CfE Higher Chemistry

Unit 3

Chemistry in Society

Pupil Number:

2

Chemical Industry

Importance of the Chemical Industry

the Chemical Industry is one of the largest British industries.

its products are indispensable to many aspects of modern life and many are used for the

benefit of society

it is the only manufacturing industry to export more than it imports and so earns a trade

balance surplus from these exports for Britain

also invisible trade balance surplus from selling licenses to use British processes abroad

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.

Stages in the manufacture of a new product

The manufacture of a new product is a step-wise process from its discovery.

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:

Fossil fuels–coal, oil and natural gas

metallic ores–e.g. aluminium extracted from bauxite(Al2O3)

minerals–chlorine from sodium chloride

water and air – water in hydration of ethene to ethanol and nitrogen in the Haber Process,

oxygen in the catalytic oxidation of ammonia

organic materials–of plant and animal or vegetable oils and starch

Crude oil is a raw material from which naphtha is obtained by fractional distillation. Naphtha is

a feedstock that can be cracked to produce ethene.

research and

development

a new potentially useful chemical is prepared and patented. Some products

are discovered by accident, but others as a result of long and expensive

research.

laboratory process small scale to review the production route

pilot study

the product is now required in larger amounts and will be manufactured in a

pilot study using the route identified by the research group but in kilogram

quantities. Product quality, health hazards, and production costs will be

discussed.

scaling up planning the scaling up from lab quantities to full scale production will have

been going on from the pilot study stage.

Production plant design, planning considerations, commissioning and start up

review this will occur at each stage. All processes are reviewed and modifications

are made.

3

Batch and Continuous Processes

In a batch process the chemicals are loaded into the reaction vessel. The reaction is monitored and

at the end of the reaction the product is separated and the reaction vessel cleaned out ready for the

next batch.

In a continuous process the reactants are continuously loaded at one end of the reaction vessel and

the products are removed at the other end. Each process has advantages and disadvantages.

Batch Process v Continuous Process

In general products that are made on a very large scale will use a continuous process eg. sulfuric acid,

ammonia, iron, ethene, poly(ethene).

Products made on a smaller scale or when a continuous process would be difficult to devise or operate

will use a batch processeg. pharmaceuticals, dyes, copper refining by electrolysis.

Factors Influencing the Choice of Synthetic Route

- Cost - availability of feedstocks

- Yield of the reaction – can unreacted starting materials be recycled, or sold?

- Difficulty and cost of waste disposal

- Energy consumption

- Emissions to the atmosphere

Process Advantages Disadvantages

Batch suited to smaller scale production

up to 100 tons per annum

more versatile than continuous as

they can be used form ore than

one reaction more suited for multi

step reactions or when reaction

time is long

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

Continuous suited to large scale production

>1000 tons per annum

suitable for fast single step

processes

more easily automated using

computer control

smaller work force operates

round the clock, 365 days per

year

tend to operate with relatively

low volumes of reactants allowing

easy removal of excess heat

energy

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

4

Economic Aspects

Operating Conditions

The conditions under which a chemical process operates are chosen to maximise economic efficiency.

We have considered these in other topics but examples are:

Raising the temperature may increase the rate of a reaction but it will increase energy costs so may

not be economic

Increasing the pressure may shift equilibrium in favour of the product but will mean using stronger

reaction vessels and more powerful compressors and may not be economic.

Costs in the Chemical Industry

Costs come under 3 main categories – capital, fixed and variable costs.

Capital Costs

These are incurred when building the plant. The life of a plant is assumed to be only about 10 years

after which it is written off. The cost of this depreciation is recovered under fixed costs.

Fixed Costs

These are costs that are the same whether 1 ton or 1000 tons of product are made. The effect of

the fixed cost decreases as the amount of product increases.

They include:

- Depreciation of the plant

- Labour

- Land purchase

- Variable Costs - These are directly related to output and include:

- Raw materials and energy

- Packaging

- Waste disposal and effluent treatment

Location of the Chemical Industry

Many locations are for historical and practical reasons. They had to be near:

Water supply

Raw materials

Good communications; near ports, roads and rail

Reliable energy supplies

Available skilled labour

5

Safety and the Environment

The chemical industry is well aware of its environmental responsibilities and is acting

accordingly.

Power stations that burn fossil fuels must remove the sulfur dioxide from the gases before release to the atmosphere. The SO2 is converted to H2SO4 , which is sold.

Waste used to be dumped in quarries, rivers, the sea or stored in containers from which it could leak into streams. These methods are no longer acceptable and are increasingly becoming illegal. Waste must be treated and discharged only when it is not harmful to the environment – it must meet requirements of pH and metal ion content.

Water containing organic waste must not be discharged into rivers or canals if it will reduce significantly the oxygen content of the water, causing fish to die.

Between 1990 and 1996 discharge of potentially harmful chemicals into UK rivers was reduced by 91%.

Plants have reduced accidents by 50% in the last decade.

Road and rail tankers that carry chemicals are constructed to withstand impact in accidents.

Plants have their own fire fighting teams on site.

Plants are designed with safety in mind. Chemicals are hazardous so the accident rate will never be zero but the aim is to learn from mistakes and reduce the rate to a minimum.

It is essential at this stage to revise all of your National 5 calculations that were based on moles and equations.

6

Calculations Based on Equations

A balanced equation is taken to give the relative number of moles of each reactant and product. Since

the mass of one mole of any substance is expressed in grams, the masses involved can then be

calculated as shown.

Worked Example

Calculate the mass of water produced on burning 1g of methane

CH4 + 2O2 CO2 + 2H2O 1 mole 2 moles

It is not necessary to calculate the masses of carbon dioxide and oxygen – these substances are not included in the question.

Step 1 Convert moles to mass for the substances involved

CH4 2H2O

12 + (4 x 1) 2 [ (2 x 1) + 16 ] 16g 36g

Step 2 By proportion; determine the mass of water produced

16g 36g

1g 36 x 1

16g

= 2.25g

The Idea of Excess

Reactants are needed for a chemical reaction to take place. When one of the reactants is used up the

reaction will stop. Any reactant, which is left unreacted, is said to be “in excess”.

Worked Example 1

8g of methane is mixed with 16g of oxygen. A spark is applied to the mixture to start the reaction.

Calculate the mass of carbon dioxide produced.

CH4 + 2O2 CO2 + 2H2O

1 mole 2 moles 1 mole

Step 1 Calculate the number of moles of each reactant

No. moles CH4 = mass = 8 = 0.5moles

gfm 16

No. moles O2 = mass = 16 = 0.5moles

gfm 32

7

Step 2 Use the balanced equation to determine the quantity of each reactant used in the

reaction.

CH4 + 2O2 1 mole 2 moles

Therefore, for every 0.5 moles of methane 1 mole of oxygen would be required. Looking at the

quantities of reactants from step 1 there is not enough oxygen to allow all of the methane to react

therefore some methane will be left over at the end. The methane is said to be in excess and the

oxygen will therefore determine the quantity of carbon dioxide produced.

Step 3 Convert moles to mass and carry out a proportion calculation to determine the mass of

carbon dioxide produced.

CH4 + 2O2 CO2 + 2H2O

1 mole 2 moles 1 mole

64g 44g

16g 16 x 44

64

= 11g

Worked Example 2

What mass of hydrogen gas is produced when 2.45g of magnesium is added to 100cm3 of dilute

hydrochloric acid, concentration 1 moll-1?

Mg + 2HCl MgCl2 + H2

1 mole 2 moles 1 mole

Step 1 Calculate the number of moles of each reactant

No. moles Mg = mass = 2.45 = 0.1 moles

gfm 24.5

No. moles HCl = C x V

1 x 0.1 = 0.1 moles

Step 2 Use the balanced equation to determine the quantity of each reactant used in the

reaction.

Mg + 2HCl

1 mole 2 moles

So 0.1 moles requires 0.2 moles

From step 1 it can be seen that there is not enough acid to allow all of the Mg to react hence the Mg

will be in excess and the acid will control the mass of hydrogen produced.

8

Experiment 3.1—Excess

Magnesium reacts with dilute sulphuric acid to produce hydrogen gas.

Mg + H2SO4 (aq) H2 + MgSO4

If excess dilute sulphuric acid is used, the number of moles of unreacted acid can be found

experimentally by titration with sodium hydroxide solution using an indicator. This can then be

compared with the theoretical number of moles of unreacted acid.

Experiment

Accurately weigh out 0.4g of magnesium ribbon. Note the exact mass.

Add the magnesium to a 250cm3 volumetric flask.

Use a burette to add 50cm3 of dilute sulphuric acid, concentration 0.5 mol l-1 to the flask.

When the fizzing has completely stopped, add water to the flask up to the 250cm3 calibration

point. Secure the lid and shake well.

Titrate 25cm3 portions of this solution with sodium hydroxide solution, concentration 0.1 mol l-1,

using phenolphthein as the indicator.

Repeat until two concordant results are obtained (within 0.1cm3)

Write Up

Complete a full write up for this experiment including the all appropriate calculations.

9

Molar Volume

The molar volume is the volume of which one mole of a gas occupies.

Worked Example 1

The volume of 8g of oxygen is 5.5 litres. Calculate the volume of 3 mol of oxygen.

gfm O2 = 2 x 16 = 32g

8g 5.5

96g 5.5 x 96

8

= 66 litres

Worked Example 2

Calculate the volume of 0.14g of nitrogen. (Assume the molar volume of the gas is 23 litres mol-1)

gfm N2 = 2 x 14 = 28g

28g 23

0.14g 23 x 0.14

28

= 0.115 litres or 115 cm3

Worked example 3

Calculate the volume of hydrogen produced when 0.2g of zinc reacts with excess sulfuric acid.

(Take the molar volume to be 22.2 litres mol-1)

Zn + H2SO4 ZnSO4 + H2

1 mole 1 mole

1mole Zn 65g Therefore, we have 0.003 moles of H2

0.2 x 1 0.2g 1 mole 22.2 Litres

65 0.003moles 0.003 x 22.2

=0.003moles 1

= 0.068 Litres

Reacting Volumes

1 mole of any gas occupies roughly the same volume (at the same temperature and pressure).

This means that equal volumes of gases contain the same number of moles.

= =

We already know that a balanced

equation gives the same number of moles of reactants and products.

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

1 mole 3 moles 2 moles

So.. 1 litre + 3 litres 2 litres

1 litre of O2 1 litre of H2 1 litre of CO2

10

NOTE: liquid or solid reactants and products are not included in this

calculation!

C(s) + O2(g) CO2(g)

1mole 1mole 1mole

So.. not a gas 1 litre 1 litre

Worked Example 1

30cm3 of methane is completely burned in 100cm3 of oxygen. What is the volume and composition of

the gas at the end of the experiment? (All volumes are measured at atmospheric pressure and room

temperature)

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

1mole 2moles 1mole

Step 1: from the balanced equation determine which reactant is in excess

1 mole CH4 reacts with 2 moles of O2

30cm3 60cm3

Since there is 100cm3 of O2, O2 is in excess by 40cm3

Step 2: since the oxygen is in excess the methane will be used up first and hence will control the

volume of product.

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

1mole 1mole

30cm3 30cm3

So the total volume and composition of gas at the end of the experiment is:

Total Volume at the end will be: 70cm3

40cm3 of O2 left unreacted and 30cm3 of CO2 produced

NOTE: since the water is a liquid it is not considered.

11

Experiment 3.2—Synthesising sodium citrate—Percentage Yield Calculation.

Sodium citrate is also known as food additive E331. It is often used as an acidity regulator in drinks,

and as an emulsifier for oils when making cheese. It allows the cheeses to melt without becoming

greasy. Sodium citrate is also the main ingredient in over the counter cystitis medication. You are

going to synthesis sodium citrate and calculate the percentage yield and atom economy achieved.

Experiment

Weigh out 3.5g of citric acid crystals, add to a beaker

Measure out 50cm3 of 1.0 mol l-1 sodium hydroxide and add to the crystals

Stir the mixture until all the crystals have dissolved

Evaporate the sodium citrate solution in an evaporating basin

Weigh dried powder with an accurate balance and calculate the percentage yield.

Write up

Complete a full write up for this experiment. The calculation involving percentage yield can only be

completed once the next page is completed.

12

Percentage Yield Calculations

The yield in a chemical reaction is the quantity of product obtained. The actual yield can be

compared, as a percentage, with the theoretical.

Worked Example 1

5g of methanol reacts with excess ethanoic acid to produce 9.6g of methyl ethanoate. Calculate the

percentage yield.

Step 1: determine the theoretical yield (the quantity expected from the balanced equation)

CH3OH + CH3COOH CH3OOCCH3 + 2 H20

1 mol 1 mol

32g 74g

5g 74 x 5

32

= 11.56g

Theoretical Yield = 11.56g

Step 2: The actual yield is always given in the question.

Actual yield = 9.6g

Step 3: Percentage yield = 9.6 x 100

11.56

= 83%

The percentage yield is a very important consideration for industrial chemists. They must take

account of cost of raw materials, plant-running costs etc. If the yield of product is not sufficient

enough to cover the costs of production then the process would not be considered to be economically

viable.

Atom Economy

Atom economy is a measure of the proportion of reactant atoms which are incorporated into the

desired product of a chemical reaction.

Calculation of atom economy therefore also gives an indication of the proportion of reactant atoms

forming waste products.

In developing an atom economical reaction pathway the industrial chemist may well prefer

rearrangement and addition reactions over less environmental friendly substitution and elimination

reactions.

% atom economy =Mass of desired product(s)

Total mass of reactantsx 100

13

Example 1: Addition reaction – halogenation of an alkene

Total mass of reactants = 56 g + 159.8 g = 215.8 g

(Note: Product mass is also 215.8 g)

Mass of desired product (2,3-dibromobutane) = 215.8 g

This process is 100% atom efficient, with all the reactant atoms included within the desired product.

Example 2: Elimination reaction

Total mass of reactants = 161 g + 74 g = 235 g

(Note: Total product mass = 235 g)

Mass of desired product ethylene oxide = 88 g

This elimination reaction is therefore only 37.4% atom efficient, with the remaining 62.6% in the

form of unwanted waste products (calcium chloride and water).

C C

Br

CH3

H

Br

CH3

HC C

CH3

H

CH3

H

Br2

(Z)-but-2-ene

C4H

8

1mol(12 x 4) + (8 x 1)

= 56g

Bromine

Br2

1mol2 x 79.9

= 159.8g

2,3-dibromobutane

C4H

8Br

2

1mol(12 x 4) + (8 x 1) + (79.9 x 2)

= 215.8g

+

% atom economy =Mass of desired product(s)

Total mass of reactantsx 100

% atom economy =215.8

x 100 = 100%215.8

C C Cl

H

H

OH

H

H

C C

O

H

HH

H

Ca(OH)2

2-chloroethanol

C2H

5OCl

2mol2[(12 x 2) + (5 x 1) + 16 + 35.5]

= 161g

Calcium hydroxide

Ca(OH)2

1mol40 + 2(16 +1)

= 74g

Water

H2O

2mol2 [(2 x 1) + 16]

= 36g

CaCl2

2H2O

2

ethylene oxide

C2H

4O

2mol2[(12 x 2) + (4 x 1) + 16]

= 88g

Calcium chloride

CaCl2

1mol40 + (2 x 35.5)

= 111g

2

Desired Product Waste Products

+

+

+C C Cl

H

H

OH

H

H

C C

O

H

HH

H

Ca(OH)2

2-chloroethanol

C2H

5OCl

2mol2[(12 x 2) + (5 x 1) + 16 + 35.5]

= 161g

Calcium hydroxide

Ca(OH)2

1mol40 + 2(16 +1)

= 74g

Water

H2O

2mol2 [(2 x 1) + 16]

= 36g

CaCl2

2H2O

2

ethylene oxide

C2H

4O

2mol2[(12 x 2) + (4 x 1) + 16]

= 88g

Calcium chloride

CaCl2

1mol40 + (2 x 35.5)

= 111g

2

Desired Product Waste Products

+

+

+

% atom economy =88

x 100 = 37.4%235

% atom economy = Mass of desired product(s) x 100

Total Mass of reactants

14

Equilibria

Reversible Reaction

Many reactions are reversible. A reversible reaction can reach equilibrium in a closed system. A

reaction reaches equilibrium when the rate of the forward reaction equals the rate of the reverse

reaction.

reactants ⇌ products

At equilibrium, the concentration of the products and the

reactants will remain constant.

The concentration of reactants will probably not equal the

concentration of the products.

Where an industrial process produces an equilibrium, costly reactants may not be completely

converted into products; chemists try to manipulate the equilibrium to achieve the best possible

conversion rate.

Catalysts increase the rate at which an equilibrium is formed but do not affect the equilibrium

position.

The Position of Equilibrium

The position of equilibrium means the extent to which the reaction has taken place.

The equilibrium in this case lies well to the right hand side. There is 80% conversion

of A into B. Sometimes this is indicated by the size of the arrows.

In this example, equilibrium would be said to lie to the left.

Can you explain what happens in the

case of water?

Le Chateliers Principle

An equilibrium will move to undo any change imposed upon it by temporarily favouring either the

forward or backward reaction until a new equilibrium position is reached.

reactants ⇌ products

If the forward reaction is favoured we say the equilibrium has moved to the right.

If the reverse reaction is favoured we say the equilibrium has moved to the left

15

Products of an Equilibrium Reaction with Time

At the start of the reaction there will be 100%

reactants. However, as the reaction proceeds the

reactants will be used up and the products will be

formed. Eventually the reaction will reach equilibrium.

At this stage the concentration of both the reactants

and products will remain constant.

In a reversible reaction when the curve levels off this indicates that the reaction has reached

equilibrium. (it does not indicate that the reaction has stopped).

Shifting the Position of Equilibrium

If, once an equilibrium has been established the reaction conditions are changed then the position of

equilibrium will be altered. If a new equilibrium is established where there is an increase in products

we say the equilibrium has shifted to the right. If the new equilibrium results in an increased amount

of reactants then we say the equilibrium has shifted to the left.

Changing reaction conditions such as concentration, temperature and pressure will cause a shift in

equilibrium position.

Changing Concentration

Increasing the concentration of the reactants, or decreasing the concentration of the products

shifts the position of equilibrium to the right hand side

Decreasing the concentration of the reactants or increasing the concentration of the products shifts

the position of equilibrium to the left.

%

0

50

100

Time

25%

%

0

50

100

Time

75%

%

0

50

100

Time

75%

25%

16

Changing Temperature

Increasing the temperature favours the endothermic reaction.

Decreasing the temperature favours the exothermic reaction.

Changing Pressure

In the gaseous state molecules have high energies and are fast moving. Pressure is the result of gas

molecules bombarding the walls of the vessel in which the gas is contained. The greater the number

of molecules in a given volume the greater the pressure.

The pressure on the right hand side is greater than the pressure on the left hand side because there

are more molecules.

NOTE:

1. Pressure only affects the equilibrium of a system that involves gases.

2. A pressure change will alter equilibrium only if there are different numbers of moles of

gases on each side.

A(g) + B(g) C(g) + D(g)

Changing pressure will not affect this equilibrium since there are the same number of moles of gas

on each side of the equation.

2A(g) + B(g) C(g) + D(g)

Changing pressure will affect this equilibrium since there are different numbers of moles of gases

on each side of the equation.

Increasing the pressure will favour the side with the lowest number of moles in the gas state.

Decreasing the pressure will favour the side with the highest number of moles in the gas state.

Low pressure High pressure

C(g) 2D(g)

17

Experiment 3.3—Dehydration of hydrated cobalt chloride

Crystals of cobalt chloride which contain water molecules of crystallisation (hydrated cobalt chloride,

CoCl2.6H2O) are pink. Anhydrous cobalt chloride, CoCl2, on the other hand is blue. The conversion of

hydrated cobalt chloride to anhydrous cobalt chloride is an example of a reversible reaction, i.e. the

reaction can go forwards or backwards.

Experiment

Add a spatula of hydrated cobalt chloride to a test tube.

Heat the crystals until the water molecules are driven off. Observe the colour change.

Allow the anhydrous cobalt chloride to cool before adding a few drops of water. Observe the

colour change.

Write up

Complete a write up for this experiment to make suitable notes on your observations.

18

The equilibrium between nitrogen dioxide and dinitrogen tetroxide

In chemistry, Le Chatelier's principle, also called

Chatelier's principle or "The Equilibrium Law", can be used

to predict the effect of a change in conditions on a

chemical equilibrium. The principle states that:

If a chemical system at equilibrium experiences a

change in concentration, temperature, volume, or

pressure, then the equilibrium shifts to counteract the

imposed change and a new equilibrium is established.

Experiment

Pressure demo:

Collect a gas syringe filled with an equilibrium mixture of brown nitrogen dioxide and colourless

dinitrogen tetroxide. Press in the plunger of the syringe as far as possible. The colour of the gas

mixture will become darker as the concentration of the gas mixture increases. Within a few seconds

it will become paler, as the position of the equilibrium responds to the increased pressure, and brown

nitrogen dioxide is converted into colourless dinitrogen tetroxide. Release the plunger and the

changes will reverse themselves. Pull the plunger out to reduce the pressure and note the colour

changes.

Temperature demo:

Fill a second syringe with 50 cm3 of air and seal with a screw clip. Clamp both syringes vertically in a

2L beaker of water so that they are immersed up to the 100cm3 mark. Note the temperature and the

readings of both syringes (which should be the same). Heat the water gently with a Bunsen burner

and record the temperature and the volume of gases every 10 °C or so. Before taking each reading,

remove the Bunsen burner and stir the water for a couple of minutes to ensure that the

temperatures of the gases are the same as that of the water. Twist the plungers of the syringes

before taking each reading to ensure that they are not sticking. Continue taking readings until the

temperature is about 70 or 80 °C. Plot graphs of volume against temperature for both gases on the

same axes. The nitrogen dioxide/dinitrogen tetroxide mixture will expand more than air as the

equilibrium responds to the increase in temperature by producing more nitrogen dioxide. If there is

time, take further readings as the water cools to check for leaks.

Notes

A full description for this experiment can be found on the Royal Society of Chemistry web page

under classic chemistry demonstrations.

http://media.rsc.org/Classic%20Chem%20Demos/CCD-81.pdf

19

Effect of a Catalyst

Consider the reversible reaction at equilibrium:

If a catalyst is added the rates of both the forward and backward reaction will be increased.

There is the same decrease in the activation energies of the forward and backward reactions and

hence there is the same increase in their rates. Therefore, a catalyst does not alter the position

of equilibrium. It does however, speed up the rate at which equilibrium is attained.

Equilibrium in Industry

The manufacture of ammonia is a very important reaction. Ammonia is made from nitrogen and

hydrogen by the Haber Process. The reaction is reversible and if the conditions were kept constant

equilibrium would be attained.

At equilibrium only 15% yield of ammonia is achieved however, the reaction conditions can be altered

to increase the rate of the forward reaction to increase the yield of ammonia.

Use of a catalyst

An iron catalyst is used in the Haber Process. The catalyst does not alter the

position of equilibrium. However, it does allow the process to be carried out

more efficiently at a lower temperature.

The use of a catalyst does not change the equilibrium it only enables the

position of equilibrium to be reached more quickly.

Reaction pathway

Potential

energy

A + B

C + D

Ea

Eb

Ec

Ed

without catalyst

with catalyst

20

(a) Concentration

As a result of economic considerations, the Haber Process does not actually ever attain

equilibrium. In a condenser, the ammonia gas is cooled and the liquid ammonia piped off.

Constantly removing the ammonia gas decreases the rate of the backward reaction. In

addition, the unreacted nitrogen and hydrogen are recycled. The increase in the

concentrations of the reactant gases increases the rate of the forward reaction. The effect

of both of these factors is a more economical production of ammonia.

(b) Pressure

Since the number of product molecules is less than the number of reactant molecules,

increasing the pressure increases the rate of the forward reaction. The pressure chosen is

about 200 atmospheres. Beyond this pressure, the relative increase in the rate cannot justify

the increased costs. SEE EXPERIMENT 3.4

(c) Temperature

Since the reaction is exothermic, increasing the temperature decreases the percentage

conversion of reactants to ammonia. However, if the temperature is too low the reaction is

too slow. As a result, the Haber Process is carried out at a moderately high temperature of

about 400oC.

Experiment 3.4—Effect of pressure on equilibrium (DEMO)

Chemical Energy

Enthalpy of Combustion

When 1 mole of a substance burns it releases a certain amount of energy

which we call the enthalpy of combustion.

Definition of Enthalpy of Combustion:

The enthalpy of combustion is the energy released to the surroundings when one mole of a substance

burns completely in oxygen.

When writing balanced equations for enthalpies of combustion it is important to ensure only 1 mole of

the substance being burned is included in the equation:

C + O2 --> CO2

H2 + ½ O2 --> H20

S + O2 --> SO2

CO + ½ O2 --> CO2

C3H8 + 5O2 --> 3CO2 + 4H2O

Take a note of the

results/observations and

explain what you see.

21

Enthalpy Calculations

There are two types of combustion calculations. In the first type the temperature difference is not

mentioned and this calculation simply requires a proportion calculation to be carried out.

WORKED EXAMPLE 1

When 1g of ethanol, gram formula mass 46g, is burned 30kJ of energy is released. Calculate the

Enthalpy of Combustion for ethanol.

1g 30kJ

46g 30 x 46

1

= 1380kJ

This means that when 1 mole of ethanol burns it will produce 1380kJ. Since combustion is always an

exothermic reaction the enthalpy of combustion is - 1380 kJmol-1

WORKED EXAMPLE 2

0.16g of methanol, CH3OH is burned in a spirit burner. The heat from this combustion causes the

temperature of 100g of water to be raised from 20oC to 27oC. Calculate the enthalpy of combustion

of methanol.

Step 1 Eh = c m ΔT

= 4.18 x 0.1 x 7

= 2.926 kJ

Step 2 0.16g 2.926kJ

32g 2.926 x 32

0.16

= 585.2kJ

So Enthalpy of Combustion, ΔH = - 585.2 kJmol-1

Remember mass is always

expressed in kg & 1cm3 of

water equals 1g

22

Experimental 3.5—Enthalpy of combustion

The enthalpy of combustion of a substance is the energy released when one mole of the substance is

completely burned in oxygen.

The aim of this experiment is to determine the enthalpy of combustion of ethanol i.e. the

enthalpy change for the reaction:

CH3CH2OH (l) + 3O2(g) 2CO2(g) + 3H2O(l)

A measured mass of ethanol is burned in a spirit burner and the heat

released is transferred to a copper can containing a known volume of

water. From the resulting temperature rise, the enthalpy of combustion of

ethanol can be calculated.

In this experiment we assume that all the heat released in the combustion

reaction is absorbed only by the water in the copper can.

Experiment

Weigh the spirit burner (already containing ~50cm3 ethanol) with its cap on and record its

mass. (The cap should be kept on to cut down the loss of ethanol through evaporation)

Using the measuring cylinder, measure out 100 cm3 of water into the copper can.

Set up the apparatus as directed by your teacher.

Measure and record the temperature of the water.

Remove the cap from the spirit burner and immediately light the burner.

Slowly and continuously stir the water with the thermometer. When the temperature has risen

by about 10 °C, recap the spirit burner and measure and record the maximum temperature of

the water.

Reweigh the spirit burner and record its mass.

Write up

Complete a full write up for this experiment. Within the experiment calculate the enthalpy of

combustion for ethanol.

23

Hess’s Law

The Law of Conservation of Energy states that energy can neither be created nor destroyed, but

that it can be changed from one form to another. This particular application of the conservation of

energy law to chemical reactions is known as Hess’s Law.

Hess’s Law: the enthalpy change in converting reactants into products is the same regardless of

the route by which the reaction takes place.

The total enthalpy change for route 1 = ∆H1

The total enthalpy change for route 2 = ∆H2 + ∆H3

The total enthalpy change for route 3 = ∆H4 + ∆H5 + ∆H6

According to Hess’s Law the total enthalpy change for 1, 2 & 3 will be identical.

∆H1 = ∆H2 + ∆H3 = ∆H4 + ∆H5 + ∆H6

The reaction can take place by two methods:

According to Hess’s Law:

∆H1 = ∆H2 + ∆H3

1

2

∆H1

3

∆H3 ∆H2

∆H4

∆H5

∆H6

A B

C

D E

NaOH(s)

+

NaOH(aq)

+ HCl(aq)

NaCl(aq)

+ HCl(aq)

Route 2

Route 1

∆H1

∆H2 ∆H3

24

Applications of Hess’s Law

Hess’s Law can be used to calculate enthalpy changes, which are difficult or impossible to

determine, by experiment.

Enthalpy of Formation

Definition: The enthalpy of formation is the quantity of heat energy taken in or given out when 1

mole of a substance is formed from its elements in their normal state.

Calculations

The information you need to tackle a calculation will either be given in the question or can be found in

the data booklet.

Worked Example 1

Calculate the enthalpy of formation of ethane given that the enthalpies of combustion of carbon,

hydrogen and ethane are -394kJmoll-1, -286kJmoll-1 and -1561kJmol-1 respectively.

The first thing that you must do is write a balanced equation for the equation that you are required

to calculate the enthalpy change for. This is called the Target Equation.

Target Equation : 2C(s) + 3H2(g) C2H6(g) ∆H = ?

Next you need to construct balanced equations for each of the substances in your target equation,

using the information given in the question.

(1) C(s) + O2(g) CO2(g) ∆H = -394

(2) H2(g) + ½O2(g) H2O(l) ∆H = -286

(3) C2H6(g) + 3½O2(g) 2CO2(g) + 3H2O(l) ∆H = -1561

These equations can now be rearranged to give the target equation. Note that whatever change you

make to the equation you must also make to the enthalpy value!

2 x (1) 2C(s) + 2O2(g) 2CO2(g) ∆H = -788

3 x (2) 3H2(g) + 1½O2(g) 3H2O(l) ∆H = -858

reverse (3) 2CO2(g) + 3H2O(l) C2H6(g) + 3½O2(g) ∆H = +1561

Once everything that appears on both the reactant side and the product side have been cancelled you

should be left with your target equation!

2C(s) + 3H2(g) C2H6(g) ∆H = -85kJmoll-1

25

Worked example 2

Use the enthalpies of combustion in your data booklet to find the enthalpy change for the reaction

between ethyne and hydrogen to produce ethane.

Target Equation : C2H2 + 2H2 C2H6 ∆H = ?

(1) C2H2 + 2½O2 2CO2 + H20 ∆H = -1301

(2) H2 + ½O2 H20 ∆H = -286

(3) C2H6 + 3½O 2CO + 3H2O ∆H = -1561

(1) C2H2 + 2½O2 2CO2 + H20 ∆H = -1301

2 x (2) 2H2 + O2 2H20 ∆H = -572

reverse (3) 2CO2(g) + 3H2O(l) C2H6(g) + 3½O2(g) ∆H = +1561

C2H2 + 2H2 C2H6 ∆H= -312kJmoll-1

Bond Enthalpies

Bond enthalpies can be divided into two categories (see the Data Booklet).

The bond dissociation enthalpy is the energy required to break one mole of bonds and form two

separate atoms, all species being in the gaseous state. These values are accurately known but it is

only possible to calculate them for diatomic molecules, e.g.

H2(g) → 2H(g) +436 kJ mol–1

N2(g) → 2N(g) +945 kJ mol–1

HCl(g) → H(g) + Cl(g) +432 kJ mol–1

For molecules with more than two atoms, such as methane, it is only possible to calculate the mean

bond enthalpy since the situation is much more complex with the C–H bonds breaking off one after

the other to form a different fragment of the original molecule.

The removal of the first hydrogen atom from CH4 does not require the same energy as the removal

of the second hydrogen from the CH3 fragment and so on. The value obtained is the average or mean

bond enthalpy for C–H, i.e.

CH4(g) → C(g) + 4H(g) ∆H = 4 x C–H = 1648 kJ mol

Therefore the mean bond enthalpy of C–H = 1648 / 4 = 412kJ mol–1

Mean bond enthalpies are quoted in data books for bonds of that type in any molecule but can also be

calculated from other enthalpy changes, as shown in the following examples.

26

Experiment 3.6—Hess’s law

Solid potassium hydroxide can be converted into potassium chloride solution by two different routes:

Route 1 is the direct route whereby potassium chloride solution is made by adding solid potassium

hydroxide directly to hydrochloric acid. Let's suppose it has an enthalpy change of ΔH1.

KOH(s) + HCl(aq) KCl(aq) + H2O(l) ΔH1

Route 2 is the indirect route and involves two steps. In the first of these solid potassium hydroxide

is dissolved in water:

KOH(s) + aq KOH(aq) ΔH2a

The resulting potassium hydroxide solution is then added to hydrochloric acid to form potassium

chloride solution:

KOH(s) + HCl(aq) KCl(aq) + H2O(l) ΔH2b

According to Hess's Law the overall enthalpy change involved in converting solid potassium hydroxide

into potassium chloride solution will be the same no matter whether the direct or indirect route is

taken.

The aim of this experiment is to confirm Hess's Law.

27

Experiment

ROUTE 1 (direct route)

Using the measuring cylinder, measure out 25cm3 of 1 mol l-1

hydrochloric acid into a plastic beaker or polystyrene cup.

Measure and record the temperature of the acid.

Weigh out accurately about 1.2g of potassium hydroxide into a

plastic beaker or polystyrene cup and record the mass. Make sure

the mass of potassium hydroxide does not exceed 1.4g.

Add the acid to the potassium hydroxide. Slowly and continuously stir the reaction mixture

with the thermometer until all the solid reacts.

Measure and record the highest temperature reached by the reaction mixture.

ROUTE 2 (indirect route)

Step A

The solution you prepare in this step is needed in step B - DON'T THROW ITAWAY!

Using the measuring cylinder, measure out 25 cm3 of water into a plastic beaker or polystyrene

cup.

Measure and record the temperature of the water.

Weigh out accurately about 1.2g of potassium hydroxide into a plastic beaker or polystyrene

cup and record the mass. Make sure the mass of potassium hydroxide does not exceed 1.4g.

Add the water to the potassium hydroxide. Slowly and continuously stir the reaction mixture

with the thermometer until all the solid dissolves.

Measure and record the highest temperature reached by the solution.

Keep the solution you have just prepared but allow it to cool down for some time before

proceeding to step B.

Step B

Using the measuring cylinder, measure out 25 cm3 of 1 mol l-1 hydrochloric acid into a plastic

beaker or polystyrene cup.

Measure and record the temperature of the acid.

Measure and record the temperature of the potassium hydroxide solution you prepared in step

A.

Add the acid to the potassium hydroxide solution and stir the reaction mixture slowly and

continuously with the thermometer.

Measure and record the highest temperature reached by the reaction mixture.

28

Example Question

Calculate the mean bond enthalpy of the C–H bond from the enthalpy of formation of methane

and any other required data from the Data Booklet.

Step 1: Write down the required equation:

CH4 → C(g) + 4H(g) ∆H = 4 x C–H

Step 2: Express all the information given in equation form:

(1) C(s) + 2H2(g) → CH4(g) ∆H = –75 kJ mol–1

(2) C(s) → C(g) ∆H = + 716 kJ mol–1

(3) H2(g) → 2H(g) ∆H = +436 kJ mol–1

Step 3: Use the three equations to obtain the required balanced equation:

(1) Reverse CH4(g) → C(s) + 2H2(g) ∆H = +75 kJ mol–1

(2) Nothing C(s) → C(g) ∆H = +716 kJ mol–1

(3) x2 2H2(g) →4H(g) ∆H = +872 kJ mol–1

Step 4: Add the three equations to give the required equation and DH:

CH4(g) → C(g) + 4H(g) ∆H = 1663 kJ mol–1

1663 kJ = 4 x C–H C–H = 1663 / 4

Mean bond enthalpy of C–H = 415.75kJ mol–1

REDOX Reactions

A REDOX reaction involves two half reactions - oxidation and reduction.

These half reactions can be written as ion-electron equations. Ion-electron equations are found on

page 12 of the Data Booklet.

Oxidation involves the LOSS of electrons (OIL):

Fe Fe2+ + 2e Mg Mg2+ + 2e

The ion-electron equation for oxidation must be written in reverse.

Reduction involves the GAIN of electrons (RIG):

Cu2+ + 2e Cu Ag+ + e Ag

Remember, oxidation and reduction are two halves of the same chemical reaction. The combined

reaction is called a Redox reaction.

To form the overall redox reaction, the ion-electron equations for the oxidation and reduction must

be combined, ensuring that the electrons cancel.

29

Oxidation: Mg(s) Mg2+ + 2e

Reduction: Ag+ + e Ag(s) (x2)

Mg(s) + 2Ag+ Mg2+ + 2Ag(s)

Oxidation: Al(s) Al3+ + 3e (x2)

Reduction: 2H+ + 2e H2(g) (x3)

Mg(s) + 2Ag+ Mg2+ + 2Ag(s)

Oxidising and Reducing Agents

In a redox reaction the species that is oxidised is described as a reducing agent - a species that

allows reduction to occur.

Similarly, a species which is reduced is described as an oxidising agent - a species that allows

oxidation to occur.

eg. Mg(s) + 2Ag+ Mg2+ + 2Ag(s)

The elements with low electronegativities (metals) tend to form ions by losing electrons (oxidation)

and so can act as reducing agents. The strongest reducing agents are found in group 1.

The elements with high electronegativities (non-metals) tend to form ions by gaining electrons

(reduction) and so can act as oxidising agents. The strongest oxidising agents are found in group 7.

Using the Data Booklet

When writing redox equations you must first identify the oxidation and reduction half reactions and

then combine them to give the redox reaction. Use the Electrochemical Series on page 12 of the data

booklet.

Example:

Iron (II) sulfate solution is reacted with potassium permanganate solution until the first appearance

of a permanent pink colour.

Step 1: What four ions are present in the two solutions?

Step 2: Write the ion-electron equation for the oxidation reaction.

Step 3: Write the ion-electron equation for the reduction reaction.

Step 4: Add the two half equations together to give the redox equation (remember to balance

the electrons first)

30

Experiment 3.7—OXIDISING AND REDUCING AGENTS

Halogen displacement

Experiments with halogen displacement can be used to compare the strength of elements as oxidising

agents. In this experiment you will be using chlorine, bromine and iodine water firstly to test their

effect on pH paper, and secondly to test their reaction with potassium halides.

Experiment

Put a piece of Universal Indicator paper onto a white tile.

Use a glass stirring rod to transfer a few drops of the chlorine water onto the indicator paper.

Repeat this with a fresh piece of paper and the bromine water then the iodine water.

In a test-tube, add some chlorine water to a solution of potassium bromide.

Add some chlorine water to a solution of potassium iodide.

Add some bromine water to a solution of potassium chloride

Add some bromine water to a solution of potassium iodide

Add some iodine water to a solution of potassium chloride

Add some iodine water to a solution of potassium bromide

Complete the table noting your observations

Results table

Effect on

indicator paper

Reaction with

potassium

chloride solution

Reaction with

potassium

bromide solution

Reaction with

potassium iodide

solution

Chlorine water

Bromine water

Iodine water

31

Experiment 3.8—OXIDISING AND REDUCING AGENTS

Blue bottle experiment

The blue bottle reaction is a chemical reaction in which in a closed bottle an aqueous solution

containing dextrose (D-glucose), sodium hydroxide and a methylene blue and some air turns from

colorless to blue upon shaking and which then decolorizes again after a while. After shaking again the

blue color returns and this cycle can be repeated several times. The reaction will work with other

reducing sugars besides glucose and also with other reducing dyes. This experiment allows you to see

that dextrose acts as a reducing agent and oxygen acts as an oxidising agent.

Blue Bottle NOT Blue Bottle

Experiment

(Note the optimum temperature for this reaction is 25-300C)

Using a measuring cylinder, measure 30cm3, 0.5mol l-1 sodium hydroxide into a 100cm3 conical

flask, volumetric flask or bottle

Add 1g of D-glucose and swirl until the solids are dissolved.

Add 1cm3 of the methylene blue solution.

The blue colour will turn colourless after about one minute.

Stopper the flask or bottle. Shake the flask vigorously so that air dissolves in the solution. The

colour will change to blue. This will fade back to colourless over about 30 seconds. The more

shaking, the longer the blue colour will take to fade. The process can be repeated for over 20

cycles. After some hours, the solution will turn yellow and the colour changes will fail to occur.

Write up

Complete a full write up for this experiment. Make a note of any observations that you see.

32

More Complicated Half-Equations

There are some ion-electron equations that are not given in the data book. You must learn the rules

to work them out for yourself.

Worked example

Write an ion-electron equation for the following reaction:

Cr2O72- Cr3+

Step 1: If necessary, balance the central atom / ion.

Cr2O72- 2Cr3+

Step 2: Add water (H2O) if it is needed to balance the oxygen atoms.

Cr2O72- 2Cr3+ + 7H2O

Step 3: Balance the hydrogen in the water by adding hydrogen ions.

Cr2O72- + 14H+ 2Cr3+ + 7H2O

Step 4: Calculate the total electrical charge on each side of the equation.

Cr2O72- + 14H+ 2Cr3+ + 7H2O

(12+) (6+)

Step 5: Add electrons to balance the electrical charges.

Cr2O72- + 14H+ + 6e- 2Cr3+ + 7H2O

(12+) + (6-) (6+)

Chemical Analysis

Chromatography

Chromatography is a method of separating and analysing a mixture of

soluble chemical substances.

There are different types of Chromatography techniques. The most

common is paper chromatography.

Uses of Chromatography

Chromatography can be used to follow the course of a chemical reaction or to

establish the products of a chemical reaction.

It can also be used to identify additives in foods, analyse fibres found at a crime scene and

fingerprinting.

33

Paper Chromatography

During paper chromatography a small sample of the mixture being

tested is spotted onto the base line (a straight line usually drawn in

pencil above the level of the solvent) on the filter paper. The filter

paper is then placed in a solvent.

By capillary action the solvent moves up the paper. This is when

different components of the mixture are separated.

Components can move quickly or slowly up the paper depending on the solvent used.

There are two stages in paper chromatography; the Mobile Phase and the Stationary Phase.

The solvent is the mobile phase in paper chromatography and the wet paper is the stationary phase.

Why does separation occur?

Separation of the mixture occurs due to the polarity and size of the molecules within the mixture.

Depending on what solvent is used depends on how the molecules behave.

For example, if a mixture contains very polar molecules and non-polar molecules and a polar solvent,

such as ethanol, is used the polar molecules move much quicker up the filter paper. This is because

the less polar molecules will be more attracted to the mobile phase.

In terms of large and small molecules smaller molecules move much quicker and further up the filter

paper compared to larger molecules.

The Retention Factor (Rf) Values

The Rf factor can be used to compare the different components found in a sample. The Rf values of a

mixture being tested can be compared with known samples.

Note: if two substances on chromatography paper have the same Rf value it is likely that they are

the same compound. If they do not have the same Rf value then they are definitely different

compounds.

The Rf value of the red, green and

blue particles can be measured and

calculated to prove that certain

molecules are present in the mixture.

34

Experiment 3.10—CHEMICAL ANALYSIS

Wool chromatography

Chromatography is the collective term for a set of laboratory techniques for the separation of

mixtures. The mixture is dissolved in a fluid called the mobile phase, which carries it through a

structure holding another material called the stationary phase. The various constituents of the

mixture travel at different speeds, causing them to separate. This experiment uses white wool as the

stationary phase and ethanol as the mobile phase. You will be using this technique to separate the

colour constituents of a felt tip pen.

Experiment

Cut a length of white wool approximately 1m long

Fill a 100cm3 beaker with approximately 20cm3 of methylated spirits

Make a 1cm long mark on one end of the wool approximately 20cm from the end with your

chosen pen. Allow the ink to dry.

Attach a crocodile clip to each end of the wool and set the beaker of methylated spirits on the

edge of a table

Carefully lower the end of wool with the ink into the beaker making sure the ink is not

submerged, drape the wool over the spout of the beaker and allow it to hang towards the floor

with an empty beaker on the floor under the wool.

Watch as the alcohol is wicked up the wool and towards the floor under the influence of

gravity. The alcohol dissolves and separates the different colour components of the ink at

different rates resulting in a series of coloured band moving towards the floor.

35

Redox Titrations

As with volumetric titrations there has to be a way of determining the end-point of the reaction. For

many reactions an indicator must be used however, for some redox reactions the end-point can be

recognised from a colour change in one of the reactants, for example, when potassium permanganate

solution (purple) reacts with iron(II) sulfate solution, the permanganate ions are reduced to

colourless manganese ions. When a colour change involving one of the reactants is used to determine

the end-point the reaction is said to be self-indicating.

Worked example:

20cm3 of iron(II)sulphate were titrated with 0.01mol-1 potassium permanganate solution until a

permanent pink colour was observed. If the volume of potassium permanganate used was 25.6cm3,

what is the concentration of the iron(II) sulfate solution?

Step 1: write the ion-electron equations and combine to give the redox equation.

Fe2+ Fe3+ + e- (x5)

MnO4- + 8H+ + 5e- Mn2+ + 4H20

5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O

Step 2: calculate the number of moles of the ‘known’ substance

Moles = c x v

MnO4- 0.01 x 0.0256

0.000256 moles

using the mole ratio:

Step 3: calculate the concentration of the iron(II) sulfate solution

C = moles / v

0.00128

0.02

= 0.064moll-1

36

Experiment 3.11—CHEMICAL ANALYSIS

Redox titration of Vitamin C

Vitamin C (ascorbic acid) is an important component of our diet. In its absence the protein, collagen,

cannot form fibres properly and this results in skin lesions and blood vessel fragility.

Although vitamin C occurs naturally in many fruits and vegetables many people take vitamin C tablets

to supplement their intake.

Vitamin C can undergo a redox reaction with iodine in which the vitamin C is oxidised and the iodine

molecules are reduced.

The aim of this experiment is to determine the mass of vitamin C in a tablet by carrying out a redox

titration using a solution of iodine of accurately known concentration and starch solution as an

indicator.

37

Experiment Add a vitamin C tablet to the beaker. Add some deionised water (approximately 50cm

3) to the beaker and stir the mixture

until the tablet has dissolved. Carefully add the resulting solution to the 250cm

3 volumetric flask. Rinse out the

beaker several times with water and add the washings to the flask. Add water to the standard flask to bring the volume of the solution up to the

graduation mark on the neck. Stopper the flask and invert it several times to make sure the solution is thoroughly

mixed. After rinsing the pipette with a little of the vitamin C solution, pipette 25cm

3 of it into

the conical flask. Add a few drops of starch solution to the vitamin C solution in the conical flask. After rinsing the burette with a little iodine solution, fill the burette with the iodine

solution. Note the initial burette reading. Since the solution has a dark colour, it is difficult to

see the bottom of the meniscus. Take the burette reading from the top of the meniscus.

Add the iodine solution slowly from the burette whilst gently swirling the solution in

the conical flask. Initially you will see a blue/black colour as the iodine reacts with the starch but this will rapidly disappear as the iodine reacts with the vitamin C.

Near the end-point of the titration the colour disappears more slowly. At this point

add the iodine solution drop by drop until the solution just turns a blue/black colour and remains so.

This is the end-point of the titration i.e. all the vitamin C has reacted. Note the final

burette reading. Wash out the conical flask. Repeat the titrations until concordant results are obtained.

Write up

Complete a full write up for this experiment. Show all calculations and determine the mass of Vitamin

C in the tablet.

38

Word Meaning Atom Economy

The proportion of reactant atoms which are incorporated into the

desired product of a chemical reaction.

% atom economy =Mass of desired product(s)

Total mass of reactantsx 100

Batch processes

These involve mixing one batch of reactants in a vessel, allowing the

reaction to complete, then removal of the products, before reloading

with a fresh batch of reactants.

Bond Dissociation

Enthalpy Energy required to break one mole of bonds to form two separate

atoms, all species being in the gaseous state.

Capital costs

Capital costs are incurred in the initial building of the plant and

associated infrastructure.

Cash flow

Cash flow is the movement of money into and out of a business, and is a

measure of profitability.

Continuous processes

These take place in a plant where reactants are fed in continuously at

one end, and products are removed continuously from the other.

Chromatography

A method of separating and analysing a mixture of soluble chemical

substances.

Dynamic equilibrium A dynamic equilibrium is achieved when the rates of two opposing

processes become equal, so that no net change results

Electrodes

The electrodes (one negative and one positive) dip into the electrolyte

and form the connection to the electric wires. Any chemical reactions

occur at their surfaces.

Electrolysis

Electrolytes are liquids containing ions which are free to move, and are

therefore capable of conducting electricity.

Electrolyte

Electrolytes are liquids containing ions which are free to move, and are

therefore capable of conducting electricity.

Electrolytic cell

An electrolytic cell is the equipment used to carry out electrolysis –

oxidation occurs at the positive electrode and reduction at the negative

electrode.

39

Equilibrium

Chemical equilibrium is the state reached by a reaction mixture when

the rates of forward and reverse reactions have become equal

Excess Reactant added which is not all used up in the reaction.

Feedstock

Feedstocks are the reactants from which other chemicals can be

extracted or synthesised.

Fixed costs

Fixed costs are incurred irrespective of whether a plant is operating at

maximum or only at partial capacity.

Hess’s law

The enthalpy change for a chemical reaction is independent of the route

taken, providing the starting point and finishing point is the same for

both routes.

Molar Volume

The volume, in litres, occupied by one mole of gas under standard

conditions of temperature and pressure.

Ion-electron equations

An ion-electron equation is a half-equation, either an oxidation or a

reduction, which in combination of the opposite type, can be part of a

complete redox equation.

Oxidation

An oxidation is a loss of electrons by a reactant in any reaction.

Oxidising agent An oxidising agent is a substance which accepts electrons.

Pilot plant

A pilot plant is a small scale chemical plant, built with similar materials

to a fullscale one, to investigate the chemical and engineering aspects

of a process before a full-scale plant is built.

Raw materials Raw materials are the initial starting materials from which the a

process’s feedstock is derived.

Reducing agent

A reducing agent is a substance which donates electrons.

Reduction

A reduction is a gain of electrons by a reactant in any reaction.

Stoichiometry

Stoichiometry means literally ’measuring elements’ and is concerned

with the quantities of substances reacting and the quantities of

substances produced as derived from the balanced equation, i.e. it is

concerned with mole relationships.

Variable costs

Variable costs are dependent on a plant’s output.

Yield Quantity of product obtained in a reaction, usually given as % yield =

actual/theoretical x100.

40

The Chemical Industry Exercise 1

1. The chemical industry will design a process to

Maximise what?

Minimise what?

2. Describe the stages in the manufacture of a new product.

3. What does the term Feedstock mean?

4. What are the main raw materials used in the chemical industry?

5. Give examples of capital, fixed and variable costs.

6. Describe a continuous process, giving advantages and disadvantages.

7. Describe a batch process, giving advantages and disadvantages.

8. Along with economic viability suggest other factors that will impact on the location and route

used to manufacture a new product.

9. Aluminium is extracted from its purified oxide by molten electrolysis. Suggest two advantages

and disadvantages of siting the aluminium smelters in the Scottish Highlands.

The Chemical Industry Exercise 2 Past Paper Questions

1. Which element is NOT a raw material in the chemical Industry?

A Nitrogen

B Oxygen

C Sodium

D Sulfur

2. The costs involved in the industrial production of a chemical are made up of fixed costs and

variable costs.

Which of the following is most likely to be classified as a variable cost?

A The cost of land rental

B The cost of plant construction

C The cost of labour

D The cost of raw materials

3. Which of the following is produced by a batch process?

A Sulfuric acid from sulfur and oxygen

B Aspirin from salicylic acid

C Iron from iron ore

D Ammonia from nitrogen and hydrogen

41

The flow chart summarises some industrial processes involving ethene.

The feedstocks for ethene in these processes are

A ethane and glycol

B ethane and ethanol

C glycol and poly(ethene)

D glycol, poly(ethene) and ethanol.

5. Which of the following compounds is a raw material in the chemical industry?

A Ammonia

B Calcium carbonate

C Hexane

D Nitric acid

6. Cerium metal is extracted from the mineral monazite.

The flow diagram for the extraction of cerium from the

mineral is shown on the previous page.

Name the type of chemical reaction taking place in Step A.

In Step B, cerium hydroxide is heated to form cerium

oxide, Ce2O3, and compound Z.

Name compound Z.

In Step C, cerium metal is obtained by electrolysis

What feature of the electrolysis can be used to reduce the

cost of cerium production?

42

7. About 2.5 million tonnes of sulfuric acid are produced each year in the UK. Sulfuric acid was

made industrially by the Chamber Process. The following chemical reactions are involved.

Sulfur is burned to produce sulfur dioxide.

Sulfur dioxide reacts with water to produce sulfurous acid.

Nitric oxide is produced by the catalytic oxidation of ammonia;

water is also a product of this reaction.

Nitric oxide reacts with oxygen to form nitrogen dioxide

Nitrogen dioxide reacts with sulfurous acid to form sulfuric acid and regenerate nitric oxide.

Copy and complete the flow diagram below of the Chamber process with the names of the chemicals

involved (2 Marks)

To show the name of a chemical

To Show a reaction

Sulfuric acid

43

The Mole Revision Exercise 3

Moles and mass

1.) Write the formula and then work out the gram formula mass (mass of 1 mole) of:

a) hydrogen sulfide (b) magnesium bromide

c) hydrogen sulfate (d) hydrogen nitrate

2.) What is the mass of the following?

a) 10 moles of calcium carbonate (b) 3 moles of sodium chloride

c) 0.5 moles of lithium fluoride (d) 0.2 moles of methane (CH4)

3.) How many moles of compound are present in:

a) 14 g of nitrogen gas (b) 400 g of copper(I) oxide

c) 85 g of lithium chloride (d) 117 g of sodium sulfide

Moles and concentration

1) Calculate the number of moles of:

a) sodium chloride in 1000 cm3 of 1 mol l-1 solution.

b) sodium chloride in 500 cm3 of 0.5 mol l-1 solution.

c) potassium nitrate in 250 cm3 of 0.5 mol l-1 solution.

2) What volume of water is required to make:

a) 3 moles of potassium hydroxide into a 1.5 mol l-1 solution?

b) 0.5 moles of sodium chloride into a 2 mol l-1 solution?

c) 0.1 moles of lithium oxide into a 0.05 mol l-1 solution?

3) What concentration of solution is obtained if we dissolve:

a) 2 moles of sodium hydroxide in 500 ml of water?

b) 1 mole of lithium chloride in 2 L of water?

c) 1.5 moles of magnesium bromide in 100 cm3 of water?

4) What concentration of solution is obtained if we dissolve:

a) 40 g of sodium hydroxide in 1 litre of water?

b) 58.5 g of sodium chloride in 100 cm3 of water?

c) 585 g of sodium chloride in 10 litres of water?

5) What volume of water is required to make:

a) 56 g of potassium hydroxide into a 1.5 mol l-1 solution?

b) 117 g of sodium chloride into a 2 mol l-1 solution?

c) 15 g of lithium oxide into a 0.05 mol l-1 solution?

6) What mass of substance is required to make:

a) 500 cm3 of 2 mol l-1 aluminium nitrate solution?

b) 2 L of 0.5 mol l-1 potassium bromide solution?

c) 100 cm3 of 3 mol l-1 calcium sulfate solution?

44

Balancing Equations and Calculations from Balanced Equations Revision

1. Balance the following equations:

a) N2 + H2 NH3

b) N2 + O2 NO

c) Na + H2O NaOH + H2

d) CH4 + O2 CO2 + H2O

2. 2Mg + O2 MgO

Calculate the mass of oxygen required to burn 6 g of magnesium.

3. CaCO3 CaO + CO2

What mass of CaCO3 is required to form 2.8 g of calcium oxide?

4. N2 + 3H2 2NH3

What mass of ammonia (NH3) is produced from 7 g of nitrogen?

Excess Calculations Exercise 4

In each of the following reactions calculate which reactant is in excess

1. a) Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

6.5 g of Zinc added to 25 cm3 of dilute sulfuric acid,

concentration 2 mol l-1

b) Mg (s) + 2HCl(aq) MgCl2(aq) + H2(g)

2.4 g of magnesium added to 100 cm3 of dilute hydrochloric acid,

concentration 1 mol l-1

c) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

3.6 g of Zinc added to 50 cm3 of dilute hydrochloric acid,

concentration 0.5 mol l-1

d) Mg (s) + H2SO4(aq) MgSO4(aq) + H2(g)

4.7 g of magnesium added to 25 cm3 of dilute sulfuric acid,

concentration 1 mol l-1

2.) Iron(II) sulfide reacts with hydrochloric acid as follows:

FeS(s) + 2HCl(aq) FeCl2(aq) + H2S(g)

If 4.4 g of iron(II) sulfide was added to 160 cm3 of 0.5 mol l-1 hydrochloric acid, show by calculation

which substance is in excess.

45

3. A student added 0.20 g of silver nitrate, AgNO3, to 25 cm3 of water. This solution was then

added to 20 cm3 of 0.0010 mol l-1 hydrochloric acid. The equation for the reaction is:

AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)

Show by calculation which reactant is in excess.

4. What mass of ammonia gas is produced when 1.32 g of ammonium sulfate is heated with 1 g of

sodium hydroxide?

(NH4)2SO4 + 2NaOH 2NH3 + Na2SO4 + 2H2O

5. What mass of sulfur dioxide is produced when 1.26 g of sodium sulphite is added to 50cm3 of

dilute hydrochloric acid , concentration 2 mol l-1

Na2SO3 + 2HCl 2NaCl + H2O + SO2

6. Calcite is a very pure form of calcium carbonate which reacts with nitric acid as follows:

CaCO3(s) + 2HNO3(aq) Ca(NO3)2(aq) + H2O(l) + CO2(g)

A 2.14 g piece of calcite was added to 50.0 cm3 of 0.200 mol l-1 nitric acid in a beaker. Calculate the

mass of calcite, in grams, left unreacted.

7. Copper(II) oxide reacts with sulfuric acid as follows:

CuO(s) + H2SO4(aq) CuSO4(aq) + H2O(l)

1.6 g of copper(II) oxide is added to a beaker containing 50 cm3 of

0.25 mol l-1 sulfuric acid. Calculate the mass of copper(II) oxide remaining after the reaction was

complete.

8. Lead reacts with hydrochloric acid as follows:

Pb(s) + 2HCl(aq) PbCl2(aq) + H2(g)

If 6.22 g of lead was added to 50 cm3 of 1 mol l-1 hydrochloric acid, calculate the mass of lead left

unreacted.

9. A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 10.00 g of

silver nitrate. The reaction that occurs is

Zn(s) + 2AgNO3(aq) 2Ag(s) + Zn(NO3)2(aq)

(a) Determine which reactant is in excess.

(b) Calculate how many grams of silver will be formed.

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10. A piece of lithium with a mass of 1.50 g is placed in an aqueous solution containing 6.00 g of

copper (II) sulfate. The reaction that occurs is:

2Li(s) + CuSO4(aq) Cu(s) + Li2SO4 (aq)

(a) Determine which reactant is in excess.

(b) Calculate how many grams of copper will be formed.

Molar Volume Exercise 5

1. Decide which has the greater volume, measurements made at the same temperature and

pressure

a) 17 g of ammonia or 17 g of methane

b) 20 g of neon or 20 g of nitrogen

c) 4 g of hydrogen or 44 g of carbon dioxide

d) 6.4 g of sulfur dioxide or 8 g of oxygen

2. The volume of 0.22 g of propene C3H6 is 118 cm3. Calculate the volume of 2 moles of propene

at this temperature and pressure .

3. Calculate the volume of 2.4 g of ethane, C2H4. (take the molar volume to be 23.6 litres mol-1)

4. The volume of 1 g of hydrogen is 11.6 litres. Calculate the volume of 4 moles of hydrogen at

this temperature and pressure.

5. 3 g of an alkane occupies a volume of 2.24 litres. What is the molecular formula of the alkane

(take the molar volume to be 22.4 litres mol-1)

6. Using the densities in the data booklet calculate the volume of 10 g of

(a) Hydrogen

(b) Argon

7. Using the densities in the data booklet calculate the mass of 10 litres of

(a) Helium

(b) Nitrogen

8. From the following data calculate the approximate formula mass of the gas X.

Mass of plastic bottle empty = 112.30 g

Mass of plastic bottle + Gas X = 113.52 g

Capacity of plastic bottle = 1 litre

Molar volume of gas X = 23.6 litres

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9. A flask, capacity 600 cm3 was used to calculate the molar volume of sulfur dioxide.

The following data was obtained:

Mass of evacuated flask = 512.97 g

Mass of flask + sulfur dioxide = 514.57 g

10. Some of liquid Z is injected into a large syringe and it evaporates. The following results were

recorded:

Mass of syringe before injection = 4.648 g

Mass of syringe after injection = 4.774 g

Syringe reading before injection = 0 cm3

Syringe reading after injection = 84 cm3

Calculate the relative formula mass of liquid Z

(Molar volume = 30.6 litre mol-1)

Balanced Equations Using Mass and Volumes Exercise 6

Molar volume 24 litres unless stated otherwise

1.) What volume (in l) of carbon dioxide would be produced by completely reacting 60 g of carbon

with oxygen?

C + O2 CO2

2.) What volume (in l) of hydrogen would be produced by completely reacting 60 cm3 of

hydrochloric acid of concentration 1.2 mol l–1 with zinc?

Zn + 2HCl ZnCl2 + H2

3.) What volume (in l) of carbon dioxide would be produced by completely reacting 10 g of calcium

carbonate with hydrochloric acid?

CaCO3 + 2HCl CaCl2 + H2O + CO2

4.) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

What volume of hydrogen gas is produced when 5 g of magnesium is added to 100 cm3 of

dilute hydrochloric acid concentration 2 mol l-1 ( Molar volume =23.6 Litres mol-1)

5.) What mass of magnesium oxide is obtained when 24 g of magnesium is ignited in 5 litres of

oxygen

Mg(s) + O2(g) MgO(s)

(Molar volume is 22.8 litres mol-1)

48

6. What mass of water is produced when 2 litres of hydrogen is burned in excess oxygen (use

the density of hydrogen given in the data book)

7. Calculate the volume of oxygen required for the complete combustion of

1 g of ethane C2H4

(Molar volume is 32.2 litres mol-1)

8. Chlorine gas can be produced by heating calcium hypochlorite, Ca(OCl)2, in dilute hydrochloric

acid.

Ca(OCl)2(s) + 2HCl(aq) Ca(OH)2(aq) + 2Cl2(g)

Calculate the mass of calcium hypochlorite that would be needed to produce 0·096 litres of

chlorine gas. (Take the molar volume of chlorine gas to be 24 litres mol–1.)

9.) Hydrogen fluoride gas is manufactured by reacting calcium fluoride with concentrated

sulfuric acid.

CaF2 + H2SO4 CaSO4 + 2HF

What volume of hydrogen fluoride gas is produced when 1.0 kg of calcium fluoride reacts

completely with concentrated sulfuric acid?

(Take the molar volume of hydrogen fluoride gas to be 24 litres mol–1.)

10.) In the lab, nitrogen dioxide gas can be prepared by heating

copper(II) nitrate.

Cu(NO3)2(s) CuO(s) + 2NO2(g) + ½O2(g)

Calculate the volume of nitrogen dioxide gas produced when 2.0 g of copper (II)

nitrate is completely decomposed on heating.

(Take the molar volume of nitrogen dioxide to be 24 litres mol–1.)

49

Reacting Volumes Exercise 7

1. In each of the following reactions decide the ratio of the volume of product(s) to the volume

of reactant(s).

a) H2(g) + F2(g) 2HF(g)

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

c) 2C(s) + O2(g) 2CO(g)

d) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)

2. N2(g) + 2O2(g) 2NO2(g)

What volume of nitrogen dioxide is produced when 100 cm3 of nitrogen is sparked in excess

oxygen?

3. When the following gases are burned calculate the volume of oxygen required and the volume

of carbon dioxide produced

a) 50 cm3 of methane

b) 200 cm3 of ethane

c) 250 cm3 of carbon monoxide

4. 10 cm3 of butane gas is mixed with 75 cm3 of oxygen and the mixture exploded. Calculate the

volume and composition of the resulting gas mixture.

C4H10(g) + 6.5 O2(g) 4CO2(g) + 5H2O(l)

5. If 100cm3 of propene is burned completely with 900 cm3 of oxygen what will be the volume

and composition of the resulting gas mixture?

6 50 cm3 of ethyne, C2H2 is burned completely in 220 cm3 of oxygen

a) What will be the volume and composition of the resulting gas mixture?

b) What will be the volume and composition of the resulting gas mixture if the

experiment was repeated at 200oC?

50

7. What volume of oxygen (in litres) would be required for the complete combustion of a gaseous

mixture containing 1 litre of carbon monoxide and 3 litres of hydrogen?

8. 20 cm3 of ammonia gas reacted with an excess of heated

copper(II) oxide.

3CuO + 2NH3 3Cu + 3H2O + N2

Assuming all measurements were made at 200 °C, what would be the volume of gaseous

products?

9. 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)

ethyne

What volume of gas would be produced by the complete combustion of 100 cm3 of ethyne gas?

All volumes were measured at atmospheric pressure and room temperature.

10. The equation for the complete combustion of propane is:

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

30 cm3 of propane is mixed with 200 cm3 of oxygen and the mixture is ignited. What is the

volume of the resulting gas mixture? (All volumes are measured at the same temperature and

pressure.)

51

Percentage Yield and Atom Economy Exercise 8

1.) CH3OH + C2H5COOH C2H5COOCH3 + H20

Methanol propanoic acid methyl propanoate

In preparation, 40.4 g of methyl propanoate is obtained from 18.3 g of methanol.

(a) Calculate the percentage yield

(b) Calculate the atom economy in regards to methyl propanoate

2.) C3H7OH oxidation C2H5CHO

Propan-1-ol Propanal

In a preparation, 3.2 g of propanal is obtained from 3.9g of propan-1-ol. Calculate the

percentage yield.

3.) C3H6 + Br2 C3H6Br2

20.4 g of 1,2-dibromopropane is obtained from 5.2 g of propene.

Calculate the percentage yield

4.) N2 + 3H2 2NH3

Under test conditions, 2 kg of hydrogen reacts with excess nitrogen to produce 1.5 kg of

ammonia.

Calculate the percentage yield

5.) 2SO2 + O2 2SO3

Under test conditions, 1 kg of sulfur dioxide reacts with excess oxygen to produce 0.8 kg of

sulfur trioxide.

Calculate the percentage yield

52

6. 2FeCl2 + Cl2 2FeCl3

5.072 g of iron (II) chloride yields 4.869 g of iron (III) chloride, calculate the percentage

yield.

7. Calculate the atom economy for the production sulfur trioxide assuming that all the reactants

are converted into products.

H2SO4 H2O + SO3

8. Which reaction below has the highest atom economy for producing water?

2C2H6 + 7O2 4CO2 + 6H2O

C3H6 + 4½O2 3CO2 + 3H2O

9. 4NH3 + 5O2 4NO + 6H2O

a) 1.68 x 103 kg of NO was produced from1.36 x103 kg calculate the percentage yield.

b) Calculate the atom economy in terms of the production of NO

10. Fe2O3 + 3CO 2Fe + 3CO2

6.65 x 104 kg of an iron ore which is impure iron (III) oxide is reacted with an excess of

carbon monoxide, producing 2.79 x 104 kg of iron.

a) Calculate the mass of pure iron (III) oxide in the ore. (Assuming that all the iron

(III) oxide is reduced to iron and that the impurities do not take part in the reaction)

b) Calculate the percentage by mass, of the iron (III) oxide in the ore.