Topic 6: Energetics

➢ Enthalpy (H):
It is the total energy associated with the materials which react.

➢ Enthalpy change (∆H):

It is the heat energy transferred between the system and surroundings at constant
pressure. The standard conditions are 100 kPa and a specified temperature, usually 298 K.

➢ Exothermic reaction:
► During the exothermic reaction, heat energy is given out from
the system (chemicals). Therefore, the temperature rises up.
► The amount of energy released during bond making (formation)
is more than the energy absorbed during bond break.
► ΔH is negative because the energy of the products is less than
the energy of the reactants.
► The products have less energy than the reactants.

 Examples for Exothermic Processes:

1- Combustion of a fuel.
2- Respiration of human beings and animals where the energy in the food is released.
3- Displacement reaction.
4- Dissolving sodium hydroxide in water.
5- Adding concentrated sulfuric acid to water.

➢ Endothermic Reactions:
► During the endothermic reaction, heat energy is gained by the
system (chemicals) from surroundings. Therefore, the
temperature falls down.
► The amount of energy released during bond making is less than
the energy absorbed during bond break.
► ΔH is positive because the energy of the products is more than
the energy of the reactants.
► The products have more energy than the reactants.

 Examples for Endothermic Processes:

1- Thermal decomposition reaction.
2- The photosynthesis process in green plants.
3- The reaction between dilute ethanoic acid and solid sodium hydrogen carbonate
4- Dissolving ammonium nitrate, potassium iodide, urea and sodium thiosulphate in water.

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➢ Measuring the Enthalpy Change for a Reaction Experimentally
“Calorimetric method”

Calculating the enthalpy change of reaction, ∆Hr from experimental data

Energy change = mass of solution x heat capacity x temperature change
Q = m x c x ΔT
energy mass specific change in
change of water heat of water temperature
(J) (g) (4.18 J g-1 K-1) (K)
𝑯𝒆𝒂𝒕 𝑬𝒏𝒆𝒓𝒈𝒚 𝑪𝒉𝒂𝒏𝒈𝒆 (𝑸)
Molar Enthalpy Change(ΔH) = KJ mol-1
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 (𝒏)
• The heat capacity of water is 4.18 J g-1K-1.
• In any reaction where the reactants are dissolved in water we assume that the heat capacity
is the same as pure water.
• Also assume that the solutions have the density of water, which is 1g cm-3.

 Enthalpy change of a Displacement Reaction:

The addition of 6.0 g (an excess) of magnesium metal to
100 cm3 of copper(II) sulphate “CuSO4(aq)” solution (2.0
mol dm-3) raised the temperature from 20.0°C to 65.0°C.
Calculate the enthalpy change of the reaction:
Mg(s) + CuSO4(aq) MgSO4(aq) + Cu(s)
● Specific heat capacity “c” of the solution = 4.18 J g-1 K-1.
● The density of the solution = 1.00 g cm-3

Answer:
1- Find the energy change:
The mass of the solution = 100 x 1.00 = 100 g
Energy gained to surroundings = m x c x ΔT
= 100 x 4.18 x (65 – 20)
= 18810 J
2- How many moles completely reacted:
Conc X V 2.00 X 100
Moles of CuSO4(aq) = = = 0.20 moles
1000 1000
3- Calculate the molar enthalpy change:

18810⁄
1000
ΔH = = – 94. 050 KJ/mol
0.20

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 Practical determination of the heat value of Ethanol:
A student carried out an experiment to determine the heat
value of ethanol. He got the following measurements:
● Mass of heated water = 670.0 g
● Initial mass of burner = 538.5 g
● Final mass of burner = 536.2 g
● Initial temp. of water = 25.0 °C
● Final temp. of water = 76.0 °C

1- Calculate the rise in temperature.

 The rise in temperature = 76 - 25 = 51°C
2- Calculate the heat gained (in kJ) by water(C = 4.2 J/°K.g).
 Heat gained = M x C x T
= 670 x 4.2 x 51
= 143514 J
= 143.514 kJ
3- Assuming that the heat produced from the fuel is completely absorbed by water.
i- Calculate the mass of the used fuel in this experiment.
 Mass of fuel = 538.5 - 536.2 = 2.3 g
ii- Calculate the ethanol moles.
2.3
 The ethanol mole = = 0.05 mol
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iii- Calculate the molar enthalpy change of ethanol measured in kJ mol-1.
143.514
 The heat value = = - 2870.28 kJ mol-1
0.05

General method
1. Calculate energy change for quantities using Q = m x cp x ∆T
2. Work out the moles of the reactants used
3. Calculate ∆H by dividing Q by the number of moles of the reactant
4. Add a sign and unit (divide by a thousand to convert Jmol-1 to kJmol-1)

Errors in measuring enthalpy change for Errors in measuring enthalpy change for
reactions combustion

• Energy transfer to or from surroundings. • Energy losses from calorimeter

• Reaction or dissolving may be incomplete or slow. • Incomplete combustion of fuel
• Density of solution is taken to be the same as • Incomplete transfer of energy
water. • Evaporation of fuel after weighing
• Approximation in specific heat capacity of solution. • Heat capacity of calorimeter not included
• The method assumes all solutions have the heat • Measurements not carried out under standard
capacity of water. conditions
• Neglecting the specific heat capacity of the (as H2O is gas, not liquid, in this experiment)
calorimeter.
• We ignore any energy absorbed by the apparatus.

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Examples
1. Calculate the enthalpy change of reaction for the reaction where 25cm3 of 0.2 M copper
sulphate was reacted with 0.01mol (excess of zinc). The temperature increased 7oC .
Solution:
Q = m x cp x ∆T = 25 x 4.18 x 7 = 731.5 J
Moles of CuSO4 = conc x volume = 0.2 x 25/1000 = 0.005 mol
∆H = Q / no of moles = 731.5 / 0.005 = 146300 J mol-1
= 146 kJ mol-1 to 3 sf
Finally add in the sign to represent the energy change: if temp increases the reaction is
exothermic and is given a minus sign e.g. –146 kJ mol-1.

2. 25cm3 of 2M HCl was neutralised by 25cm3 of 2M NaOH. The Temperature increased 13.5oC
What was the energy change per mole of HCl?
Solution:
Q = m x cp x ∆T = 50 x 4.18 x 13.5 = 2821.5 J
moles of HCl = conc x volume = 2 x 25/1000 = 0. 05 mol
∆H = Q / no of moles = 2821.5 / 0.05 = 564300 J mol-1
= - 56.4 kJ mol-1

3. Calculate the enthalpy change of combustion for the reaction where 0.65g of propan-1-ol was
completely combusted and used to heat up 150g of water from 20.1 to 45.5oC
Solution:
Q = m x cp x ∆T = 150 x 4.18 x 25.4 = 15925.8 J
moles of propan-1-ol = mass / Mr = 0.65 / 60 = 0.01083 mol
∆H = Q/ no of moles = 15925.8/0.01083 = 1470073 J mol-1
= 1470 kJ mol-1 to 3 sf
= –1470 kJ mol-1

4. Methane is burned in presence of oxygen as fuel according to the equation:

CH4 + 2 O2 CO2 + 2 H2O

Decide whether the reaction is exothermic or endothermic and calculate the overall energy
change for this reaction. The table gives some average bond dissociation energies.
Bond C−H O=O C=O O−H
Average bond dissociation energy (KJ/mol) 410 498 740 460

• Energy absorbed during bond break = ………………………………………………………………………………………………. KJ

• Energy released during bond formation = ……………………………………………………………………………………………….KJ
• The overall energy change = ……………………………………………………………………………………………….KJ
• The reaction is ……………………………………………………………… as ……………………………………………………………………………………………
…..………………………………………………………………………..……………………………………………………………………………………………………………………

4
Calorimetric method
One type of experiment is one in which substances are mixed in an insulated container and the
temperature rise measured as reaction between two solutions.

General method
• Washes the equipment (cup and pipettes etc) with the solutions to be used.
• Dry the cup after washing.
• Put polystyrene cup in a beaker for insulation and support.
• Measure out desired volumes of solutions with volumetric pipettes and transfer to insulated cup.
• Clamp thermometer into place making sure the thermometer bulb is immersed in solution measure
the initial temperatures of the solution or both solutions if 2 are used
• Do this every minute for 2-3 minutes
• At minute 3 transfer second reagent to cup.
 If a solid reagent is used then add the solution to the cup first and then add the solid weighed
out on a balance.
 If using a solid reagent then use ‘before and after’ weighing method
• Stirs mixture (ensures that all of the solution is at the same temperature)
• Record temperature every minute after addition for several minutes.

 If the reaction is slow then the exact temperature

rise can be difficult to obtain as cooling occurs
simultaneously with the reaction.
 To counteract this we take readings at regular time
intervals and extrapolate the temperature
curve/line back to the time the reactants were
added together.
 We also take the temperature of the reactants for
a few minutes before they are added together to
get a better average temperature.
 If the two reactants are solutions then the
temperature of both solutions need to be
measured before addition and an average
temperature is used.

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➢ Standard Enthalpy Changes of Reaction ( ΔHrӨ):
It is the enthalpy change which occurs when one mole of a substance is formed from its
elements in their the quantities of substances mentioned in a balanced equation react
under standard conditions, and with everything in its standard state.

➢ Standard conditions are:

• 100 kPa pressure (= 105Pa nearly equal 1atm )
• 298 K (room temperature or 25oC)
• Solutions at 1mol dm-3
• All substances should have their normal state at 298K

➢ Standard Enthalpy Change of Formation (ΔH𝑓Ө )

It is the enthalpy change when 1 mole of a compound is formed from its constituent
elements in their standard states and under standard conditions.
2Fe (s) + 1.5O2 (g) → Fe2O3 (s) 𝚫𝐇𝒇Ө = -824.2 KJ mol-1
The standard enthalpy of formation of an element = 0 kJ mol-1

➢ Standard Enthalpy Change of Combustion (ΔH𝑐Ө )

It is the enthalpy change which occurs when one mole of the compound is burned
completely in excess of oxygen under standard conditions.
CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (l) ∆ 𝐇𝐜𝛉 = -890.3 KJ mol-1
 Incomplete combustion will lead to soot (carbon), carbon monoxide and water. It will be less
exothermic than complete combustion.

➢Standard Enthalpy Change of Neutralisation (ΔH𝑛Ө )

It is the enthalpy change when an acid neutralises an alkali under standard conditions to
produce 1 mole of water.
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l) ∆ 𝐇𝐧𝛉 = -57.1 KJ mol-1
 Enthalpy changes of neutralisation for strong acids and alkalis, the values are similar,
with values between -56 and -58 kJ mol-1 and decreases for the neutralisation between
weak acids and weak bases.

➢Standard Enthalpy Change of Atomisation (ΔH𝑎𝑡

Ө
)
It is the enthalpy change when 1 mole of gaseous atoms is formed from its element under
standard conditions.
𝛉
Na (s) → Na (g) ∆ 𝐇𝐚𝐭 = +107 KJ mol-1
 Enthalpy change of atomisation is always endothermic (has positive value) as it involves
increasing the separation between atoms which requires energy.

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➢ Hess’s Law
Hess’s law states that total enthalpy change for a reaction is independent of the route by
which the chemical change takes place as long as the initial and final conditions are the same.
It is used to calculate the ∆Hf for substance difficult to measure and to determine the
stability of compounds.

This means that the enthalpy to go from A to B

doesn’t depend on the route taken.
As formula: ∆Hr = ∆H1 + ∆H2

Example 1

Example 2

7
Using Hess’s law to determine enthalpy changes from enthalpy
changes of formation.

8
Using Hess’s law to determine enthalpy changes from enthalpy
changes of combustion.

9
➢ Bond enthalpy and Mean Bond enthalpy
 The bond enthalpy ΔHB is the enthalpy change when one mole of a bond in the
gaseous state is broken.
 The mean bond enthalpy is the enthalpy change to break one mole of a bond
averaged out over many different molecules is broken in the gaseous state.

• These values are positive because energy is required to break a bond.

• The definition only applies when the substances start and end in the gaseous state.
• We use values of mean bond energies because every single bond in a compound has a slightly
different bond energy.
• 'Average' refers to the fact that bond enthalpy is different in different molecules.

Example:
In CH4 , there are 4 C-H bonds. Breaking each one will require a different amount of
energy. However, we use an average value for the C-H bond for all hydrocarbons.

For diatomic molecules as : Cl2(g) → 2Cl(g) ΔBH = +243 kJmol-1

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Remember:
Calculated values of enthalpy of combustions from enthalpy of formation data will be more
accurate than if calculated from average bond enthalpies.
Because average bond enthalpy values are averaged values of the bond enthalpies from
various compounds.

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Summary
➢Using Hess’s law to determine enthalpy changes from enthalpy changes of
formation.

➢Using Hess’s law to determine enthalpy changes from enthalpy changes of

combustion.

➢Using Hess’s law to determine enthalpy changes from enthalpy changes for bond
energies.

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 Topic 7: Intermolecular Forces
➢ A compound containing elements of a large difference in electronegativity (> 1.7)
will be ionic compound.
➢ A molecule containing elements of similar or a small electronegativity difference
will be purely covalent molecule.
➢ A polar covalent bond forms when the elements in the bond have different
electronegativity’s of around 0.3 to 1.7.

Intermolecular forces:
• Intermolecular forces occur between covalent molecular substances and noble
gases. It do not occur in ionic substances.
• Types of intermolecular force
1. Permanent dipole-permanent dipole interactions.
2. London forces (dispersion force / instantaneous dipole-induced dipole force).
3. Hydrogen bonds.

1. Permanent dipole-permanent dipole interactions.

Polar molecules have permanent dipole-permanent dipole interaction in which the
negative end of one molecule is attracted towards the positive end of another.

2. London Forces (or Dispersion force or Instantaneous dipole induced dipole force)
 They occur between all simple covalent molecules and the separate atoms in noble
gases.
 Simple molecular substances as: N2, P4, O2, S8, F2, Cl2, Br2, I2 have only dispersion
force as they are non-polar.
 In any molecule the electrons are moving constantly and randomly so parts of the
molecule become more or less negative i.e. small temporary or transient dipoles
which can induced a dipole in the molecule next to it and so on.

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 The induced dipole is always the opposite sign to the original one.

 London force increases by:

• Increasing the number of electrons (and protons) in the molecule.
• Increasing the number of contact points between the molecules.

 Examples
 The boiling point of Pentane is 36°C and 2,2-dimethylpropane is 10°C although
both are non-polar and having same number of electrons.
As the molecule in pentane can line up beside each other so there are a large
number of contact points and stronger dispersion force that needed more energy to
break and higher boiling point.

 The boiling point of noble gases increase as the number of electrons increased
As London force (dispersion force) between the atoms increased by increasing
number of electrons.

 Iodine is solid while chlorine is gas at room temperature

As London force (dispersion force) between the molecules increased by
increasing number of electrons (Iodine has 106 electrons but Chlorine has 34 electrons).

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3. Hydrogen Bonds
 It occurs in compounds that have a hydrogen atom attached to one of the three
most electronegative atoms of fluorine, oxygen and nitrogen, which must have
an available lone pair of electrons.

 There is a large electronegativity difference between the H and the F,O,N.

 Hydrogen bonding occurs in addition to London forces.
 The bond angle is often 180° around the H atom.
 Hydrogen bonds can be formed between hydrogen and F,O,N not Cl. Explain
 As the + hydrogen atom is small so can form a strong interaction with small -
atoms as F,O,N but Cl is too large to get close enough to Hydrogen atom to form
hydrogen bond.
 Hydrogen bonding is stronger than the other two types of intermolecular
bonding.

 Water can form two hydrogen bonds per molecule, because the electronegative
oxygen atom has two lone pairs of electrons on it.
 Water has a higher boiling point than expected as it can form stronger hydrogen
bonding ( 2H-bonds per molecule) in addition to their London forces so needs
more energy to break the bonds.(Water boils at 100°C, Hydrogen fluoride at 20°C
and Ammonia at -33°C).

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  The density of ice at 0°C is less than that of water at 0°C which is very unusual
as the molecule in ice are arranged in rings of six, held together by hydrogen
bonds creating large areas of open space inside the rings.

 A single Hydrogen bond between HF molecules is stronger than that between

H2O molecules as Fluorine is more electronegative than Oxygen.
 Boiling point of HF is less than that of water as each molecule of water can form
two hydrogen bonds while each HF molecule can form only one.
 Not all of the hydrogen bonds in HF are broken on vaporization as HF is
polymerized even in the gas phase.
 A single hydrogen bond in NH3 as nitrogen has only one lone pair electrons and it
is weaker than that in H2O and HF as Nitrogen is less electronegative than
Oxygen and Fluorine.

The trends in boiling temperatures of the hydrogen halides HF

to HI

• The boiling temperature increases from HCl to HI due to increasing the number
of electrons in the halogen atom down the group so dispersion force become
stronger and require more energy to break.
• The boiling temperature of HF is much higher due to the stronger hydrogen bond
between the HF molecules in addition to the dispersion force.

Note: Permanent dipole interaction doesn’t support the increase in boiling temperature as
electronegativity decreases down the group so the partial charges become smaller
and permanent dipole forces become weaker.
16
The trends in boiling temperatures of main group hydrides

• The anomalously high boiling points of H2O, NH3 and HF are caused by the
hydrogen bonding between these molecules in addition to their London forces that
require more energy to break.
• The general increase in boiling point from H2S to H2Te or from PH3 to SbH3 is
caused by increasing London forces between molecules due to an increasing
number of electrons.
The trends in boiling temperatures of alkanes with increasing
chain length

• The boiling temperature increased with increasing molecular mass as dispersion

force increase by increasing number of electrons per molecule also increasing
points of contact between adjacent molecules.
• Branched alkanes have lower boiling temperature than their unbranched
isomers due to fewer points of contact between adjacent molecules that
decreases the overall intermolecular force of attraction between molecules.

17
The trends in boiling temperatures of alcohols
• Alcohols form hydrogen bonds in addition to dispersion forces. This means
alcohols have higher boiling points and relatively low volatility compared to
alkanes with a similar number of electrons.

• Ethanol and acetone both are polar having permanent dipole-dipole interaction,
but ethanol has a higher boiling point as ethanol has also hydrogen bonds
between molecules.
• Methoxymethane is non-polar having only London force between molecules
• Alcohols, carboxylic acids, proteins, amides all can form hydrogen bonds.

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 Solubility
~~~~~~~~~~
 Like dissolve like.

➢ Dissolving of ionic solid in water:

• Highly polar solids such as ionic salts, e.g. sodium chloride dissolve in water
(polar solvent) but not in hexane (non-polar solvent).
• When an ionic lattice dissolves in water it required energy to break down the
bonds in the lattice to separate the ions “lattice dissociation energy” and
forming new bonds between the metal ions and water molecules “hydration
energy” which released energy.
• The negative ions are attracted to the δ+ hydrogens on the polar water
molecules and the positive ions are attracted to the δ- oxygen on the polar
water molecules.
• Ionic compounds dissolve in water when:
Lattice dissociation energy < Hydration energy.

➢ Solubility of simple alcohols:

• The smaller alcohols are soluble in water because they can form hydrogen bonds
with water.

19
 • The longer the hydrocarbon chain the less soluble the alcohol as London forces
predominate between the alcohol molecules.
• -OH, -NH and – C=O groups can form hydrogen bonds with water.
➢ Insolubility of compounds in water:
Polar molecules such as halogenoalkanes or non-polar substances like hexane are
insoluble in water as they cannot form hydrogen bonds with water molecules.

➢ Solubility in non-aqueous solvents:

• Hexane and octane are two non-polar liquids that mix completely as both
contain weak London forces which extend throughout their mixture.
• Iodine which has only London forces between its molecules will dissolve in a
non- polar solvent such as hexane which also only has London forces.
• Propanone is a useful solvent because it has
both polar and non-polar characteristics. It
can form London forces with some non-polar
substances such as octane with its CH3
groups. It’s polar C=O bond can also
hydrogen bond with water.

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Topic 8: Redox Chemistry and Groups 1, 2 and 7

Topic 8(A): Redox Chemistry

➢ Redox reaction:
It is the reaction that involves both reduction and oxidation.
➢ Background to oxidation and reduction:
• Oxidation is the process of addition of oxygen or removing of hydrogen.
• Reduction is the process of removing oxygen or addition of hydrogen.

➢ Oxidation and reduction in terms of electron loss and gain:

• Oxidation is the process of electron loss (OIL):
Zn → Zn2+ + 2e-
It involves an increase in oxidation number.
• Reduction is the process of electron gain (RIG):
Cl2 + 2e- → 2Cl-
It involves a decrease in oxidation number.

➢ Oxidising agent and reducing agent:

• Oxidising agent: increases the oxidation number of another atom.
It is the substance which gets reduced and gains electrons.
• Reducing agent: decreases the oxidation number of another atom.
It is the substance which gets oxidised and loses electrons.

➢ Oxidation number:
 It is a number given to each atom or ion in a compound which show us its
degree of oxidation.
 It can be positive, negative or zero.

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➢ Rules for determining the oxidation number:
1. The oxidation number of an uncombined element is Zero.
2. The sum of the oxidation numbers of all the elements in a neutral compound is
Zero.
3. The sum of the oxidation numbers of all the elements in an ion is equal to the
charge on the ions.
4. The more electronegative element in a substance is given a negative oxidation
number.
5. In compounds many atoms or ions have fixed oxidation numbers:
• Group I element are always +1.
• Group II element are always +2.
• Fluorine is always -1.
• Hydrogen is +1 (Except metal hydrides as NaH where it is -1).
• Oxygen is -2 (Except in Peroxides as H2O2 where it is -1, Superoxides as
KO2 where it is -½ and in F2O where it is +2).

➢ Appling the oxidation number rules:

Calculate the oxidation number for the underlined atoms in the following:
KMnO4 HNO3 H2S
K + Mn +4O = zero H + N + 3 O = zero 2H + S= zero
+1 + Mn + 4(-2) = zero +1 + N +3 (-2) = zero 2(+1) S = zero
+1 + 1Mn - 8 =zero +1 + N – 6 = zero +2 + S = zero
Mn= -1 + 8 N =-1 + 6 S = -2
Mn= +7 N = +5

S2O3-2 H2SO4 SO2

2S + 3 O = -2 2H + S + 4 O = zero S + 2O = zero
2S +3 (-2) = -2 2(+1) + S + 4(-2) = zero S + 2(-2) = zero
2S - 6 = -2 +2 + S - 8 = zero S + 2(-2) = zero
2S = + 4 S=+6 S - 4 = zero
S = +2 S = +4

➢ Systematic names:
When an element can have more than one oxidation state, the names of its compounds
and its ions often include the oxidation number of the element using a Roman numeral in
bracket and often referred as the “systematic name”.

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➢ Disproportionation:
It is the simultaneous oxidation and reduction of an element in a single reaction.
Examples:
• Cl2(aq) + H2O(l) → HClO(aq) + HCl (aq)
0 +1 -1 Chlorine disproportionates

• 2H2O2(aq) → 2H2O (l) + O2(g)

-1 -2 0 Oxygen disproportionates

• Cu2O(s) + H2SO4(aq) → Cu(s) + CuSO4(aq) + H2O(l)

+1 0 +2 Copper disproportionates

➢ Using oxidation number to classify reactions:

➢ Writing full equation from ionic half-equations:

A full equation is written by adding two half equations together after balancing the
number of electrons in both to cancel them out.
• Displacement reaction between zinc and copper (II) sulphate:
Oxidation half-equation : Zn → Zn2+ +2e-
Reduction half-equation: Cu2+ + 2e- → Cu
Ionic equation : Zn + Cu2+ → Cu + Zn2+

• Reaction between Iron (II) with Chlorine

Oxidation half-equation : Fe2+ → Fe3+ + e- (equation X2)
Reduction half-equation: Cl2 + 2e- → 2Cl−
Ionic equation : 2Fe2+ + Cl2 → 2Fe3+ + 2Cl−

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• Reaction between acidified potassium manganate (VII) with aqueous solution
of iron (II) sulfate.
Oxidation half-equation : Fe2+ → Fe3+ + e- (equation X5)
Reduction half-equation: MnO4− + 8H+ + 5e- → Mn2+ + 4H2O
Ionic equation : 2Fe2+ + MnO4− + 8H+ → 5Fe3+ + Mn2+ + 4H2O

• Reaction between acidified potassium manganate (VII) with acidified

potassium iodide solution.
Oxidation half-equation : 2I− → I2 + 2e-
Reduction half-equation: MnO4− + 8H+ + 5e- → Mn2+ + 4H2O
Ionic equation :

• Reaction between acidified dichromate (VI) with hydrogen peroxide solution.

Oxidation half-equation : H2O2 → O2 + 2H + 2e-
Reduction half-equation: Cr2O72− + 14H+ + 6e- →2Cr3+ + 7H2O
Ionic equation :

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 Topic 8(B): The elements of Groups 1 and 2

➢ General properties of groups (I) & (II) elements:

• They are all metals.
• They are good conductors of electricity.
• Their compounds are all white or colourless.
• They react with acids to give hydrogen gas.
• Group (1) elements have an oxidation number of +1 and group (II) elements is +2 in
their compounds.
• Compared with group (I) elements group (II) are:
 Harder and denser.
 Of higher melting point.
 Exhibit stronger metallic bonding due to 2 outer shell electrons.

➢ Ionisation of groups (I) & (II) elements:

• The First ionisation energy is the energy required to remove an electron from one
mole of atom in the gaseous state.
M(g) → M+(g) + e−
• The second ionisation energy is the energy required to remove an electron from one
mole of positive ion in the gaseous state.
M+(g) → M2+(g) + e−
• The ionisation energy of groups (I) &(II) elements decreases down the group as :
▪ As the atomic radius increases down the group so the effect of nuclear charge on
the outermost electrons decreases.
▪ The outer shell electrons become more shielded from the attraction of the nucleus
by the repulsive force of inner shell electrons so less energy needed to remove the
electrons.
• The ionisation energy of Magnesium is more than that of Sodium [Same Period]
As the atomic radius decreases across the period and the effect of the nuclear charge
increases so the ionisation energy increased.

➢ Reactivity of groups (I) & (II) elements:

• Reactivity increased downward the group as the atomic radii increase there is more
shielding the nuclear attraction decreases and it is easier to remove (outer) electrons
and so cations form more easily.
• Group 1 elements are more active than group 2 due to increasing the atomic radii so
ionisation energy decreased and less energy needed to remove an electron.
• Group 1 elements and Barium from group 2 are stored under oil due to their reactivity.
25
➢ Reactions of the elements of Group 1 (Li to K) and
Group 2 (Mg to Ba)
• The group 1 and 2 metals form ionic compounds.
• The group 1 and 2 metals act as reducing agents.
• The group 1 and 2 metals get more reactive going down the group.

 Reaction with Oxygen:

• Group 1& 2 metals react with oxygen in air forming white layer of metal oxide.
• They burn in air and more rapidly with oxygen giving white solid (Flame test).
• Example:
▪ Lithium burns in air with red flame giving white solid powder.
4Li(s) + O2(g) → 2Li2O(s)
▪ Magnesium ribbon burns in air with bright white flame giving white solid powder.
2Mg(s) + O2(g) → 2MgO(s)

 Reaction with Chlorine:

• Group 1& 2 metals react with chlorine gas forming solid metal chloride.
• The reaction with chlorine becomes more vigorous down the group.
• Example:
2Na(s) + Cl2(g) → 2NaCl(s)
Mg(s) + Cl2(g) → MgCl2(s)

 Reaction with Water:

Group (I) elements
• All Group 1 metals react with water forming metal hydroxide and hydrogen.
• Reactivity increased down the group.
• Example:
▪ Lithium floats, darts, bubbles of gas and gets smaller then disappears.
2Li(s) + 2H2O (g) → 2LiOH(aq) + H2(g)
▪ Sodium floats, darts, bubbles of gas, gets smaller forming sphere then
disappears and may catch fire.
2Na(s) + 2H2O (g) → 2NaOH(aq) + H2(g)
▪ Potassium floats, darts, bubbles of gas gets smaller forming sphere then
disappears and catch fire with lilac flame.
2K(s) + 2H2O (g) → 2KOH(aq) + H2(g)
Group (II) elements
• Most of Group 2 metals react with water forming metal oxides or hydroxides and
hydrogen.
• Reactivity increased down the group.
• Example:
▪ Beryllium doesn’t react with water due to a thick oxide layer on the surface of
the metal.
26
 ▪ Magnesium reacts very slowly with cold water and rapidly with steam.
Mg(s) + H2O (g) → MgO(s) + H2(g)
Mg(s) + H2O (l) → Mg(OH)2(aq) + H2(g)

▪ The other group 2 metals will react with cold water faster down the group to
form hydroxides.
Ca(s) + 2H2O (l) → Ca(OH)2 (aq) + H2 (g)
The solution turns cloudy as Ca(OH)2 is slightly soluble in water
Ba(s) + 2H2O (l) → Ba(OH)2 (aq) + H2 (g)
Ba(OH)2 is soluble in water

➢ Reactions of the elements of Group 1 and Group 2

Oxides and Hydroxides
 Reaction with Water:
Group (I) oxides and hydroxides:
• All group 1 oxides react with water giving an alkaline solution.
Example : Li2O(s) + H2O(l) → 2LiOH(aq)
• All group 1 hydroxides are soluble in water.
Group (II) oxides and hydroxides:
• Solubility of oxides and hydroxides increases down the group.
• Beryllium oxide does not react with water.
• Magnesium oxide reacts slightly with water giving Mg(OH)2 which is slightly
soluble in water so fewer free OH- ions are produced giving low pH.
MgO(s) + H2O(l) → Mg(OH)2(aq)
• Calcium oxide is known as “Quick lime” as it reacts quickly with water forming
“Slaked lime” and its aqueous solution known as “Lime Water” which used to test
for Carbon dioxide gas.
CaO(s) + H2O(l) → Ca(OH)2(aq)
𝒔𝒉𝒐𝒓𝒕 𝒕𝒊𝒎𝒆 + 𝐂𝐎𝟐 𝐟𝐨𝐫 𝐥𝐨𝐧𝐠 𝐭𝐢𝐦𝐞
Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l) → Ca(HCO3)2(aq)
• Strontium oxide and Barium oxide react similarly as Calcium oxide.

 Reaction with Acids:

Group (I) oxides and hydroxides:
• All are basic oxides so they react with dilute acids giving soluble salt and water.
Example: Li2O(s) + 2HCl(aq) → 2LiCl(aq) + H2O(l)
• The hydroxide react with dilute acids in neutralisation reaction giving salt and
water.
Example: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

27
 Group (II) oxides and hydroxides:
• All are basic oxides except (Beryllium oxide which is amphoteric) so they react
with dilute acids giving soluble salt and water.
Examples: CaO(s) + 2HCl(aq) → CaCl2(aq) + H2O(l)
CaO(s) + H2SO4(aq) → CaSO4(s) + H2O(l)
• The hydroxide react with dilute acids in neutralisation reaction giving salt and
water.
Examples: Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + H2O(l)
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(aq) + H2O(l)

➢ The trends in solubility of the hydroxides and sulfates

of Group 2 elements
• Group II hydroxides become more soluble down the group as the cation radius
increased so the ionic attraction decreases (less lattice enthalpy) and the polarizing
power decreases so attraction with water decreases (less hydration enthalpy).
• Be(OH)2 is the least soluble.
• All Group II hydroxides when not soluble appear as white precipitates.

• Group II sulfates become less soluble down the group.

• BaSO4 is the least soluble.

 Why Calcium stops reacting with dilute sulfuric acid after few seconds although it
reacted initially? Ca(s) + H2SO4(aq) → CaSO4(s) +H2(g)
As calcium reacts with sulfuric acid forming an impermeable solid CaSO4 layer on the top
of the calcium which is insoluble even in sulfuric acid so calcium stop reacting.

➢The trends in thermal stability of the nitrates and the

carbonates of the elements in Groups 1 and 2
• Thermal decomposition is the breaking down of a compound into their elements
or to simpler compounds by heat.
• Thermal stability depends on the polarizing power of the cation which depends
on size and charge of the cation.
• As the charge increases and radius decreases, the charge density increases and
polarizing power increases so thermal stability decreases.
• Down Group (I):
 The radius increases with same charge
 The charge density decreases
 The polarizing power decreases
 So the thermal stability increases.
 Except Lithium as it has small radius and high charge density so it has similar stability
as group (II)

28
➢Thermal stability of the Groups (I) & (II) Carbonates:
• Thermal stability of the Groups (II) Carbonates increases down the group as the cation
radius increased (charge density decreases) so its polarizing effect decreases and distort
the carbonate ion less (Weakening of the C-O bond in the carbonate ion decreases).

• Groups (I) Carbonates don’t decompose as the cation have only +1 charge so the charge
density is not enough to polarize the carbonate ion, EXCEPT Lithium as its ion is small
enough to have a polarizing effect.

 Group (I) Carbonate → doesn’t decompose  (Except Lithium carbonate)

Example: Na2CO3(s) → doesn’t decompose

 Group (II) Carbonate → Metal Oxide + Carbon dioxide

Example: MgCO3(s) → MgO(s) + CO2(g)
• Carbon dioxide gas produced turbid colourless lime water.

➢Thermal stability of the Groups (I) & (II) nitrates:

• Thermal stability of the Groups (II) Nitrates increases down the group as for Carbonates,
Magnesium nitrate decomposes the easiest as Magnesium cation is small with greater
charge density that causes more polarisation effect on nitrate ion and weakness the N-O
bond)

• Groups (I) Nitrates decompose but not in the same way as Group (II) EXCEPT Lithium
Nitrate that decomposes as Group (II) Nitrates.

 Group (I) Nitrate → Metal nitrite + Oxygen  (Except Lithium nitrate)

Example: 2NaNO3(s) → 2NaNO2(s) + O2(g)
• Oxygen gas produced relight a glowing splint.

 Group (II) Nitrate → Metal Oxide + nitrogen dioxide + Oxygen

Example: 2Mg(NO3)2(s) → 2MgO(s) + 4NO2(g) + O2(g)
• NO2 is brown fumes
• Oxygen gas produced relight a glowing splint.

➢Thermal stability of the Groups (I) & (II) Hydrogen carbonates:

 Group (I)&(II) Hydrogen carbonate → Metal carbonate + Water + Carbon dioxide
Example: 2Ca(HCO3)2(s) → CaCO3(s) + H2O(l) + CO2(g)
• Carbon dioxide gas produced turbid colourless lime water.

29
➢ Flame test
Method
• Dip a platinum or nichrome wire ( they are unreactive metals so will not give out any
flame colour and of high melting point) in concentrated hydrochloric acid (to clean
the wire and reacting with salt forming chlorides are more volatile than other salts)
• Dip the wire into the sample of metal salt and put it with the sample in the blue part of
the flame.
• Observe the flame colour change:
Cation Colour Cation Colour
Li+ Red Be2+ No colour
Na+ Yellow Mg2+ No colour
K+ Lilac / pale violet Ca2+ Orange
Rb+ Red Sr2+ Red
Cs+ Blue Ba2+ Apple green

Explanation for occurrence of flame

• In a flame test the heat is absorbed by the electron to move to a higher energy
level.
• The electron is unstable at the higher energy level and so drops back down.
• As it drops back down from the higher to a lower energy level, energy is
emitted in the form of visible light energy with the wavelength of the observed
light within the range of visible light.
Note: Mg2+ doesn’t produce change in flame colour as the light emitted is not within the
visible range.

➢ Test for some Ions

Ion Test Result
1. Carbonate ions Add dilute acid Effervescence and CO2 gas produced
[CO32-] e.g. dilute hydrochloric acid that turns lime water milky.
2. Aq. Sulphate ions Acidify (using dil. HCl acid) White precipitate of BaSO4.
[SO42-] then add aq. Barium chloride Ba2+ + SO42-→ BaSO4
(OR Acidify using dil. HNO3 acid
then add aq. Barium nitrate).
3. Ammonium ions Add aq. sodium hydroxide and Ammonia gas produced with pungent
[NH4+] warm smell that turns damp red litmus paper
blue.
And forms white solid smoke with HCl
fumes.

30
Titrations:
➢ The method for carrying out the titration
• Rinse equipment (burette with acid, pipette with alkali, conical flask with distilled water)
• Fill pipette by 25 cm3 of alkali into conical flask
• Adds acid solution from burette
• Make sure the jet space in the burette is filled with acid.
• Add a few drops of indicator and refer to colour change at end point
 phenolphthalein [pink (alkali) to colourless (acid): end point pink colour just disappears] [use
if NaOH is used]
 methyl orange [yellow (alkali) to red (acid): end point orange][use if HCl is used]
• Use a white tile underneath the flask to help observe the colour change
• Add acid to alkali whilst swirling the mixture and add acid drop wise at end point
• Repeats titration until at least 2 concordant results are obtained- two readings within 0.2
of each other

➢ Working out average titre results

Only make an average of the concordant titre results

➢ Recording results
• Results should be clearly recorded in a table
• Result should be recorded in full (i.e. both initial and final readings)
• Record titre volumes to 2dp (0.05 cm3)

➢ Safety precautions
• Acids and alkalis are corrosive (at low concentrations acids are irritants) so Wear eye
protection and gloves
• If spilled immediately wash affected parts after spillage
• If substance is unknown treat it as potentially toxic and wear gloves.

➢ A conical flask is used in preference to a beaker because it is easier to swirl the mixture in a
conical flask without spilling the contents.
➢ Distilled water can be added to the conical flask during a titration to wash the sides of the
flask so that all the acid on the side is washed into the reaction mixture to react with the
alkali.
➢ It does not affect the titration reading as water does not react with the reagents or change
the number of moles of acid added.
➢ Only distilled water should be used to wash out conical flasks between titrations because it
does not add an extra moles of reagents.

31
 Errors:
Calculating apparatus errors
• Each type of apparatus has a sensitivity error
 Balance ± 0.001 g
 Volumetric flask ± 0.1 cm3
 25 cm3 pipette ± 0.1 cm3
• Burette ± 0.05 cm3
• Calculate the percentage error for each piece of equipment used by:
% error = ± sensitivity error x 100
• Measurement made on apparatus e.g. for pipette
% error = 0.05/ 25 x100
• To calculate the maximum percentage apparatus error in the final result add all the individual
equipment errors together.

Reducing errors in a titration

• Replacing measuring cylinders with pipettes or burettes which have lower apparatus
sensitivity error will lower the error
• To reduce the error in a burette reading it is necessary to make the titre a larger volume. This
could be done by: increasing the volume and concentration of the substance in the conical
flask or by decreasing the concentration of the substance in the burette.

➢ If the burette used in the titration had an uncertainty for each reading of +/– 0.05 cm3 then during
a titration two reading would be taken so the error on the titre volume would be +/– 0.10 cm3.

Reducing errors in measuring mass

• Using a more accurate balance or a larger mass will reduce the error in weighing a solid
• Weighing sample before and after addition and then calculating difference will ensure a more
accurate measurement of the mass added.

Example:
A 25 cm3 sample of vinegar was diluted in a 250 cm3 volumetric flask. This was then put in a
burette and 23.10 cm3 of the diluted vinegar neutralised 25 cm3 of 0.1 M NaOH.
What is the concentration of the vinegar in gdm-3 ?
CH3CO2H + NaOH → CH3CO2 -Na+ + H2O
Step 1: work out amount, in mol, of sodium hydroxide amount = conc x vol = 0.10 x 0.025
= 0. 00250 mol
Step 2: use balanced equation to give moles of CH3CO2H
1 moles NaOH : 1 moles CH3CO2H
So 0.00250 NaOH : 0.00250 moles CH3CO2H
Step 3: work out concentration of diluted CH3CO2H in 23.1 (and 250 cm3) in mol.dm-3
Conc = amount/Volume = 0.00250 / 0.0231 = 0.108 mol.dm-3
Step 4: work out concentration of original concentrated CH3CO2H in 25cm3 in mol.dm-3
Conc = 0.108 x 10 = 1.08 mol.dm-3
Step 5: work out concentration of CH3CO2H in original concentrated 25 cm3 in g.dm-3
Conc in g.dm-3 = conc in mol.dm-3 x Mr = 1.08 x 60 = 64.8 g dm-3
32
Making a solution from a solid
• Weigh required mass of solute in a weighing bottle.
• Tip contents into a beaker and add 100 cm3 of distilled water.
• Use a glass rod to stir to help dissolve the solid.
• Sometimes the substance may not dissolve well in cold water so the
beaker and its contents could be heated gently until all the solid had
dissolved.
• Pour solution into a 250 cm3 graduated flask via a funnel.
• Rinse beaker and funnel and add washings from the beaker and glass
rod to the volumetric flask.
• make up to the mark with distilled water using a dropping pipette for
last few drops.
• Invert flask several times to ensure uniform solution.

Diluting a solution
• Pippette 25 cm3 of original solution into volumetric flask.
• Make up to the mark with distilled water using a dropping pipette for last few drops.
• Invert flask several times to ensure uniform solution.

33
 Topic 8(C): The elements of Group 7

• Fluorine (F2): Pale yellow gas, highly reactive and very strong oxidising agent.
• Chlorine (Cl2): greenish, reactive gas, poisonous in high concentrations
• Bromine (Br2) : red liquid, that gives off dense brown/orange poisonous fumes
• Iodine (I2): shiny grey solid sublimes to purple gas.

➢ General properties of group VII elements

• They behave in a similar chemical manner.
• Non-metals.
• They exist as diatomic molecules.
• Melting and boiling points increase down the group As the molecules become
larger they have more electrons and so have stronger dispersion forces between the
molecules required more energy has to break the force this increases the melting and
boiling points.
• Colour deepens down the group.
• Electronegativity decrease down the group as the effect of nuclear charge
decreases, the atomic radii increases due to the increasing number of shells and
the shielding effect of electrons in inner energy levels increases therefore the
attraction forces between the nucleus and the outer electrons becomes weaker.
• Reactivity decreases down the group as the atoms get bigger with more
shielding so they less easily attract and accept electrons.
• It is more difficult to gain electrons down the group so the oxidation power
decrease.
• They can exist in covalent and ionic compounds
• They exhibit a range of oxidation numbers.

➢ Solubility of group (VII) elements [Halogens]:

 Solubility in Water:
• The solubility of the halogens in water decreases down the group.
• Chlorine reacts in water forming Hydrochloric acid and Hypochlorous acid
(bleaching material) in disproportionation reaction.
Cl2(g) + H2O → HCl(aq) + HClO(aq)
• Chlorine is used in water treatment as it is disinfectant.
• Bromine dissolves and reacts with water in a similar way but to a lesser extent.
• Iodine is almost insoluble in water.
34
  Solubility in Hexane:
• Halogens are non-polar so they are more soluble in hydrocarbon solvents as
hexane than water.
• Fluorine dissolves giving pale yellow solution, Chlorine gives pale yellow
solution, bromine give pale orange solution and iodine gives violet solution.

➢ Oxidation reactions with Group 1 and 2 metals:

• The halogens react strongly with all of the more electropositive elements and
become reduced to negative halide ions.
• 2Na(s) + Cl2(g) → 2NaCl(s)
• Mg(s) + Cl2(g) → MgCl2(s)
• 2Fe(s) + 3 Cl2(g) → 2FeCl3(s)

➢ The disproportionation reaction of chlorine with cold and hot

dilute aqueous sodium hydroxide:
 Chlorine with cold Sodium Hydroxide (15°C):
2NaOH(aq) + Cl2(aq) → NaCl(aq) + NaClO (aq) + H2O(l)
Sodium Chlorate(I)
0 -1 +1
Chlorine oxidised in NaOCl and reduced in NaCl

 Chlorine with hot Sodium Hydroxide (70°C):

6NaOH(aq) + 3Cl2(aq) → 5NaCl(aq) + NaClO3 (aq) + 3H2O(l)
0 -1 +5
Chlorine oxidised in NaOCl3 and reduced in NaCl

 Bromine and iodine react similarly but less reactivity

35
➢ Solid Group 1 halides with concentrated sulfuric acid
• The reducing power of the halides increases down group 7.
• This is because as the ions get bigger so it is easier for the outer electrons to be given
away as the effect of the nuclear charge on them becomes smaller.
• This can be clearly demonstrated in the various reactions of the solid halides with
concentrated sulfuric acid.

 With Fluoride and Chloride

The H2SO4 is not strong enough an oxidising reagent to oxidise the chloride and fluoride ions.
No redox reactions occur.
NaF(s) + H2SO4(aq) → NaHSO4(s) + HF(g)
Observations: White steamy (misty) fumes of HF are evolved.
NaCl(s) + H2SO4(aq) → NaHSO4(s) + HCl(g)
Observations: White steamy fumes of HCl are evolved.

 With Bromide
H2SO4 plays the role of acid in the first step producing HBr and then acts as an oxidising agent in
the second redox step.
Acid- base step : NaBr(s) + H2SO4(aq) → NaHSO4(s) + HBr(g)
Redox step : 2HBr + H2SO4 → Br2(g) + SO2(g) + 2H2O(l)
Observations: White steamy fumes of HBr are evolved.
Red fumes of Bromine are also evolved.
A colourless, acidic gas SO2 with choking smell.
Oxidation half equation: 2Br −→ Br2 + 2e-
Reduction half equation: H2SO4 + 2 H+ + 2e− → SO2 + 2 H2O
Br− ions are stronger reducing agents than Cl− and F− and after the initial acid-base reaction it
reduces the Sulphur in H2SO4 from +6 to + 4 in SO2.

 With Iodide
H2SO4 plays the role of acid in the first step producing HI and then acts as an oxidising agent in
the three redox steps
NaI(s) + H2SO4(aq) → NaHSO4(s) + HI(g)
2HI + H2SO4(aq) → I2(s) + SO2(g) + 2H2O(l)
6HI + H2SO4(aq) → 3I2(s) + S (s) + 4 H2O (l)
8HI + H2SO4(aq) → 4I2(s) + H2S(g) + 4H2O(l)

Observations:  White steamy fumes of HI are evolved.

 Black solid and purple fumes of Iodine are also evolved
 A colourless, acidic gas SO2
 A yellow solid of Sulphur
 H2S (Hydrogen Sulphide), a gas with a bad egg smell.
I− ions are the strongest halide reducing agents. They can reduce the Sulfur from +6 in H2SO4 to + 4 in
SO2, to 0 in S and -2 in H2S.
36
➢ Reaction of hydrogen halides with water (to produce acids)
The hydrogen halides are all soluble in water to form acidic solution (H3O+).
HCl(g) + H2O(l) → H3O+ + Cl−

➢ Reaction of hydrogen halides with ammonia gas (to produce

ammonium halides)

All the hydrogen halides react readily with ammonia to give the white smoke of the
ammonium halide
HCl(g) + NH3 (g) → NH4Cl (s)
HBr(g) + NH3 (g) → NH4Br (s)
HI(g) + NH3 (g) → NH4I (s)
This can be used as a test for the presence of hydrogen halides

➢ Test for halides

Anion Test Observation
Halides Ions: Add drop of dil
nitric acid then
Silver nitrate
a) Chloride ion solution. a) White Precipitate [AgCl]
[Cl-1] Add dilute Soluble in dilute and concentrated
Ammonia then Ammonia.
concentrated
b) Bromide ion b) Cream Precipitate [AgBr]
Ammonia.
[Br-1] Insoluble in in dilute Ammonia but
Soluble in conc. Ammonia.

c) Iodide ion [I-1] c) Yellow Precipitate [AgI]

Insoluble in in dilute Ammonia
Insoluble in conc. Ammonia.

AgX Leave it in The precipitate darkness or turns grey as

Sunlight silver is formed:
2AgX → 2Ag + X2

The role of nitric acid is to react with any carbonates present to prevent formation of the
precipitate Ag2CO3.

 Silver chloride dissolves in dilute ammonia to form a Colourless complex ion

AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+ (aq) + Cl− (aq)

37
 Topic 9: Introduction to Kinetics and Equilibia

9A: Kinetics
➢ Rate of reaction:
• The rate of reaction is defined as the change in concentration of a substance per unit
time, it’s unit is mol dm-3s-1.
change in concentration
• Rate of reaction =
time for change to happen
• So, in order to measure the rate of a reaction, we need to find out:
1. How fast one of the reactants is being used up, or
2. How fast one of the products is being formed
• In a graph of concentration of a reactant against time, the gradient (slope) of the graph
indicates the rate of the reaction.

➢ The factors which affect the rate of reaction:

The reaction goes faster when any of the following factors is increased.
1- Temperature : for all reactions.
2- Surface Area : for solid reactant.
3- Concentration : for aqueous (solutions) reactant.
4- Pressure : for gas reactant.
5- Catalyst : for some specific reactions.
6- Light Intensity : for some specific reactions.
7- Nature of Reagents.

38
➢ The collision theory of reactivity
For a reaction to occur:
1. There should be collisions between the molecules of reactants
2. These collisions should be energetic enough
3. 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, which increases
the rate of the reaction
• Increasing the temperature increases the proportion of molecules with sufficient
energy to react, which increases the rate of reaction.

➢ The Activation Energy

It is the minimum energy which particles need to collide to start a reaction.
• It is the energy needed to break the bonds between
reactants.
• A low activation energy means a fast reaction that will
occur at room temperature.
• A high activation energy means slow reaction that will
require heat energy or a catalyst as it will not occur
spontaneously at room temperature.

39
➢ Maxwell Boltzmann Distribution
The Maxwell-Boltzmann energy distribution shows the spread of energies that molecules of a
gas or liquid have at a particular temperature

How can a reaction go to completion if few particles have energy greater than Ea?
As particles can gain energy through collisions.

By Increasing Temperature
• As the temperature increases the distribution shifts towards having more molecules with
higher energies.
• At higher temps both the Emp and mean energy shift to high energy values although the
number of molecules with those energies decrease.

40
➢ Effect of Increasing Temperature
• At higher temperatures the energy of the
particles increases. They collide more
frequently and more often with energy
greater than the activation energy.
• More collisions result in a reaction
• As the temperature increases, the graph
shows that a significantly bigger proportion
of particles have energy greater than the
activation energy, so the frequency of
successful collisions increases.

➢ Effect of Increasing Surface area

Increasing surface area of solids will cause collisions to occur more frequently between
the reactant particles and this increases the rate of the reaction.

➢ Effect of Catalysts
• Catalysts is a chemical substance that increases reaction rates without getting used up
and chemically unchanged.
• As they provide an alternative route or
mechanism with a lower activation energy.
• Catalyst weaken the bond of the reactants, the
reaction take place on the catalyst surface.
• Enzymes are proteins that act as biological
catalysts.
• Catalysts are used to make processes more
sustainable by using less energy to give higher atom economy.

If the activation energy is lower, more particles will have energy > EA, so there will be a
higher frequency of effective collisions. The reaction will be faster.

41
➢ Effect of Increasing Concentration and Increasing Pressure
At higher concentrations (for dissolved reactants) and pressures (for gaseous reactants)
there are more particles per unit volume and so the particles collide with a greater
frequency and there will be a higher frequency of effective collisions.

Note:
If a question mentions a doubling of concentration/rate then make sure you mention
double the number of particles per unit volume and double the frequency of effective
collisions.

If concentration increases, the shape of the

energy distribution curves do not change (i.e.
the peak is at the same energy) so the Emp
and mean energy do not change
The curves will be higher, and the area under
the curves will be greater because there are
more particles.

42
 9B: Equilibria

• Many reactions are reversible as: N2 + 3H2 2NH3

• A case of dynamic equilibrium is reached when:-
1- The rate of forward reaction equals the rate of backward reaction.
2- The concentration of any reactant and/or product remains
constant (does not change at equilibrium).
3- All reactants and products are present at any moment.

➢ Le Chatelier’s Principle

Le Chatelier’s principle states that if an external condition is changed the equilibrium will
shift to oppose the change (and try to reverse it).

➢Factors which affect equilibrium:

 Effect of pressure:
On increasing the pressure, the equilibrium shifts in the direction which produces less
number of moles of gases (less volume) and vice versa.

 Effect of temperature:
• Increasing the temperature makes the equilibrium shifts into the direction that takes in
heat (the endothermic direction).
• Decreasing the temperature makes the equilibrium shifts into the exothermic direction.

 Effect of concentration:
Increasing or decreasing the concentration in aqueous solutions makes the equilibrium
shifts into the direction of less concentration.

 A catalyst doesn’t affect the position of equilibrium in a reversible reaction as it

speeds up both forward and backward reactions equally.

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Examples:
 Effect of Pressure on equilibrium:
CO (g) + 2H2(g) CH3OH (g)
• If pressure is increased the equilibrium will shift to oppose this and move towards the
side with fewer moles of gas to try to reduce the pressure to the right side, giving a
higher yield of methanol.
• Decreasing pressure will cause the equilibrium to shift towards the side with more
moles of gas to oppose the change and thereby increase the pressure.
• If the number of moles of gas is the same on both sides of the equation then changing
pressure will have no effect on the position of equilibrium
H2(g) + Cl2(g) 2HCl(g)
• Increasing pressure may give a higher yield of product and will produce a faster rate.
Industrially high pressures are expensive to produce and of high risk.

 Effect of Temperature on equilibrium:

N2 (g)+ 3H2(g) 2NH3 (g)
• If temperature is increased the equilibrium will shift to oppose this and move in the
endothermic, backwards direction to try to decrease temperature. The position of
equilibrium will shift towards the left, giving a lower yield of ammonia.
• If temperature is decreased the equilibrium will shift to oppose this and move in the
exothermic direction to try and increase the temperature by giving out heat.
• Low temperatures may give a higher yield of product but will also result in slow rates of
reaction.

 Effect of Concentration on equilibrium:

I2 + 2OH- I- + IO- + H2O
Brown colourless
• Increasing the concentration OH- ions causes the equilibrium to shift to oppose this and
move in the forward direction to remove OH- ions so the position of equilibrium will
shift towards the right, giving a higher yield of I- and IO-. (The colour would change from
brown to colourless)
• Adding H+ ions reacts with the OH- ions and reduces their concentration so the
equilibrium shifts back to the left giving brown colour.
•

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Iodine (I) chloride and Iodine (III) chloride:
I2(s) + Cl2(g) → 2ICl(l)
ICl(l) + Cl2(g) ICl3(s)
Brown Green Yellow

• If chlorine gas is passed through a U-tube containing solid iodine, a brown liquid of
Iodine (I) chloride ICl is formed with brown vapour above it and the bottom of tube gets
hot as it is an exothermic reaction.
• If more chlorine gas is passed through the U-tube a yellow solid Iodine (III) chloride ICl3 is
formed.
• If the chlorine supply is removed and the U-tube is tipped horizontal, the yellow crystals
disappear and a brown gas is seen.
• If the chlorine supply is attached again the yellow solid forms as the equilibrium moves
to the right and producing Iodine (III) chloride again.

Stored methane in equilibrium:

Methane Hydrated (s) Methane (g) + Water (l) ∆H= +ve
• There are huge amount of methane trapped as solid methane hydrate I ice structure
deep in the oceans which is in equilibrium wit gaseous methane and water through
endothermic reaction.
• If the Earth’s temperature rises, the equilibrium will move to the right to oppose this
change releasing more methane into the atmosphere.
• If there is an increase in pressure, equilibrium will move to the left turns the methane
gas back into solid methane hydrate.

Nitrogen dioxide and dinitrogen tetroxide:

2NO2 N 2O 4 ∆H= -58 kj/mol
Brown pale yellow
• If temperature increases equilibrium shifts to the left side as backward reaction is
endothermic and the mixture gets darker.
• If the pressure decrease the equilibrium shifts to the left to the side of more number
of moles and the mixture gets darker.

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➢ Importance of equilibrium to industrial processes:

Haber process:
N2 (g) + 3H2(g) 2NH3 (g) ∆H = -91 kJ mol-1

Temperature = 450oC, Pressure= 200 – 1000 atm, Catalyst = iron

Low temp gives good yield but slow rate:
High pressure gives good yield and high rate (too high energy costs)
Catalyst speed up the reaction.

Contact process:
Stage 1: S (s) + O2 (g) → SO2 (g)
Stage 2: SO2 (g) + ½O2 (g) SO3 (g) ∆H = -98 kJ mol-1

Temperature = 450oC, Pressure = 10 atm, Catalyst = V2O5

Low temp gives good yield but slow rate.
High pressure gives slightly better yield and high rate cost much.
Catalyst speed up the reaction.

Production of methanol from CO

CO (g) + 2H2(g) CH3OH (g) ∆H = -ve (exothermic reaction)

Temperature = 400oC, Pressure = 50 atm, Catalyst = chromium and zinc oxides

Low temp gives good yield but slow rate:
High pressure gives good yield and high rate but coat much.

 In all cases catalysts speeds up the rate allowing lower temp to be used (and hence lower
energy costs) but have no effect on equilibrium

 In all cases high pressure leads to too high energy costs for pumps to produce the pressure
and too high equipment costs to have equipment that can withstand high pressures.

 Recycling unreacted reactants back into the reactor can improve the overall yields of all
these processes

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Topic 10: Organic Chemistry: Halogenoalkanes, Alcohols and Spectra

10A: General principles

➢ Types of chemical reactions of organic compounds:

➢ Addition reaction: which involve two molecules combining to make only one new
molecule.
CH2=CH2 (g) + H2O (g) → CH3CH2OH (g)
➢ Elimination reaction: which involve removal of small molecule (often water) from the
organic molecule using alcoholic alkali (ethanolic KOH) where the OH− acts as base. .

➢ Substitution reaction: which involves replacing an atom by another atom.

CH3CH2Br (l) + OH− (aq) → CH3CH2OH (aq) + Br− (aq)
➢ Oxidation reaction: which involves addition of oxygen or removal of hydrogen using
an oxidizing agent as KMnO4, K2Cr2O7.
CH3CH2OH (aq) + O2 (g) → CH3COOH (aq) + H2O (l)
➢ Reduction reaction: which involves removal of oxygen or addition of hydrogen.
➢ Hydrolysis reaction: which involves breaking down of a compound using water in
presence of acid or enzyme as catalyst.
➢ Polymerisation: which many monomers joined together to form one very large
polymer.

➢ Types of reagents:
➢ Nucleophile: is an electron-pair donor which used to form a covalent bond, e.g.
H2O:, :NH3, OH−, CN−,…….

➢ Electrophile: is an electron pair acceptor and is attracted to high electron density

molecules, e.g. Polar HCl, HBr, HI ,Cl2, Br2,,……

➢ Free radical is a specie that contains an unpaired electron, e.g. Cl, Br, NO, NO2 ,……
it can be formed by hemolytic fission in presence of UV light or heterolytic fission.

Homolytic fission:

Heterolytic fission:

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 10B: Halogenoalkanes

➢ Naming Halogenoalkanes
Based on original alkane, with a prefix indicating halogen atom: Fluoro for F; Chloro for Cl;
Bromo for Br; Iodo for I.

➢ Classifying Halogenoalkanes:
Halogenoalkanes can be classified as primary, secondary or tertiary depending on the number of
carbon atoms attached to the C-X functional group.

➢ Reactions of Halogenoalkanes:
Halogenoalkanes undergo either nucleophilic substitution or elimination reactions.

• Elimination: removal of small molecule (often water) from the organic molecule.

• Nucleophilic substitution reactions:

Substitution: swapping a halogen atom by another atom or groups of atoms.
Nucleophile: electron pair donator e.g. :OH-, :NH3, CN−, H2O:

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 Elimination reaction:
• Using alcoholic alkali as ethanolic KOH as The OH− acts as a base.
• With Primary halogenoalkane:

• With unsymmetrical secondary and tertiary halogenoalkanes two (or sometimes three)
different structural isomers can be formed.

• The structure of the halogenoalkane also has an effect on the degree to which
substitution or elimination occurs in this reaction.
• Primary tends towards substitution Tertiary tends towards elimination.

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 Nucleophilic Substitution reaction:

The rate of these substitution reactions depends on the strength of the C-X bond,
the weaker the bond (as its length increased), the easier it is to break and the faster
the reaction.
➢The iodoalkanes are the fastest to substitute and the fluoroalkanes are the slowest. So the
fluoroalkanes are very unreactive.

 Nucleophilic substitution with aqueous hydroxide ions

• Reagent: Potassium (or sodium) hydroxide
• Conditions: In aqueous solution; Heat under reflux
• Mechanism: Nucleophilic Substitution
• Role of reagent: Nucleophile, OH−
• Change in functional group: halogenoalkane →alcohol

 The OH– is a stronger nucleophile than water as it has a full negative charge and so is
more strongly attracted to the Cδ+
 The aqueous conditions needed is an important point. If the solvent is changed to
ethanol an elimination reaction occurs.

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SN2 nucleophilic substitution mechanism for halogenoalkanes
This mechanism occurs with primary halogenoalkanes.
( SN2 =S “substitution”, “N” nucleophilic, “2”two reactants at the beginning of the mechanism)

SN1 nucleophilic substitution mechanism for tertiary halogenoalkanes

Tertiary haloalkanes undergo nucleophilic substitution in a different way.

 Tertiary halogenoalkanes undergo this mechanism as the tertiary carbocation is made

stabilised by the electron releasing methyl groups around it (see alkenes topic for
another example of this).
 Also the bulky methyl groups prevent the hydroxide ion from attacking the
halogenoalkane in the same way as the mechanism above.
 Primary halogenoalkanes don’t do the SN1 mechanism because they would only form
an unstable primary carbocation.

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 Nucleophilic substitution with water (Hydrolysis)
• Hydrolysis is defined as the splitting of a molecule (in this case a halogenoalkane) by a
reaction with water.
• Water is a poor nucleophile but it can react slowly with halogenoalkanes in a
substitution reaction.

 Aqueous silver nitrate is added to a halogenoalkane and the halide leaving group
combines with a silver ion to form a SILVER HALIDE PRECIPITATE.
 The precipitate only forms when the halide ion has left the halogenoalkane and so the
rate of formation of the precipitate can be used to compare the reactivity of the
different halogenoalkanes.

• The iodoalkane forms a precipitate with the silver nitrate first as the C-I bond is
weakest and so it hydrolyses the quickest.
• The quicker the precipitate is formed, the faster the substitution reaction and the
more reactive the haloalkane.
• The rate of these substitution reactions
depends on the strength of the C-X
bond, the weaker the bond, the easier it
is to break and the faster the reaction.

➢ This reaction can be speeded up by using AgNO3 with NaOH as it is a stronger

nucleophile than water.
➢ Sometimes ethanol is added in addition to water as a solvent to dissolve
halogenoalkane and AgNO3 mixture.

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 Nucleophilic substitution with ammonia
• Reagent: NH3 dissolved in ethanol
• Conditions: Heating under pressure in a sealed tube
• Mechanism: Nucleophilic Substitution
• Type of reagent: Nucleophile, :NH3
• Change in functional group: halogenoalkane → amine

Reaction Mechanism:

Further substitution reactions can occur between the halogenoalkane and the amines
formed leading to a lower yield of the amine as secondary and tertiary amines may be
formed. Using excess ammonia helps minimize this.

Further reactions

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 Nucleophilic substitution with alcoholic Potassium Cyanide
• Reagent: KCN dissolved in ethanol
• Conditions: Heating under reflux
• Mechanism: Nucleophilic Substitution
• Type of reagent: Nucleophile, :CN−
• Change in functional group: halogenoalkane → nitrile

Reaction Mechanism:

 This reaction increases the length of the carbon chain and form nitriles as
Bromoethane forms propanenitrile
 Potassium cyanide (KCN) is highly toxic.

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 10C: Alcohols
• General formula of alcohols is CnH2n+1OH
• They are compounds containing OH group.

➢ Naming of Alcohols
• These have the ending -ol and if necessary
the position number for the OH group is
added between the name stem and the –ol.

• If the compound has an –OH group in

addition to other functional groups that need
a suffix ending then the OH can be named
with the prefix hydroxy-.

• If there are two or more -OH groups then di, tri

are used.
Add the ‘e’ on to the stem name though

➢ Types of Alcohols

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➢ The reactions of Alcohols
 Combustion:
Alcohols combust with a clean flame

 Elimination of Alcohol to alkene by heating with conc.

phosphoric acid:
CH3CH(OH)CH2CH3 → H2O + CH2=CHCH2CH3 OR CH3CH=CHCH3 “cis”
OR CH3CH=CHCH3 “trans”

 Substitution reactions of Alcohols to form Halogenoalkanes

“Halogenation”
Various halogenating compounds can be used to substitute the –OH group for a halogen

 Reaction with phosphorous (V) halide:

• This reaction with PCl5 (phosphorous (v) chloride) can be used as a test for alcohols.
• You would observe misty fumes of HCl produced.
• For Br and I it is best to use PI5, PI3 and Br equivalents

 Reaction with hydrogen bromide:

KBr + H2SO4 → KHSO4 + HBr
C2H5OH + HBr → C2H5Br + H2O
• This reaction KBr mixed with 50% H2SO4 and heat under reflex.
• If conc. H2SO4 used : HBr will be oxidized to Br2 and alcohol will be dehydrated to
alkene.

 Reaction with phosphorous (III) halide:

P4 + I2 → 4PI3

• PI3 is produced by heating under reflux red Phosphorous and Iodine.

• Concentrated or diluted H2SO4 are not used as HI will be oxidized to I2 and
phosphorous acid is formed.
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 Reaction of Alcohols with sodium:

Observations: • Effervescence,
• The mixture gets hot,
• Sodium dissolves,
• A white solid is produced.
This reaction can be used as a test for alcohols

 Oxidation of Alcohols:
• Potassium dichromate K2Cr2O7 is an oxidising agent that causes alcohols to
oxidise.
• The reaction depends on: the type of alcohol, i.e. whether it is primary,
secondary, or tertiary, and on the conditions.

➢Partial Oxidation of Primary Alcohols

Reaction: primary alcohol → aldehyde
Reagent: potassium dichromate (VI) solution and dilute sulphuric acid.
Conditions: (use a limited amount of dichromate) warm gently and distil out
the aldehyde as it forms.
Observation: the orange dichromate ion (Cr2O7 2-) reduces to the green Cr3+ ion

➢Full Oxidation of Primary Alcohols

Reaction: primary alcohol → carboxylic acid
Reagent: potassium dichromate(VI) solution and dilute sulphuric acid
Conditions: use an excess of dichromate, and heat under reflux: (distill off
product after the reaction has finished)
Observation: the orange dichromate ion (Cr2O72-) reduces to the green Cr3+ ion.

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 Oxidation of Secondary Alcohols
Reaction: secondary alcohol → ketone
Reagent: potassium dichromate(VI) solution and dilute sulphuric acid.
Conditions: heat under reflux
Observation: the orange dichromate ion (Cr2O72-) reduces to the green Cr3+ ion.
There is no further oxidation of the ketone under these conditions.

Tertiary alcohols cannot be oxidised at all by potassium dichromate: This is

because there is no hydrogen atom bonded to the carbon with the OH group.

➢ Distinguishing between Aldehydes and Ketones

It depends on the fact that aldehydes can be further oxidised to carboxylic acids whereas
ketones cannot be further oxidised.
Fehling’s (Benedict’s) solution
Reagent: Fehling’s Solution containing blue Cu2+ ions.
Conditions: heat gently
Reaction: aldehydes only are oxidised by Fehling’s solution into a carboxylic acid and
the copper ions are reduced to copper (I) oxide
Observation: With Aldehydes : Blue Cu2+ ions in solution change to a red precipitate
of Cu2O but with Ketones do not react.

➢ Distinguishing between alcohols and carboxylic acids

Add sodium carbonate, it will fizz and produce carbon dioxide gas with carboxylic acids
only.
Add blue litmus paper, it changes red with carboxylic acids and unchanged with alcohols.

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Summary:

59
Organic Techniques
➢ Distillation
In general used to separate an organic product from its reacting mixture.

Example
Reaction: primary alcohol → aldehyde
Reagent: potassium dichromate (VI) solution and
dilute sulfuric acid.
Conditions: use a limited amount of dichromate
and warm gently and distil out the aldehyde
as it forms [This prevents further oxidation to
the carboxylic acid]
Observation: Orange dichromate solution
changes to green colour of Cr3+ ions.

➢ Reflux
Reflux is used when heating organic reaction mixtures for long periods. The
condenser prevents organic vapours from escaping by
condensing them back to liquids.

Example
Reaction: primary alcohol → carboxylic acid
Secondary or tertiary alcohol → Ketone
Reagent: potassium dichromate(VI) solution and dilute sulfuric
acid
Conditions: use an excess of dichromate, and heat under reflux:
(distill off product after the reaction has finished using
distillation set up)
CH3CH2CH2OH + 2[O] → CH3CH2CO2H + H2O

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 Observation:
• Orange dichromate solution changes to green colour of Cr3+ ions.
• Reflux is used when heating organic reaction mixtures for long periods.
• The condenser prevents organic vapours from escaping by condensing them back to
liquids.
• Never seal the end of the condenser as the build up of gas pressure could cause the
apparatus to explode. This is true of any apparatus where volatile liquids are heated
including the distillation set up Water in Water out Liebig condenser thermometer.
• Heat Anti-bumping granules are added to the flask in both distillation and reflux to
prevent vigorous, uneven boiling (smooth boiling).
• Note:
The bulb of the thermometer should be at the T junction connecting to the condenser
to measure the correct boiling point.

Purifying an organic liquid

• Put the distillate of impure product into a separating funnel.
• Wash product by adding either
 Sodium hydrogencarbonate solution , shaking and releasing the
pressure from CO2 produced.
 Saturated sodium chloride solution.
• Allow the layers to separate in the funnel, and then run and
discard the aqueous layer.
• Run the organic layer into a clean, dry conical flask and add three spatula loads of drying
agent (anhydrous sodium sulphate) to dry the organic liquid.
• Carefully decant the liquid into the distillation flask
• Distill to collect pure product.

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 10D: Mass spectra and IR
➢ Mass spectrometry
Measuring the Mr of an organic molecule
• In a mass spectrometer an electron is removed from an organic molecule forming a
positive ion called “The molecular ion” which tend to be break up into fragments.
• The peak with the largest m/z, however, will be due to the molecular ion or the parent
ion and will be equal to the Mr of the molecule.
• A very small peak to the right of the molecular ion peak may be formed due to the
presence of one atom of C-13.
• The tallest peak is the peak with the greatest abundance and called the base peak.

Example of fragmentation:
• Molecular ion of ethane: (CH3-CH3)+ → CH3+ + CH3
Free radicals as CH3 are not detected in a mass spectrometer,
But CH3+ has a peak at m/z = 15
• Molecular ion of propane: (CH3-CH2-CH3)+ → CH3+ + CH2-CH3
OR (CH3-CH2-CH3)+ → CH3 + (CH2-CH3)+
We can expect to see peaks at m/z =15 (the methyl cation) and m/z = 29 (the ethyl cation)
• This is the mass spectra of butane:

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• This is the mass spectra of Propan-1-ol:

CH3CH2CH2OH+ (m/z =60) C3H7O+ (m/z = 59) + H

C3H6+ (m/z = 42) + H2O
CH2OH+ (m/z = 31) + CH3CH2
CH3CH2+ (m/z = 29) + CH2OH

• Some common m/z values and possible ions responsible for these peaks:

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Mass spectra of Propanal and Propanone (C3H6O):

Propanal Propanone

• Both mass spectra show a peak at m/z =58 which is corresponding to the molecular ion
CH3CH2CHO+.
• Both mass spectra show a peak at m/z =59 which is corresponding to C-13 isotope.
• Propanal (CH3CH2COH) mass spectra shows:
 A peak at m/z =57 which is corresponding to the molecular ion CH3CH2CO+ (losing H).
 A peak at m/z =29 which is corresponding to the CHO+ or C2H5+.
 A peak at m/z =15 which is corresponding to the CH3+.
Propanal doesn’t have a peak at 43 as the molecule can’t fragment to give CH2CO+ or C3H7+
• Propanone (CH3COCH3) mass spectra shows:
 A peak at m/z =43 which is corresponding to the molecular ion CH3CO+.
 A peak at m/z =15 which is corresponding to the CH3+.
Propanone doesn’t have a peak at 29 as the molecule can’t fragment to give CHO+ or C2H5+.

64
➢Infrared spectroscopy
• The electromagnetic spectrum of radiation includes infrared radiation.

• Polar bond absorbs infrared which vibrate (stretch) the bond while non-polar bonds
don’t.
• Polar bonds in a molecule absorb infra-red radiation at characteristic frequencies
causing the covalent bonds to vibrate by two possible effects:
 Stretching: where the bond length increases and decreases.
 Bending where the bond angle increases and decreases.

• When a molecule absorbs infrared radiation, the amount of energy absorbed

depends on: *The length of the bond
*The strength of the bond
*The mass of each atom involved in the bond

• Compounds containing different functional groups will show different IR spectra as

each bond absorbs infrared at a unique wavenumber.
• A computer will compare the IR spectra against a database of known pure
compounds to identify the compound.

65
• The spectrum is normally shown with vertical axis labelled transmittance with a
percentage from 0 to 100.
• 100% transmittance means that all the radiation is transmitted and none is absorbed.
• The horizontal axis could be labelled either frequency or wavelength with wavenumber as
unit.
• The absorption of infra-red radiation by bonds in this type of spectroscopy is the same
absorption that bonds in CO2, methane and water vapour in the atmosphere do that
cause them to be greenhouse gases.
• H2O, CO2, CH4 and NO molecules absorb IR radiation and are greenhouse gases, whilst O2
and N2 are not.
• The familiar functional groups including:
i. C–H stretching absorptions in alkanes, alkenes and aldehydes
ii. C=C stretching absorption in alkenes
iii. O–H stretching absorptions in alcohols and carboxylic acids
iv. C=O stretching absorptions in aldehydes, ketones and carboxylic acids
v. C–X stretching absorption in halogenoalkanes
vi. N-H stretching absorption in amines

66
The oxidation reaction of butan-2-ol to butanone using IR Spectroscopy:
• Initially: OH broad peak (having a wide wavenumber range) is present and no C=O peak.
• During the reaction: Both peaks are present as some alcohol has been converted to
ketone.
• Finally: The OH peak disappears as the alcohol is used up and only the C=O peak for
ketone is visible.

67