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Unit 2: Application of Core Principles of Chemistry (Edexcel International Advanced Level Chemistry)
Shapes of Molecules and Ions • The electron pairs around the central atom of a molecule dictate the shape of the
molecule.
• Steps in determining the shape of molecule:
1. Decide which atom is the center of a molecule.
2. Determine the no. of electron pairs around the central atom.
i. Look up valence electrons in central atom
ii. Add one electron for each atom joined to the central atom.
iii. Add electron if particle is negatively charged or subtract electron if the
particle is positively charged.
3. Determine the no. of bond pairs and lone pairs.
Valance Shell Electron Pair Repulsion (VSEPR) stated that:
– The electron pairs (either bond pair or lone pair) repel each other and move as far
apart as possible.
lone-pair lone-pair repulsion > lone-pair bond-pair repulsion> bond-pair bond-pair repulsion
• The repulsion between electron pair increased by an increase in electronegativity of the
central atom.
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Molecule Electron dot diagram No. of bonding
pair
No. of lone pair Shape Geometry
BeCl2 Linear
H2O Bent
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BF3 Trigonal planar
NH3 Pyramidal
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CH4 Tetrahedral
PF5 Trigonal bipyramid
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SF6 Octahedral
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No. of
bonding pair
No. of
lone pair
Shape of molecule Example
2 0 Linear BeCl2
3 0 Trigonal planar BF3
4 0 Tetrahedral CH4
3 1 Trigonal pyramidal NH3
2 2 Bent H2O
5 0 Trigonal bipyramid PF5
6 0 Octahedral SF6
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Bond Energy and Bond Length
• Bond energy is defined as the standard enthalpy change for breaking the bond in 1 mol
gaseous molecules.
• Bond length is the distance between the nuclei of two bonded atoms. The shorter the bond
length, the higher it is the bond energy.
Bond Polarity
• Covalent bonds may have some ionic character which resulting in a polar covalent bond.
• A polar covalent bond is formed between two atoms of different electronegativities.
• For example, Hydrogen fluoride consists of a polar covalent bond. This is due to the high
electronegativity of fluorine atom which tends to exert a stronger attraction on the
bonding electrons as compared to hydrogen atom.
• This unequal sharing of electrons is known as polarisation and the covalent bond is said
to be polarised.
• Therefore, a molecule is said to be polar (or has dipole moment) when its bonds are
polarised and it is not symmetrical.
• Ionic bond also contain some covalent character due to polarisation (cation attracts the
negative charge of the anion.
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Intermolecular Forces
• High degree of covalency in ionic bond exist when,
– The cation is small
– The anion is large
– The charge on both ions is large
• There are two types of intermolecular forces:
– Van der Waals’ forces
– Hydrogen bonding
– Van der Waals’ forces are forces of attraction can be divided into two,
– Dipole-dipole forces
– Temporary dipole-induced dipole forces
• Dipole-dipole forces exist between polar molecules.
• The positive end of the dipole of one molecule will attract the negative end of the dipole
of another molecule.
• As for non-polar molecules such as oxygen and nitrogen, it is suggested that there are
force of attraction between molecules since they can be liquefied and solidified.
• Temporary dipole-induced dipole attraction is due to the temporary fluctuations in the
electron density of a molecule.
Factor affecting the strength of Van der Waals’ forces
• No. of electrons - The greater the no. of electrons in the molecules, the stronger is the van
der Waals’ forces of attraction.
• Shape of molecule - For instance, strength of van der Waals’ is reduced when there is a
branching because smaller surface area of contact for van der Waals’ forces.
• Hydrogen bond can exists between molecules that contain a hydrogen atom covalently
bonded to a small and highly electronegative atom, for instance fluorine, oxygen and
nitrogen. This is also known as intermolecular hydrogen bonding
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• The highly electronegative atom will attract electron density towards itself and causes a
dipole moment.
• Hence, the hydrogen atom of one molecule will possess a partial positive charge and will
be attracted to the fluorine, oxygen or nitrogen atom of another molecule that carries a
partial negative charge.
Intermolecular & Intramolecular Hydrogen Bonds
• In 2-nitrophenol, there is possibility of an intramolecular hydrogen bond forming
between hydrogen atom of hydroxyl group (-OH) and oxygen atom of nitro group (-NO2)
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• However, in 4-nitrophenol, the functional groups are on opposite sides of benzene ring.
Hydroxyl group is too far away from nitro group to form intramolecular hydrogen bond.
So, intermolecular hydrogen bonds are formed between molecules
• Intermolecular bonding in 4-nitrophenol is far stronger than in 2-nitrophenol, so boiling
point 4-nitrophenol is higher than 2-nitrophenol
Physical Properties of Hydrogen Bonding
• Hydrogen bonding is responsible for the high boiling point of water and low density of
ice.
• Water might exist as gas at room temperature without existence of hydrogen bonding.
• In the presence of hydrogen bonding, ice has an open structure which account for the
lower density of ice as compared to water.
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Giant Covalent Molecules
Diamond
• Carbon has an electronic arrangement of 1s2 2s2 2p2. In diamond, each carbon shares
electrons with four other carbon atoms - forming four single bonds.
The physical properties of diamond
• has a very high melting point (almost 4000°C). Very strong carbon-carbon covalent
bonds have to be broken throughout the structure before melting occurs.
• is very hard. This is again due to the need to break very strong covalent bonds operating
in 3-dimensions.
• doesn't conduct electricity. All the electrons are held tightly between the atoms, and aren't
free to move.
• is insoluble in water and organic solvents. There are no possible attractions which could
occur between solvent molecules and carbon atoms which could outweigh the attractions
between the covalently bound carbon atoms.
Graphite
• Graphite has a layer structure.
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• Each carbon atom in graphite uses three of its electrons to form simple bonds to its three
close neighbours. That leaves a fourth electron in the bonding level. These "spare"
electrons in each carbon atom become delocalised over the whole of the sheet of atoms
in one layer. They are no longer associated directly with any particular atom or pair of
atoms, but are free to wander throughout the whole sheet.
The physical properties of graphite
• It has a high melting point, similar to that of diamond. In order to melt graphite, it isn't
enough to loosen one sheet from another. You have to break the covalent bonding
throughout the whole structure.
• It has a soft, slippery feel, and is used in pencils and as a dry lubricant for things like
locks. You can think of graphite rather like a pack of cards - each card is strong, but the
cards will slide over each other, or even fall off the pack altogether. When you use a
pencil, sheets are rubbed off and stick to the paper.
• It has a lower density than diamond. This is because of the relatively large amount of
space that is "wasted" between the sheets.
• Insoluble in water and organic solvents - for the same reason that diamond is insoluble.
Attractions between solvent molecules and carbon atoms will never be strong enough to
overcome the strong covalent bonds in graphite.
• Conducts electricity. The delocalised electrons are free to move throughout the sheets. If
a piece of graphite is connected into a circuit, electrons can fall off one end of the sheet
and be replaced with new ones at the other end.
Silicon Dioxide
• Silicon dioxide is also known as silicon (IV) oxide.
• Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all
you need to do is to modify the silicon structure by including some oxygen atoms.
The physical properties of silicon dioxide
• It has a high melting point - varying depending on what the particular structure is
(remember that the structure given is only one of three possible structures), but around
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1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the
structure before melting occurs.
• It is hard. This is due to the need to break the very strong covalent bonds.
• It doesn't conduct electricity. There aren't any delocalised electrons. All the electrons are
held tightly between the
Fullerene
• a spherical molecule and contain 60 carbon atoms
• it consists of hexagonal rings of carbon atoms and alternating pentagonal and heptagonal
carbon rings to allow curvature of the surface producing molecules that have a complete
hollow shape
• They are classed as simple covalent molecules
• Fullerene molecules can be used for drug delivery into the body
Graphene
• A new substance that is a one layer of graphite
• Three covalent bonds per atom and the 4th outer electron per atom is delocalised
• Have very high tensile strength because of the strong structure of many strong covalent
bonds
• Can conduct electricity due to the presence of free moving electrons
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Carbon Nanotubes
• Have very high tensile strength because of the strong structure of many strong covalent
bond
• Can conduct electricity well along the tube due to the presence of free and delocalised
electrons
• Can be used as a vehicle to deliver drugs to cells
Metallic Bonding
• It is defined as the electrostatic attraction between the positively charged metal ions and
the ‘cloud’ of delocalised electrons
The physical properties of metal
• Metals can be deformed since the electron ‘cloud’ prevent the repulsion among the
cations.
• Since it electrons are free to move throughout the metal piece, hence metal is a good
electrical and thermal conductors.
• Owing to these typical properties, metal such as aluminum, copper can be used for
packaging, drawn into wires, and their alloys are strong enough for usage in aeroplane,
car, train and building.
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Redox Redox Process
o Redox reaction – a reaction in which the reduction of one reactant and the oxidation of
another reactant occurs simultaneously.
o Oxidising agent – a reactant that causes oxidation and itself reduced in the reaction.
o Reducing agent – a reactant that causes reduction and itself oxidised in the reaction.
Redox in terms of Electron Transfer
o Oxidation is defined as loss of electrons whereas reduction is defined as gain of electrons.
o e.g. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
o Zn(s) is a reducing agent and itself oxidised to Zn2+(aq).
o Cu2+(aq) is an oxidixing agent and itself reduced to Cu(s).
o Disproportionation is a redox reaction in which both oxidation and reduction of the same
element occurs.
o e.g. Cu2O(s) + H2SO4(aq) Cu(s) + CuSO4(aq)
+1 0 +2
Rules for working out Oxidation Numbers
o All free atoms in elements have an oxidation number of zero.
E.g. Mg Fe O2 Cl2
o For simple ions, the oxidation number is the same as the charge on the ion.
E.g. Na+ Al3+ Mg2+
o For covalent compound, the covalent bond is changed into ‘ionic bond’ by assuming that the
bonded electrons are on the more electronegative atom. The oxidation numbers are the
charge on the ‘ions’.
E.g. H2O SO3 HCl
o In a neutral molecule, the sum of the oxidation numbers of all the atoms equals zero.
o e.g. Al2O3 2(3+) + 3(-2) = 0
o In an ion, the sum of the oxidation numbers of all the atoms in the formula unit equals the
charge on the ion.
o e.g. MnO4- 1(+7) + 4(-2) = -1
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Construct Redox Equations
o Redox equation can be constructed as follow:
• Copy out the two half equations from data booklet.
▪ Write the half-equation with more positive Eө on top.
▪ Write the half-equation with more negative Eө underneath and reversed
so that all the reactants are on the left.
• Multiply the half-equations with appropriate factors so that the number of electrons
involved in each is the same.
• Add two equations together, canceling out the ions or molecules that appear on both
sides of the equations
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The Periodic Table – Groups 2 and 7 Group 2
o Atomic radius
- Atomic radium increases down the group
- Atoms have more shells of electrons making the atom bigger
o Melting point
- Melting point decreases
- Metallic bonding weakens as the atomic size increases
- Distance between the positive ions and delocalized increases
- Weaker electrostatic forces between the positive ions and delocalized electrons
o First ionization energy
- First IE decreases
- The outermost electrons are held more weakly as they are successively further from the
nucleus in additional shells
- The outer shell electrons become more shielded from the attraction of the nucleus
o Reactivity
- Increases down the group
- As the atomic radii increase, there is more shielding
- Nuclear attraction decreases
- Easier to remove electron
- Cations form more easily
- Must be stored under oil to protect it from reaction with water and oxygen
- Mg and Be form a protective oxide layer to avoid further corrosion
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o Reaction with oxygen
Group 2 element Equation Property
Be 2Be(s) + O2(g)2BeO(s)
Amphoteric (ionic compound
with covalent character)
Mg 2Mg(s) + O2(g)2MgO(s)
Basic
Ca 2Ca(s) + O2(g)2CaO(s)
Basic
Sr 2Sr(s) + O2(g)2SrO(s)
Basic
Ba 2Ba(s) + O2(g)2BaO(s)
Basic
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o Reaction with water
Group 2 element Property Equation
Be Has no reaction with either
cold water or steam
-
Mg Has no reaction with cold
water but it reacts with
steam
MgO dissolves slightly in
water to give weakly
alkaline solution
Mg(s) + H2O(g)MgO(s) + H2(g)
MgO(s) + H2O(l)Mg(OH)2(aq)
Ca React vigorously with cold
water
Ca(s) + 2H2O(l)Ca(OH)2(s) +
H2(g)
Sr React vigorously with cold
water
Sr(s) + 2H2O(l)Sr(OH)2 + H2(g)
Ba React vigorously with cold
water
Ba(s) + 2H2O(l)Ba(OH)2 + H2(g)
- Solubility of group II hydroxides increases down the group
o Solubility of sulphates
- Less soluble down the group
- BaSO4 is the least soluble
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o Thermal decomposition of group 2 carbonate
- MgCO3(s)MgO(s) + CO2(g) (at 540oC)
CaCO3(s)CaO(s) + CO2(g) (at 900oC)
SrCO3(s)SrO(s) + CO2(g) (at 1290oC)
BaCO3(s)BaO(s) + CO2(g) (at 1360oC)
- Thermal stability increases down the group
- Cationincrease in size
- Charge density decreases
- Polarizing power of the cationdecreases
- Anion electron cloud is less distorted
o Thermal decomposition of group 2 nitrate
- Mg(NO3)2(s)MgO(s) + 2NO2(g) + ½ O2(g)
Ca(NO3)2(s)CaO(s) + 2NO2(g) + ½ O2(g)
Sr(NO3)2(s)SrO(s) + 2NO2(g) + ½ O2(g)
Ba(NO3)2(s)BaO(s) + 2NO2(g) + ½ O2(g)
- Thermal stability increases down the group
- Cationincrease in size
- Charge density decreases
- Polarizing power of the cationdecreases
- Anion electron cloud is less distorted
Group 7
Element Colour and state
at room
temperature
Boiling point/ ͦC Solubility
In water In organic
solvent
Fluorine Pale yellow gas -188 - -
Chlorine Yellow-green gas -35 Moderately
soluble
Yellow solution
Bromine Dark red liquid 58 Slightly soluble Orange-brown
solution
Iodine Black solid 183 Insoluble Violet/purple
solution
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o Redox behavior
- Oxidizing power of halogens decreases down the group
F2 Cl2 Br2 I2
- Reducing power of halide ions increases down the group
F-<Cl-< Br-< I-
- Feasible halogen displacement redox reactions
F2 + 2Cl- Cl2 + 2F-
F2 + 2Br- Br2 + 2F-
F2 + 2I- l2 + 2F-
Cl2 + 2Br- 2Cl- + Br2
Cl2 + 2I- l2 + 2Cl-
Br2 + 2I- 2Br- + l2
o Volatility
- Decreases down the group
- Halogen has simple molecular structure consisting of diatomic molecule
- Held together by weak VDW forces
- Down the group, the molecular size increases
- VDW increases
- Boiling point of halogen increases and halogen becomes less volatile
o Reaction of halogen with hydrogen
- H2 + F22HF
H2 + Cl22HCl
H2 + Br22HBr
H2 + I22HI
- Thermal stability of hydrides decreases down the group
- Due to decrease in strength of the H-X bond
- As the size of halogen increases
- Size of halogen increases increase in bond length decrease in bond strength
- HF>HCl>HBr> HI (thermal stability)
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- The acid strength of hydrides increases down the group
- Due to decrease in the strength of H-X bond
- HI>HBr>HCl> HF (acid strength)
- Oxidation of hydrides (HX H2 and X2)
HF Electrolysis
HCl KMnO4 (strong oxidizing agent)
HBr H2O2 (weaker oxidizing agent)
HI Readily oxidized at atm O2
o Reaction of halide ions with silver followed by ammonia
- The function of ammonia
✓ Ag+ (aq) + 2NH3 (aq) [Ag(NH3)2]+ (aq)
✓ NH3 acts as a ligand with Ag+ ion
✓ A ligand is ion or molecule that donates its lone pair electron through
coordinate bond/dative bond to metal ion
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o Reaction of halide ions with concentrated sulphuric acid
Ion Observation Equation
F-
White fumes of HF
NaF + H2SO4 NaHSO4 + HF
Cl-
White fumes of HCl
NaCl + H2SO4 NaHSO4 + HCl
Br-
White fumes of HBr + red brown
bromine vapor
NaBr + H2SO4 NaHSO4 + HBr
2HBr + H2SO4 Br2 + SO2 + 2H2O
I-
White fumes of HI, black solid
iodine + purple iodine vapor +
bad egg smells hydrogen
sulphide
NaI + H2SO4 NaHSO4 + HI
2HI + H2SO4 I2 + SO2 + 2H2O
8HI + H2SO4 4I2 + H2S + 4H2O
o Reaction of chlorine with NaOH
- With cold NaOH, Cl2 undergoes disproportionation.
0 -1 +1
Cl2 + 2NaOH(aq)NaCl(aq) + NaClO(aq) + H2O(l)
- With hot NaOH, Cl2 undergoes disproportionation.
0 -1 +5
2Cl2 + 6NaOH(aq)5NaCl(aq) + NaClO3(aq) + 3H2O(l)
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Kinetics Activation Energy
o Reaction involves collision between more than two species
o Orientation of collision refers to the molecules hit each other in the right way to decide
whether a particular collision will result in a reaction
o Energy of collision is refer to activation energy
o Even if the species are orientated properly, a reaction will not happen unless the particles
collide with a certain minimum energy called activation energy of the reaction
o Activation energy is the minimum energy required before a reaction can occur
Maxwell-Boltzmann Distribution
• In any system, the particles present will have a very wide range of energies
• For gases, this can be shown on a graph called Maxwell-Boltzmann distribution which is
a plot of the number of particles possess each particular energy to react when they collide
• The graph is asymmetry/not normal distribution
• The area under the graph represents the total number of molecules
• At any given temperature, most molecules have an average energy, a small number of
molecules have very high or very low energies
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Effect of Temperature On Reaction Rate
• The effect of temperature on reaction rates is illustrated in the Maxwell-Boltzmann
distribution
• An increase in temperature increases the average kinetic energy of molecules and a
greater molecules has higher kinetic energy
• At higher temperature, the number of molecules with energy same or greater than
activation energy increased, so reaction rate increases because the number of frequency
of effective collisions increase
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Effect of Catalyst on Reaction Rate
• Adding a catalyst will speed up rate of reaction
• A catalyst provides an alternative route for the reaction to happen with a lower activation
energy
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• Properties of a catalyst are as follow:
– Only a small quantity is required to catalyse a reaction
– It does not undergo any chemical changes at the end of reaction, but it can
undergo physical changes
– It is specific in its action and does not catalyse all reactions
– It alters the reaction mechanism that will lower the activation energy
– It does not change the reaction enthalpy or equilibrium constant, but it can change
the order of reaction
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Chemical Equilibria Reversible Reactions
o Reversible reaction is one which proceeds in both directions.
o The reactants are not completely converted to products.
o Equilibrium is reached where both reactants and products are present.
o e.g. H2(g) + I2(g) 2HI(g)
o Equilibrium can only occur in a closed system.
o e.g. When few cm3 of water are placed in a flask which is corked, liquid is changing to the
gas and vice versa continuously and the total amount remain the same, hence, equilibrium is
reached.
H2O(l) H2O(g)
Irreversible Reactions
o If the activation energy of the reverse reaction (Eb) is exceptionally high, then the reaction
will be unfavourable known as irreversible reaction.
H2O(l) H2O(g) (open system)
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Dynamic Equilibrium
o A reversible reaction can reach dynamic equilibrium when the rate of the forward reaction
becomes equal to the rate of backward reaction.
o e.g. 3H2(g) + N2(g) 2NH3(g)
o As soon as NH3 molecules are formed, they decomposed back to N2 and H2. At this stage, the
total amount of each substance remains the same since the production rate of NH3 equals to
the decomposing rate of H2 and N2.
Le Chatelier’s Principle
o Le chatelier’s Principle states that if a change is made to a system in equilibrium, the system
reacts in such a way as to tend to oppose the change and a new equilibrium is formed.
o e.g. If something is added to the system at equilibrium, then the system will behave so as to
remove it and vice versa.
o It summarises the effect of external factors such as temperature, concentration and pressure
on a system at equilibrium.
Changes in Concentration
• e.g. Fe3+(aq) + SCN-(aq) [Fe(SCN)]2+(aq)
• When extra SCN-(aq) is added, by Le chatelier’s Principle, the position of equilibrium
shifts to the right so as to remove the extra SCN-(aq). Hence, more [Fe(SCN)]2+(aq) is
produced.
• However, when some SCN-(aq) is removed, the position of equilibrium will shifts to the
left so as to replace some SCN-(aq). Therefore, less [Fe(SCN)]2+(aq) is produced.
Changes in Pressure
• Pressure changes can only affect the reaction equilibrium when gases are involved.
• e.g. 2SO2(g) + O2(g) 2SO3(g)
• When pressure increased, the position of equilibrium shifts to the right so as to reduce the
pressure by reducing the number of molecules of gas. The new equilibrium contains more
SO3(g).
• When pressure decreased, the position of equilibrium shifts to the left so as to increase
the pressure by increasing the number of molecules of gas. The new equilibrium contains
more SO2(g) and O2(g).
Changes in Temperature
• e.g. 2SO2(g) + O2(g) 2SO3(g) ∆H= -94.9kJ/mol
• The reaction above is an exothermic reaction, hence, when temperature increased, by Le
chatelier’s Principle, the position of equilibrium will move to a direction that can reduce
the temperature, that is, shifts to the left. So more SO2(g) and O2(g) produced.
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• Whereas when temperature decreased, the position of equilibrium shifts to the right in
order to increase the temperature, hence, more SO3(g) is produced.
Equilibrium Constant, Kc & Kp
o The equilibrium constant is a measure of the extent to which the reactants are converted into
products before equilibrium is reached.
o The equilibrium constant is only affected by temperature.
o Changes in concentration, pressure or the presence of catalyst will not affect the equilibrium
constant.
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Q: Given the equilibrium constant for following equation is 4.0,
CH3CH2OH + CH3COOH CH3COOCH2CH3(l) + H2O(l)
Calculate the mass of ethanoic acid that must be mixed with 2.0 mol of ethanol to produce 1.5
mol of ethyl ethanoate at equilibrium.
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Haber Process
o It is a process in manufacturing ammonia, NH3.
3H2(g) + N2(g) 2NH3(g) ∆H = -92kJ/mol
o Typical conditions are as follow:
– A pressure of 200 atm.
• Le chatelier’s Principle predicts that high pressure gives a higher yield of
ammonia.
• But when pressure is too high, it would results in higher cost and maintanance
of equipment. Hence, a pressure of 200 atm is used.
– A moderate temperature of 500oC.
• Le chatelier’s Principle predicts that low temperature gives a higher yield of
ammonia.
• However, if temperature is too low, the rate of reaction is too slow. Hence, a
moderate temperature of 500oC is used.
– Iron catalyst (finely divided) is also used to increase the rate of reaction so that the
equilibrium is reached faster.
– Small amount of promoters such as K2O or Al2O3 also needed to enhance the
efficiency of the catalyst.
– N2 and H2 are used in a proportion of 1:3. N2 H2 is obtained from fractional
distillation of liquid air and natural gas, respectively.
Contact Process
o This is the main chemical reaction occurring in the Contact process. It is a reversible,
exothermic reaction accompanied by a decrease in volume. Applying Le Chatelier's Principle,
to get a good yield of sulphur trioxide,
o The following points are to be borne in mind: A low temperature is necessary. An optimum
temperature of
o 410-450oC is found suitable. High pressure would favour the forward reaction. It is not
possible to build acid resistant towers which can withstand high pressures. A pressure of 760
- 1520 mm of Hg is used.
o Excess of oxygen is necessary as it would favour the forward reaction. A suitable catalyst is
required.
o Platinum is a more efficient catalyst than vanadium pentoxide. But it is more expensive. It
also gets poisoned by impurities like arsenic (III) oxide, Vanadium pentoxide though less
efficient is cheaper and not affected by impurities. K2O may be used as a promoter to
enhance the activity of the catalyst.
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o 410-450oC is found suitable. High pressure would favour the forward reaction. It is not
possible to build acid resistant towers which can withstand high pressures. A pressure of 760
- 1520 mm of Hg is used.
o Excess of oxygen is necessary as it would favour the forward reaction. A suitable catalyst is
required. Platinum is a more efficient catalyst than vanadium pentoxide. But it is more
expensive. It also gets poisoned by impurities like arsenic (III) oxide, Vanadium pentoxide
though less efficient is cheaper and not affected by impurities. K2O may be used as a
promoter to enhance the activity of the catalyst.
Bronsted-Lowry Theory of Acids and Bases
o It states that an acid is a species which donates proton (H+) (proton donor) to a base; a base
is a species that accepts proton (H+) (proton acceptor) from an acid.
e.g. HA + H2O H3O+ + A-
acid base conjugate conjugate
acid base
Strong and Weak Acids
• The relative strengths of acids depend on the extent of dissociation of the acid solution.
• The strongest acid is the one with greatest degree of dissociation.
• Therefore, strong acids are strong electrolytes while weak acids are weak electrolytes.
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Organic Chemistry
Alcohols
o Contain –OH group
o Can form H-bond with other alcohol molecules and with water
o Has high MP and BP
o Soluble in water
o Can be used as a solvent
Manufacture of Alcohol
i. Fermentation of sugars
C6H12O6 2C2H5OH + 2CO2
This process has a number of advantages:
o it is a low-technology process, which means it can be used anywhere
o it does not use much energy
o it uses sugar cane as a raw material, which is a renewable resource
There are, however, a few disadvantages associated with this process:
o it is a batch process, which means that once the reaction has finished the vessel needs to
emptied before the reaction can be started again
o it is a relatively slow process
o it produces fairly impure ethanol
ii. Hydration of ethene
o Temperature: 300 oC
Pressure: 60 atm
Catalyst: H3PO4
C2H4 + H2O C2H5OH
This process has a number of advantages:
- it is a relatively fast process
- it is a continuous flow process, which means that ethene can be entered into the vessel
continuously and the reaction never has to be stopped
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- it produces pure ethanol
There are also a number of disadvantages associated with this process:
- it requires fairly high technology
- it uses a lot of energy
- the ethene comes from crude oil, which is a non-renewable resource
Primary, Secondary and Tertiary Alcohols
Primary
C C C
H
H
H H
H
HOH
H
The carbon attached to the OH is attached to 0 or 1 other carbon atom
Secondary
H C
H
HC
H
C
H
H
H
OH
The carbon attached to the OH is attached to 2 other carbon atoms.
Tertiary
H C
H
HC
H
C
H
H
OH
H H
C
H
The carbon atom attached to the OH is attached to 3 other carbon
atoms.
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Reaction of Alcohols
o Alcohol molecules are saturated and polar, containing a +ve carbon. Thus alcohols
tend to undergo nucleophilic substitution reactions.
o The OH can combine with an adjacent H atom to form a stable H2O molecule. Thus
alcohols can also undergo elimination reactions.
o Alcohols can lose hydrogen and undergo a variety of oxidation reactions.
i. Nucleophilic Substitution
o Alcohol Halogenoalkane
a. Condition: HBr (mixture of NaBr/KBr and conc. H2SO4), heat and distil
HI (mixture of NaI/KI and conc. H3PO4)
o Rate of reaction: tertiary alcohol > secondary alcohol > primary alcohol
o Nucleophilic substitution of tertiary alcohol reacts with conc. HCl at room
temperature
b. Condition: PCl3/PCl5
3 CH3-CH2-CH2-OH + PCl3 3 CH3-CH2-CH2-Cl + H3PO3
CH3-CH2-CH2-OH + PCl5CH3-CH2-CH2-Cl + POCl3 + HCl
o Can be used a test for –OH groups as white fumes of HCl produced
ii. Reaction between Sodium and Ethanol
CH3-CH2-CH2-OH + Na CH3-CH2-CH2-O-Na+ + H2
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iii. Elimination
o Condition: Al2O3, heat/ conc. H3PO4
CH3-CH2-CH2-OH CH3-CH2=CH2+ H2O
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iv. Oxidation
a. Partial oxidation of primary and secondary alcohol
Condition: acidified K2Cr2O7, heat under distillation
• If a primary alcohol is mixed with an oxidising agent, two hydrogen atoms
can be removed and an aldehyde will be formed:
E.g. CH3CH2OH + [O] CH3CHO + H2O
H H
HH
CC OHH + [O] + H2OC
O
H
H
H
CH
Ethanol Ethanal
• If a secondary alcohol is mixed with an oxidising agent, two hydrogen atoms
can be removed and a ketone will be formed:
Eg CH3CH(OH)CH3 + [O] CH3COCH3 + H2O
C
O
CH3
CH3CH
OH
CH3CH3
+ [O] + H2O
Propan-2-ol Propanone
• Tertiary alcohols are not readily oxidised since they do not have available
H atoms to give up.
b. Complete oxidation of primary alcohol
Condition: acidified K2Cr2O7, heat under reflux
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Summary of oxidation reactions of alcohols and carbonyls
primaryalcohols
secondaryalcohols
tertiaryalcohols
aldehydes ketones
carboxylicacids
X
[O]
[O]
[O]
v. Esterification
Condition: Conc. H2SO4 (catalyst)
vi. Iodoform
Condition: Sodium hydroxide and iodine
Positive result: pale yellow precipitate (triiodomethane)
• Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.
• If "R" is a hydrocarbon group, then you have a secondary alcohol. Lots of secondary
alcohols give this reaction, but those that do all have a methyl group attached to the
carbon with the -OH group.
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• No tertiary alcohols can contain this group because no tertiary alcohols can have a
hydrogen atom attached to the carbon with the -OH group. No tertiary alcohols give the
triiodomethane (iodoform) reaction.
Halogenoalkane
o Contain a halogen atom e.g. F, Cl, Br, I attached to an alkyl group
o Also called as alkyl halides
o Different kinds of halogenoalkanes
Primary The carbon which carries the halogen atom
is only attached to one other alkyl group
Secondary The carbon which carries the halogen atom
is attached to two other alkyl group
Tertiary The carbon which carries the halogen atom
is attached to three other alkyl group
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Physical Properties
o Boiling points below room temperature
o Gases at room temperature
o BP: CH3X < CH3CH2X < CH3CH2CH2X VDW (permanent dipole) gets stronger as
the molecule gets bigger or has more electrons
o BP: CH3F < CH3Cl < CH3Br < CH3I increase in size from F to I (increase in number
of electrons), stronger VDW (permanent dipole)
o BP: tertiary halogenoalkane < secondary < primary attractions are stronger if the
molecules can lie closely together, more VDW forces can be formed
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Chemical Reactivity
o Bond strength falls as from C-F to C-I
o Bond strength as bong length
o Size of halogen atom: F < Cl < Br < I
o C-F bond length is the shortest, strongest bond strength
o Rate of reaction: C-F < C-Cl < C-Br < C-I
o Chemical Reactions:
- Nucleophilic Substitution
▪ Nucleophile: a species which is strongly attracted to a region of positive
charge e.g. negative ions or have a strongly - ▪ Nucleophilic substitution can be divided into SN2 & SN1 mechanism
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SN2 SN1
Primary & Secondary Secondary & Tertiary
S: substitution N: nucleophilic 2: initial stage of
the reaction involves 2 species
S: substitution N: nucleophilic 1: initial
stage of the reaction involves 1 species
Primary halogenoalkane did not use the SN1
mechanism because primary carbocation is
unstable than the tertiary
Tertiary halogenoalkane can form stable
tertiary carbocation due to the presence of
electron-donating alkyl groups
1. Hydrolysis
Reagent: Aqueous NaOH/KOH
Condition: Heat under reflux
Product: Alcohol
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2. Elimination
Reagent: NaOH/KOH in ethanol
Condition: Heat under reflux
Product: Alkene
3. Reagent: NaCN/KCN in ethanol
Condition: Heat under reflux
Product: Nitrile
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4. Reagent: ammonia in ethanol
Condition: sealed tube
Product: amine
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Mass Spectra and IR
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Green Chemistry
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