principle predicts that if a product (or reactant) is added (or
removed), the system will undergo an equilibrium shift to restore the equilibrium. (either
more reactants will be produced, or more products will be produced).
example;
a pop bottle has an equilibrium between CO2(g) and CO2(aq), when the cap
is removed CO2(g) escapes, but when the cap is put back on the system reestablishes
equilibrium and the concentrations of CO2(g) and CO2(aq) are lower.
Variables Affecting Chemical Equilibrium:
Concentration
if more reactant is added: the system will use some of that added reactant and
change it into product to re-establish the equilibrium
if more product is added: the system will produce more reactant to re-establish
equilibrium
if reactant is taken away: system shifts left to produce more reactants
if product is take away: systems shifts right to produce more products
Temperature
the energy in a chemical equilibrium can be treated as a reactant or product
endothermic reaction: reactants + energy ⇌ products
exothermic reaction: reactants ⇌ products + energy
energy can be added or removed from a system by heating or cooling the
container. in either situation, the equilibrium shifts to minimize the change. if the system
is cooled, the system tries to “warm” itself and the equilibrium shifts the direction that
produces heat. if heat is added, the equilibrium shifts in the direction that absorbs heat.
Volume
increasing volume (decreasing pressure): shift happens toward the side with the
larger total amount of gaseous entities
decreasing volume (increasing pressure): shift happens toward the side with the
smaller total amount of gaseous entities
Variables That Do Not Affect Equilibria:
Catalysts - alter the activation energy required, do not affect concentrations of entities.
Inert Gases - gases that cannot react, do not change the concentrations of other entities.
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The Reaction Quotient:
this tells us that the larger the value of K, the more the reaction (as written) favours
the products. a very small K means the reactants are favoured.
if we know the concentrations of the substances in a closed chemical system, we
may want to determine wether the system is at equilibrium — and if not, in which
direction it will shift to reach equilibrium.
if a chemical system begins with only reactants, it is obvious that the system will
shift right to produce products. if both reactants and products are present, the direction is
less obvious. in such case, we can substitute concentrations into the equilibrium law
expression to produce a trial value called the reaction quotient (Q).
Reaction Quotient (Q): a test calculation using measured concentration values of a
system on the equilibrium law expression
Interpreting Q
Q=K
system is at equilibrium
Q>K
too much product
so the system shifts left to produce more reactant
Q<K
too much reactant
so the systems shifts right to produce more product
example 1;
the following reaction occurs in a closed container at 445℃. the equilibrium
constant, K, is 0.020.
2HI(g) ⇌ H2(g) + I2(g)
if the concentration of hydrogen iodide is 0.20 mol/L, hydrogen is 0.15 mol/L
and iodine is 0.09 mol/L, determine if the system is at equilibrium and if not, the
direction the reaction will proceed
K = 0.020 ∴ reactant favoured
Q = [H2(g)][I2(g)]
Q = (0.15)(0.09)
—————
—————
[HI(aq)]2
(0.20)2
Q = 0.3375
K = 0.020
Q > K too much product ∴the system will shift left to produce more reactants
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Equilibrium Calculations:
example 2;
carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. at
900℃, K is 4.200. calculate the concentration of all entities at equilibrium if 4.000 mol of each
entity is initially placed in a 1.000-L closed container.
Step 1: write a balanced chemical equation and list known values
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
V = 1.000L
K = 4.200
nall entities = 4.000mol
Step 2: calculate concentrations if necessary
[all] = 4.000mol ÷ 1.000L
[all] = 4.000mol/L
Step 3: calculate Q and determine the direction of shift
Q = [CO2(g)][H2(g)]
——————
[CO(g)][H2O(g)]
Q = (4.000)(4.000)
——————
(4.000)(4.000)
Q = 1.000
K = 4.000
K > Q too much reactant ∴ the system will shift right to produce more product
Step 4: ICE table
CO(g)
H2O(g)
CO2(g)
H2(g)
I
4.000
4.000
4.000
4.000
C
-x
-x
+x
+x
E
4.000 - x
4.000 - x
4.000 + x
4.000 + x
Step 5: sub into equilibrium law expression and find x
4.200 = (4.000 + x)(4.000 + x)
—————————
(4.000 - x)(4.000 - x)
→ 8.198 - √4.200 - x = 4.000 + x
→ 4.198 = 3.049x
→→ x ≐ 1.38
√4.200 = √(4.000 + x)2
√4.200 = 4.000 + x
→
————— → ——— ————
√(4.000 - x)2
1
4.000 - x
Step 6: use x value to solve for the equilibrium concentrations
[CO(g)] = [H2O(g)] = 4.000 - x
[CO2(g)] = [H2(g)] = 4.000 + x
= 4.000 - 1.38
= 4.000 + 1.38
= 2.623
mol/L
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The Solubility Product:
this is a special case of equilibrium where excess solute is in equilibrium with its aqueous
solution.
Solubility: the concentration of a saturated solution of a solute in a particular solvent at
a particular temperature; a specifi c maximum concentration
Ksp: the solubility product constant
the value obtained from the equilibrium law applied to a saturated solution
copper (I) chloride is dissolved in water. (has a low solubility, so even a small amount results in a saturated solution)
Balanced Equation:
CuCl(s) ⇌ Cu(aq)+ + Cl(aq)Solubility Equilibrium Law Expression:
Ksp = [Cu(aq)+] [Cl(aq)-]
Calculating Solubility Using Ksp Values:
example; calculate the Ksp for magnesium fluoride at 20℃, given a solubility of 0.00172
g/100mL
MgF2(s) ⇌ Mg(aq)2+ + 2F(aq)1. n = m÷M
= 0.00172g ÷ 62.31g/mol
= 2.76 x 10-5 mol
2. 100mL x 1L
——
1000mL
= 0.1L
3. C = 2.76 x 10-5 mol ÷ 0.100mL
= 2.76 x 10-4 mol/L
5. Ksp = (2.76 x 10-4)(5.52 x 10-4)2
= 8.3 x 10 -17
4.
MgF2(g)
Mg2+(aq)
2F-(aq)
I
—
0
0
C
—
+x
+ 2x
E
—
x
x
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Q>K
• supersaturated
• ∴ precipitate formed
Predicting Precipitation:
find Q (trial in product) and compare to K
example 1;
CuCl(s) ⇌ Cu+(aq) + Cl-(aq)
Q=K
• saturated solution
• ∴ at equilibrium
Ksp = [Cu+ (aq)][Cl-(aq)]
Q = (____)(____)
= _____
Q<K
• unsaturated
• ∴ no precipitate formed
example 2;
The Ksp of AgCl(s) is 1.8 x 10-10 at 25℃ in each case determine the type
of solution and whether a predicate forms.
[Ag+(aq)] = 1.6 x 10-7
a) [Cl-(aq)] = 1.3 x 10-5
Q = (1.3 x 10-5) (1.6 x 10-7)
Q = 2.08 x 10-12
Q < Ksp ∴ unsaturated
b) [Cl-(aq) ] = 1.5 x 10-5
Q = 3 x 10-9
[Ag+(aq)] = 2.0 x 10-4
Q > Ksp ∴ supersaturated
Common Ion Effect:
- when equilibrium exists in a solution, the equilibrium can be shifted by dissolving
into the solution a compound that adds a common ion
example;
- saturated solution of sodium chloride
- in equilibrium with undissolved sodium
NaCl(aq) ⇌ Na+(aq) + Cl- (aq)
- if a few drops of HCL is added, additional crystals of sodium chloride will form
- use Le Chatelier’s principal to explain the HCL releases large numbers of
chloride ions into the solution
HCl(aq) ⇌ H+ + Cl-(aq)
- those additional ions increase the concentration of chalice ions, shifting the NaCl
equilibrium to the left
- causing NaCl to precipitate out
- a similar result would happen if a solution containing Na+(aq) ions were added
instead
- in this case, the common ion would be the sodium ions
Common Ion Effect - a reduction in the solubility of a salt caused by the
presence of another salt having a common ion
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Acids & Bases:
- Acids and bases are electrolytes that for aqueous solutions with unique properties:
Properties of Acids & Bases
Acids
Bases
Property
sour
bitter
taste
<7
7 - 14
pH
forms H+ in solution
forms OH-(aq)
turns litmus red
turns litmus blue
litmus
used for cleaners
Recall the Arrhenius definition of acids and bases:
Acids - solutes that produce H+(aq)
Bases - solutes that produce OH-(aq)
Problem - does not explain how NH3 has basic properties
In 1923 a chemist in Denmark and a chemist in England (independantly) recognized that
in most acid-base interactions, a proton (H+ ion) is transferred from one reactant to
another
A Bronsted-Lowry ACID is a proton-donor
A Bronsted-Lowry BASE is a proton-acceptor
Bronsted-Lowry Theory:
Example;
H2O(l)
+
base
HCl(aq)
acid
⇌
H3O+(aq)
+
Cl-(aq)
hydronium ions
Amphoteric (amphiprotic) - a substance capable of acting as an acid or base in
different chemical scenarios
In proton transfer at equilibrium, both forward and reverse reactions involve BronstedLowry acids and bases.
Conjugate acid-base pair - two substance whose formulas only differ by a proton
example; identify the conjugate acid-base pair in the following reactions:
CH3COOH(aq)
acid
+
H2O(l)
base
⇌
CH3COO-(aq)
conjugate base
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+
H3O+(aq)
conjugate acid
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Properties of Strong & Weak Acids & Bases:
Properties
Acids
Bases
Strong
Weak
Strong
Weak
Attraction for
protons
very weak
strong attraction
strong attraction
weak attraction
% ionization
100% ionization
low (less than
50%)
100% ionization
low (less than
50%)
Nature of
conjugate
partner
weak conjugate
base
strong conjugate weak conjugate
base
acid
strong conjugate
acid
pH
<< 7 (lesser)
<7
> 7 (greater)
>> 7
Auto-ionization of Water:
H2O(l)
+
HCl(aq)
base
acid
Kw = 1.0 x 10-14
Kw = [H3O+(aq)][OH-(aq)]
1.0 x 10-14 = [H+(aq)][OH-(aq)]
⇌
H3O+(aq)
+
conjugate acid
[product ÷ reactant]
Cl-(aq)
conjugate base
example;
if we had pure water, what would be the concentration of H+ and
OH-?
pH = -log [H+(aq)]
[H+(aq)] = 5.3 x 10 -6
[H+(aq)] = 6.2 x 10 -3
[H+(aq)] = 7.8 x 10 -6
1.0 x 10-14 = [ x ][ x ]
√1.0 x 10-14 = √x2
1.0 x 10-7
=x
pH = 5.28
pH = 2.21
pH = 5.11
Dissolving an Acid - dissolving an acid increases the concentrations
Dissolving a Base - dissolving a base increase the concentration of OH-(aq)
- in neutral solutions [H+(aq)] = [OH-(aq)]
- in acidic solutions [H+(aq)] > [OH-(aq)]
- in basic solutions [H+(aq)] < [OH-(aq)]
- since Kw = [H+(aq)][OH-(aq)], if you know one of the concentrations, you can
find the other
example;
a 0.15 mol/L solution
of HCl is at SATP, calculate
the concentration of
hydroxide ions.
H2O(l)
+
HCl(aq)
⇌
H3O+(aq)
+
Cl-(aq)
C = 0.15 mol/L
[H+(aq)] = 1.0 x 10-7 mol/L
Kw = [H+(aq)][OH-(aq)]
1.0 x 10-7
(0.15)x
———— = ———
6.67 x 10-7 mol/L = x
0.15
0.15
** strong acid ∴ 100% ionization
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pH:
pH = -log [H+(aq)]
[H+(aq)] = 10-pH
0 - 2 → strong acid ; 3 - 6 → weak acid ; 8 - 11 → weak base ; 12 - 14 → strong base
example;
a sample of sodium hydrozide of 1.0 mol/L, determine the concentration of H+
ions
+
OH-(aq)
NaOH
→
Na+(aq)
c = 1.0 mol/L
c = 1.0 mol/L
+
1.0 x 10-14 = [H+]1.0
Kw = [H (aq)][OH (aq)]
———— ————
[H+] = 1.0 x 10-14
Terms:
• strong - 100% ionization/dissociation (NO equilibrium)
• weak - <50% ionization/dissociation (equilibrium formed)
• concentrated - high molarity
• diluted - low molarity
pOH:
pOH = -log [OH-(aq)]
[OH-(aq)] = 10-
0 - 2 → strong base ;3 - 6 → weak base ; 8 - 11 → weak acid ; 12 - 14 → strong acid
** this is true if all solutions have the same concentrations
example;
a)
calculate the pOH of a solution that has a hydroxide ion concentration of 2.5 x
-13
10 mol/L
pOH = -log -2.5 x 10(-13)
pOH = 12.60
b)
calculate the pH of this solution
12.60 + 1.40 = 14.00
Kw = [H+(aq)][OH-(aq)]
-13
-14
+
**pH and pOH always add to 14
1.0 x 10 = [H (aq)]2.5 x 10
-2
+
4.0 x 10 mol/L =[H (aq)]
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Percent Ionization of Weak Acids:
weak acids - only a little bit ionizes (∴ equilibrium forms)
% ionization = [H+(aq)]
———— x 100%
[OH-(aq)]
example;
calculate the % ionization of methanol acid if a 0.10M solution has a pH of 2.38
HCO2H(aq)
⇌
+
H+(aq)
CO2H-(aq)
0.10 mol/L
pH = 2.38
[H+] = 10-2.38 = 4.17 x 10-3 mol/L
% ionization = (4.17 x 10-3 ÷ 0.10) x 100%
= 4.2%
** ∴ reactant favoured
Acid Ionization Constant, Ka:
Ka - equilibrium constant for a weak acid
example;
calculate the Ka of acetic acid if a 0.1000M solution at equilibrium at SATP, has a
percent ionization of 1.3%
CH3COOH(aq)
⇌
Ka = [H+][CH3COOH-]
———————
[CH3COOH]
0.013 = [H+]
——
0.1000
+
H+(aq)
CH3COOH-(aq)
% ionization =
[H+]
—————
[CH3COOH]
** ∴ reactant favoured
= 1.3 x 10-3 mol/L
Ionization Constant for Weak Bases:
B - weak base (a base will have at least one lone pair of electrons)
B(aq)
base
+
H2O(l)
acid
⇌
BH+(aq)
conjugate acid
+
OH-(aq)
conjugate base
Kb = [BH(aq)][OH-(aq)]
———————
[B(aq)]
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Relationship Between Ka & Kb:
Kw = 1.0 x 10-14
Kw = Ka · Kb
← for an acid base conjugate pair (each reacting with water separately)
NH3(aq)
+
base
⇌
H2O(l)
acid
NH4+(aq)
conjugate acid
OH-(aq)
conjugate base
Kb = [NH4(aq)][H2O(l)]
———————
[NH3+(aq)]
NH4+(aq)
acid
+
⇌
H2O(l)
base
NH3(aq)
conjugate base
H3O(aq)
conjugate acid
Ka = [NH3(aq)][H3O(aq)+]
———————
[NH4+(aq)]
Kw = Ka · Kb
LS = Kw
= [H3O+][OH-]
RS = [H3O+][NH3] [NH4+][OH-]
—————— · —————
[NH4]
[NH3]
= [H3O+][OH-]
example;
find the value of Kb for the acetate ion C2C3O2, if the Ka of acetic acid is 1.8 x 10-5
Kw = Ka · Kb
1.0 x 10-14 = 1.8 x 10-5 · Kb
5.6 x 10-10 = Kb
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