57135296 Chemistry 9701 Complete Book for a Levels

8/6/2019 57135296 Chemistry 9701 Complete Book for a Levels

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A Level. Chemistry (9701)www.revision-notes.co.cc

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Chapter 1 (AS-Level)Atomic Structure

Discovering the electron

A. Study of Cathode raysAt normal pressures, gases are poor conductors, but

at lower pressure they become better conductors of

electricity. Scientists studied the effects of passing

electricity through gases at low pressures.

A scientist named William Crookes saw that the glass

of the containing vessel opposite the cathodeglowed when the voltage applied was high.

A solid object placed between the cathode and the

glass causes a shadow. They proposed that the glow

was caused by rays coming from the cathode, called

cathode rays.

There was some argument about whether the rays

are waves or particles. The most important evidenceis that they are deflected by magnetic fields, which is

best explained by assuming that these are electrically

charged particles.

The direction of deflection, which is towards the positive pole shows that they are negatively

charged.

B. J.J. Thomsons e/m experiment

J.J. Thomson measured the deflection of a narrow beam of cathode rays in both magnetic andelectrical fields. His results allowed him to calculate the charge-to-mass ratio (e/m) of the particles.

Their charge-to-mass ratio was found to be exactly the same, whatever gas or types of electrodes

were used in the experiment.

The cathode ray particles had a very small mass, 1/2000 of a hydrogen atom. He decided to call

them the electrons.

The name was suggested by George Johnstone Stoney for the units of electricity.

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C. Millikans Oil drop experimentThe electron charge was first measured by the American Physicist Robert Millikan using his Oil

drop experiment.

He gave the oil drops

negative charge by spraying

them into air ionized by X-

rays. He adjusted the charge

on the plates so that the

upward force of attraction

equaled the downwards

force of gravity, and so the

drop would stay stationary.

Calculations on the forces allowed him to find the charges on the drops. These were multiples ofthe charge on an electron.

He found that the electron had a charge of 1.602 x 10-19

C (coulombs). They had a mass of 9.109 x

10-31

kg, which is 1/1837 of the mass of a hydrogen atom.

Discovering the protons and neutrons

The new atomic models; the Plum-Pudding, and the Nuclear AtomNew discoveries needed new atomic

models. If there are negatively charged

electrons in all neutral atoms, then there

must be a positively charged particle to

cancel them out.

For some time, the most favored was

the J.J. Thomsons Plum-Pudding

model, in which the electrons wereembedded in a pudding of positive

charge.

Then an experiment in 1909, by Hans Geiger and Ernest Rutherford, changed everything. They

were investigating how alpha particles were scattered when they were fired at very thin sheets of

different metals.

They detected alpha particles by flashes of light that were caused by the impact with a fluorescent

screen. Some, however, were deflected with large angles.



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The Plum-Pudding model couldnt explain the surprising observations. However, Rutherford

suggested that atoms consist of largely empty space and that the mass is largely concentrated into

a very small, positively charged nucleus.

Most alpha particles pass through the empty space in the atom with very little deflection. When

the alpha particle approaches on a path close to the nucleus, however, the positive charges

strongly repel each other, and the alpha particle is deflected through a large angle.

Particles in the nucleusThe protonRutherford reasoned that there must be particles that are responsible for the positive charge. He

and Marsden fired alpha particles through hydrogen, nitrogen and other elements.

They detected new particles with positive charge and the approximate mass of a hydrogen atom.

He called these particlesprotons.

The proton has a positive charge of 1.602 x 10-19

C, equal in size but different in charge of that of

an electron. It has a mass of 1.673 x 10-27

kg.

Each electrically neutral atom has the same number of protons as the number of electrons outside

the nucleus.

The neutron

The mass of the atom, which is concentrated in its nucleus, cannot depend only on protons, as

they usually provide around half of the atomic mass.

Rutherford then proposed that there is another particle in the nucleus which has the same mass of

the proton but with no electrical charge. As the particle is not charged, its detection was very

difficult, not until 1932.

One of his co-workers, James Chadwick, produced sufficient evidence for the existence of the

neutron.



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Using the apparatus Chadwick

bombarded a block of beryllium

with alpha particles. No charged

particles were detected on the

other side of the block. However,

when a block of paraffin wax was

placed near the beryllium, charged

particles were detected and

identified as protons.

The alpha particles had knocked out neutrons from the beryllium, and in turn these knocked out

protons out of the wax.

Order of discovery of subatomic particles

1. Electrons (cathode ray)2. Protons3. Neutrons (1932)

Atomic and Mass numbers

Atomic Number (Z)The difference between atoms of different elements is the number of protons in the nucleus of

each atom.

The atomic number shows:

The number of protons in the nucleus The number of electrons in a neutral atom of that element The position of the element in the periodic table

Mass number (A)It is useful to have a measure for the total number of particles in the nucleus of an atom. It is

called the mass number. For any atom:

The mass number is the sum of the number of protons and the number of neutrons in thenucleus



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Behavior of protons, neutrons and electrons in electric fieldsThe three particles behave differently in electric fields due to a different charge on them and their

relative masses.

Protons are attracted to the negative pole and electrons are attracted to the positive pole.

Because electrons are much lighter, they are deflected more.Neutrons are not deflected as they

have no charge on them.

Summary:Particle Name Relative Mass Relative Charge

Proton 1 +1

Neutron 1 0

Electron negligible -1

Isotopes

In Rutherfords atomic model, the nucleus consists of protons and neutrons, each with a mass of

one atomic unit. The relative atomic masses of elements should then be whole numbers. It was a

puzzle why chlorine has a relative atomic mass of 35.5.

The answer is that atoms of the same element are not all identical.

These atoms that have the same proton number but a different mass number are calledisotopes.

For example, hydrogen has three isotopes:

Isotope Protium Deuterium Tritium

Symbol

Protons 1 1 1

Neutrons 0 1 2

Electrons in atoms

The electrons are involved in the changes that happen in chemical reactions, so if we knew

everything about the arrangements of the electrons in atoms and molecules, we could predict

most of the ways that chemicals behave, purely from mathematics.



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There are some models that show how the electrons are arranged around the nucleus. The first

and most simple one is that thy orbit the nucleus which was rejected.

This is because the electrons would soon lose energy and fall down into the nucleus.

The second model is that they orbit around the nucleus, in shells or energy levels.

Arrangement of electrons: Energy levels and shellsIn 1913, a Danish physicist Niels Bohr proposed his ideas about arrangements of the electrons in

atoms.

The German physicist Max Planck had proposed in his Quantum Theory, that energy is atomic, and

can only be transferred in packets, or quanta.

Bohr applied this theory to the energy of electrons. So he suggested that electrons can only have

energy in quanta, and so they can only exist in quantized levels of energy.

If an electron gains or loses energy, it will travel up and down the energy levels, respectively, but

cannot stay somewhere in between.

These energy levels are most commonly called as shells.

Shells are numbered 1, 2, 3, 4, etc. These numbers are known as the principal quantum numbers,

which are given the symbol n. Such numbers correspond to the periods in the periodic table.

For a given element, electrons are added to the shells as follows:

Up to 2 in shell 1 Up to 8 in shell 2 Up to 18 in shell 3



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Ionisation energy

When an atom loses an electron it becomes a positive ion. We say that it has been ionized. Energy

is needed to remove electrons and this is generally called Ionisation Energy.

The first ionisation energy of an element is the amount of energy needed to remove one electron

from each atom in a mole of atoms of an element in the gaseous state.

The general symbol for ionisation energy is Hi and for the first ionisation energy it is Hi1.

X(g) X+

(g) + e-

1M 1M

2nd

Ionization Energy/Enthalpy/Potential

X+(g) X2+(g) + e-

1M 1M

3rd

Ionization Energy/Enthalpy/Potential

X2+

(g) X3+

(g) + e-

1M 1M

Examples of ionisation energies of the element Nitrogen:

Electrons

removed

1 2 3 4 5 6 7

N 1400 2860 4580 7480 9450 53300 64400

Trends in Hi (Ionization energy):

1) HIincreases as more and more electrons are removed i.e. H I2> HI1.2) There are one or more particularly large sizes of HIwithin the successive HI of each

element (except for H and He), due to the jump from one energy level to another.

Factors affecting HI:

1) The size of the positive nuclear charge; the greater the positive nuclear charge the greaterthe HI.

2) The distance of electrons from nucleus; the further the e- is from the nucleus, the weakerthe attraction and therefore the smaller the ionization energy needed.

3) The shielding effect of inner electrons; as the number of inner shells increases, the HIdecreases.



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In general:

1. As we go from left to right across a period, the nuclear charge increases and atomic sizedecreases, HIincreases.

2. As we go down a group the shielding effect and atomic size increases and HIdecreases.

HI increases (Periodic table)

HI decreases

3. Electrons are in discrete energy levels (quantized) (or shells) characterized by theirprincipal quantum numbers (n= 1,2,3,4, etc)

4. The nearest to the nucleus is shell K with n=1, the next is L with n=2. The furthest awayfrom the nucleus has the highest energy (energy becomes higher)(less negative)

5. The maximum number of electrons in an energy level (or shell) is given by the equation#e = 2n2where n is principal quantum number

Need for a more complex model

The electronic configurations are not so simple. For some elements, like lithium to neon, their first

ionization energies dont increase evenly.

This, and other complications, shows the need for a more complex model of electronic

configurations, other than the Bohrs model.



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Bohrs model as shown by quantum mechanics:

1) Energy levels consist of one or more sub-levels which contains orbitals at different

energies.

2) Sublevels are s, p, d, f types S contains one orbital P contains three orbitals D contains fve orbitals F contains seven orbitals.3) An orbital is a region in space around the

nucleus where there is a high probability

of finding a particular electron.

4) Orbitals have different 3D shapes, no exact boundaries. They are fuzzy like clouds, oftencalled electronic clouds, drawn with boundaries to represent 90-95% of the space in which

an electron exists.

5)

S orbitals are spherical

6) While P orbitals are dumbbell shaped with 2 lubes extending along X, Y and Z axes.

Filling the orbitals and electronic configurationsThree important rules:

1) Pauli Exclusion Principle.- Any orbital cannot hold more than 2

electrons, in this case they should be

paired (spin-paired)

2) In filling shells and orbitals, the orbitals ofthe lowest energies are filled first. (i.e. the

electronic configuration [distribution of

electrons] should be given in its lowest

energy state.

3) Orbitals of equal energies (degenerateorbitals) are filled singly first. Then you

should double up with more electrons.

(Lowest energy state = Ground state)



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Examples of electronic configuration of some elements:

Electronic configuration

for some of the

elements from 19 to 36:

19 K [Ar] 4s1

20 Ca [Ar] 4s2

21 Sc [Ar] 3d1

4s2

24 Cr [Ar] 3d5

4s1

25 Mn [Ar] 3d5 4s2

29 Cu [Ar] 3d10

4s1

30 Zn [Ar] 3d10

4s2

31 Ga [Ar] 3d10

4s2

4p1

35 Br [Ar] 3d10

4s2

4p5

36 Kr [Ar] 3d10

4s2

4p6

First 18 elements:

1 H 1s1

1 He 1s2

3 Li 1s2

2s1

4 Be 1s2

2s2

5 B 1s2

2s2

2p1

6 C 1s2

2s2

2p2

7 N 1s2

2s2

2p3

8 O 1s2

2s2

2p4

9 F 1s2

2s2

2p5

10 Ne 1s2

2s2

2p6

11 Na 1s2

2s2

2p6

3s1

12 Mg 1s2

2s2

2p6

3s2

13 Al 1s2 2s2 2p6 3s2 3p1

14 Si 1s2

2s2

2p6

3s2

3p2

15 P 1s2

2s2

2p6

3s2

3p3

16 S 1s2

2s2

2p6

3s2

3p4

17 Cl 1s2

2s2

2p6

3s2

3p5

18 Ar 1s2

2s2

2p6

3s2

3p6



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Some electronic configurations arrow in box form:

1s 2s 2p

H

He

Li

F

Ne

4s 3d 4p

K [Ar]

Sc [Ar]

Mn [Ar]

Ga [Ar]

Kr [Ar]

END OF LESSON



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Chapter 2Atoms, molecules and Stoichiometry

Counting atoms and molecules

There are two important definitions to remember in this chapter:

Relative Atomic Mass, Ar, or an element:o Average mass of one atom relative to the mass of one atom of C12 which is

considered to be 12 (atomic mass unit A.M.U)

Relative Isotopic Mass of an Isotope of an element: The mass of one atom of the isotope relative to that of one atom of C12.

To calculate the Ar of an element we have to consider all the isotopes of the element and their

abundance.

= ( %)Example, to find the relative atomic mass of chlorine:

Isotopes:

Chlorine-35, abundance = 75.5%

Chlorine-37, abundance = 24.5%Therefore:

= (

)

=

The mass of different molecules are compared in a similar fashion. The relative formula mass(Mr)

of a compound, is the mass of a molecule of the compound relative to the mass of an atom ofcarbon-12.

To find the relative Mr of a compound, we add up all the Ars of the elements in the compound.

Example for CH4:

= ( )

=



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Determination of Ar from mass spectra

Ar is determined using an instrument called the mass spectrometer. The instrument is shown

below:

Knowledge of the working of the mass spectrometer is not required by CIE.

The results of the mass spectrometer would be shown on a computer screen, as a chart of

abundance against mass. For example, for zirconium:

0

10

20

30

40

50

60

90 91 92 93 94 95 96

%A

geAbundance

Mass/charge ratios (m/e)

ex: Zirconium

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Counting chemical substances in bulk

The mole and Avogadros constant

A mole of atoms is a quantity that containsAvogadros number(61023) of atoms.

Similarly,

A Mole of molecules:o It is a quantity of the substance that

contains Avogadros number of

molecules.

(e.g. : a mole of ions a mole of

electrons)

In terms of mass,

A mole of atoms is a quantity in grams equal to the relative atomic mass.

For example, 1mol of S atoms weighs 32 grams.

Relative Molecular Mass (Mr) is the sum of atomic masses of all atoms in the molecule.

Examples are found in the book.

The empirical (simplest) formula & molecular formula:

The Empirical Formula:Of a compound shows the simplest whole-number ratio of the elements in the compound

The Molecular Formula:Of a compound shows the real number of each element in a molecule of a compound.

Example 1: SAQ 2.10 pg21

Q: Copper oxide has the following composition by mass: Cu = 0.635g ; O = 0.08g.

Calculate empirical formula of the oxide:

ANS:

Ar (Cu) =63.5 Ar (O) =16

Cu O

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2 1

Cu2O

Combustion analysis

The composition by mass of organic compounds can be found by combustion analysis. This

involves the complete combustion in oxygen of a sample of a known mass.

In combustion analysis, all the carbon is converted to carbon dioxide and all the hydrogen into

water.

These produced are carefully collected and weighed. Calculation gives the mass of carbon and

hydrogen present.

If oxygen is also present, its mass is found by subtraction (elimination). Other elements require

other methods.

- mass of C in a sample = mass of CO2

- mass of H in a sample = mass of H2O

Example:

SAQ 2.11 pg22

Q: On complete combustion of 0.4g of a hydrocarbon (only H and C), 1.257g of CO2 and 0.514g of H2O were

produced.

a) Find the Empirical formula of the hydrocarbonANS: Find C: 1.257

Find H: 0.514

C = 0.3428g H = 0.0571g

= 0.02856 = 0.0571

= 1 = 2

CH2

b) If relative molecular mass of the hydrocarbon is 84, what is its molecular formula

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ANS: mass of CH2 = 14

= 6

So, the molecular formula is C6H12

Calculations involving reacting masses:Fe2O3 + 3CO 2Fe + 3CO2

- molar mass of Fe2O3 = (256) + (316) = 160 g/mol

- one mole of Fe2O3 gives 2 moles of Fe.

160g of Fe2O3 gives (256) = 112g of iron

1000g of Fe2O3 gives 112

= 700g of iron


QUE: Calculate the mass of iron produced from 1000 tons of Fe2O3. How many tons of Fe2O3 would

be needed to produce 1 ton of iron? If the iron ore contains 12% of Fe2O3, how many tons of ore

are needed to produce 1 ton of iron?

ANS:

1) 1,000,000,000g of Fe2O3 gives 112

= 700,000,000g = 700 tons

2) 112

= 1,000,000, x = 1.43 tons3) 1.43

= 11.9 tons

Calculations involving concentration:Concentration is how much solute is available in a specific volume of solution.

Concentration by Mass:o how many grams of solute in 1 dm3 solution. (unit is g/dm3)(m/v)

Concentration by Moles (Molar concentration):o How many moles of solute in 1 dm3 solution (unit is moles/dm3(n/V)

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Example 1:

QUE: What amount of NaOH is present in 24.0cm3

of an aqueous 0.010 mol/dm3?

ANS:

Convert the volume to dm3

1dm3

= 10 10 10cm3

= 1000cm3

24.0 cm3

=

dm

3

Amount of NaOH in 24.0cm3

=

0.010mol

= 2.40 10-4

mol

Calculations involving gas volumes:

Equal volumes of different gases contain the same number of molecules under same conditions oftemperature and pressure, and this number is Avogadros Number.

The opposite is also true, equal numbers of molecules of different gases, under same conditions of

temperature and pressure occupy the same volume.

At room temperature and pressure (r.t.p), one mole of any gas occupies approximately 24dm3

(at

s.t.p, this is 22.5dm3).Reacting volumes of gases under same conditions of temperature and

pressure can be used to determine the formula and stoichiometry of reaction.

Example 1:

QUE: 10cm3

of hydrocarbon burned completely in 50cm3

of oxygen produced 30cm3

of CO2 at r.t.p.

Determine the formula of hydrocarbon and write a balanced equation of the reaction.

ANS:

HC(g) + O2 (g) CO2 (g) + H2O(l)

Volume: 10cm3

/ 10 50cm3/10 30cm

3/10 -

Gas Volume Ratio: 1 : 5 : 3

Gas Mole Ratio: 1 : 5 : 3

3 moles of C come from 3 moles of O2 react with 3 moles of CO2

5-3 = 2 moles of O2 which react with hydrogen

4H2 + 2O2 4H2O

2 moles of O2 react with 8 moles of H atoms, which gives C3H8

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O

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QUE: 20cm3of gaseous hydrocarbon Y burned completely in 60cm

3of oxygen to produce water

and 40cm3

of CO2 (@ r.t.p)

a) What is formula of hydrocarbon Yb) Write a balanced equation for the reaction.

ANS:

HC(g) + O2 (g) CO2 (g) + H2O(l)

Volume: 20cm3

60 cm3

40 cm3

--

Gas volume ratio: 1 : 3 : 2

Gas mole ratio: 1 : 3 : 2

4C + 4O2 4 moles of Co2 6-4= 2 moles to react with H2

8H + 2O2 4H2O C4H8/2 gives C2H4 + 3O2 2CO2 + 2H2O

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Summary:

Relative Atomic Mass, Ar, or an element is the average mass of one atom relative to the mass of

one atom of C12 which is considered to be 12 (atomic mass unit A.M.U).

Relative Isotopic Mass of an Isotope of an element the mass of one atom of the isotope relative to

that of one atom of C12.

= ( %)

A mole of atoms is a quantity that containsAvogadros number(61023) of atoms.

=

Relative Molecular Mass (Mr) is the sum of atomic masses of all atoms in the molecule

The Empirical Formula of a compound shows the simplest whole-number ratio of the elements in

the compound

The Molecular Formula of a compound shows the real number of each element in a molecule of a

compound.

=

=

=

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Chapter 3 (AS-Level)Chemical bonding and structure

A. Covalent bondingSimple molecular

- Liquids and gases- Low melting solids- H2O, NH2, C2H5OH, Sucrose

Properties:

A. Low boiling point and melting pointB. Nonconductors of electricityC. May be insoluble in H2O, may dissolve in organic solvents

Giant molecular- Solids of high melting point and boiling point- Quartz

Properties:

A. High boiling point and melting pointB. NonconductorsC. Insoluble in H2O and other solventsIn covalent bonds, electron pairs are shared between atoms. The electron pairs lying between

the two nuclei are attracted by both nuclei, thus bonding them and thus overcoming the

repulsion between them.

In covalent compounds, the shared electron pairs are in molecular orbitals rather than atomic

orbitals

Molecular Orbitals arise from the overlap of atomic orbitals.

Atomic Orbitals Molecular Orbitals

s, p, d , ,

H H

C = C

H H

Examples of simple covalent molecules:



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Molecular

formula

Structure Bonds

Cl2 Cl-Cl

CH4

H2O

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NH3

C2H4

CO2

O = C = O

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Dative covalent (coordinate) bond:

A lone pair from one atom overlaps with an empty orbital in another atom.

Ex: +

NH3 + [H] + H

H N H

H

+

H

H N H

H

Ex:

Al2Cl6

Cl Cl Cl

Al Al

Cl Cl Cl

Dative covalent

bond

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B. Bonds of intermediate(in between covalent & ionic) characterIonic, of covalent character:

Polarization of Ions:

These are ionic compounds that show some properties which are more characteristic of covalent

compounds. These ionic compounds contain anions which have become polarized.

1) Polarization of anion is the distortion of electronic cloud by the cation.2) Polarization brings more electron charge between the two ionic nuclei producing a certain

degree of covalent bonding.

3) Distortion:

+ _ purely ionic

+ _ partial ionization

+ _ extensive Ionic Molecule of Covalent Character

4) Cations with smaller radius and greater charge (i.e. greater charge density) have greaterpolarization on anions.

5) Anions with larger radius and greater charge are more easily polarized by cations.

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Covalent, of ionic character

Polarity of Covalent Bond

1) Covalent bonds in molecules become polar if there is difference in electronegativitybetween the elements. EN is the ability to attract the shared e-pairs.

EN increases Most EN

Least EN

Periodic Table

(F>O>N>Cl)

Most electro-negative

2) The more electronegative atom attracts the shared pair(s) more to itself leading toformation of a dipole. (i.e. ionic character).

3) In polyatomic molecules, the shape of the molecule must be taken into account.Diatomic and polyatomic molecules

Diatomic

To predict the polarity in a molecule, the molecular shape should be known

Polyatomic

Non-polar, the 4 polar bonds cancel

each other out.

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Shapes of simple molecules:

We apply VSEPR (Valency Shell Electron-Pair Repulsion Theory).

Bonding and non-bonding e-pairs repel each other. The repulsion causes these pairs to move as far

apart as possible. The orientation in space of these pairs determines the shape of the molecule.

CH4 (tetrahedron)

NH3 (trigonal pyramidal)

SO42-

ion

Lone pair

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H2O (V-shaped)

CO2 (linear)

BF3 (trigonal planar)

SF6 (octahedral)

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Summary:1) Simple Molecules, central atom has no LPs

Formula Shape

AB2 linear

AB3 trigonal planar

AB4 tetrahedral

AB5 trigonal bipyramidal

AB6 octahedral

2) Simple Molecules, central atom has LPsFormula # of LPs Shape

AB2 1 V-shaped

AB2 2 V-shaped

AB3 1 trigonal pyramidal

Lone pairs, bonding-pairs and bond angles:

Lone-pairs , Bonding-pairs , Bond Angles

LPs occupy bigger space than BPs

LP-LP > LP-BP > BP-BP

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Bond length and bond enthalpy:

Single Bond Length > Double Bond Length > Triple Bond Length

Bond Enthalpy: energy required to break 1 mol of given bond in 1 mol of gaseous molecules.

Single Bond Enthalpy < Double Bond Enthalpy < Triple Bond Enthalpy

C. Metallic bonding

Properties of Metals:

Shiny Good conductor of electricity & heat in solid state

Sonorous Ductile Malleable

Bonding: In metallic bonding (lattice), the atoms lose their electrons (outer-shell), which extend

throughout the lattice, thus forming a sea of electrons surrounding a lattice of positive ions.

Metallic bonding explains the properties of metals.

Intermolecular forces:

Instantaneous dipole, induced dipole (van der waals forces)

Present in atoms and non-polar molecules.

+

-

+

-

He

attraction of + to - charge

Ex:noble gases, alkanes, polymers like LDPE & HDPE.., graphite

Gases: Cl2 Liquids: Br2 Solids: I2

He

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Permanent dipole, dipole

Present in polar molecules.

Ex: H2O, NH3, HCl, CHCl3

Hydrogen bonding (H-bonding)

H-bonding is present in molecules with possibility of H-bonding. This is by far the strongest type ofintermolecular force.

i.e. with: F-H, O-H, N-H bonds

Ex: H2O, NH3, proteins, C2H5OH (carboxylic acids)

H

H H O

O H

H H O

Hydrogen bonding

Hydrogen bonding

END OF LESSON

Hydrogen bonding is responsible

for the liquid state of water, and

water is responsible for the

presence of life.

+

+

+

+

+

+

-

-

-

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Chapter 4 (AS-Level)States of Matter

The three states:

Intermolecular forces tend to bring order to molecules, while kinetic energy brings disorder.

Brownian motion occurs in only gases and liquids.

How much order is there in liquid?There is short-range order and long-range disorder.

Arrangement of Particles in Solids

There are different types of solids:

Giant Metallic (Metals) Giant Ionic (Ionic Compounds) Giant Covalent (Graphite, SiO2 diamond) Simple Covalent (I2)

Solid Liquid Gas

Impurity

Kinetic Energy Winning

Intermolecular Forces Winning

Order Disorder

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Difference in properties of solids, liquids, and gases:

They are due to differences in spacing and speed of particles.

Spacing

In gases, particles are much further apart than in liquids and solids. There is little difference

between liquids and solids.

Both (solids and liquids) are hard to compress Gases are poor conductors, because of the large distances between particles. Liquids and solids are better conductors

For heat to be transferred by particles, movement of energy of particles must be passed from one

to another (by collision in gases & liquids and vibration in solids).

Metals are unique in thermal and electrical conduction due to the presence of free electrons.

Solids Liquids Gases

Amount of order of

arrangement of

particles

Very orderly Short-range order;

longer-range disorder

Almost complete

disorder

Shape Fixed Takes shape of the

container

No Shape

Position of particles Fixed; no movement;

vibration in place

Some movement Always moving rapidly

Spacing of particles Close (10-10

m) Close (10-10

m) Far apart (10-8

m)

Compressibility Very low Very low High

Conduction of heat Metals & graphite

conduct; others poor

Metals very good;

others poor

Very poor

Why gases liquefy and solids melt?

By bringing molecules closer (applying pressure) and slower (reducing their temperature),

intermolecular forces become sizable and overcome repulsive forces. Molecules stick together

leading to liquid state.

Each gas has its characteristic T at which intermolecular forces are strong enough to win (Tc critical

temperature).

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For melting, heating is necessary, i.e. thermal energy to increase kinetic energy of molecules must

be supplied to overcome intermolecular forces to allow molecules to be free to move.

Explaining changes of state

Experiment to measure the vapor pressure of a liquid:

Read page 57. Not required for examination.

Remarkable Substances:

Liquid Crystals

These are liquids which have sufficient long-range order to behave like solids (at certain range of

temperature). Usually molecules are thin, long, and not very symmetrical. Arrangement of

molecules can be upset by slight changes in the surroundings.

At equilibrium, the rate at

which molecules leave the

liquid equals the rate at which

molecules join the liquid.

Solid LiquidFusion - Melting

Freezing - Solidification

GasLiquid

Evaporation - Boiling

Condensation -

Liquefaction

A liquid boils when its vapor

pressure equals atmospheric

pressure.

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Behavior of Real Gases

Real Molecules Occupy Space Less volume than ideal volume

Real Molecules Attract each other Less pressure than ideal value

Real Gas Equation

Real gases approach ideal gas behavior quite away from conditions of their liquefaction.

Opposite of Liquification= Low pressure (small intermolecular forces) and high temperature (large

distances between molecules)

END OF LESSON

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Chapter 5a (AS-Level)Chemical Energecs

Chemical reactions are accompanied by enthalpy changes, in the form of heat

energy.

Enthalpy, given the symbol H, is the total energy content of thee reacting materials.

Enthalpy changeis the change in the total energy content of the reacting materials, given

the symbol H (Delta H).

When H is negative, the reaction is exothermic; when its positive the reaction is

endothermic.

Some chemical reactions are:

Exothermic:Examples:

- Combustion of fuel- Respiration in our bodies- Reaction of metals with acids- Reaction of alkalis with acids- Reaction of quicklime with water:

CaO(s) + H2O(l)Ca(OH)2 (aq) + E

Endothermic:Examples:

Decomposition of limestone:CaCO3(S) + E CaO(s) + CO2 (g)

Photosynthesis:CO2 (g) + H2O + E Carbohydrates (Glucose)

Dissolving NH4Cl:NH4Cl(s) + H2O + ENH4+ (aq) + Cl- (aq)

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Energy is conserved

Energy is neither created nor destroyed, in either exothermic or endothermic reactions. In

endothermic reactions, the energy is transferred to the reactants (the system) from the

surroundings. In exothermic reactions it is transferred to the surroundings, all in the form ofheat.

Enthalpy and Enthalpy changes

Mentioned earlier are the definitions of enthalpy and enthalpy change.

H has a unit of kilojoules per mole (kJ mol-).

We can illustrate enthalpy changes using Enthalpy Profiles (or Enthalpy graphs). Shown

below are some examples:

Example:

() () () ()

In which Delta H = -890 kJ/mol.

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Standard Enthalpy Changes (H)

When we want to compare the enthalpy changes of reactions, we must use standard conditions.

A standard enthalpy change for a reaction takes place under these conditions, which are known as

standard:

Pressure = 1 atm = 100kPa Temperature = 298K Mole = 1mol Concentration = 1M =1

of solutions

Reactants and Products in normal physical and standard condition

Standard enthalpy change of reaction (Hr)

Hr is the enthalpy change when the amounts of reactants given in reaction equation react under

standard conditions to give products in the standard conditions.

Example:

2H2 (g) +2O2 (g) 2H2O(l)

HR

= -572kJ

BUT

H2 (g) +

O2(g) H2O(l)

HR = -286 kJ

Standard enthalpy change of formation of

compounds (Hf)

HF is the enthalpy change when 1 mol of the

compound is formed from its elements under

standard conditions.

Example:

HF (H2O) = -286 kJ/mol

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Standard enthalpy change of combustion (Hc)

HC is the enthalpy change when 1 mol of the substance is completely burned in oxygen under

standard conditions.

Example:

HC (H2) = -286 kJ/mol

Other enthalpy changes

We can define other standard enthalpy changes in a similar manner:

The standard enthalpy change of hydration is the enthalpy change when one mole ofa gaseous ion dissolves in water to give an infinitely dilute solution.

The standard enthalpy change of solution is the enthalpy change when one mole of asolute dissolves in a solvent to give an infinitely dilute solution.

The standard enthalpy change of neutralization is the enthalpy change when onemole of H

+ions from an acid is completely neutralized by an alkali to give one mole of

water.

The standard enthalpy change of atomization of an element is the enthalpy changewhen one mole of gaseous atoms is formed from one mole of the element in its

standard state.

Bond making, breaking and enthalpy change

Bond Making is Exothermic

Bond Breaking is Endothermic

Bond Enthalpy (BE) is the energy needed to break 1 mole of a particular covalent bond.

This indicates the strength of the bond, the higher the energy needed, the higher bond strength

and vice versa.

Bond enthalpies

It is useful to measure the amount of energy need to break a covalent bond, as it indicates

the strength of the bond. It is called Bond Enthalpy.

They are also:

All are positive quantities

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All are average values Compared from bonds in gaseous compounds Difficult to measure directly They are given in tables (such as data booklets which are given in AS/A level exams)

Example:

Bond H-H C-C C=C C-H O=O O-H C-O C=O

Enthalpy

Change

+436 +347 +612 +413 +498 +464 +358 +805

Calculating Enthalpy change of a reaction from bond enthalpies:

CH4 (g) + 2O2 (g) Co2 (g) + 2H2O(l)

O=C=O

Form 2 C=O bonds

H O=O

H C H O=O O

H H H

Break 4 C-H bonds Break 2 O=O bonds O

H H

Form 3 O-H bonds

Hr= (sum of all BEs from broken bonds) (sum of all BEs from formed bonds)Calculation:

HC= Hr = [4BE(C-H) + 2BE(O=O)] [2BE(C=O) +4BE(O-H)]

= 4(413) +2(498) 2(805) 4(464)

= -890kJ

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Measuring energy transfers and enthalpy changes

Enthalpy change of combustion

Measurements ofHr are important as they help to compare the energy available from the

oxidation of different flammable liquids, which may be used as fuels.

The apparatus is shown below:

Measurements that are taken:

Mass of cold water The temperature rise of the water The loss in mass of the fuel

The specific heat capacity of water is also needed, which is 4.2 Jg-1

K-1

.

Specific heat capacity of a substance is the heat required to raise the temperature of a unit

mass of a substance (one gram or one kilogram) one degree Celsius or Kelvin.

Therefore:

While heat capacity of a substance or apparatus, is the heat required to raise the

temperature of the whole substance or apparatus one degree Celsius or kelvin.

Heat capacity is determined in a different experiment.

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See page 77 & 78 for diagram and explanation + solved example.

H of neutralization of acid and alkali

Use known amounts of acid and alkali (measured volumes and masses in case of solid Mix quickly and measure temperature change up to 0.1 degree Celsius to 0.2 degree

accuracy

Use the equation : Calculate for 1 mol of H+

Summary:

() () ()

Mass of liquid in

calorimeter in

grams

Enthalpy change

of combustion in

kJ/mol

Heat Capacity of

the liquid in

Jg-1

K-1

Change in

temperature in

K

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Enthalpy change of solution of NaOH

Method:

Use polystyrene cup to hold the solution. Weigh and add 100g distilled water.

Determine initial temperature. Add few pellets of solid NaOH quickly to the water. Stir mixture immediately and start a timer to take temperature readings at regular

intervals.

Temperature rises to a max and then drops, readings should be taken for enoughtime.

Weigh the cup and find mass of NaOH added. Plot temperature versus time and extrapolate to find T as in the figure.

Heat evolved = m x 4.18 x T

Then calculate for 1 moleH solution

Example:

Mass of cup = 8g Mass of cup + water = 108.15g Mass of distilled water = 100.15g Mass of NaOH + cup + water = 114.35g Mass of NaOH = 6.20g Initial temperature of cup = 15 C

From the graph, T = 16C

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Calculation:

E = m.c.T

E = 100.15 x 4.18 x 16 = 6700J released when dissolving 6.2g of NaOH.

For 1 mole of NaOH 40 grams should be used (or calculated) to get the enthalpy change ofsolution of NaOH.

Enthalpy change by different routes (Hess law)

In chemical reactions:

Hess law states that:

The total enthalpy change for a chemical reaction is independentof the route

followed from reactants to the products, provided the initial and final conditions

are the same.

So we deduce the following (using the diagram):

Hess law is another form of the law of conservation ofenergy, which is the first law inthermodynamics.

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Using Hess law in enthalpy change calculations

A. Calculating Hr from bond enthalpyExample:

In Habers process:

N2 (g) + 3 H2 (g) 2 NH3 (g)

Hr = Hf = Broken bonds Formed bondsHr = Hf = BE (NN) +3 BE (H-H) 6 BE (N-H)

B. Calculating Hr from enthalpy changes of formation HfExample (given the Hf of reactants and products):

CaO (s) + H2O (l) Ca(OH)2 (s)

O2 + Ca H2 +

O2

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Hf (CaO) + Hf (H2O) + Hr = Hf [Ca(OH)2]

(Figures given by the examiner)

(-6351) + (-285.8) + Hr = (-984.1)

Hr = -65.2 kJ/mol of Ca(OH)2

Therefore:

Hr = *Hf (Products)+ *Hf (Reactants)+C. Calculating HFormation from HCombustion

Example:

Find Hf (CH4), given HCombustion

C (s) 2 H2 (g) CH4 (g)

Hf (CH4) + HC (CH4) = HC (C) + 2HC (H)

Hf (CH4) = (-394) + 2(-286) (-890)Hf (CH4) = -76 kJ/mol

END OF LESSON

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Chapter 6 (AS-Level)Electrochemistry

Elements have different combining power, which is different ratios of atoms that combine with each other

is called valency (Strength).

Example:

MgO which has a ratio of 1:1 Al2O3 which has a ratio of 2:3

At present a more useful measure is used, Oxidation state or Oxidation number.

The charge an atom would have in a molecule or ion, if electrons were completely transferred

(to the more electronegative atoms) in covalent molecules or ions.

Rules for assigning oxidation numbers:

1. Oxidation numbers are calculated as the number of electrons that an atom loses, gains or shares informing ionic or covalent bonds.

2. The oxidation state of free element is zero. E.g. H2, Br2, O2, etc.3. The oxidation number of a simple ion (monatomic) is its charge. E.g. Na+ (+1)4. The oxidation number of hydrogen in compounds is (+1), except in metal hydrides, where it is (-1)5. The oxidation number of oxidation in compounds is (-2), except in peroxides and in OF2 where it is

(-1) and (+2) respectively.

6. In the molecules and ions, the more electronegative atom in given the negative oxidation numbers.7. The sum of oxidation numbers of all atoms in a neutral molecule is zero.8. The sum of oxidation numbers of all atom in a complex ion is equal to the charge of the ion.

Oxidation Numbers Calculation Examples:

The oxidation number of Carbon in CO2 is:(-2) x 2 = (+4)

The oxidation number of Carbon in CH4 is:X + 4 x(+1) = 0, therefore X = (-4)

of Mn in KMnO4 is:1 + X + 4(-2) = 0

X = +7

of Cl in KClO3 is:1+ X + 3(-2) = 0

X = +5

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Redox Reactions:Reduction Oxidation Reactions:

The term redoxis used for the simultaneous processes of reduction and oxidation, in which one element is

reduced and another oxidized.

Examples:

Both are redox reactions.

Fe is oxidized, it is the reducing agent.

O is reduced it is the oxidizing agent.

From the example:

Fe had an oxidation number of 0. After the reaction, the Fe lost 3 electrons, so its oxidation number increased.Similarly, the oxygen had an oxidation number of 0 before the reaction and gained 4 electrons, so its oxidation

number decreased from 0 to -2.

Oxidaon is: The loss of electrons. Increase in oxidaon number.

Reducon is: The gain of electrons. The decrease in oxidaon number.

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Electrolysis

It is a redox Chemical reaction that takes place when a direct current is passed in an aqueous solution or

molten ionic compound.

Oxidation takes place at the anode.

Reduction takes place at the cathode.

Example:

Electrolysis of H2O

4 OH- O2 + 2 H2O + 4 e

-

4 H+

+ 4 e- 2 H2

Net reaction:

4 OH-+ 4 e

- 2 H2O + 2 H2 + O2

More information available in the book.

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Electrolysis of Brine using Diaphragm cell

At the anode (oxidation):

2 Cl- 2 e

-+ Cl2

At the cathode (reduction):

2 H+

+ 2 e- H2

Net reaction:

H2O H+

+ OH-

OH-+ Na+ NaOH

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Extraction of Aluminium from Alumina

The method used to extract the metal form its ore depends on the position of the metal in the reactivity series. If the

metal is high up in the series, its ores are stable and can be only be obtained by electrolysis.

Aluminium is extracted from bauxite (Al2O3) by electrolysis. However, bauxite has a very high melting point but can

be dissolved in molten cryolite at 900 degrees Celsius. In other words, the cryolite is used to lower the melting point

of bauxite.

Cathode reaction:

4 Al3+

+ 12 e-= 4 Al

Anode reaction:

6 O2-= 3O2 + 12 e

-

Carbon dioxide is also produced from this reaction. It is produced from the carbon electrodes burning in the heat

and oxygen produced.

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Electrolytic purification of copper

Metals can be refined or purified by electrolysis. The impure metal forms the anode and the pure metal forms the

cathode. The electrolyte is an aqueous metal salt.

In the purification of copper, impure copper is used as the anode and a thin sheet of pure copper is used as the

cathode.

The following reaction occurs:

At anode, the copper is ionized (becomes an ion):

Cu -2e-= Cu

2+

At the cathode, the copper ion is unionized, which produces solid copper on the cathode:

Cu2+

+2e-= Cu

As electrolysis takes place the pure copper sheet gains mass and the anode (impure copper) loses mass and the

impurities are deposited under the anode in the container.

This means that the copper ions had left the impure copper side and joined the pure copper side.

The colour of the copper (II) sulphate solution does not change as the concentration of copper ions in the solution

does not change.

END OF LESSON

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Chapter 7a (AS-Level)Equlilibria

Reversible Changes:

1. Physical changes, examples: Melting of ice, freezing of H2O

H2O H20 Dissolving CO2 in H2O like in fizzing drinks

CO2 (aq) CO2 (g)2. Chemical changes, examples:

Formation of ozone and its decomposition to oxygen3 O2 + UV Light + CFCs 2 O3

Thermal decomposition of CaCO3 to CaO and CO2, and the formation of CaCO3 fromCaO left in an atmosphere of CO2

CaCO3 (s) CaO (s) + CO2 (g)

Equilibrium, a state of balanced change:

Examples:

Dissolve salt in H2O until no more dissolves (saturated solution + non-dissolved salt) =equilibrium between solution and solid, which is concentration of saturated solution

stays constant (at a constant temperature).

Although the ions are in constant motion, moving from solid to solution and to

solution to solid.

NaCl (s) Na+ Cl- (aq)

Similar equilibrium in butane cylinder.Molecules in liquid are in equilibrium with those in the gas phase

C4H10 (l) C2H10 (g)

Equilibrium and chemical changesFor example:

CaCO3 + Heat CaO + CO3

There is no equilibrium in an open container.

CaO + CO2 CaCO3

In an atmosphere of carbon dioxide in a closed container.

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In general, increasing the pressure shifts the equilibrium to the side of fewer gaseous molecules

and vice versa.

Effect of catalyst on equilibrium

Catalysts have no effect on equilibrium, but allow the equilibrium to be achieved faster.

They do this by decreasing the activation energy, and hence speed it up. More catalyst could mean

a faster reaction rate, but doesnt change the equilibrium concentration of reactants or products.

Law of chemical equilibriumFor the general equation:

aA +bBcC + dD

The following is always true:

KC is the equilibrium constant;

A, B, C and D are the concentration of the corresponding substances in the general equation;

The superscript, a, b, c and d are moles of the corresponding substances in the general equation.

For example:

H2 + I2 2HI

Calculating KC

For example:

OH

CH3COCH3 + HCN CH3CCH3

CN

Initial concentration 0.05 mol/dm3

0.05 mol/dm3

0.00 mol/dm3

Change -0.0233 mol/dm3

-0.0233 mol/dm3

+0.0233 mol/dm3

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KCand Le Chateliers principle

For example:

2 HI H2 + I2 Forward reaction is endothermic

When the temperature is raised:

The concentration of products increase, while

the concentration of reactants decrease, so KC

has to increase.

Example 2:

2 SO2 + O2 2 SO2 Forward reaction is exothermic

The concentration of the product decrease as the temperature is increased, while the

concentration of the reactants increases, so KC has to decrease.

KC and concentration changes

For example:

2 HI H2 + I2

At 500K,

If the concentration of HI is increased, KC decreases. So to maintain equilibrium, the concentration

of H2 and I2 has to increase so the value of KC reaches 6.25x10-3.

Equilibrium constants and pressure changesIn reactions involving gases, KP (of pressures) is used.

E.g.

2 HI H2 + I2

p(H2) indicates the equilibrium partial pressures of H2 in a closed system containing the equilibrium

mixture at constant temperature.

The partial pressure of a gas in a mix of gases is the pressure exerted by that gas alone, if it

occupies the container alone.

Calculating partial pressures

For example, air at 500kPa has 1 mole oxygen and 4 moles nitrogen.

(1)Mole fraction =

Of oxygen = , and of nitrogen =

(2)Partial pressure = mole fraction x total pressure

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p(O2) = x 500 = 100kPa

p(N2) = x 500 = 400kPa

Haber process and KPThe Haber process is a process in which ammonia is produced from reacting hydrogen and

nitrogen together.

The reaction is:

N2 (g) + 3 H2 (g) 2 NH3 (g)

The conditions are:

A pressure between 2.5 and 15 MPa A temperature between 670K to 770K

In terms of partial pressures of nitrogen, hydrogen and ammonia, the equilibrium constant

expression is:

Using KC and KPExample 1:

CH3COOH(l) + C2H5OH(l) CH3COOC2H5 (l) + H2O(l)

KC =

Example 2:

H2(g) + I2 (g) 2HI(g)

KC =

Example 3:

3H2(g) + N2(g) 2NH3(g)

KC =

Example 4:

Mixture of 0.500 mol/dm3

H2 and 0.500 mol/dm3

I2 was placed in a 1.00 dm3

stainless-steel flask at

430 degree Celsius. Calculate the equilibrium concentration of H2, I2 and HI.

KC of the reaction is 54.3 at this temperature.

2 HI H2 + I2

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Initial 0.500 mol/dm3

0.500 mol/dm3

0.00 mol/dm3

Change -X mol/dm3

-X mol/dm3

2x mol/dm3

Equilibrium (0.500 x) mol/dm3

(0.500 x) mol/dm3

2x mol/dm3

()

() ()

So the equilibrium concentrations are:

Equilibrium 0.091 mol/dm3

0.091 mol/dm3

0.818 mol/dm3

The Haber process3 H2(g) + N2(g)

2NH3(g) (H = -93 kJ/mol)

Due to the unreactive nature of nitrogen, a high amount of energy is needed. Increasing thetemperature would drive the equilibrium to the left as the forward reaction is exothermic.

Pressure is also used to increase the reaction rate, which would drive the equilibrium to theright. Very high pressures increase the cost of the plant, and low pressures make the

reaction very slow.

A catalyst can be used to overcome the problem of the low rate of reaction. An iron catalystis used, with small amounts of K, Mg, Al, and Si oxides that improve the efficiency of the

catalyst.

The rate of the reaction can be increased by increasing the temperature and by accepting alower equilibrium percentage of ammonia in the mixture.

Ammonia is removed as it is formed, so the reaction mixture is not left to reach equilibrium. Rapid expansion is used to cool the mixture. Ammonia liquefies, while unreacted nitrogen

and hydrogen are recycled into the reaction vessel.

Conditions are:

Between 2.5 and 15MPa pressure 670 to 770K temperature Finely divided or porous iron catalyst with metal oxide promoters

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The contact processIn this process, sulphuric acid is manufactured from sulphur.

(1)Sulphur is burned in air at 1000C to make SO2S + O2 SO2

(2)Sulphur dioxide is converted into Sulphur trioxide by using vanadium (v) oxide catalyst.2 SO3 + O2 2 SO3 (H = -197 kJ/mol)

(3)Sulphur trioxide is dissolved in sulphuric acid to make oleum (H2S2O7) which is then dilutedinto sulphuric acid.

SO3 + H2SO4 H2S2O7

H2S2O7 + H2O 2 H2SO4

Conditions are:

V2O5 catalyst Temperature of 400 to 600C Pressure just above atmospheric pressure

These conditions used to convert sulphur dioxide into sulphur trioxide are a compromise. Higher

yields of SO3 are produced when using air in excess, forcing equilibrium to the right.

A temperature of 400C is used because the catalyst is inactivated below it.

Uses of sulphuric acid include:

The manufacture of paints The manufacture of detergents and soaps The manufacture of phosphate fertilizers The manufacture of dyestuff

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With metal carbonates

2 H+

+ 2 Cl-+ CuCO3 Cu

2++ 2 Cl

-+ H2O + CO2

Net ionic equation:

2 H+

+ CuCO3 Cu2+

+ H2O + CO2

Definition of acids and basesArrhenius (1884):

An acid is a substance that produces an excess of H+ in aqueous solutions

A base is a substance that produces excess of OH+ in aqueous solutions.

Brnsted & Lowry

An acid is proton donor.

A base is a proton acceptor.

Examples:

When hydrochloric acid dissolves in water:HCl + H2O H3O+ + Cl-

Proton Donor Proton Acceptor Conjugate acid of H2O Conjugate base of HCl

When ammonia dissolves in water:

Ammonia accepts a proton from the water and so its a base. The ammonium ion is its conjugate

acid. Water donates a proton, so its an acid. The hydroxide ion is its conjugate base.

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The role of waterPure water conducts electricity. This is a fact. This means that pure water contain ions of itself.

Pure water can also be electrolyzed by a direct current. Every now and then, one water molecule

reacts with another water molecule to form ions. A proton leaves one water molecule and is

accepted by the other. These ions transfer electrons during electrolysis.

2 H2O (l) H3O+

(aq) + OH

-

(aq)

Or more simply as:

H2O (l) H+

(aq) + OH-(aq)

Base behavior and neutralizationWhen acid and bases react, they are said to neutralize each other. For example, in the acid HCl,

there are 2 ions, H+

and Cl-and a water molecule. In the base NaOH, there are Na

+and OH

-ions

and a water molecule. When these are mixed, the protons and hydroxide ions react to form water.

H+

(aq) + OH-(aq) H2O (l) (H= -57 kJ/mol)

This is what neutralization is the formation of water by the exothermic forward reaction. The

ions remaining stay dissolved in water.

Acids and bases of varying strengthStrong acid and bases are those which are totally ionized when dissolved in water. The strong acids

include hydrogen halides and strong bases include the group I hydroxides.

Weak acids and bases dont ionize totally when they dissolve in water. They even might not ionize

at all. Examples for weak acids include Ethanoic acid. They hardly donate protons at all.Organic

acids like Ethanoic acid and citric acids are typical weak acids. Weak acids are similarly weak in

accepting protons. These include the conjugate bases of strong acids, such as sulphate and

chloride ions.

Acid Base

Strongest acid

Weakest acid

Hydrochloric HCl H+

+ Cl- Chloride Weakest base

Strongest base

Benzoic C6H5COOHH+

+

C6H5COO-

Benzoate

Ethanoic CH3COOHH+

+

CH3COO-

Ethanoate

Ammonium NH4+H+ + NH3 Ammonia

Phenol C6H5OHH+

+ C6H5CO- Phenoxide

Hydrogen carbonate HCO3-H+ + CO3

2- Carbonate

Water H2OH+

+ OH- hydroxide

END OF LESSON

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Chapter 8a (AS-Level)Reacon kinecs

Rates of reaction why bother?There are many reasons why chemists study the rate of reaction, for example to:

Gain insight into the reaction mechanism (how the reaction proceeds). Understand the chemical processes taking place in our bodies and environment. Improve the rate of the production in chemical industries.

When a reaction proceeds through several steps, the slowest step in the mechanism determines

the overall rate of the reaction, and it is called the rate-determining steps.

Chemical reactions proceed at different rates

The rate of reactions tells us how fast the reaction occurs. It is measured by amount of a reactant

used up, or the amount of product produced, in a given time.

OR:

The reacon rate is a posive quanty.

Example:

H2O2 + 2 I +2 H 2 H2O + I2

If concentration of I2 is changed from 0 to 10-5

mole in 10 seconds, then the reaction rate is:

The unit is always mol.dm-3

.s-1

When t approaches 0,

, which is the instantaneous reaction rate at time t.

In practice, it is the slope of the tangent at time t, to the concentration-time curve.

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4. In some reactions, the intensity of lightIncreasing the light intensity of radiation (visible or UV light) increases the rate.

For example, in the free radical substation of methane by chlorine, an increase in the intensity of

UV light increases the rate of the reaction.

5. CatalystsCatalysts usually speed up the reaction by lowering the activation of the energy of the reaction.

For example, nickel is used in the hydrogenation of vegetable oils to make margarine.

Monitoring and measuring reaction ratesWe can measure reaction rates in a variety of ways to study the factors which affect the chemical

reaction rates. To study the effect of temperature on a reactions rate, all the other variables must

be kept constant, while the temperature may be changed. A fair test is required.

We must also know the mole ratios of the reactants and products as shown by the balanced

chemical equation. No side reactions must be taking place, as these will affect the measurements.

Now the way in which the reaction progress can be monitored must be decided. There are 2 types

of method accessible to us:

1. Destructive methodThis based on chemical analysis, for example, using titration. Samples are taken from reaction

mixture at suitable intervals, quenched (to stop the reaction, using ice for example) and analyzed

(and the concentration is determined).

2. Non-destructive methodsThis is done using physical methods to measure a property that changes as reaction proceeds.

Properties like mass, volume, pressure, pH, colour, electric conductivity can be used at regular

time intervals (quenching not included).

Monitoring reaction rate using mass loss

When a gas is evolved during a

reaction, monitoring mass loss

may provide a suitable method of

measuring the reaction rate.

There needs to be sufficient loss

in mass to be followed using a

reasonably accurate balance. For

example, 2.00 g of small marble

chips will give a satisfactory loss

in mass when treated with 150

cm3

of hydrochloric acid.

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Monitoring reaction rate using colour intensity

Using colour intensity, concentration of a specific substance can be measured. For example, the

concentration of copper ions in the following experiment can be estimated using a

spectrophotometer.

The copper ions replace the atoms of iron. After several minutes, the blue colour of (a) has

become paler and a red-brown deposit has formed on the iron wool (b).

The spectrophotometer is a device that measures how much light of a particular wave-length can

pass through a sample, liquid or gas.

The beaker in (a) contains 1 dm3

of 1.00 mol dm-3

aqueous copper sulphate. Iron wool reacts with

the copper ions in the solution, displacing them and changing the colour of the solution as a result.

Cu2+

(aq) + Fe (s) Fe2+

(aq) + Cu (s)

Solutions appear coloured because they absorb radiation in the visible region of the spectrum.

Aqueous copper sulphate absorbs radiation in the yellow, orange and red regions, which only

allows blue light to pass through and so the solution looks blue.

Measuring reaction rates of gases using pressure changes

Measurements of pressure changes at a given

temperature can be used to calculate concentration

change as a reaction proceeds. For example, this

method can be used to monitor the production of

carbon dioxide from limestone in a sealed container.

CaCO3 (s) CaO (s) + CO2

The figure shows a notional system to investigate the

effect of heat on the decomposition of limestone.

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Monitoring reaction rates of solutions using chemical analysis

If there is a change in acidity or basicity as a reaction proceeds, suitable titrations can be made to

follow the rate. The rate of formation of sulphurous acid, H2SO3 (a component of acid rain formed

by the reaction of SO2 with water), could be followed by measuring the increase in concentration

of hydrogen ions produced.

This is monitored by titrating samples of the increasingly acidic solution against a basic solution of

known concentration, for example, 0.001 mol dm-3

aqueous sodium hydroxide. The more sodium

hydroxide that is needed to neutralize the sample, the more sulphuric acid is present.

The collision theory of reactivityFor a reaction to occur:

1. There should be collisions between the molecules of reactants2. These collisions should be energetic enough3. They should occur with the right orientation of molecules

Important points:

Molecules will react only if they collide with each other And if there is enough energy in the collision Increasing the concentration increases the probability of collision of collision, which

increases the rate of the reaction

Increasing the temperature increases the proportion of molecules with sufficient energy toreact, which increases the rate of reaction

The Boltzmann distribution of molecular energiesThe Boltzmann distribution represents the number of particles with particular energies.

Billions and billions of particles in a gas are in constant random motion. A few are almost

motionless. A minority have momentarily speeds far above the average. The majority of particles

have speeds around the average. This is shown by the graph below. The difference between the

energy of the molecules is only due to speed (as they have the same mass).

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With a catalyst:

A catalyst is a substance added to reactions in the purpose of increasing the rate. It doesnt change

at the end of the reaction. It lowers the energy of activation which is achieved by allowing the

reaction to take place by a different mechanism or by a different pathway.

The reaction rate increases because the catalysed reaction pathway has lower activation energy

than that of the uncatalysed reaction. The Boltzmann distribution below shows how the lower

activation energy increases the number of molecules that will react on collision.

EnzymesThese are proteins that act as biological catalysts. Enzymes:

Show great specificity Are extremely sensitive to changes in pH and temperature Are extremely sensitive to molecules that are inhibitors and cofactors Are far more efficient than the inorganic catalysts used in the chemical industry

END OF LESSON

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Chapter 9 (AS-Level)Chemical Periodicity

Versions of the periodic tableThe most common versions of the periodic table include vertical groups I, II, VII, and horizontal

periods from 1 to 7 and blocks of elements, and are named as follows:

S-block, which contains groups I &II P-block, which contains groups III to VII & 0 D-block, which contains transition metals F-block, which contains lanthanides & actinides

Periodic patterns of physical properties of elements (the first 36 elements

only)

Electronic configuration:

In S-block, the outermost electrons are in s-orbital, in p-block the outermost electrons are inthe p-orbitals.

Elements in same group have the same number of electrons in their outer shell For the elements in groups I 0 the number of outer shell electrons is the same as the

group numbers

Group 8 (noble gases) have 8 es in their outermost shell.Atomic radiiAtomic radius is 2 types:

Covalent radius, which is half the distance between nuclei in neighboring atoms in amolecule.

Metallic radius, which is half the distance between neighbouring nuclei in metallic crystal(lattice).

Trends in atomic radii:

Increase down a group Decrease across a period Relatively constant in T.M energies Opposite trends to ionization energies

These trends are due to the combined effect of:

Nuclear charge Distance of outer electronic shell from nucleus (atomic size) Shielding effects of inner electronic shells

Ionic radii

Ionic radii decrease across a period The radii of positive ions (cations) are smaller than those of corresponding atoms

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F: 1s2, 2s

2, 2p

5 1680 kJ/mol

Ne: 1s2, 2s

2, 2p

6 2080 kJ/mol

Na: 1s2, 2s

2, 2p

6, 3s

1 494 kJ/mol

This is due to the combined effect of these factors:

Nuclear charge (protons in nucleus) Atomic size (distance of outer electrons from nucleus) Shielding effect of inner shells

First ionization energies in periods

Considering Li, Be and B:

Li: 1s2, 2s1 IE = 519 kJ/mol

Be: 1s2, 2s2 IE= 900 kJ/mol (Has the same shielding effect but greater nuclear charge and

smaller size).

B: 1s2, 2s2, 2p1 IE = 799 kJ/mol (The electron in the p-orbital is of higher energy level, so is

easier to remove).

From the above graph, a general trend can be seen across periods, but this trend is uneven. For

example look the above elements, Lithium, Beryllium and Boron. Someone might predict that

Boron has the highest first ionization energy, but Beryllium does.

From experimental evidence like this, it is known that it is easier to remove electrons from a p

orbital than from a s orbital in the same shell. P orbitals are higher energy levels than the s orbitals

for a given quantum number. So an electron is easier to remove from the p orbital of the Boron

atom than from the s orbital.

First ionization energies in Groups

Elements are placed in groups in the periodic table, as they show similar physical and chemical

properties. The first ionization energies decrease down a group. This is due to:

Nuclear charge increasing Atomic radii increases, so there is less attraction between electrons and protons Shielding effect of inner electrons

The last two factors decrease the effect of the first factor.

First ionization energies and reactivity

The lower the first ionization energy of an element in the periodic table, the higher the reactivity

of that particular element with acids or bases, with metals or non-metals.

In reactions between metals and non-metals, the metal has to form a positive ion for areaction to occur. The metals with lower first ionization energies react faster and morevigorously.

Group I metals have the least first ionization energies because they have the lowest firstionization energies. Reactivity increases down the group.

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Reactions of period 3 elements

MgO has a very high melting point and withstands corrosive chemicals. It is also used as anadditive to cattle feed. Without it cattle become ill very fast.

Al2O3 layer forms over aluminium metal, which prevents it from further reacting. SO2 used to produce H2SO4 which is used in the manufacture of paints, dyestuff, etc. NaCl is used a table salt, to remove snow from roads, etc.

MgCl2 is used as a source for Mg and Cl2 CCl4 is used as a grease dissolver. SiCl4 is used as a source of pure silicon for the manufacture of Integrated circuits.

Preparation of period 3 oxides

Na2O:Sodium is burned in oxygen, which makes sodium oxide. It burns with a yellow flame.

4 Na + O2 2 Na2O

MgOMagnesium is burned in air. It burns vigorously with a white flame.

2 Mg + O2 2 MgO

Al2O3Aluminium forms Al2O3 when powdered and burned. It is also a layer over the metal.

4 Al + 3 O2 2 Al2O3

P4O10Phosphorous is burned in oxygen (excess).

P4 + 5 O2 P4O10

SO2 and SO3Sulphur is burned in air.

S + O2

SO2

SO2 can be turned into SO3 by using V2O5 as a catalyst with heat an air.

2 SO2 + O2 2 SO3

Preparation of period 3 chlorides

NaClSodium burns in chlorine gas to make NaCl.

2 N

57135296 Chemistry 9701 Complete Book for a Levels

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