22Al + FeAl + Fe22OO3 3 22Fe + AlFe + Al22OO33

Calculate the mass of iron produced from 40g of iron Calculate the mass of iron produced from 40g of iron oxide.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 1Which reactant would be in excess if 6.5g of zinc was added to 25cm3 of dilute sulphuric acid ofconcentration 2mol l-1

n = no of mole (moles)

c = concentration (mol l-1)

v = volume (litres)

Worked Example 2a) 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 3a) 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

Volume = mass density

Molar Volume

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.

Molar volume = formula mass density

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.

Batch Pros (advantages)

• 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

Continuous Pros (advantages)

• 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 1When 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 2Excess 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 310kg 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 reactantsx100

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 EquilibriumThe 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. Use of a catalyst2. Concentration3. Pressure (of gases)4. Temperature

1. Use of a catalystA 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 tothe 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

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 _________.

C(s) + H2O(g) CO(g) + H2(g)

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

(your calculator won’t always do this for you, look at the temp. change!)

ΔH is always positive

(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 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

Enthalpy of Neutralisation

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 kgΔT = Temperature rise of water in °C

Worked example 1 Enthalpy of combustion of methanol

Methanol mass before 53.65g Methanol mass after 53.46g Mass of water heated 100g Temp rise in water 10oC

Worked example 2 Enthalpy of solution of NH4NO3

Mass of solute (NH4NO3) 1.00g

Mass of water 75g Temp of water initially 20.4oC Temp of solution 18.8oC Temp change 1.6oC

Worked example 3 Enthalpy of neutralisation of HCl by NaOH

HCl + NaOH NaCl + H2O

Temp of acid before mix 19.5 oCTemp of base before mix 18.5 oCTemp of solution after mix 32.5 oC

Solutions used = 20cm3 2moll-1 HCl20cm3 2moll-1 NaOH

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;

Use the enthalpies of combustion of carbon and hydrogen to calculation the enthalpy of formation of enthanol.

2C + 3H2 + ½O2 C2H5OH

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 bondsof 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 1Calculate the enthalpy change, using bond enthalpies, for

the following reaction;

H2 (g) + Cl2 (g) 2HCl (g)

Calculate the enthalpy change, using bond enthalpies, for the

following reaction;

CH4 (g) + O2 (g) CO2 (g) + 2H2O (g)

Worked Example 2

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-)

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 1

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.