2Al
+
Fe2O3
2Fe
+
Al2O3
Calculate the mass of iron produced from 40g of iron oxide.
C9H2O
+
14 O2
9 CO2
+
10 H2O
Calculate the mass of water produced when 6.4 grams of
.
nonane is burned. Show your working
clearly.
The idea of excess
When a chemical reaction involving two
reactants is carried out, usually one of the
reactant gets completely used up and some of
the other reactant is left over.
The reactant which is left over is said to be
in excess.
Worked Example 1
Which reactant would be in excess if 6.5g of zinc
was added to 25cm3 of dilute sulphuric acid of
concentration 2mol l-1
n = no of mole (moles)
c = concentration (mol l-1)
v = volume (litres)
Worked Example 2
a) Which reactant would be in excess if 0.972g of
magnesium was added to 50cm3 of 0.10 mol l-1
hydrochloric acid?
b) Calculate the mass of hydrogen produced
Worked Example 3
a) Which reactant is in excess when 0.506g
magnesium carbonate is added to 100cm3 of
0.5moll-1 nitric acid.
b) Calculate the mass of carbon dioxide produced
Molar Volume
Volume = mass
density
Molar volume = formula mass
density
When dealing with gases its more appropriate to use
volume instead of it’s mass.
The molar volume of a gas (measured at 0oC @ 1atm)
can be worked out using the middle equation.
The molar volume is the volume that one mole of the
gas takes up.
Molar volume allows us to relate gases to one another.
Density Worked example
What is the density of butane (C4H10) in gl–1 if
its molar volume is 22 l mol–1?
As the temperature and pressure change, the
volume a gas takes up changes.
Provided the molar volume of a gas is known,
the volume of a gas can be calculated from
the number of moles of the gas using the
following relationship.
where;
V
= volume of gas (litre)
n
= no of moles (moles)
Vm = molar volume (l/mol.)
Molar Volume Worked example 1
What volume of carbon dioxide is produced by
roasting 25g of calcium carbonate? (molar volume is
24 lmol –1)
Molar Volume Worked example 2
If 300cm3 of a gas weigh 0.55g, what is the formula
mass of the gas? (molar volume 24 lmol–1)
Gas Mixtures (common in multiple choice)
If 25cm3 of ethene are burned in 100cm3 of
oxygen, what would be the composition of the
resulting gas mixture?
Balanced Equation + Gases example
What volume of oxygen is required for the
complete combustion of 4 litres of methane?
• the chemical industry is one of the largest
industries in Britain.
• its products are vital to many aspects of
modern life and many are used for the
benefit of society.
• the chemical industry involves the
investment of large sums of money but
employs relatively few people making it a
capital intensive and not a labour intensive
industry.
Top 5 categories of ‘chemicals’ made
1) Basic inorganics and fertilisers
2) Dyestuffs, paint and pigments
3) Petrochemicals and polymers
4) Pharmaceuticals
5) Specialities (e.g. explosives)
Stages in the manufacture of a new product
Raw Materials and Feedstocks
• A feedstock is a chemical from which other
chemicals are manufactured.
• Feedstocks are made from raw materials (the basic
resources that the Earth supplies to us)
They are:
•
•
•
•
•
•
Water – (used in hydration of ethene to ethanol)
Air – (N2 used in Haber process)
Fossil fuels – coal, crude oil and natural gas
Metallic ores – e.g. aluminium extracted from bauxite (Al2O3)
Minerals – e.g. chlorine from sodium chloride
Organic materials – from plant and animal origin e.g. veg oils
Batch and Continuous Processing
• There are two main types of chemical processing.
Batch and continuous.
• In batch processing the chemicals are loaded into
the reaction vessel and the reaction is monitored.
At the end of the reaction the product is collected
and the reaction vessel is cleaned out ready for the
next batch.
• In continuous processing the reactants are
continuously added at one end of the reaction vessel
and the products are removed at the other end.
• Each process has advantages and disadvantages.
Pros (advantages)
Batch
• suited to smaller scale production (up to 100 tons per year)
• more versatile than continuous as they can be used for more
than one reaction
• more suited for multi step reactions or when reaction time is
long
Cons (disadvantages)
• possibility of contamination from one batch to the next
• filling and emptying takes time during which no product, and
hence no money, is being made
• safety – relatively large amounts of reactants may not be
controllable in the event of an exothermic reaction going
wrong.
An example of a batch process is used in the making of
pharmaceuticals
Pros (advantages)
Continuous
• suited to large scale production (>1000 tons per year)
• suitable for fast single step processes
• more easily automated using computer control
• smaller workforce operates round the clock 365 days per year
• tend to operate with relatively low volumes of reactants
allowing easy removal of excess heat energy
Cons (disadvantages)
• very much higher capital cost before any production can occur
• not versatile, can make only one product
• not cost effective when run below full capacity
Some examples of a continuous process is the making of
sulphuric acid, iron and poly(ethene) and ammonia.
What is percentage yield and why do we use it?
In chemical reactions we rarely, if ever, get the
amount/quantity of products we calculate from a (balanced)
chemical equation.
The reasons for this can be:
• at the end of the reaction there may be reactant left
unconverted to product (see excess)
• some reactant may be converted into a by-product
• the isolation of the product may be difficult
Therefore chemists like to calculate the percentage yield of
a reaction (on a small scale) to see if it makes ‘economical
sense’. (good ‘money’ example on textbook pages 146/147)
Worked Example 1
When 5g of methanol reacts with excess ethanoic acid 9.6g
of methyl ethanoate is produced.
What is the percentage yield in this reaction?
Worked Example 2
Excess ethyne was reacted with 0.1 moles of hydrogen
chloride and 4.1g of the product - 1,1 dichloroethane – were
obtained.
Calculate the percentage yield.
Worked Example 3
10kg of nitrogen reacts with excess hydrogen producing 1kg of
ammonia. Calculate the percentage yield.
N2 + 3H2
2NH3
Atom Economy
Although % yield can be used to calculate the overall
efficiency of a chemical reaction, it does not take
into account how much of the reactants are changed
into (unwanted) by-products.
Atom economy allows chemists to examine the
proportion of reactants that are converted into the
desired product (i.e. the chemical they want.)
Atom Economy =
Mass of desired product(s)
Total mass of reactants
x100
Worked Example
Calculate the atom economy for the production of
ethyl propanoate, assuming that all reactants are
converted into products.
C2H5OH + C2H5COOH
C2H5OOCC2H5 + H2O
Reversible Reactions and Equilibrium
In one way reactions (example 1) the reactants change
completely into products. These products do not change
back into the reactants.
Example 1
However, there are many reactions (example 2) in which the
products can react to reform the reactants. These are
called reversible reactions.
Example 2
Reversible reactions give rise to equilibrium.
In general terms…
At the start with A and B, the rate of the forward reaction is
high because the concentrations of A and B are high.
The rate of the back reaction is initially 0 because there are no
products (C and D) yet.
As the reaction proceeds the concentrations of A and B
decrease while the concentrations of C and D increase. This
continues until the two rates become equal.
At this point the concentration of A, B, C and D are constant and
the (closed) system is at chemical equilibrium.
Equilibrium is only possible in a closed system – i.e. no substances
are added or removed.
At equilibrium, the forward and backward reactions are
continuing and their rates are equal.
Therefore the concentrations of A, B, C + D remain
constant. This is known as dynamic equilibrium
Equilibrium does not mean there are 50% reactants 50%
products (i.e. equal amounts)
Position of Equilibrium
The position of equilibrium varies from reaction to
reaction. Sometimes it occurs when the forward
reaction is almost complete whilst other times it occurs
when the forward reaction has barely started.
We use the terms;
‘equilibrium lies to the left’ when the conc. of reactants
is greater than the conc. of products
and
‘equilibrium lies to the right’ when the conc. of products
is greater than the conc. of reactants
Factors that affect the Position of Equilibrium
As many reactions are at equilibrium, it is
important we understand how to alter its position
as this has a bearing on the yield of reactants to
products.
Factors include;
1.
2.
3.
4.
Use of a catalyst
Concentration
Pressure (of gases)
Temperature
1. Use of a catalyst
A catalyst lowers the activation energy between reactants and
products by providing an alternative reaction path.
The activation energy is lowered by the same amount for both the
forward reaction and the back reaction.
As a catalyst speeds up both the forward and back reactions,
equilibrium is reached more quickly.
However a catalyst has no effect on the position of equilibrium.
Le Chatelier’s Principle
The effect of changes in concentration,
pressure and temperature on an equilibrium
can be predicted using Le Chatelier’s
Principle;
“If a system at equilibrium is subjected to
a change, the system will adjust to oppose
the effect of the change.”
2. Concentration
Increasing the concentration of A or B will speed up
the forward reaction. The equilibrium position moves
to the right to counteract this change i.e. more C
and D are produced.
Decreasing the concentration of C or D will slow down
the back reaction This means the equilibrium will
move to the right in order to counteract the changes.
i.e. more C and D are produced.
Increasing the concentration of C and D or
decreasing the concentration of A and B moves the
equilibrium to the left.
Increasing the conc. of either Fe3+(aq) or CNS-(aq) will result in the
equilibrium position moving to the ______, using up some of the
additional reactants and producing more FeCNS2+(aq).
The solution will become more ____. The reverse is true i.e.
decreasing the concentration of either reactant will result in less
red.
Increasing the concentration of FeCNS2+(aq) will result in the
equilibrium position moving to the ______ to use up some of the
additional product by making more Fe3+(aq) and CNS-(aq).
The solution will become less red. The reverse is true i.e. decreasing
the concentration of FeSCN2+(aq) will result in more red.
3. Pressure (of gases)
If the pressure on an equilibrium system is
increased, then the equilibrium position shifts
to reduce the pressure.
In other words, increasing the pressure on an
equilibrium system will result in the equilibrium
shifting to reduce the pressure i.e. moving to
the side that has the smallest number of gas
particles.
Pressure Example 1
There is 1 mole of gas on the left hand side of the reaction
and 2 moles of gas on the right hand side of the reaction.
Increasing the pressure on this system would result in the
equilibrium position moving to the ______ i.e. consuming
NO2(g) and producing more N2O4(g).
The system will become a lighter in colour.
Decreasing the pressure on this equilibrium system will result
in the equilibrium position moving to the right i.e. the side
that has the most gas particles, in order to increase the
pressure.
The brown colour of the system becomes darker.
Pressure Example 2
There are ____ moles of gas on the left hand side of
the reaction and ______ mole(s) of gas on the right
hand side of the reaction.
Increasing the pressure on this system results in the
equilibrium position moving to the _________.
Decreasing the pressure on this equilibrium system will
result in the equilibrium position moving to the
_________.
Pressure Example 3
C(s) + H2O(g)
CO(g) + H2(g)
There are ____ mole(s) of gas on the left hand side of
the reaction and ______ mole(s) of gas on the right
hand side of the reaction.
Increasing the pressure on this system results in the
equilibrium position moving to the _________.
Reducing the pressure on this equilibrium system will
result in the equilibrium position moving to the
_________.
4. Effect of Temperature on the Position of Equilibrium
In a system at equilibrium, if the forward reaction is
exothermic the back reaction must be endothermic, and vice
versa.
In an endothermic reaction, energy can be considered as a
reactant of the reaction.
If the temperature of an endothermic equilibrium system is
increased, the equilibrium position shifts to use up the heat by
producing more products. (moves to right)
In an exothermic reaction, energy can be considered as a
product of the reaction.
If the temperature of an exothermic equilibrium system is
increased, the equilibrium position shifts to use up the heat by
producing more reactants. (moves to left)
Temperature
Example 1
If the temperature is increased the position of equilibrium
moves to the _________.
If the temperature is decreased the position of equilibrium
moves to the ________.
An increase in temperature favours the back reaction
and therefore _______ the concentration of
methanol.
This suggests that to get a high yield of methanol we
should carry out the reaction at low temperature.
However, low temperature means a low rate and a long
time to establish equilibrium.
In industry a compromise is reached at a moderately
high temperature (200 to 300°C) which gives a
worthwhile rate but a reduced yield of methanol.
Reminder - Potential Energy Diagrams
Exothermic
Endothermic
ΔH is always negative
ΔH is always positive
(your calculator won’t always do this
for you, look at the temp. change!)
(remember you must write in the sign!)
Enthalpy of Combustion
Enthalpy of combustion (heat of combustion) is
the heat energy given out when one mole of a
substance burns completely in oxygen.
For example;
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −892 kJ mol−1
Enthalpy of Solution
Enthalpy of solution is the enthalpy change
when 1 mole of solute is completely dissolved.
For example;
NaCl(s)
Na+(aq) + Cl−(aq)
ΔH = +5 kJ mol−1
Enthalpy of Neutralisation
Enthalpy of neutralisation is the heat energy
released when 1 mole of water is formed by
neutralisation of an acid with a base.
If more than one mole of water is produced
this energy must be multiplied accordingly.
For example
H+(aq) + OH−(aq)
H2O(l)
ΔH = −57.3 kJ mol−1
Calculating Enthalpy Changes Via Experiment
Energy absorbed (kJ) = c x m x ΔT
where:
c = Specific heat of water, 4.18 kJ kg−1 °C−1
m = Mass of water in kg
ΔT = Temperature rise of water in °C
Worked example 1
Enthalpy of combustion of
methanol
Methanol mass before
Methanol mass after
Mass of water heated
Temp rise in water
53.65g
53.46g
100g
10oC
Worked example 2 Enthalpy of solution of NH4NO3
Mass of solute (NH4NO3)
Mass of water
Temp of water initially
Temp of solution
Temp change
1.00g
75g
20.4oC
18.8oC
1.6oC
Worked example 3 Enthalpy of neutralisation of
HCl by NaOH
HCl
+
NaOH
NaCl
+
Temp of acid before mix
19.5 oC
Temp of base before mix
18.5 oC
Temp of solution after mix 32.5 oC
Solutions used = 20cm3 2moll-1 HCl
20cm3 2moll-1 NaOH
H2O
The Definition
Hess's law states that the enthalpy change for converting
reactants into products is the same regardless of the route
taken.
Therefore the enthalpy change for two routes from the
same reactants to the same products will be equal.
Hess’s Law in ‘real life’ context
a) A man works hard and saves throughout his whole working life. At age
60 he retires with personal wealth of £3million.
b) Another man (a stock broker) lives the life of luxury, making and
losing a fortune several times through his life. At the age of 60 he
has a personal wealth of £3million.
c) Another man lives in near poverty until the age of 60 when he wins
£3million on the ‘Lotto’.
The three men are age 60 and have very different stories, but ultimately
they all end up with £3million. Their wealth is therefore independent of
how their lives were lived.
Hess’s Law Calculation Example 1
Calculate the enthalpy change for the formation of 1 mole of
methane from its elements (using the enthalpies of combustion of
carbon, hydrogen and methane from the data book.)
Hess’s Law Calculation Example 2
The enthalpy of formation of propan-1-ol from carbon, hydrogen and
oxygen can be represented by the equation;
Use the enthalpies of combustion of carbon and hydrogen to calculation
the enthalpy of formation of propan-1-ol.
Hess’s Law Calculation Example 3
The enthalpy of formation of ethanol from carbon, hydrogen and oxygen
can be represented by the equation;
2C + 3H2 + ½O2
C2H5OH
Use the enthalpies of combustion of carbon and hydrogen to calculation
the enthalpy of formation of enthanol.
Hess’s Law Calculation Example 4
Calculate the enthalpy of formation of silane.
Bond Enthalpies
The molar bond enthalpy is the enthalpy change when a bond in
gaseous molecule is broken. (Note – mean bond enthalpies are
averages but used in the same way)
Bond enthalpy data is in the data booklet
E.g.
Cl2(g)
2Cl-(g)
ΔH = +243 kJmol-1
In other words it takes 243 kJ of energy to break all bonds
of the chlorine molecules into chlorine atoms.
Therefore;
2Cl-(g)
Cl2(g)
ΔH = -243 kJmol-1
In other words making one molecule of chlorine from its atoms
releases 243kJ of energy.
Worked Example 1
Calculate the enthalpy change, using bond enthalpies, for the
following reaction;
H2 (g) + Cl2 (g)
2HCl (g)
Worked Example 2
Calculate the enthalpy change, using bond enthalpies, for the
following reaction;
CH4 (g) + O2 (g)
CO2 (g) + 2H2O (g)
Oxidation, Reduction and Displacement Reaction Revision
Displacement reactions are examples of redox reactions.
E.g. Zinc metal displaces silver from silver (I) nitrate solution.
This is because zinc is higher than silver in the electrochemical
series.
The zinc metal is the reducing agent. It donates electrons and it
is oxidised.
The silver ion is the oxidising agent. It accepts electrons and it is
reduced.
Oxidation cannot happened without reduction and vice versa.
Combining the two equations produces a REDOX equation.
The number of electrons lost in the oxidation must balance
the number of electrons gained in the reduction.
NB -The nitrate ions do not appear in the redox equation.
This is because they are spectator ions and are not directly
involved in the electron transfer.
Steps to writing a balanced ion electron half equation;
•
Write down the main chemical in its two forms
(oxidised and reduced) and balance the main
atom(s)
•
Balance oxygen by adding water molecules.
•
Balance hydrogen by adding hydrogen ions
•
Balance the charge by adding electrons. (1-)
Worked Example 1
Potassium permanganate (KMnO4) oxidises Fe2+ ions
to Fe3+ ions. During this reaction the MnO4– ions are
reduced to Mn2+ ions.
a) Complete the reduction half equation
MnO4-
Mn2+
b) Write the redox equation for the reaction.
Worked Example 2
Write the redox equation for copper reacting
with nitric acid
Reduction; NO3-
NO
Worked Example 3
Write the redox equation for the reaction between
bromine and sodium sulphite. In this reaction sodium
(Na+) is a spectator ion.