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SCH 4U REVIEW 1 of 39

SCH 4U REVIEW

CHAPTER 1: ORGANIC COMPOUNDS

organic compound – a compound that contains carbon and usually hydrogencatenation – the property of carbon to form a covalent bond with another carbon atom, forming

long chains or ringsfunctional 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 NamingNUMBER PREFIX FOR

NAMINGALKYL GROUP

NAMEFUNCTIONAL

GROUPPREFIX FOR

NAMING

1 meth- methyl- - F fluoro2 eth- ethyl- - Cl chloro3 prop- propyl- - Br bromo4 but- butyl- - I iodo5 pent- pentyl- - OH hydroxy6 hex- hexyl- - NO2 nitro7 hept- heptyl- - NH2 amino8 oct- octyl-9 non- nonyl-10 dec- decyl-

Isomers of Alkyl Groups

butyl (or n-butyl) CH3 – CH2 – CH2 – CH2

isobutyl (or i-butyl) CH3 – CH – CH3

CH2

s-butyl (secondary butyl) CH3 – CH – CH2 – CH3

t-butyl (tertiary butyl) CH3


CH3 – C – CH3

SYMBOLSR, R΄, R˝ alkyl groupX halogen atom(O) oxidizing agent (like KMnO4 or Cr2O7

2- in H2SO4)PRIORITY FOR NAMING (FROM HIGHEST TO LOWEST)OH hydroxylNH2 aminoF, Cl, Br, I fluoro, chloro, bromo, iodoCH2CH2CH3 propylCH2CH3 ethylCH3 methyl

ALKANES

DEFINITIONa hydrocarbon with only single bonds between carbon atoms; general formula, CnH2n+2

NAMING

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 –

EXAMPLE(S)

propane CH3 – CH2 – CH3

cyclohexane CH2

CH2 CH2

| | CH2 CH2

CH2

POLARITY non-polarFORCES dispersion forces

BOILING POINT

relatively low increases with chain length straight chains have higher b.p.s than branched chains

SOLUBILITY immiscible in water and other polar solvents


REACTIONS

combustion C3H8 + 5 O2 3 CO2 + 4 H2O

substitution (halogenation)1) CH4 + Cl2 CH3Cl + HCl2) CH3Cl + Cl2 CH2Cl2 + HCl

ALKENES

DEFINITIONa hydrocarbon that contains at least one carbon-carbon double bond; general formula, CnH2n

NAMING

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

| | – C = C –

EXAMPLE(S) propene CH2 = CH – CH3

POLARITY non-polarFORCES dispersion forces

BOILING POINT relatively lowSOLUBILITY immiscible in water and other polar solventsREACTIONS halogenation (with Br2, Cl2, etc.)

ethene + bromine 1,2 – dibromoethane Br Br

CH2 = CH2 + Br2 CH2 – CH2

hydrogenation (with H2)ethyne + hydrogen ethane H H

CH2 = CH2 + 2 H2 CH2 – CH2

H Hhydrohalogenation (with hydrogen halides) propene + hydrogen bromine 2-bromopropane H Br | |H – C = CH – CH3 + HBr H – C – CH – CH3

| | H H

room temperature

catalyst, heat, pressure

room temperature


hydration (with H2O) 1-butene + water 2-butanol

CH3 – CH2 – CH = CH2 + HOH CH3 – CH2 – CH – CH2

OH H

MARKOVNIKOV’S RULE

“The rich get richer” … when a hydrogen halide or water is added to an alkene or alkyne, the hydrogen atom bonds to the carbon atom within the double bond that already has more hydrogen bonds.

GEOMETRIC ISOMERS

a cis isomer has both alkyl groups on the same side of the molecular strucutre

CH3 CH3

C = C cis-2-butene

H H

a trans isomer has alkyl groups on the opposite side of the molecular strucutre

CH3 H C = C trans-2-butene

H CH3

ALKYNES

DEFINITIONa hydrocarbon that contains at least one carbon-carbon triple bond; general formula, CnH2n-2

NAMING

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 – C ≡ C –

EXAMPLE(S) propyne CH ≡ C – CH3

POLARITY non-polarFORCES dispersion forcesBOILING POINT

relatively low

SOLUBILITY immiscible in water and other polar solventsREACTIONS Same as Alkenes (resulting in double bonds rather than single)

AROMATIC HYDROCARBONS

H2SO4 catalystPREPARATION REACTION FOR ALCOHOLS


DEFINITION a compound with a structure based on benzene (a ring of six carbons)

NAMING

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

GENERAL FORMULA

EXAMPLE(S)

methylbenzene

POLARITY non-polarFORCES dispersion forcesBOILING POINT

relatively low

SOLUBILITY immiscible in water and other polar solventsREACTIONS Same as Alkanes (performs substitution reactions)

ORGANIC HALIDES

DEFINITIONa compound of carbon and hydrogen in which one or more hydrogen atoms have been replaced by halogen atoms

NAMINGhalogen atoms are considered to be attachments to the parent chain and are numbered and named with a prefix as such

GENERAL FORMULA

R – X

EXAMPLE(S)

H H | |1,2-dichloroethane H – C – C – H | | Cl Cl

Cl1,2-dichlorobenzene Cl

POLARITY polar (due to halogen atom)FORCES dispersion forces (increased strength b/c of carbon-halogen bonds)BOILING POINT

higher than the corresponding hydrocarbons

SOLUBILITY more soluble in polar solvents

CH3


REACTIONS

addition ethylene + hydrogen idodie 1-iodoethene H I | | H – C ≡ C – H + H – I H – C = C – H

substitution ethane + chlorine 1-chloroethane + hydrogen chloride H Cl CH3 – CH3 + Cl – Cl H – C – C – H + H – Cl

H Helimination 2-bromopropane + hydroxyl group propene + bromine ion + water H Br H H H H | | | H – C – C – C – H + OH- H – C – C = C – H + Br- + H2O

H H H H

ALCOHOLS

DEFINITION

an organic compound characterized by the presence of a hydroxyl functional group (OH-)

NAMING

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

EXAMPLE(S)

propanol CH3 – CH2 – CH2 – OH

phenol

POLARITY polar (due to hydroxyl group)FORCES hydrogen bonding and dispersion forcesBOILING high (due to capacity for hydrogen bonding)

O

< 109.5°

O

H S+

S-

OH


POINTSOLUBILITY very soluble in polar solvents and nonpolar solvents (due to OH- group)

hydration (preparation) 1-butene + water 2-butanol

CH3 – CH2 – CH = CH2 + HOH CH3 – CH2 – CH – CH2

OH H dehydration (elimination)

propanol propene + water

CH3 – CH – CH2 CH3 – CH = CH2 + HOH

OH H

1°, 2°, AND 3° ALCOHOLS

Primary Alcohol (1°) CH3 – CH2 – OH 1 other carbon group attached to carbon with OH OH Secondary Alcohol (2°) CH3 – CH – CH3

2 other carbon groups attached to carbon with OH OH

Tertiary Alcohol (3°) CH3 – C – CH3

3 other carbon groups attached to carbon with OH CH3

ETHERS

DEFINITION

an organic compound with two alkyl groups (the same or different) attached to an oxygen atom

NAMING

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

ethoxypropaneGENERAL FORMULA

R – O – R΄

H2SO4 catalyst

H2SO4 catalyst

O

116°

O

C S+

S-C

C

C


EXAMPLE(S) methoxyethane CH3 – O – CH2 – CH3

POLARITY polar (due to the V-shape and C - O bonds)FORCES dispersion forcesBOILING POINT

medium (higher than hydrocarbons, lower than alcohols of similar length)

SOLUBILITY soluble in polar solvents and nonpolar solventscondensation (preparation) * addition of two alcohols (same or different) *

methanol + methanol methoxymethane + water

CH3 – OH + CH3 – OH CH3 – O – CH3 + HOHALDEHYDES

DEFINITIONan organic compound that contains the carbonyl group (–C=O) on the end carbon of a chain

NAMING the “e” ending of the parent hydrocarbon is changed to “al” to indicate the presence of the R-C=O

GENERAL FORMULA

O ||

R[H] – C – H

EXAMPLE(S)

O ||propanal CH3 – CH2 – C – H

POLARITY polar (due to carbonyl group)FORCES dispersion forcesBOILING POINT

medium (higher than ethers, lower than alcohols of similar length)

SOLUBILITY similar solubility to alcoholsREACTIONS oxidation (preparation)

1° alcohol + mild oxidizing agent aldehyde + water

ethanol + (O) ethanal + water

OH O | ||CH3 – C – H + (O) CH3 – C – H + HOH | H

H2SO4 catalyst


reduction (hydrogenation)

aldehyde + hydrogen 1° alcohol

proponal + hydrogen 1-propanol

O OH CH3 – CH2 – C – H + H2 CH3 – CH2 – C – H

KETONES

DEFINITIONan 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)

NAMING the “e” ending of the parent hydrocarbon is changed to “one”

GENERAL FORMULA

O ||

R – C – R΄

EXAMPLE(S)

O ||propanone (acetone) CH3 – C – CH3

POLARITY polar (due to carbonyl group)FORCES dispersion forcesBOILING POINT

medium (higher than ethers, lower than alcohols of similar length)

SOLUBILITY similar solubility to alcohols

REACTIONS

oxidation (preparation)

2° alcohol + oxidizing agent ketone + water

2-propanol + (O) propanone + water

OH O || CH3 – C – CH3 + (O) CH3 – C – CH3 + HOH H

reduction (hydrogenation)

ketone + hydrogen 2° alcohol

butanone + hydrogen 2-butanal

O 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: – C O – H

NAMING the “e” ending of the parent alkane is changed to “oic acid” numbering starts with the C of the carboxyl group

GENERAL FORMULA

O ||

R[H] – C – OH

EXAMPLE(S)

CH2CH3

4-ethyl-3-methylhexanoic acid CH3 – CH2 – CH – CH – CH2 – COOH

CH3

POLARITY polar (due to carboxyl group)FORCES dispersion forces and hydrogen bondingBOILING POINT

very high (higher than hydrocarbons of similar length due to –CO and –OH groups)

SOLUBILITY similar solubility to alcohols

REACTIONS

dissolution in water (proof of acidity)

ethanoic acid + water carboxylate ion + hydronium ion

CH3COOH + H2O CH3COO- + H3O+

O – CH3 – C * shares the double bond Ooxidation (with weak oxidizer)

1) 1° alcohol + weak oxidizing agent aldehyde + water2) aldehyde + weak oxidizing agent carboxylic acid + water O 1) CH3 – CH2 – OH + Cr2O7

2- + 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 + 4 MnO4

- + 4 H+ 3 CH3 – C – OH + 5 H2O + 4 MnO2

O


NOTE: Ketones are not readily oxidizing, as in the oxidation of aldehydes.

ESTERS

DEFINITIONan organic compound characterized by the presence of a carbonyl group bonded to an oxygen atom

NAMING 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

GENERAL FORMULA

O ||

R[H] – C – O – R΄

EXAMPLE(S)

2-methylbutyl propanoate O CH3

CH3 – CH2 – C – O – CH2 – CH – CH2 – CH3

POLARITY less polar than carboxylic acids (loss of OH- group)FORCES dispersion forcesBOILING POINT

medium (lower than carboxylic acids, higher than aldehydes / ketones of similar length due to extra O)

SOLUBILITY less soluble than acids

REACTIONS

condensation (formation)

carboxylic acid + alcohol ester + water

ethanoic acid + methanol methyl ethanoate + water O O || || CH3 – C – OH + CH3 – OH CH3 – C – O – CH3 + H2O

hydrolysis (saponification)

ester + NaOH sodium salt of acid + alcohol

O O

R – C – O – R΄ + Na+ + OH- R – C – O- + Na+ + R΄ – OH

NOTE: Esters generally have nice odours and are used to create artificial flavours.

AMINES

DEFINITIONan ammonia molecule in which one or more H atoms are substituted by alkyl or aromatic groups

NAMING1) nitrogen group is named as a substituent group using “amino-”2) alkyl group is named as a substituent group from “-amine”

GENERAL FORMULA R΄ [H]

| R – N – R΄΄ [H]


EXAMPLE(S)

1) 1-aminopropane NH2

2) propylamine CH2 – CH2 – CH3

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)FORCES dispersion forces and N-H bondsBOILING POINT

medium (lower than alcohols of similar length, higher than hydrocarbons)

SOLUBILITY soluble in water

REACTIONS

formation

alkyl halide + ammonia amine + ammonium halide

iodoethane + ammonia aminoethane + ammonium iodide

CH3 – CH2 – I + 2 NH3 CH3 – CH2 – NH2 + NH4I

1°, 2°, AND 3° AMINES

Primary Amine (1°) CH3 – N – H 1 alkyl group attached to N | H methylamine

Secondary Amine (2°) CH3 – N – CH3

2 alkyl groups attached to N | H dimethylamine Tertiary Amine (3°) CH3 – N – CH3

3 alkyl groups attached to N | CH3 trimethylamine

AMIDES

DEFINITIONan organic compound characterized by the presence of a carbonyl functional group (C=O) bonded to a nitrogen atom

NAMINGalkyl group attached to double-bonded O is considered to be the substituent group, attached to the parent “-amide”

GENERAL FORMULA

O R΄΄ [H] || |

R[H] – C – N – R΄ [H]

EXAMPLE(S) O ||


propanamide CH3 – CH2 – C – NH2

POLARITY slightly more polar than amine of similar length (extra O)FORCES dispersion forces and N-H bondsBOILING POINT

same as amines

SOLUBILITY soluble in water

REACTIONS

formation

carboxylic acid + amine amide + water

ethanoic acid + ammonia ethanamide + water O H O || | || CH3 – C – OH + H – N – H CH3 – C – NH2 + HOH

EXAMPLE OF A STEPPED SYNTHESIS REACTIONWrite 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 + hydrogen chloride CH3 – CH2 HOH CH3 – CH2 + H – Cl

| | Cl OH

3. ethanol + strong oxidizer ethanoic acid + water + manganese dioxide CH3 – CH2 + MnO4

- CH3 – C – OH + HOH + MnO2

| || OH O4. chloroethane + ammonia aminoethane + ammonium chlorine CH3 – CH2 + 2 H – N – H CH3 – CH2 – NH2 + NH4Cl | | Cl H5. ethanoic acid + aminoethane N-ethylethanamide + water O H || | CH3 – C – OH + CH3 – CH2 – NH2 CH3 – C – N – CH2 – CH3 + HOH || O

COMMON NAMES

COMMON NAME IUPAC NAME

COMMON NAME IUPAC NAME

ethylene ethene formic acid carboxyl group (COOH)propylene propene acetic acid ethanoic acidacetylene ethyne toluene / phenyl methane methyl benzene

formaldehyde methanal acetate ethanoateacetaldehyde ethanal acetamide ethanamide

acetone propanone

H2SO4 catalyst


FLOW CHART OF ORGANIC REACTIONS

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

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

monomer


polyester – a polymer formed by condensation reactions resulting in ester linkages between monomers

dialcohol + dicarboxylic acid polyester (dimer) + water O O O O || || || || HO – CH2 – CH2 – OH + C – C – O – CH2 CH2 – O – C – C – + 2 H2O

| | n

HO OH

polyamide – a polymer formed by condensation reactions resulting in amide linkages between monomers; also known as a nylon

diamine + dicarboxylic acid polyamide (dimer) + water H H O O H H O O | | || || | | || || N – CH2 – N + C – C – N – CH2 – N – C – C – + 2 H2O

| | | | n

H H HO OH

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 + fatty acids fat (triglyceride) H O O

| 3 || ||H – C – OH HO – C – R H – C – O – C – R | |H – C – OH H O | ||H – C – OH H – C – O – C – R’

|

SCH 4U REVIEW 16 of 39 H O ||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 phosphate base

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 orbitorbital – 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 levell – 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

NUMBER OF ORBITALS

MAX. NUMBER OF ELECTRONS

PRESENT AT n =

0 s 1 2 1-71 p 3 6 2-72 d 5 10 3-73 f 7 14 4-7

6p


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

e.g., 9F

2p

2s

1s

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:

32 e- 6s5d

4f

18 e- 5s

5p

18 e- 4s

8 e- 3s

8 e- 2s

2 e- 1s

4p

3p

2p

4d

3d

n (primary quantum number or energy level)

l (secondary quantum number or sublevel)

m1 (one orbital is distinguished from another at the same sublevel)m2 (spin)


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 4s 3dELECTRON CONFIGURATION electron configuration – a method for communicating the location and number of electrons in

electron energy levels

e.g., O: 1s2 2s2 2p4

3p5 S2-: 1s2 2s2 2p6 3s2 3p6

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 mechanicsHeisenberg uncertainty princple – it is impossible to simultaneously know exact position and

speed of a particle

principal quantum number

orbital

number of electrons in orbital(s)


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. Arrange atoms symmetrically around the central atom (usually listed first in the formula, not usually oxygen and never hydrogen).

2. 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.

3. 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.


4. Place a bonding pair of electrons between the central atom and each of the surrounding atoms.

5. 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.

6. 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.

7. Draw the Lewis structure and enclose polyatomic ions within square brackets showing the ion chare.

e.g., SO3

S O3

has 6 3 (6) 6 + 18 = 24 e-

wants 8 3 (8) 8 + 24 = 32 e-

difference = 8 e-

= 4 bonds

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. Draw the Lewis structure for the molecule, including the e- pairs around the central atom.2. Count the total number of bonding pairs and lone pairs of electrons around the central atom.3. Use the table below to predict the shape of the molecule.# OF e- PAIRS

AROUND CENTRAL

ATOM

ORIENTATION OF

ELECTRON PAIRS

NUMBER OF BONDING AND LONE

PAIRS

BOND ANGLES

SHAPE EXAMPLEMOLECULAR GEOMETRY

2 linear 2 BP, 0 LP 180° linear BeF2 X – A – X

S

O O

O

. . .


3 triangular planar 3 BP, 0 LP 120° trigonal

planar BF3

X|A

X X

4

tetrahedral 4 BP, 0 LP 109.5° tetrahedral CH4

X|A

X X X

tetrahedral 3 BP, 1 LP < 109.5° trigonal pyramidal

NH3

PCl3A

X X X

tetrahedral 2 BP, 2 LP < 109.5° v-shaped or bent

H2OOF2

A

X X tetrahedral 1 BP, 3 LP 180° linear HCl A – X

5 trigonal bipyramidal 5 BP, 0 LP 120° &

90°trigonal

bipyramidal PCl5

X X | X

A | X

X

6 octahedral 6 BP, 0 LP 90° octahedral SF6

X X | X

A X | X X

e.g., HOF(l) . . H : O : F : 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

2.1 H H 2.1 therefore, it is a polar molecule H 2.1 INTERMOLECULAR FORCES intermolecular force – the force of attraction and repulsion between molecules; much weaker

than covalent bonds

∂-

∂+∂+

∂+


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 PARTICLESFORCE/BOND

PROPERTIES

EXAMPLESMELTING POINT

CONDUC-TIVITY

SOLUBILITY IN LIQUID/SOLUTION

Ionic

ions(+ and -) ionic

medium (higher as

ionic charge

increases, size of ion decreases)

yes yes

NaCl, Na3PO4, CaF2,

MgO, CuSO4•5H2O

Metallic

cations

metallic (fixed

nuclei with mobile

delocalized electrons,

allowing for conduction of heat and electricity)

high(higher as number of

valence electrons increases)

yes noPb, Fe,

Cu, Al, Ag, Au


Polar Molecular

molecules

London, dipole-dipole,

hydrogen

low(higher as number of electrons, polarity of molecules,

and presence

of hydrogen

bonds increases)

no

yesPF3, ICl,

CHCl3, H2O, NH3, SO3

Nonpolar Molecularmolecules or single

atomsno

inert gases, diatomic elements,

CO2, CCl4, SF6, BF3

Covalent Network Crystal

atoms

covalent (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 no

diamond, SiC, AlN, BeO,

SiO2, CuCl2, Mg2S

NETWORK SOLIDS (AKA “SUPERMOLECULES”) ALLOTROPES OF CARBON:

a) diamond – 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) 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

a) quartz – a 3-D network solidb) mica – a 2-D network solidc) 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

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 EXAMPLE OF CHANGEsolution (ΔHsol) NaBr(s) Na+

(aq) + Br-(aq)

combustion (ΔHcomb) CH4(g) + 2 O2(g) CO2(g) + H2O(l)

vaporization (ΔHvap) CH3OH(l) CH3OH(g)

freezing (ΔHfr) H2O(l) H2O(s)

neutralization (ΔHneut) 2 NaOH(aq) + H2SO4(aq) 2 Na2SO4(aq) + 2 H2O(l)

formation (ΔHf) C(s) + 2 H2(g) + ½ O2(g) CH3OH(l)

ΔH = n ΔHvap or sol = mcΔT ΔH = q n

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: 1.41 g Mass of Water Heated: 97.00 g Mass of Water Vapourized: 0.18 g Mass of Pop Can: 16.07 g

mn M

E p (k

J)


Temperature Change of Water: 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: ΔH = qtotal

n = 9.72 kJ 0.0235 mol = 414 kJ/mol

% error: 100% – 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 TERMSe.g., C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2802.7 kJ

METHOD 2: THERMOCHEMICAL EQUATIONS WITH ΔH VALUESe.g., C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) ΔH = -2802.7 kJ

METHOD 3: MOLAR ENTHALPIES OF REACTIONe.g., ΔHrespiration = -2802.7 kJ/mol glucose

METHOD 4: POTENTIAL ENERGY DIAGRAMe.g.,


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) ΔH°1 = +34 kJ

(2) NO(g) + ½ O2(g) NO2(g) ΔH°2 = -56 kJ

2 x (1): N2(g) + 2 O2(g) 2 NO2(g) ΔH°1 = 2(+34) kJ

-2 x (2): 2 NO2(g) 2 NO(g) + O2(g) ΔH°2 = -2(-56) kJ

N2(g) + O2(g) 2 NO(g) ΔH° = +68 kJ + 112 kJ = +180 kJUSING 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 gmH2O = 4.0 kgcH2O = 4.18 J/(g•°C)

ΔH = Σ nΔH°f (products) – Σ nΔH°

f (reactants)

= 3 mol x -393.5 kJ + 4 mol x -285.8 kJ – 1 mol x -104.7 kJ + 5 mol x 0.0 kJ 1 mol 1 mol 1 mol 1 mol = -2219 kJ

ΔHc (propane) = qwater

n ΔHc = mcΔT

ΔT = n Δ H c

mc

SCH 4U REVIEW 27 of 39 = 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 reactionrate of reaction – the speed at which a chemical change occurs, generally expressed in

concentration per unit time, such as mol/(L•s)rate = Δc Δtaverage 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 equatione.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

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]

SCH 4U REVIEW 28 of 39e.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) INITIAL RATE OF REACTION (MOL/(L•S)

1 0.001 0.004 0.0022 0.002 0.004 0.0083 0.003 0.004 0.0184 0.004 0.001 0.0085 0.004 0.002 0.0166 0.004 0.003 0.024

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)

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. 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 required for molecules to react

activated complex – an unstable chemical species containing partially broken and partially formed bonds representing the maximum potential energy point in the change; also called the transition state

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)

activation

net potential energychange, ΔH


reaction mechanism – a series of elementary steps that makes up an overall reactionelementary step – a step in a reaction mechanism that only involves one-, two-, or three-particle

collisionsrate-determining step – the slowest step in a reaction mechanismreaction intermediates – molecules formed as short-lived products in reaction mechanisms

e.g., HBr(g) + O2(g) HOOBr(g) (slow)HOOBr(g) + HBr(g) 2 HOBr(g) (fast)2 HOBr(g) + HBr(g) H 2O(g) + Br2(g) (fast)4 HBr(g) + O2(g) 2 H2O(g) + 2 Br2(g)

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 collisions3. CONCENTRATION: an increase in reactant concentration increases the rate of reaction 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 space4. 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

reactionmechanism

elementary step

rate-determining step

reaction intermediate


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 solutionphase 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

reactionEQUILIBRIUM LAW EQUATION For the general chemical reaction: aA + bB ↔ cC + dD

K = [C] c [D] d where: • A, B, C, D are chemical entities in gas or aqueous phases (liquids and [A]a [B]b 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 changeequilibrium shift – movement of a system at equilibrium resulting in a change in the

concentrations of reactants and productsVariables Affecting Chemical Equilibria:

VARIABLE TYPE OF CHANGE RESPONSE OF SYSTEMconcentration increase shifts to consume added reactant

decrease shifts to replace removed reactanttemperature increase shifts to consume added thermal energy

(away from heat term)decrease shifts to replace removed thermal energy

(towards heat term)volume increase (decrease in

pressure)shifts towards side with larger total number

of gaseous entitiesdecrease (increase in

pressure)shifts towards side with smaller total

number of gaseous entitiescommon ion effect dissolving a compound into

solution that adds a common ion

shifts away from common ion to consume the added reactant

VARIABLES THAT DO NOT AFFECT CHEMICAL EQUILIBRIAcatalysts - no effect

adding inert gases - no effectSOLVING EQUILIBRIUM PROBLEMS


1. Write a balanced equation for the reaction and list the known values.2. 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) + I2(g) ↔ 2 HI(g) K = 49.7

[H2(g)] = 4.00 mol [I2(g)] = 2.00 mol [HI(g)] = 0.00 mol/L 2.00 L 2.00 L = 2.00 mol/L = 1.00 mol/L

H2(g) + I2(g) ↔ 2 HI(g)

Initial concentration (mol/L) 2.00 1.00 0.00Change in concentration (mol/L) – x – x +2 xEquilibrium concentration (mol/L) 2.00 – x 1.00 – x 2x

K = [HI] 2 [H2] [I2]49.7 = (2x) 2 (2.00 – x) (1.00 – x) 4x2 = 49.7 (2.00 – x) (1.00 – x)0.92x2 – 3.00x + 2.00 = 0

x = –b ± √ b 2 – 4ac 2a = 3.00 ± √ 9.00 – 7.36 1.84 = 2.33 or 0.932.33 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

[H2(g)] = 2.00 mol/L - x = 2.00 mol/L - (0.93 mol/L)

= 1.07 mol/L[I2(g)] = 1.00 mol/L - x = 1.00 mol/L - (0.93 mol/L)

= 0.07 mol/L[HI(g)] = 2x = 2(0.93 mol/L)

= 1.87 mol/L


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. Write a balanced equation and list the known values.2. Use the solid product to write an equilibrium equation for it dissolving into ions.3. Find the Ksp value for the solid product and write it next to the equilibrium equation.4. 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) + SO4

2-(aq) Ksp = 1.08 x 10-10

nBa2NO3 = c v nNH4NO3 = c v = (0.50 mol/L) (0.130 L) = (0.20 mol/L) (0.020 L) = 0.065 mol = 0.0040 mol = nBa = nSO4

[Ba2+(aq)] = n [SO4

2-(aq)] = n

v v = 0.065 mol = 0.004 mol

0.150 L 0.150 L = 0.43 mol/L = 0.027 mol/L

Ksp = [Ba2+] [SO42-]

= (0.43) (0.027) = 0.011 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 assistanceentropy, 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)

Choose the solid to write the equilibrium equation (remember any NO3

- molecule is aqueous)

Divide by the “new” volume (i.e., 20 mL and 130 mL equals 150 mL)


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

Endothermic (ΔH > 0) Exothermic (ΔH < 0)

Entropy increases (ΔS > 0)spontaneous at high temps.

nonspontaneous at low temps.H2O(s) H2O(l)

spontaneous C(s) + O2(g) CO2(g)

Entropy decreases (ΔS < 0)nonspontaneous3 O2(g) 2 O3(g)

spontaneous at low temps.nonspontaneous at high temps.

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 donorBrønsted-Lowry base – proton acceptoramphoteric (amphiprotic) – a substance capable of acting as an acid or a base in different

chemical reactions; a substance that may accept or donate a proton e.g.,

HC2H3O2(aq) + H2O(l) ↔ C2H3O2-(aq) + H3O+

(aq)

strong acid – an acid with a very weak attraction for protons and easily donates it to a basestrong 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 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)] [H+(aq)] = Kw [OH-

(aq)] = Kw = (1.0 x 10-7 mol/L) (1.0 x 10-7 mol/L) [OH-

(aq)] [H+(aq)]

= 1.0 x 10-14

In neutral solutions [H+(aq)] = [OH-

(aq)]

In acidic solutions [H+(aq)] > [OH-

(aq)]In basic solutions [H+

(aq)] < [OH-

(aq)]PH pH – the negative of the logarithm to the base ten of the concentration of hydrogen (hydronium)

ions in a solutionpOH – the negative of the logarithm to the base ten of the concentration of hydroxide ions in a

solution

pH = –log [H+(aq)] pOH = –log [OH-

(aq)] pH + pOH = 14.00

conjugate pair

acid

base


[H+(aq)] = 10–pH [OH-

(aq)] = 10–pOH

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) 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/LpOH = –log [OH-

(aq)] = –log (3.3 x 10-2)

= 1.47pH = 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% [H+(aq)] = p x [HA(aq)]

concentration of acid solute 100

acid ionization constant (Ka) – equilibrium constant for the ionization of an acide.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) 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

HF(aq) ↔ H+(aq) + F-

(aq)

Initial concentration (mol/L) 0.100 0.000 0.000Change in concentration (mol/L) – x + x + x Equilibrium concentration (mol/L) 0.100 – 0.0078 0.0078 0.0078

= 0.0922

100%

7.8%

SCH 4U REVIEW 35 of 39Ka = [H + (aq)] [F - (aq)] pH = –log [H+

(aq)] [HF(aq)] = –log (0.0078)

= 0.0078 2 = 2.1 0.0922 = 6.6 x 10-4

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 NUMBERS1. 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-)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 oxidation-reduction 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 0 +4 -2 +1 -2


H2S(g) + O2(g) SO2(g) + H2O(g)

reductionBALANCING REDOX EQUATIONS USING OXIDATION NUMBERS this method is most appropriate when dealing with covalent compounds1. 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) + Cr2O7

2-(aq) Ag+

(aq) + Cr3+(aq)

oxidation: lost 1 e- (x 6)

6 Ag(s) + Cr2O72-

(aq) + 14 H+(aq) 6 Ag+

(aq) + 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

processes1. 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.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.e.g., MnO4

– + N2H4 MnO2 + N2

4 [MnO4– + 2 H2O + 3 e- MnO2 + 4 OH–]

3 [N2H4 + 4 OH– N2 + 4 H2O + 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–

0 +6 -2 +1 +3


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 seriesvoltage – the potential energy difference per unit charge; measured in volts (V), 1 J/Celectric current – the rate of flow of charge past a point; measured in amperes (A), 1 C/s

TYPENAME OF

CELLHALF-REACTIONS

CHARACTERISTICS AND USES

primary cell – electric cell that cannot

be recharged

dry cell (1.5 V)

2 MnO2 + 2 NH4+ + 2 e- Mn2O3 + 2 NH3 + H2OZn Zn2+ + 2 e-

inexpensive, portable flashlights, radios

alkaline dry cell (1.5 V)

2 MnO2 + H2O + 2 e- Mn2O3 + 2 OH-

Zn + 2 OH- ZnO + H2O + 2 e-

longer shelf life, higher currents for longer periods

same uses as dry cell

mercury cell (1.35 V)

HgO + H2O + 2 e- Hg + 2 OH-

Zn + 2 OH- ZnO + H2O + 2 e-

small cell, constant voltage during life

hearing aids, watches

secondary cell – electric cell that can be recharged

Ni-Cad cell (1.25 V)

2 NiO(OH) + 2 H2O + 2 e- 2 Ni(OH)2 + 2 OH-

Cd + 2 OH- Cd(OH)2 + 2 e-

completely sealed, lightweight

power tools, shavers, portable computers

lead-acid cell (2.0 V)

PbO2 + 4 H+ + SO42- + 2 e- PbSO4 + 2 H2O

Pb + SO42- PbSO4 + 2 e-

large currents, reliable for recharges

all vehicles

fuel cell – electric cell

that produces electricity by a continually supplied fuel

aluminum-air cell (2 V)

3 O2 + 6 H2O + 12 e- 12 OH-

4 Al 4 Al3+ + 12 e-

high energy density, readily available aluminum alloys

electric cars

hydrogen-oxygen cell

(1.2 V)

O2 + 2 H2O + 4 e- 4 OH-

2 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 conditionsreference 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

standard oxidation potential (ΔEo°) – 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

ΔE° = ΔEr° – ΔEr

°

cell cathode anode

A positive standard cell potential (ΔE° > 0) indicates that the overall cell reaction is spontaneous.

e.g., SOA OAAg(s) | Ag+

(aq) || Cu2+(aq) | Cu(s)

RA SRA

2 [Ag+(aq) + e- Ag(s) ] Cu(s) Cu2+

(aq) + 2 e-

Cu(s) + 2 Ag+(aq) Cu2+

(aq) + 2 Ag(s)

ΔE° = ΔEr° – ΔEr

°

cell cathode anode

ΔE° = ΔE°Ag+ – ΔE°

Cu2+

= 0.80 – 0.34 = 0.46 V

organic compounds summary - fsmith …fsmith.pbworks.com/f/sch 4u review.doc · web viewone of a...

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