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HSC CHEMISTRY STUDENT WORKBOOK
 0402890724

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TEACHER EDITION

HSC CHEMISTRY STUDENT WORKBOOK CORE MODULE 1: PRODUCTION OF MATERIALS COLLATED BY:

BRONWEN HEGARTY 1

PRODUCTION OF MATERIALS How to Use this Workbook This workbook is designed for use both by students and teachers.  It provides a guided pathway through the syllabus content for Core Module 1 of the NSW Chemistry Syllabus.  Worksheets direct students and teachers to the most significant parts of each section of the syllabus, including the first-hand investigations and research topics.  The workbook is structured so that it can form the basis of a teaching sequence and of lesson plans for teachers.  Revision questions provide practice for students at the completion of each syllabus section and allow students to test their understanding of each section.  Questions similar to those asked in the HSC are included in each section to allow students to gain experience in the answering techniques needed for success in the final HSC examination.  The Notes on the Syllabus allow students and teachers to recognise the context of this Core Module and to relate the learning and teaching to the relevant issues for society and the environment.  Students and teachers can use the checklist provided to ensure that they have covered all syllabus “dot-points” in their learning or teaching.  A Periodic Table and Data Sheet, including a Table of Reduction Potentials, are provided. 2

Notes on the Syllabus It is essential that students be thoroughly familiar with the syllabus and they should download it from the NSW Board of Studies website (www.boardofstudies.nsw.edu.au). Students should recognise the context of this Core module. HSC questions are often framed around the following points:  The needs of humans for food, clothing and shelter have traditionally come from the natural environment.  The twentieth century saw an increase in world population, an explosion in the use of traditional materials and a reduction in supplies of traditional resources. Research was needed to develop a wider range of materials to satisfy technological developments.  Chemists and chemical engineers continue to search for new sources of traditional materials such as those from the petrochemical industry. Fossil organic reserves are dwindling and new sources of the organic chemicals presently used have to be found. Chemists are continually searching for compounds to be used in the design and production of new materials.  Students should take into account the implications of chemistry for society and the environment and the current issues, research and developments in chemistry.

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Student Summaries. Students should prepare their own summaries for each of the syllabus “dot-points”, remembering that recent HSC examination papers have included questions closely following the syllabus wording. These questions may be worth up to 7 or 8 marks. In particular, students should ensure that their summaries include analysis, assessment, evaluation etc when it is specified in the syllabus. Following is a checklist to allow you to prepare your summaries and to tick them off when completed. Syllabus statement

Summary completed

Fossil fuels provide both energy and raw materials such as ethylene, for the production of other substances Identify the industrial source of ethylene from the cracking of some of the fractions from the refining of petroleum Identify that ethylene, because of the high reactivity of its double bond, is readily transformed into many useful products Identify data, plan and perform a first-hand investigation to compare the reactivities of appropriate alkenes with the corresponding alkanes in bromine water Identify that ethylene serves as a monomer from which polymers are made Identify polyethylene as an addition polymer and explain the meaning of this term Outline the steps in the production of polyethylene as an example of a commercially and industrially important polymer Identify the following as commercially significant monomers: - vinyl chloride - styrene by both their systematic and common names Describe the uses of the polymers made from the above monomers in terms of their properties Analyse information from secondary sources such as computer simulations, molecular model kits or multimedia resources to model the polymerisation process Gather and present information from first-hand or secondary sources to write equations to represent all chemical reactions encountered in the HSC course Some scientists research the extraction of materials and energy from biomass to reduce our dependence on fossil fuels Discuss the need for alternative sources of the compounds presently obtained from the petrochemical industry Explain what is meant by a condensation polymer Describe the reaction involved when a condensation polymer is formed Describe the structure of cellulose and identify it as an example of a condensation polymer found as a major component of biomass Identify that cellulose contains the basic carbon-chain structures needed to build petrochemicals and discuss its potential as a raw material Use available evidence to gather and present data from secondary sources and analyse progress in the development and use of a named biopolymer. This analysis should name the specific enzyme(s) used or organism used to synthesise the material and an evaluation of the use or potential use of the polymer produced related to its properties 4

Syllabus statement (cont.)

Summary completed

Other resources, such as ethanol, are readily available from renewable resources such as plants Describe the dehydration of ethanol to ethylene and identify the need for a catalyst in this process and the catalyst used Describe the addition of water to ethylene resulting in the production of ethanol and identify the need for a catalyst in this process and the catalyst used Process information from secondary sources such as molecular model kits, digital technologies or computer simulations to model: - the addition of water to ethylene - the dehydration of ethanol Process information from secondary sources to summarise the processes involved in the industrial production of ethanol from sugar cane Process information from secondary sources to summarise the use of ethanol as an alternative car fuel, evaluating the success of current usage Solve problems, plan and perform a first-hand investigation to carry out the fermentation of glucose and monitor mass changes Present information from secondary sources by writing a balanced equation for the fermentation of glucose to ethanol Describe and account for the many uses of ethanol as a solvent for polar and non-polar substances Outline the use of ethanol as a fuel and explain why it can be called a renewable resource Describe conditions under which fermentation of sugars is promoted Summarise the chemistry of the fermentation process Define the molar heat of combustion of a compound and calculate the value for ethanol from first-hand data Identify data sources, choose resources and perform a first-hand investigation to determine and compare heats of combustion of at least three liquid alkanols per gram and per mole Assess the potential of ethanol as an alternative fuel and discuss the advantages and disadvantages of its use Identify the IUPAC nomenclature for straight-chained alkanols from C1 to C8 Oxidation-reduction reactions are increasingly important as a source of energy Explain the displacement of metals from solution in terms of transfer of electrons Identify the relationship between displacement of metal ions in solution by other metals to the relative activity of metals Account for changes in the oxidation state of species in terms of their loss or gain of electrons Describe and explain galvanic cells in terms of oxidation-reduction reactions Outline the construction of galvanic cells and trace the direction of electron flow Define the terms anode, cathode, electrode and electrolyte to describe galvanic cells Perform a first-hand investigation to identify the conditions under which a galvanic cell is produced 5

Syllabus statement (cont.)

Summary completed

Perform a first-hand investigation and gather first-hand information to measure the difference in potential of different combinations of metals in an electrolyte solution Gather and present information on the structure and chemistry of a dry cell or lead-acid cell and evaluate it in comparison to one of the following: - button cell - fuel cell - vanadium redox cell - lithium cell - liquid junction photovoltaic device (eg the Gratzel cell) in terms of:  chemistry  cost and practicality  impact on society  environmental impact Solve problems and analyse information to calculate the potential EO requirement of named electrochemical processes using tables of standard potentials and half-equations Nuclear chemistry provides a range of materials Distinguish between stable and radioactive isotopes and describe the conditions under which a nucleus is unstable Describe how transuranic elements are produced Process information from secondary sources to describe recent discoveries of elements Describe how commercial radioisotopes are produced Identify instruments and processes that can be used to detect radiation Identify one use of a named radioisotope: - in industry - in medicine Describe the way in which the above named radioisotope is used and explain its use in terms of its chemical properties Use available evidence to analyse benefits and problems associated with the use of radioactive isotopes in identified industries and medicine

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Section 1 FOSSIL FUELS AS SOURCES OF MATERIALS AND ENERGY 1.

Ethylene (ethene) is an alkene produced from crude oil (petroleum). The chief components of crude oil are alkanes. Explain the terms alkanes and alkenes and the differences between these families of compounds. Alkanes – a homologous series of saturated hydrocarbon compounds, unreactive except combustion. General Formula = CnH2n+2 Alkenes – a homologous series of unsaturated hydrocarbon compounds, reactive because of double C=C bond. General Formula = CnH2n

2.

How are the various fractions of crude oil separated? Each fraction consists of compounds with similar chain length and boiling point. Fractions are separated by fractional distillation. The fractions are given identifying names (petrol, diesel, kerosene, etc).

3.

Draw up a table, listing the chief components of crude oil. For each fraction, give the range of number of carbons, examples and approximate boiling points of the alkanes in that fraction. No of carbons

Name of fraction

1 to 4 5 to 7 5 to 12 12 to 16

LPG Petroleum ether Petrol Kerosene

Boiling point range (°C) 25

Motor oils, greases Asphalt and tar

> 350 residue

Example

Use

propane hexane octane

Fuel gas Solvents Motor fuel Aviation fuel, heating Motor fuel, heating oil Lubricating oil Roads, roofing

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4.

What is meant by “catalytic cracking of a long chain alkane”? Describe how this is achieved and why the process is significant. Catalytic cracking involves use of a catalyst and heat to bring about the conversion of longchain hydrocarbons into smaller molecules. Mixtures of alkanes and alkenes are formed. By further cracking the alkenes, ethylene may be formed. The process uses surface catalysts, called zeolites, to provide a lower energy pathway for the reaction. The process is significant as useful unsaturated hydrocarbons and short chain saturated hydrocarbons can be produced from the less useful long chain saturated components. C20H42  C10 H20 + C10H22 decene

decane

C10 H20  2C5H10 pentene

C5H10  C2H4 + C3H6 ethylene

5.

propene

Only some fractions of crude oil (LPG and chains with about 10 carbons) are used for obtaining ethylene from cracking. What is “LPG”, what is it used for and what compounds are in this mixture? LPG = liquified petroleum gas; used as a gaseous fuel (gas cylinders); components are methane, ethane, propane and butane.

6.

Using structural formulae, write the equation for the catalytic cracking of decane (10 carbons) to form ethylene and 3-methylheptane. CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3  CH2CH2 + CH3CH2CH(CH3)CH2CH2CH2CH3 decane

7.

ethylene

3-methylheptane

Alkenes are more reactive than alkanes. Explain why, in terms of the bonding in these compounds. Alkenes contain 1 reactive C=C double bond, whereas the alkanes have only unreactive C-C single bonds.

8.

Alkenes often undergo “addition” reactions. What is an addition reaction? Addition reactions involve adding atoms onto the carbon chain, at the position of the double bond. For example, bromine, Br2, adds onto the carbon chain as 2 bromine atoms, 1 onto each of the 2 carbons previously held together by the double bond. 2-pentene reacts with bromine to form 2, 3-dibromopentane. 8

9.

Ethylene undergoes reaction with halogens without need for a catalyst. Give an equation for the reaction of ethene with bromine (Br2). Why is this described as an addition reaction? CH2=CH2 (g) + Br2 (aq)  CH2BrCH2Br (l) It is an addition reaction because the bromine has been added into the molecule, a double bond has been converted to a single bond and no other product is formed.

10.

Alkanes can undergo substitution reactions in the presence of UV light. Write a balanced equation for the substitution of ethane with bromine (Br2). UV

C2H6 (g) + Br2 (aq)

11.

 C2H5Br (l) + HBr (g)

Explain why HBr (g) is always formed when alkanes undergo substitution reactions with bromine. Substitution reactions always involve the removal of a hydrogen atom by the bromine molecule to form hydrogen bromide gas. A range of other products may be formed. Explain why, using as an example the reaction between ethane and bromine in the presence of UV light. Ethane has 6 hydrogen atoms that can react with the bromine molecule. If only 1 hydrogen is removed, then only 1 product, bromoethane (apart from the hydrogen bromide gas) could form. However, if 2 molecules of bromine react with each molecule of ethane, then 2 different bromoalkanes could be formed (1,1-dibromoethane and 1,2-dibromoethane). 2 molecules of HBr (g) would also form. With longer chain alkanes, the number of different bromoalkanes increases significantly. By-products like hydrogen (H2) and a 4-carbon alkane could also form in small quantities.

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12.

Explain why the UV light cannot be described as a catalyst. Use an energy diagram to clarify the role of a catalyst versus the role of UV light. The diagram shows the energy changes during an endothermic reaction. Reactions which involve UV light are endothermic, with the UV light providing the activation energy for the reaction.

The UV light provides the activation energy shown as Y on the diagram above. Catalysts allow a reaction to proceed by a different pathway. In this example, a catalyst would speed up the reaction because the value of the activation energy would be smaller than Y (the activation energy is lowered). UV light is NOT a catalyst; it is the energy needed to get the reaction to its highest energy state but it is not a chemical substance added to make the activation energy smaller. 13.

Describe an experiment to show how you distinguished between alkanes and the corresponding alkenes in a school laboratory. Use either hexane and 1-hexene or cyclohexane and cyclohexane. Equal volumes (1 mL) of the hexane (alkane) and 1-hexene (alkene) were placed in 2 identical test tubes placed in a rack in a fume cupboard. 10 drops of bromine solution were added to each test tube simultaneously. The test tubes were shaken to observe the different rates of decolorisation of the reddish bromine solution. The test tube containing 1-hexene decolorised rapidly (10 minutes.

14.

What safety precautions did you need to take in carrying out this experiment? The reaction needs to be done in a fume cupboard as both bromine and the hydrocarbons are volatile. The vapours, if inhaled may cause respiratory problems and, in the case of bromine, burning of the respiratory tissues, as it forms an acid in water. Hydrocarbons are flammable, so the reaction should not be performed near a naked flame. Safety glasses, gloves and lab coats should be worn. 10

15.

Explain why alkanes and alkenes have similar physical properties yet different chemical properties. Give examples of the similar physical properties and of different chemical properties. Alkanes and alkenes are both hydrocarbons and both non-polar molecules. They have similar and weak intermolecular forces (dispersion forces). Alkanes and alkenes of similar mass and chain length have similar dispersion forces and these determine the physical properties such as melting point, boiling point, density. Alkanes and the corresponding alkenes have different chemical properties because of the difference in covalent bonding. Alkenes, because of the double C=C bond, react with water in the presence of conc. sulfuric acid catalyst, to form alkanols, whereas alkanes do not react with water.

16.

Would you expect hydrocarbons such as ethylene and hexane to dissolve in water? Why/why not? No, they are non-polar molecules and have little force of attraction for water molecules, which are held strongly to their neighbours by H-bonding. As a result, the hydrocarbons cannot get between the water molecules; i.e. they don’t dissolve in water.

17.

List the following compounds in increasing order of their boiling points (then check your answers from the SI Data book): methane, pentane, 1-butene, decane, propene. What determines the boiling points of these compounds? methane (-160°C), propene (-47°C), 1-butene (-6.2°C), pentane (37.8°C), decane (174°C). The boiling points are determined by the strength of the dispersion forces and the weights of the molecules. The greater the length of the chain and the greater the weight, the higher the boiling point.

18.

Ethylene is a common monomer used in polymerisation. Define the terms monomer and polymerisation. The monomer is the basic unit of the polymer. Monomers join together to form a chain of repeating units, the polymer. Polymerisation is the process which converts the monomers into the chain.

11

19.

Write an equation for the reaction of ethylene to form polyethylene.

20.

Polyethylene is an “addition polymer”. What is an addition polymer? Addition polymers are formed by “addition reactions”. The double bonds are converted to single bonds and the 2 electrons coming from the double bond join each ethylene unit onto the next.

21.

Outline the steps in the production of low density polyethylene (LDPE) from ethylene. The steps involve:  the use of an initiation reaction, which pulls electrons out of the double bond and creates a free radical (an atom or group of atoms with an unpaired electron). 

a propagation step, which adds extra 2-carbon units onto the chain.

See the 2 steps in the equations in 19 (above). R is the initiator. The polymer is LDPE. The density is low because this mechanism, controlled by this initiator, allows branching of the chain. Branched chains ensure that the chains cannot pack tightly together and hence the polymer has low density.

22.

HDPE (high density polyethylene) has different properties to LDPE and is made by a different industrial process. Explain the different properties in terms of the different structures and explain how the method of production determines the type of polyethylene formed. HDPE uses a different catalyst which ensures that the chains do not branch during the polymerisation process. Long straight chains can get close to each other and can form stronger dispersion forces than branched structures. As a result, HDPE has higher density, is less flexible and the materials made from it are more rigid (e.g. lunch boxes) than materials made from LDPE (e.g. food wrap).

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23.

Vinyl chloride and styrene are commercially significant monomers. Complete the table below.

Monomer

Systematic Name

Vinyl chloride

chloroethene

Styrene

ethenylbenzene or phenylethene

24.

Structural Formula

Name of Polymer produced PVC Polyvinyl chloride

Uses of Polymer (at least 3 for each polymer) Sewerage and drainage pipes Electrical insulation Appliance leads

polystyrene

Foam drink cups, foam packing, tool handles.

Write equations (using structural formulae) for each of the polymerisation reactions above, showing 3 repeating units in the polymer formed.

13

Other Questions relating to Section 1 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1.

Petroleum or crude oil (A) is derived from coal (B) is made in oil refineries (C) is a natural product derived from fossilised marine organisms (D) contains large quantities of ethylene

2.

There is a need to increase the proportion of the most valuable fractions of crude oil. To achieve this, oil refineries (A) join large molecules together in a process called catalytic cracking (B) break down larger molecules into smaller parts by a process called catalytic cracking (C) heat the larger molecules quickly with steam and then cool them quickly to break the bonds (D) take a large molecule like ethylene and break it up into smaller parts like octane

3.

Which of the following substances is a catalyst used in catalytic cracking? (A) iron (B) platinum (C) zeolite (D) hydrogen

4.

Pentene undergoes cracking to produce two products. Possible products could be (A) pentane and hydrogen (B) ethane and propane (C) ethylene and propane (D) propene and ethylene

5.

The catalytic cracking of alkanes in the petroleum industry is an example of “heterogeneous catalysis”. This process involves the (A) catalyst supplying a large solid surface area on which the gaseous reaction proceeds (B) reactants being a mixture of solids and liquids, hence the term “heterogeneous” (C) reactants and the catalyst being in the same state (D) catalyst being a liquid while the reactants and products are gases

6.

The alkanes (methane, ethane, propane and butane) and the alkenes (ethylene, propene and butene) are all gases at room temperature (25oC). These properties indicate (A) alkanes and alkenes only have weak dispersion forces between the molecules (B) the molecules of alkanes and alkenes are attracted to each other by both dipole-dipole and dispersion forces (C) “hydrogen bonding” is present in each molecule because of the large numbers of hydrogen atoms (D) the presence of both hydrogen bonding and dispersion forces 14

7.

Which of the following organic molecules would be the most reactive? (A) 2-methylpropane (B) 2,3-dimethylpentane (C) methane (D) 2-butene

8.

When propene reacts with bromine, the product would be (A) (B) (C) (D)

1,1- dibromopropane 1,2- dibromopropane 2,2-dibromopropane 1, 3-dibromopropane

9.

When propene reacts with water in the presence of H2SO4 catalyst, the product formed is (A) 2-propanol (B) polypropylene (C) butanol (D) propane

10.

Addition polymerisation is (A) a chemical reaction that produces monomers (B) a chemical reaction that produces long chains with single carbon-carbon bonds (C) a catalytic process that always involves an acid (D) where polymers are joined to produce long chain monomers

11.

Polyethylene is a polymer formed when (A) the ethylene molecules join without loss of atoms (B) the ethylene chain adds methyl groups onto the side during polymerisation (C) addition of a large number of polyethylene units in a chain structure occurs (D) the single bond in ethylene opens up and adds the next molecule to the chain

12.

A property common to almost all polymers is that they (A) are all low density (B) have a natural rigidity (C) are good thermal and electrical insulators (D) are chemically active

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16

Section 2 BIOMASS AS A SOURCE OF MATERIALS AND ENERGY 1.

Why do we need to discover alternatives to compounds currently obtained from petrochemicals? The supplies of petrochemicals (fossil fuels, non-renewable) are dwindling and prices are rising. Products made from petrochemicals are not biodegradable and the emission of carbon dioxide from fossil fuels impacts on the atmosphere, probably causing climate change. The impact of carbon dioxide is diminished by a change to renewable fuels and the products formed are biodegradable.

2.

Cellulose is the major component of biomass. What is biomass? Biological material from living or recently living things which can be used as a source of energy or of raw materials.

3.

Cellulose is a condensation polymer. What is a condensation polymer? A long chain molecule made by the joining of monomer units by condensation (water is produced) reactions. The -OH group from one monomer joins with an –H atom of another to form water.

4.

Glucose is the monomer for the formation of the polymer cellulose. Write an equation (using structural formulae) for this condensation polymerisation reaction. Draw 3 repeating units of the polymer chain. Equation: HO – C6H10O4 – OH + HO – C6H10O4 – OH + HO – C6H10O4 – OH  HO – C6H10O4 – O – C6H10O4 – O – C6H10O4 – OH + 2H2O 3 repeating units:





17

5.

Describe the structure of cellulose. Include in your answer:  the structure of the polymer chain  why the chains are straight (rather than coiled)  how neighbouring chains are strongly held together to form a stable structure which is difficult to break down. The polymer chain is made by the condensation reaction of β-glucose units (see question 4 above). One molecule of water is formed each time neighbouring glucose units join together. The chains are straight because the geometry of the repeating inverted rings and the C-O-C bond angles make the chain linear. Neighbouring chains are held together by the fact that they can get very close together (straight strands) AND by the fact that the neighbouring strands are attracted strongly by the hydrogen bonding between the –OH groups on glucose monomers from adjacent strands.

6.

Cellulose contains the basic carbon chain structures that are needed to build compounds that are presently obtained from petrochemicals. What does this statement mean and why is it significant? Cellulose contains repeated glucose monomers, which contain the atoms carbon, hydrogen and oxygen. Petrochemicals are currently used to make ethylene (from catalytic cracking of the long hydrocarbon chains). Ethylene (C2H4) can be converted into ethanol (C2H6O) and polymers such as polyethylene, PVC and polystyrene. These compounds all contain carbon and hydrogen; ethanol also contains oxygen. Hence theoretically cellulose could be used to make the same chemicals we currently get from crude oil (petrochemicals). Cellulose is regarded as a potential source of these materials in the future.

18

7.

What are the problems associated with using biomass as fuels? Biomass, largely cellulose, is found in the cell walls of all plants. It is the chief component of wood and as such can be used as a fuel to produce energy. However, wood is solid and the most convenient state for fuels at room temperature is liquid or (less conveniently) gas. Methane can be obtained by anaerobic decay of biomass. Since it is a gas, it is in a more convenient form for use as a transport fuel.

8.

What are the problems associated with the use of biomass as a source of other materials, including polymers? Biomass (cellulose) has such a strong structure that it is difficult to convert it back into useable 2-carbon units, which could act as the raw materials for chemical industries. Biomass is naturally broken down by aerobic bacteria but the final product is carbon dioxide, not glucose or ethanol, or other 2 carbon starting points for chemical reactions. Because there is currently no simple or efficient way of breaking cellulose into glucose, biomass is not yet being used on a large industrial scale to make other materials. If it were possible to convert cellulose into glucose on a large scale, fermentation would result in ethanol (albeit inefficiently). The ethanol could be used to make ethylene and, in turn, our current polymers. A more efficient use would be the conversion of cellulose into other polymers to replace the existing polymers based on crude oil. Biomass which contains starch can be readily broken down to form sugars, which are, in turn, used to make ethanol. Ethanol is a fuel and a source of many other chemical substances. However, all fermentation reactions are inefficient.

9.

Partially synthetic biopolymers such as rayon, cellulose acetate and CMC have been used for many years. Use ONE of these examples to describe how cellulose forms these useful biopolymers. Cellulose acetate is made from cellulose chains which have been modified by reaction of the –OH groups to attach –CH3COO, acetate groups. This polymer has been used to make OHP foils.

10.

Explain the term biopolymer and distinguish between partially synthetic biopolymers such as rayon and synthetic (industrially produced) biopolymers such as PHB. Biopolymers are polymers made totally by living things. Partially synthetic polymers are structures like cellulose acetate (above) or rayon or CMC. The cellulose strand is a natural biopolymer which has been synthetically changed by reaction of the -OH groups. Synthetic biopolymers are made from living organisms but the process is not a natural one. The process is an industrial process, using micro-organisms, which are forced to produce a polymer. 19

11.

What is the structure of PHB? How is it produced? What is it used for? Why is its production important for society and for the environment? Why is its production of limited viability in the short term? PHAs are polyhydroxyalkanoates. Certain bacteria such as Clostridium or Alcaligenes eutrophus can synthesise natural polyesters. One of these is PHB (poly-3-hydroxybutanoate). The monomer has the structure:

The polymer has the structure:

The micro-organism is fed nutrients and multiples rapidly. One of the components of the nutrient is then removed and the bacterium defends itself by producing the PHB polymer as a food storage mechanism. The bacterium is then harvested and the polymer extracted. The polymer, PHB, has similar properties to the addition polymer, polypropylene. It can be used in the production of plastic films, wrapping, bottles, disposable nappies, medical and hospital supplies. It is significant that this polymer is biodegradable. The disposal of the items produced from PHB will not cause long-term environmental problems, whereas similar products made from polypropylene will cause long-term pollution. The cost of producing biopolymers currently is prohibitive.

12.

Research another biopolymer in current production. How is it produced? Name the organism or enzyme used to synthesise the polymer. What is the biopolymer used for? Why is its production important for society and for the environment? Are there limitations to its commercial viability in the short term? No answer provided – for student research. Start with the Wikipedia website and follow the links. The range of biopolymers can be seen. Most are used for biodegradable packaging materials. Check out the water soluble products for medical uses. 20

Other Questions relating to Section 2 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1.

Scientists are looking for alternative sources of hydrocarbons because (A) new sources will be cheaper (B) new sources will not produce as much carbon dioxide gas (C) current supplies of crude oil are predicted to run out in several decades (D) supplies of cellulose are limited

2.

Two natural carbon compounds which have potential to replace crude oil as basic products to produce monomers are (A) ethanol, cellulose (B) carbon dioxide, sugar (C) butene, propanol (D) cellulose, octane

3.

Which of the following reactions explains the formation of most condensation polymers? (A) monomer  polymer + O2 (B) monomer  polymer + CO2 + H2O (C) monomer  polymer + H2O (D) monomer  polymer + CH3CH2OH

4.

Which of the following lists contains all of the elements in cellulose? (A) C, H (B) C, N, O, H (C) C, H, O (D) C, H, O, N, S, P

5.

The main agent currently used to break down cellulose into simpler compounds is (A) the acid in the stomach of mammals (B) saliva (C) bacteria (D) sunlight and water

6.

Biopolymers are polymers (A) derived from crude oil (B) produced from sugar cane by fermentation (C) made from materials produced by living things (D) made from synthetic materials

21

22

Section 3 ETHANOL AS A SOURCE OF MATERIALS AND ENERGY 1.

Draw the structural formula of ethanol.

2.

Explain why ethanol dissolves in both water and hexane, whereas hydrocarbons are not soluble in water. “Like dissolves like”. The hydrophilic –OH group is attracted to water by hydrogen bonding, so ethanol will dissolve in water. The hydrocarbon chain in ethanol is structurally like the hydrocarbon chain in hexane, so ethanol will mix with the hexane and layers do not form. Hydrocarbons are non-polar hydrophobic molecules, so do not mix with polar water molecules. Layers of the hydrocarbon (on top) and water form.

3.

Explain why the boiling points of alkanols are greater than the boiling points of the corresponding hydrocarbons. Alkanols all contain the –OH group, which forms a strong hydrogen bond to a neighbouring molecule. Hydrocarbons only possess weak dispersion forces, so neighbouring molecules are not strongly attracted and separate easily. As a result, the alkanols boil at a higher temperature.

4.

Ethanol can be dehydrated to ethylene in the presence of a catalyst. (a) Name the catalyst. (b) Why is a catalyst necessary? (c) Give the equation for the reaction. (a) Concentrated sulfuric acid. (b) A catalyst is used to speed up the reaction by providing a lower activation energy pathway. Sulfuric acid is a good dehydrating agent (it pulls the –H and –OH from the molecule of ethanol to form water). (c) conc. H2SO4 C2H6O (g) C2H4 (g) + H2O (g)

5.

Ethanol can be formed from ethylene. Explain why this reaction is described as “hydration”. Write the equation for the reaction, showing the catalyst used. The reaction is called “hydration” as ethylene chemically reacts with water to form the ethanol. conc. H2SO4

C2H4 (g) + H2O (g) 6.

C2H6O (g)

Explain why ethanol can be used as a fuel. It burns with oxygen to release energy. 23

7.

Explain why ethanol can be considered both as a renewable resource and as a chemical produced from petrochemicals. Ethanol produced from fermentation is said to be renewable as it is produced from carbohydrate-rich crops such as sugarcane and thus does not depend on a supply of crude oil. The carbohydrate-rich crops are grown in order to produce the energy in the short term. The process uses energy taken in during photosynthesis. However, over past decades (and still to a large extent in many countries) the source of ethanol, for uses other than for human consumption, has been mainly the hydration of ethylene. The ethylene was manufactured from petroleum, using the processes of fractional distillation and catalytic cracking. The chemical energy was originally derived from photosynthesis - millions of years ago.

8.

Write the equation for the production of ethanol from glucose. How can the progress of fermentation be monitored in a school laboratory? Yeast

C6H12O6 (aq)

2C2H6O (aq) + 2CO2 (g)

The progress is monitored (to some extent) by measuring the mass changes during the reaction. As the reaction proceeds, the mass of the reaction flask falls because carbon dioxide leaves. Alternatively, if the carbon dioxide is bubbled into limewater, the mass of the limewater flask increases as the carbon dioxide reacts with the limewater to produce calcium carbonate. (The experiment is invalid as:  air can enter the flask to replace carbon dioxide  removing the carbon dioxide as it forms, forces the equilibrium to the right and hence changes the yield of carbon dioxide.)

9.

Describe the conditions under which this fermentation reaction is carried out. Fermentation is carried out in anaerobic conditions (otherwise acetic acid forms), at a relatively low temperature for an industrial process (between 25°C and 37°C – otherwise the enzyme is destroyed). The solution of glucose must be very dilute (as concentrated solutions produce concentrated ethanol, which kills the yeast enzyme and stops the reaction).

24

10.

Discuss the problems/limitations associated with the use of ethanol as an alternate to hydrocarbon fuels. The production of ethanol by fermentation is an energy inefficient process. The dilute solution of ethanol formed must be fractionally distilled to remove the water. The production of sugar cane requires large land areas and is only possible in limited climatic regions. There are many wastes to be removed and the agricultural and processing steps of ploughing, planting, cutting, transporting and crushing the sugar cane use energy. Engines must be modified to burn ethanol if it is used pure. However, ethanol can be blended with petrol (up to 10%) without engine modification.

11.

Increasingly, ethanol is blended with petrol for use as a transport fuel. Assess the benefits associated with the use of ethanol in this way. The blending of ethanol with petrol is extending the supplies of petrol from crude oil and hence deferring the time when crude oil supplies will be unavailable or prohibitively expensive. However, ethanol has a lower calorific value (energy per gram) than petrol, so it is less energy efficient. Ethanol burns more cleanly, so less soot and carbon monoxide are formed. It burns at a lower temperature, so engines will not wear out as rapidly and less pollution from nitrogen and oxygen combining to form nitrogen oxides will form. Australia’s sugarcane industry had become unprofitable prior to the legislation introduced to mix ethanol with petrol, so at political, economic and social levels there have been benefits to Australians by the process of blending ethanol with petrol. Because ethanol is a renewable fuel, the nett impact on the climate of the production of carbon dioxide is less. The “greenhouse effect” is smaller, as only slightly more carbon dioxide is returned to the atmosphere than is taken out by photosynthesis. Note: Since this question involves assessment, an overall assessment statement must be included. e.g. The positive effects of using ethanol blended with petrol as a transport fuel (the renewable nature of the fuel, the reduced nett greenhouse emissions on the current atmosphere, the political, economic and social benefits in today’s society, the reduced pollution from nitrogen oxides, the lower operating temperature of engines) outweigh the negative effects of inefficiency – at least in the short term.

12.

What is meant by “the heat of combustion of ethanol”? The energy given out when one mole of ethanol undergoes complete combustion to form carbon dioxide and water, at standard conditions.

25

13.

The heat of combustion of ethanol is 1350 kJ/mole, while the enthalpy change for the combustion of ethanol is quoted as -1350 kJ/mole. Explain this difference. The heat of combustion is defined as the absolute value of the amount of heat released and hence is quoted as a positive value. The enthalpy change = H = chemical energy of products – chemical energy of the reactants. As this measure is a negative quantity (the products store less chemical energy than the reactants), the overall change is negative. It is an exothermic reaction.

14.

Explain why the heat of combustion of alkanols increases as the length of the carbon chain increases. As the length of the carbon chain of alkanols increases, there are more carbons and hydrogens present. As a result, when these carbon and hydrogen atoms recombine with oxygen atoms to form carbon dioxide and water, more bonds are formed. Note: It is the bond forming steps (not the bond breaking steps) which must be stressed in this answer.

15.

List the steps involved and the measurements taken in the determination of the heat of combustion of ethanol in the laboratory. Give a diagram of the apparatus used.

Steps in the investigation:  Ethanol was placed in spirit burner  Known mass of water was placed in beaker  Initial temperature of water recorded  Initial mass of spirit burner and ethanol recorded  Spirit burner lit  Water stirred carefully with thermometer  Spirit burner extinguished  Mass of spirit burner recorded  Highest temperature of water recorded  Changes in temperature and in mass of burner calculated  Heat of combustion determined The measurements made are:  The change in the temperature of the water  The mass of water heated  The change in mass of the spirit burner 26

16.

The calculation, in Question 15 above, is based on an incorrect assumption. As a result, the experiment cannot be regarded as valid, even though the method used may be reliable and measurements may be accurate. Assess these statements with regard to this experiment. In your answer ensure that you clarify the meanings of validity, reliability and accuracy. Also ensure that you discuss the reasons why this experiment does not give valid results and ways the validity may be improved. The experiment is invalid, as the method used cannot achieve the desired outcome – to determine the heat of combustion of the fuel. There are 2 reasons why the assumption “the heat released by the fuel = the heat absorbed by the water” is incorrect:  The heat given out by the fuel heats the surroundings, including the beaker, the stand, the air – so only a proportion is transferred to the water.  The experiment was designed to measure the heat of combustion – when complete combustion of ethanol (to form carbon dioxide and water) occurs. In this experiment, complete combustion cannot occur, as the spirit burner does not allow the maximum amount of oxygen to combine with the fuel available. It is obvious from the amount of soot which forms on the bottom of the beaker that incomplete, rather than complete, combustion has occurred. The experiment could be considered reliable i.e. reproducible (as repeating the experiment, using the same method can give the same results). It may also have been done accurately, with correct measurements using the thermometer and of the masses of water used and of the fuel burned. The experimental method cannot give valid results while heat is being lost to the surroundings. Using insulated containers, stopping air currents around the experiment, reducing the distance between the flame and the beaker, will all improve the validity, as less heat will be lost. To improve the validity with respect to the products of the combustion (to make the reaction as complete as possible), the ratio of oxygen to fuel must be increased to have excess oxygen to fuel – almost impossible with a spirit burner.

17.

A student wished to determine the heat of combustion of ethanol. He used a spirit burner containing ethanol to heat 250 g of water in a beaker. The temperature rose from 15oC to 31oC. During combustion, the burner lost 0.90 g in mass, due to the ethanol burning. Calculate the heat of combustion of ethanol, in kJ mol-1 and in kJ g-1. Heat absorbed by water = mass water x T x 4.18 Joules = 250 x 16 x 4.18 = 16,720 J = 17 kJ Assume this heat is given out by 0.90 g ethanol Heat of combustion ethanol = 17/0.9 kJ/g = 19 kJ g-1 Molar heat combustion = 19 x 46 kJ/mol = 874 kJ mol-1 Only 2 significant figures are justified, so Heat of combustion = 8.7 x 102 kJ mol-1 27

Other Questions relating to Section 3 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1.

The term dehydration of ethanol refers to the (A) addition of water (B) addition of oxygen (C) removal of water (D) removal of oxygen X

2.

CH3CH2OH



CH2 = CH2 + Y

In this reaction ethanol is converted to ethylene. Substances X and Y would be

(A) (B) (C) (D)

X concentrated HCl concentrated H2SO4 dilute H2SO4 concentrated H2SO4

Y H2O O2 H2O H2O

3.

The formula of ethanol is (A) CH3CHOH (B) CH2CH2OH (C) CH3CH2OH (D) CH3OHCH3

4.

Ethanol is totally miscible in water because (A) they both contain hydrogen atoms (B) they both have hydrogen bonding (C) they are both non-polar (D) they both have weak dispersion forces

5.

Ethanol is classified as a fuel because it (A) contains carbon (B) contains hydrogen and oxygen (C) can release energy (D) has a low BP

6.

The formula for glucose is (A) C6H10O6 (B) C6H14O4 (C) C6H12O6 (D) C4H12O6 28

7.

8.

CH3CH2OH (l) + X  2CO2 (g) + Y This is a combustion reaction. Which of the following lists correctly represents X and Y at room temperature?

(A) (B) (C) (D)

X 3O2 (g) 3H2O (l) 3O2 (g) 4O2 (g)

X



Y 3H2O (g) 3O2 (g) 3H2O (l) 5H2O (l)

ethanol + Y

During the above fermentation reaction, enzymes (A) act as catalysts and help break up the glucose molecules into smaller units (B) react with the carbon dioxide produced (C) react with limewater (D) allow the reaction to occur above 37oC 9.

A student was investigating the heat of combustion of ethanol. She used an ethanol burner that had an initial mass of 28.0 g. She then lit the burner and placed it under a beaker containing 500 mL (500 g) of water. After a few minutes she noticed that the water temperature had risen from 24oC to 38oC and the burner now weighed 26.5 g. Assuming that only the water was heated in this experiment, what would be the heat of combustion for ethanol? (A) 1360 kJ/mol (B) 2040 kJ/mol (C) 897 kJ/mol (D) 29.26 kJ/mol

10.

Ethanol is derived from plant material by the process of (A) condensation polymerisation (B) addition polymerisation (C) photosynthesis (D) fermentation

11.

The product formed when ethylene undergoes addition of water is (A) ethanoic acid (B) carbonic acid (C) ethanol (D) ethane

12.

During the fermentation of glucose the conditions must be monitored to ensure that (A) the temperature never drops below 50 °C (B) hydrogen ions are present to act as catalysts (C) an ample supply of oxygen is present (D) the solution of glucose is very dilute 29

OVERVIEW OF SECTIONS 1, 2 AND 3. Your Core 1 Chemistry syllabus (sections 1, 2 and 3) focuses on the following carbon compounds as sources of both materials and energy. Use the diagram to answer the following questions. CRUDE OIL

POLYMERS

ETHYLENE

GLUCOSE

ETHANOL

CELLULOSE 1.

Which of the above are classified as fossil fuels? Crude oil (ethylene is a petrochemical when made from crude oil – it isn’t normally used as a fuel).

2.

Which of the above compounds are classified as renewable resources? Cellulose, glucose, ethanol (when made from glucose), ethylene (when made from ethanol).

3.

Which compounds can be manufactured both from fossil fuels and from renewable resources? Ethanol, ethylene, theoretically polymers if the ethylene comes from ethanol after glucose fermentation.

4.

Why is the reversible reaction between ethylene and ethanol of such significance in the production and availability of fuels and other materials? The reversible reaction links the compounds derived normally from crude oil (ethylene and ethanol) with those classified as renewable resources. Ethanol is the pivotal point. Historically it was made from crude oil; now increasing amounts are made from fermentation of glucose, or other carbohydrate rich crop. As the supplies of crude oil dwindle, ethanol and ethylene and the current polymers could be produced from renewable resources. However, it is more advantageous to produce new polymers based on renewable resources.

30

5.

Cellulose and crude oil are both classified as natural resources. Both can be considered as fuels and as sources of other chemicals. Analyse these statements, taking into consideration current resources, technologies and potential for the future. “Cellulose and crude oil are both classified as natural resources.” Cellulose is the structural carbohydrate of all plant cells (found in the cell wall). Hence it is a natural resource. Crude oil is a fossil fuel derived from the remains of marine organisms which lived millions of years ago. Since it is derived from once living organisms, it is a natural resource. “Both can be considered as fuels.” Cellulose is found in wood, which is burnt to provide energy. The various fractions of crude oil, such as petrol, diesel, kerosene are used as transport fuels, to provide energy. Hence cellulose and crude oil can both be classified as fuels. “Both can be considered as sources of other chemicals.” Cellulose has been converted into materials such as rayon, CMC, cellulose nitrate, cellulose acetate for use as fabrics, food thickener, film and OHP foils respectively. Crude oil can be converted into ethylene, ethanol, polymers and a wide range of other chemicals. Hence both cellulose and crude oil are sources of other chemicals. Hence, the statements are correct at the present time. However, the current supplies of crude oil are dwindling and increasing in cost. Political and scientific impetus is increasing the significance of renewable resources and this trend is likely to increase in the future. However, there are currently limitations to the use of cellulose as a fuel and as a source of polymers to replace the polymers made from crude oil. Current technologies do not allow the large scale conversion of cellulose into glucose or ethanol or any other 2-carbon compound which can be converted into the wide range of materials now produced from crude oil. There is enormous potential for these technologies to change in the future, as cellulose is available as a renewable resource from every plant, and it possesses a structure based on carbon, hydrogen and oxygen atoms. This means that industry should be able to develop large scale conversion of cellulose into 6-carbon or 2-carbon units in the future and the current materials used as fuels and sources of other chemicals should be able to be derived from a readily available renewable source.

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Multiple Choice Questions relating to Sections 9.2.1, 9.2.2, 9.2.3 (similar to past HSC questions) 1.

Which equation represents cracking of a petroleum fraction? (A) (B) (C) (D)

C14H30 (l)  C14H30 (g) C7H16 (l) + 3C2H4 (g) + C3H6 (g)  C16H34 (l) C8H18 (l)  2 C4H9 (l) C16H34 (l)  C7H16 (l) + 3C2H4 (g) + C3H6 (g)

2.

Which are the best conditions for fermentation of sugars by yeast? (A) Low oxygen concentration, dilute solutions and a temperature between 25°C and 35°C (B) High oxygen concentration, dilute solutions and a temperature between 25°C and 35°C (C) Low oxygen concentration, concentrated solutions and a temperature above 45°C (D) High oxygen concentration, dilute solutions and a temperature between 45°C and 60°C

3.

The heat of combustion of 1-butanol is 2676 kJ mol-1. The value in kJ g-1 is (A) 30.4 (B) 36.2 (C) 44.6 (D) 47.8

4.

Research into synthetic biopolymers is important for society because synthetic biopolymers (A) decompose more easily than traditional synthetic polymers. (B) can be produced more cheaply than polymers based on crude oil. (C) have superior physical properties compared to polymers based on crude oil. (D) have superior chemical properties compared to polymers based on crude oil.

5.

Cellulose consists of long straight chains, formed by condensation of glucose molecules. Cellulose is best represented: (B)

32

Short Response Questions relating to Sections 9.2.1, 9.2.2, 9.2.3 (similar to past HSC questions) 1.

(a)

What type of polymerisation is shown in the following reaction?

(1 mark)

Condensation (b)

Assess the impact of the use of biopolymers on society and the environment. Include an example of a named biopolymer, describe how it is formed and how it is used. (5 marks) The use of biopolymers will, in time, have a significant impact on society and on the environment. However, at the present time, the supplies of ethylene (derived from crude oil) are still adequate to produce polymers such as polyethylene, PVC and polystyrene and society has become dependant on these materials derived from crude oil. However, in the future, when supplies of crude oil (being a fossil fuel) are no longer available or are prohibitively expensive, society will need to invest in the development, on a large scale, of synthetic biopolymers to replace the products from crude oil. As a bonus, biopolymers are biodegradable, so the detrimental environmental impacts of the UV- and bacteria-resistant polymers, based on crude oil, will be avoided. Cellulose is a natural biopolymer. At the present time, it has great potential as a source of other chemicals because it contains the correct atoms (carbon, hydrogen and oxygen). However, it is such a stable structure that it is difficult to break into useful smaller units such as glucose, ethanol or ethylene. It is converted by bacterial action to carbon dioxide. Experimentation is happening on a small scale to convert cellulose to glucose and hence to ethanol. When this bacterial hydrolysis can be achieved efficiently, cellulose will have an enormous impact on society and the environment. PHB, polyhydroxybutanoate, is a synthetic biopolymer manufactured by a bacterial process using Clostridium or Alcaligenes eutrophus which can synthesise natural polyesters. The micro-organism is fed nutrients and multiples rapidly. One of the components of the nutrient is then removed and the bacterium defends itself by producing the PHB polymer as a food storage mechanism. The bacterium is then harvested and the polymer extracted. The polymer, PHB, has similar properties to the addition polymer, polypropylene. It can be used in the production of plastic films, wrapping, bottles, disposable nappies, medical and hospital supplies. It is significant that this polymer is biodegradable. The disposal of the items produced from PHB will not cause long-term environmental problems. 33

2.

You performed a first-hand investigation that monitored mass changes during the fermentation of glucose to ethanol. (a) Outline the procedure you used. (2 marks) The fermentation flask was set up containing a dilute solution of glucose, yeast and yeast nutrient. It was fitted with a cork and delivery tube, which was placed in a conical flask of limewater. The progress is monitored (to some extent) by measuring the mass changes during the reaction. As the reaction progressed, the mass of the reaction flask was measured daily. (Alternatively, the mass of the limewater flask was measured daily.) The loss in mass of the fermentation flask or the gain in mass of the limewater flask was recorded and assumed to be due to the production of carbon dioxide, which left the fermentation flask and reacted with the limewater to form white calcium carbonate. The moles of carbon dioxide formed and hence the moles of glucose reacting were calculated. (b)

Write a balanced equation for this reaction.

(1 mark)

Yeast

C6H12O6 (aq) 3.

2C2H6O (aq) + 2CO2 (g)

The flowchart shows steps in the production of polyethylene.

(a)

Name compound X.

Ethanol

(1 mark)

(b)

The following graph shows the distribution of molecular weights of polymer molecules in the sample.

Explain the shape of the graph. (2 marks) The molecular weights of the polymer molecules vary according to the number of monomer units in the chain. The majority of molecules have short chain length, but a minority have long chain length. Why do the molecular weights of the polymer molecules vary? (1 mark) Because they have different chain lengths, with different numbers of monomer units.

34

4.

Consider the graphs shown below.

(a)

Describe the trends or patterns shown in the graph. (2 marks) The boiling points of alkanes, alkanols and alkanoic acids all increase as the molecular weight increases. The boiling points of alkanoic acids are higher than the boiling points of alkanols and are both greater than the boiling points of alkanes of the same molecular weight.

(b)

Explain these trends in terms of the structures of the compounds involved. (3 marks) Alkanoic acids are able to form, on average, 2 hydrogen bonds per pair of molecules while alkanols can only form 1 hydrogen bond per pair of molecules. (The geometry of the molecules allows stronger hydrogen bonding in alkanoic acids than in alkanols.) Alkanes are non-polar molecules, with only weak dispersion forces, so have lower boiling points than either alkanols or alkanoic acids.

5.

The heat of combustion of ethanol is 1367 kJ mol–1. A student carried out an investigation to determine the heat of combustion of ethanol. The experimental value he determined differed from the theoretical value. (a)

Identify ONE reason for this difference in values.

(1 mark)

The method is invalid. Combustion is assumed to be complete and heat is assumed to be completely transferred to the water. (b)

Calculate the theoretical mass of ethanol required to heat 50 mL of water from 21.0°C to 30.0°C. (2 marks) H = mwater x 4.18 x T = 50 x 4.18 x 9.0 = 1881 Joules = 1.88 kJ Theoretically, 46 g (1 mole) ethanol releases 1367 kJ Hence 1.88 kJ would be released from 1.88/1367 x 46 g ethanol = 0.063g (2 s.f.)

35

6.

The structures of two commercially significant monomers are shown.

(a)

Identify the preferred (common) names of both monomers. vinyl chloride and styrene

(2 marks)

(b)

Discuss how the uses of polymers depend on their properties, with reference to ONE of the above monomers. (2 marks) PVC (polyvinyl chloride) is used for plumbing pipes. Because PVC is a slightly flexible solid, which is insoluble in water and is very unreactive, it is suitable for use in drainage and sewerage pipes.

(c)

Draw the structure of ONE of these polymers showing three repeating units. (1 mark)

Note: The answer only requires the structure of the final polymer, PVC.

7.

A condensation polymer can be represented as follows:

Draw the structure of 2 monomers which combine to form this polymer.

(2 marks)

36

8.

Consider the table below:

(a) (b) (c) (d)

Which of the fuels is most efficient as a source of energy per gram? (1 mark) methane Which of the fuels is/are liquids at room temperature and pressure? (1 mark) octane, ethanol Which of the fuels is/are classified as “renewable”? (1 mark) ethanol (when it is made from glucose) Assess the potential of ethanol as an alternate fuel, making use of the data from the table. (4 marks)

Ethanol is already being used as a petrol extender, by mixing ethanol (10%) and petrol (90%). In the future, ethanol may be used as a liquid transport fuel to replace petrol. However, engines will need to be redeveloped to use pure ethanol, as the fuel:oxygen ratio for ethanol differs from the fuel: oxygen ratio for petrol. The potential of ethanol is limited by the inefficiency of the fermentation process. Fermentation requires anaerobic conditions, dilute aqueous solutions of the sugar being fermented and temperatures not far above room temperature (low for an industrial process). Fermentation is catalysed by a living organism, yeast. As a result, the temperature must be kept below 37°C and the alcohol must be dilute, otherwise the yeast will be destroyed. Because ethanol can be produced from both ethylene (by hydration) and by fermentation, it has flexibility as a fuel source. It could theoretically be derived from whichever source is economically viable at any given time. Ethanol vaporises at a lower temperature than petrol. As a result engines can operate at a lower temperature and there is less pollution from oxides of nitrogen and less wear and tear on the engines than for petrol. However, ethanol does not produce as much energy per gram as petrol (because it contains oxygen) – so it is inefficient when compared with the same mass of petrol. Overall assessment: Overriding all else: ethanol can be produced from biomass; it is classified as renewable. Since the supplies of crude oil are limited and fossil fuels have greater impact on climate change (since they release carbon dioxide into the atmosphere which was taken out by photosynthesis millions of years ago), ethanol has viability as a replacement or supplement to petrol now and has enormous potential in the coming decade(s) until a new source of transport fuel, perhaps based on cellulose or hydrogen or batteries, is readily available and economically viable.

37

9.

You have been asked to carry out an experiment to compare the heats of combustion of ethanol and 1-butanol. (a)

Identify the measurements you would need to make.

(3 marks)

The mass of water in the beaker The temperature change of this water The change in mass of the spirit burner (b)

Justify factors you would include in a risk assessment.

(3 marks)

Risk of fire – alcohols are flammable Risk of being burnt – beaker becomes hot, spirit burner is hot Risk of inhaling volatile organic vapours – organic vapours are irritating to some people who have respiratory problems 10.

Ethylene is used as a raw material in the production of other substances. (a)

Construct a flow chart to show the current industrial source of ethylene, the process by which ethylene is formed from that industrial source, and 4 useful chemicals formed from ethylene. (4 marks) Crude oil Catalytic cracking Ethylene

Ethanol

(b)

Polyethylene Styrene

Vinyl chloride

Write a balanced equation to describe the formation of ONE of these chemicals. (1 mark) conc. H2SO4

C2H4 (g) + H2O (g) C2H6O (g) The reaction is called “hydration” as ethylene chemically reacts with water to form the ethanol. Note: the catalyst is often referred to as “dilute” sulfuric acid. The water is a reactant; the concentrated sulfuric acid is the catalyst. Obviously, the combination of water and concentrated acid ends up with dilute acid, but it is important to differentiate the different roles. Water = reactant Concentrated sulfuric acid = catalyst 38

Section 4 OXIDATION AND REDUCTION REACTIONS AS SOURCES OF ENERGY

1.

What is meant by electrochemistry? The branch of science which associates electric current and changes in matter.

2.

Give an example of the production of electricity from chemical reactions. Any galvanic cell, e.g. a dry cell battery.

3.

What is the “activity series of metals”? The series of metals listed in order of their reactivity. e.g. potassium, sodium, magnesium, zinc, lead, copper, silver … are listed in decreasing order of activity.

4.

How is it determined? It is determined by reacting the metals with reagents such as water, dilute acid and oxygen and comparing the rates of reaction. It can also be determined using displacement reactions, using the rule that “an active metal will displace a less active metal from a solution containing the ions of the less active metal”.

5.

What are displacement reactions? Reactions of active metals with the ions of a less active metal. Zinc will displace copper from a solution of copper sulfate. Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)

6.

What happens in every oxidation-reduction reaction? A transfer of electrons.

7.

What is oxidation?

Loss of electrons

8.

What is reduction?

Gain of electrons

9.

Why do these reactions always occur together? Electrons will only move from the reducing agent if there is an oxidising agent available to pick them up.

10.

What is a reducing agent (also reductant)? A reducing agent causes something else to be reduced and is itself oxidised (loses electrons).

11.

What is an oxidising agent (also oxidant)? An oxidising agent causes something else to be oxidised and is itself reduced (gains electrons). 39

Refer to the Redox Table / Table of Standard Electrode Potentials provided at the back of this book. The format of this table is the same as that on recent HSC papers. Note that the format in text books varies from book to book. 12.

Where, in the table, are the best reducing agents? Top of table, on the RHS.

13.

Where, in the table, are the best oxidising agents?

14.

Which combination of chemical reagents listed on the table will have the greatest tendency to react together? Potassium metal, K, (top, right) with fluorine, F2, (bottom left).

15.

Which of the following gives up electrons most readily? Zn, Na, Cu, Ag, Mg, Fe.

Bottom of the table, on LHS.

Why? How do you know? Na is closest to the top, on RHS Which of the above metals is the most active? Why? Na; closest to the top, on RHS. Activity of metals is determined by the tendency to lose electrons. Which is the least active? Why? Ag; lowest, on RHS. Ag has the smallest tendency (of those listed) to lose electrons. Eoxidation = -0.80 V 16.

Which of the following gains electrons most readily? Cl2, F2, Cu2+, Na+, Zn2+, K+ Fluorine (F2) has the greatest tendency to gain electrons of all elements Ereduction = + 2.89 V Which of these species gains electrons least readily? K+ How do you know? The potassium ion is located at the top left of the redox table. It has the smallest tendency of all oxidising agents to gain electrons.

17.

When iron is placed in a copper II sulfate solution, solid copper forms on the iron. Explain what is occurring in terms of the transfer of electrons. Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s) Zinc metal is losing electrons and these are being gained by the copper ions.

18.

In general terms, which metal ions will be displaced by solid metals when they come into contact. Metals ions will be displaced by metals which are more active. Cu2+ will be displaced when they react with magnesium, and Cu (s), the metal, will form. Cu2+ will not be displaced when with silver, as silver is a less active metal than copper.

40

19.

Write balanced half-equations and full equations for the following oxidation-reduction reactions (using the redox table). If no reaction occurs, indicate this by writing “no reaction”. Magnesium + copper nitrate solution Mg (s) Mg2+ (aq) + 2e2+ Cu (aq) + 2eCu (s) Overall:

Mg (s) + Cu2+ (aq)

Mg2+ (aq) + Cu (s)

Copper + magnesium nitrate solution No reaction

Zinc + silver nitrate solution Zn (s) Zn2+ (aq) + 2e+ Ag (aq) + eAg (s) Overall:

Zn (s) + 2Ag+ (aq)

Zn2+ (aq) + 2Ag (s)

Zinc + dilute hydrochloric acid solution Zn (s) Zn2+ (aq) + 2e2H+ (aq) + 2eH2 (g) Overall:

Zn (s) + 2H+ (aq)

Zn2+ (aq) + H2 (g)

Sodium + chlorine Na (s) Na+ + eCl2 (g) + 2e2ClOverall:

20.

2Na (s) + Cl2 (g)

2NaCl (s) Solid would form if the elements were heated.

Write half-equations for the oxidation and reduction reactions which combine to give the following overall equations. 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) 5Fe3+ (aq) + Mn2+ (aq) + 4H2O (l) 5Fe2+ (aq) 5Fe3+ (aq) + 5eMnO4- (aq) + 8H+ (aq) + 5e-

Mn2+ (aq) + 4H2O (l)

Cr2O72- (aq) + 14H+ (aq) + 6I- (aq) ⇌ 2Cr3+ (aq) + 3I2 (aq) + 7H2O (l) Cr2O72- (aq) + 14H+ (aq) + 6e6I- (aq) 3I2 (aq) + 6e-

2Cr3+ (aq) + 7H2O (l)

41

Oxidation Number is a measure of the number of electrons which have been gained or lost by an atom in a species (by comparison with the number of electrons possessed by that atom in its elemental form).  The O.N. of an atom in the element is therefore always zero.  The charge on a simple ion is equal to its oxidation number.  The total of all the oxidation numbers of the atoms making up a species equals the charge on that species, e.g. the oxidation number for:  Na+ +1 2+  Ba +2  Copper in Cu2O +1  Copper in CuO +2  Manganese in MnO4 +7  Chlorine in Cl2 0  Chlorine in Cl-1  Oxygen in O2-2 The oxidation number changes when redox reactions (transfer of electrons) occur.  Oxidation is therefore increase in oxidation number.  Reduction is decrease in oxidation number. 21.

Give the oxidation state of iron in FeO and in Fe2O3. Explain the oxidation states in terms of loss of electrons. Oxidation state (number) of Fe in FeO = +2 (the charge on the ion = the number of electrons it has lost when it was formed from the element, Fe) Oxidation state (number) of Fe in Fe2O3 = +3 (the charge on the ion = the number of electrons it has lost when it was formed from the element, Fe)

22.

Give the oxidation state of the metal atoms in each of the following: MnO4-, CuSO4, MnCl2 and Cr2O72-. MnO4-

+7

CuSO4

+2

MnCl2

+2

Cr2O72-

+6

42

23.

Give the oxidation state (number) for each of the atoms involved in the following equations 2Na + Cl2 2Na+ + 2Cl0

0

+1

2K (s) + H2O (l) 0

24.

H = +1 O = -2

-1 2K+ (aq) + +1

2OH- (aq) + O= -2

H = +1

H2 (g) 0

Draw a diagram for the galvanic cell (known as the Daniell Cell) formed using Zn and Cu electrodes.  Label the anode, cathode, direction of electron flow, direction of ion movement in the salt bridge.  Write equations for the half-cell reactions and for the overall reaction.  Predict the voltage of this cell under standard conditions.  What changes would you expect to see after a galvanic cell had operated for a few minutes? The Daniell Cell (named after J. Daniell, 1836) is set up as shown in the diagram.

The anode half-reaction is: Zn (s) Zn2+ (aq) + 2e-

+0.76 V

The cathode half-reaction is: Cu 2+ (aq) + 2eCu (s)

+0.34 V

The overall equation is: Zn (s) + Cu 2+ (aq)

Zn2+ (aq) + Cu (s)

E = + 1.10 V

The solution in the cathode half-cell would become lighter blue. The zinc electrode would get smaller. The copper electrode would get bigger.

43

25.

Draw a diagram for the galvanic cell formed using Ag and Cu electrodes.  Label the anode, cathode, direction of electron flow, direction of ion movement in the salt bridge.  Write equations for the half cell reactions and for the overall reaction.  Predict the voltage of this cell under standard conditions.  What changes would you expect to see after a galvanic cell had operated for a few minutes?

electrons NO3

-

K+

copper anode

silver cathode KNO3 (aq)

Cu(NO3)2 (aq) AgNO3(aq) electrolytes

The anode half-reaction is: Cu (s) Cu2+ (aq) + 2e-

-0.34 V

The cathode half-reaction is: Ag + (aq) + eAg (s)

+0.80 V

The overall equation is: Cu (s) + 2Ag+ (aq)

Cu2+ (aq) + 2Ag (s)

E = + 0.46 V

The solution in the anode half-cell would become darker blue. The copper electrode would get smaller. The silver electrode would get bigger. 26.

What is the purpose of the salt bridge in a galvanic cell? To complete the circuit, allowing the flow of ions between the half-cells, so that there is no build-up of charge in either half-cell. A build-up of charge would impede or stop the flow of electrons.

27.

In which direction would electrons and ions flow in all galvanic cells? Electrons flow through the external wire from the anode to the cathode. Negative ions flow through the salt bridge from the cathode half-cell towards the anode halfcell. Positive ions flow through the salt bridge from the anode half-cell towards the cathode half-cell. 44

28.

A galvanic cell is constructed as follows: Br2(l)/Br- //Cl2(g)/Cl-. Draw a diagram of this setup and label  the anode  the cathode  direction of electron flow  direction of ion flow in the salt bridge  suitable electrodes Use the table of standard potentials to determine the voltage of this cell under standard conditions. electrons -

NO3

K+

inert (platinum) anode

inert (platinum) cathode KNO3 (aq)

bromine Br2 (l)

chlorine Cl2 (g) KBr (aq)

Br2 (l)e

electrolytes

The anode half-reaction is: 2Br- (aq) Br2 (l) + 2e-

-1.08 V

The cathode half-reaction is: Cl2 (g) + 2e2Cl- (aq)

+1.36 V

The overall voltage of the cell is 29.

KCl (aq)

Eoverall = + 0.28 V

Use the table of standard potentials to determine the standard cell potential for each of the following galvanic cells: (Necessary to use the SI Chemical Data Book)  Co  Co2+  Ag+  Ag Eoverall = 0.28 V + 0.80 V = 1.08 V How will a decrease in Co2+ concentration affect the cell potential? A decrease in the concentration of Co2+, according to Le Chatelier’s Principle, favours the oxidation of cobalt to cobalt ions. As a result, the forward reaction is favoured and the voltage will increase.  Pt, H2  OH-  Cu2+  Cu Eoverall = +0.83 V + 0.34 V = 1.17 V How will a decrease in Cu2+ affect the cell potential? A decrease in the concentration of Cu2+, according to Le Chatelier’s Principle, favours the reverse reaction. As a result, the voltage will decrease. 45

30.

The following chemicals are mixed. Use the table of standard potentials to determine whether a spontaneous reaction will occur. If a spontaneous reaction occurs, determine the overall cell potential.  Silver metal is added to a cobalt nitrate solution. No spontaneous reaction. (Eoverall is negative)  Iron (II) sulphate solution is added to bromine water. 2+ Fe (aq) Fe3+ (aq) + e-0.77 V Br2 (aq) + 2e-

2Br- (aq)

+1.10 V

Eoverall = +0.33 V  Fluorine is bubbled through water. F2 (aq) + 2e2F- (aq) 2H2O (l)

+2.89 V

O2 (g) + 4H+ (aq) + 4e-

-1.23 V

Eoverall = +1.66 V

31.

Research the properties of the following cells and complete the table below. (Note that the HSC syllabus requires students to study 1 of dry cell and lead acid and 1 only of the other cells. Button cells fulfil the syllabus requirements and are easy to understand.)

CELL TYPES DRY CELL

CHEMISTRY

Anode:

COST AND PRACTICALITY

Inexpensive

Zn (s)  Zn2+ + 2e-

Cathode: NH4+ + MnO2 + H2O  Mn(OH)2 + NH3

Electrolyte: NH4Cl, ZnCl2 paste

Best for infrequent use, as voltage drops as used

IMPACT ON SOCIETY

ENVIRONMENT IMPACT

Widely used for low energy, low cost, portable applications

No negative impact

Used as car battery

Sulfuric acid is corrosive, so care must be taken with disposal. Lead compounds have environmental and health risks.

Maximum voltage = 1.5 V LEAD-ACID BATTERY

Anode: Pb (s) + SO42- (aq)  PbSO4 (s) + 2e-

Cathode: PbO2 (s) + 4H+ + SO42- + 2e PbSO4 (s) + 2H2O

Electrolyte: Conc H2SO4

Expensive but long lasting, as rechargeable. Heavy, so adds to weight of vehicle and running costs. 6 cells in series Max each cell = 2V

Extensive application

46

CELL TYPES BUTTON CELL

CHEMISTRY

Anode: Zn (s) + 2OH-  ZnO (s) + H2O (l) + 2e-

Cathode: Ag2O (s) + H2O + 2e-  2Ag (s) + 2OH-

Electrolyte: KOH paste

32.

COST AND PRACTICALITY

IMPACT ON SOCIETY

ENVIRONMENT IMPACT

Small. Stable voltage. More expensive than dry cell, but much longer lasting. Delivers 1.6 V

Useful when small batteries needed. Watches, camera, hearing aids.

No environmental or disposal problems

Draw labelled diagrams of: (a)

a dry cell OR a lead-acid battery

(b)

a button cell

47

Other Questions relating to Section 4 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1.

Zn(s) + Cu2+(aq)



Cu(s) + Zn2+(aq)

Metal

Dark blue solution

Light blue solution

In the above displacement reaction, electrons are transferred (A) from Cu2+ to Zn (B) from Zn to Cu2+ (C) from Zn to Zn2+ (D) to the water molecules 2.

The light blue solution in Question 1 above will contain the following dissolved species (A) Zn (s), Cu2+ and Zn2+ (B) Zn (s), Cu2+, Zn2+ and Cu (s) (C) Zn (s) and Cu (s) (D) Cu2+ and Zn2+

3.

Metal Activity Series K Na Li Ba Ca Mg Al Zn Fe Sn Pb X Y Pt Au Cu (s) + 2AgNO3 (aq)  Cu(NO3)2 (aq) + 2Ag (s) This displacement equation could be used to identify substances X and Y in the activity series shown above. Which of the following would be the most correct?

(A) (B) (C) (D)

X

Y

Cl Ag Cu Cu

Ag Cu Cl Ag

48

4.

Sn4+ + 2e-  Sn2+ In this equation, the tin IV ion (A) has been reduced (B) has been oxidized (C) changed to an oxidation number of 4 (D) was oxidized by the electrons

For questions 5 to 8: V Zn

Cu Salt bridge

Zn2+ Beaker A

Cu2+ Beaker B

5.

The most active metal in the above galvanic cell is (A) Zn (B) Zn2+ (C) Cu (D) Cu2+

6.

In the above cell, the oxidant or oxidising agent is (A) Zn (B) Zn2+ (C) Cu (D) Cu2+

7.

When the above cell is operating electrons will move through the (A) solution from Zn to Cu (B) solution from Cu to Zn (C) wire from Zn to Cu (D) wire from Cu to Zn

8.

As the above cell continues to operate the (A) concentration of positive ions in beaker A will decrease (B) concentration of positive ions in beaker B will decrease and negative ions will also decrease (C) salt bridge will transfer Cu2+ ions from beaker B to beaker A (D) overall concentration of positive and negative ions remains unchanged 49

For questions 9 to 11: V Cu

Ag Salt bridge

Beaker A

Beaker B

9.

An appropriate electrolyte for beaker A would be (A) Cu(NO3)2 (aq) (B) AgNO3 (aq) (C) Cu (s) (D) CuO (s)

10.

Within the above cell, reduction is occurring (A) in the salt bridge (B) in the voltmeter (C) at the Cu electrode (D) at the Ag electrode

11.

Which of the following most accurately describes the situation at each of the electrodes in the above cell?

(A) (B) (C) (D) 12.

Cu oxidant anode anode cathode

Ag cathode cathode reductant anode

Which is the correct statement relating to the following reaction? Mg (s) + CuSO4 (aq)  MgSO4(aq) + Cu (s) (A) (B) (C) (D)

Magnesium is reduced and copper ions are the oxidant. Copper ions are reduced and magnesium is the reductant. Sulfate ions are oxidised and magnesium is the oxidant. Magnesium ions are the reduction product and copper is the reductant.

50

Short Response Questions relating to Sections 9.2.4 (similar to past HSC questions) 1.

Four metals (Pb, x, y and z) were connected in pairs in a galvanic cell and the voltage recorded.

The following results were obtained:

List the 4 metals in decreasing order of activity.

(1 mark)

z, y, Pb, x

2.

A student investigated the following electrochemical cell.

List THREE correct observations the student could make as the cell operated. The aluminium electrode would get smaller. The copper electrode would get bigger. The solution in Beaker 2 would become a lighter blue.

51

3.

A galvanic cell was made by connecting two half-cells. One half-cell was made by putting a copper electrode in a copper(II) nitrate solution. The other half-cell was made by putting a silver electrode in a silver nitrate solution. The electrodes were connected to the voltmeter as shown in the diagram.

- ions

(a)

+ ions

Complete the above diagram by drawing a salt bridge and show the direction of movement of positive and negative ions in the salt bridge. (2 marks) See above

(b)

Using the standard potentials table (at the back of book), calculate the theoretical voltage (at standard conditions) of this galvanic cell. (2 marks) The anode half-reaction is: Cu (s) Cu2+ (aq) + 2e-

-0.34 V

The cathode half-reaction is: Ag + (aq) + eAg (s)

+0.80 V

The overall equation is: Cu (s) + 2Ag+ (aq)

(c)

Cu2+ (aq) + 2Ag (s)

E = +0.46 V

Explain what is meant by “standard conditions” with respect to measurement of reduction potentials. (3 marks) Standard conditions are at 25°C and at 1 atm pressure Electrolyte concentrations must be 1 mol/L

52

4.

(a)

Identify the anode in this diagram. Lead

(b)

Write the equation for the overall cell reaction and calculate the overall cell voltage (assume standard conditions). (2 marks) The anode half-reaction is: Pb (s) Pb2+ (aq) + 2e- +0.13 V The cathode half-reaction is: Ag + (aq) + eAg (s)

(1 mark)

+0.80 V

The overall equation is: Pb (s) + 2Ag+ (aq)

Pb2+ (aq) + 2Ag (s)

E = +0.93 V

5.

Cell 1 Cell 2 Choose ONE of the cells shown above and for that cell (a) Identify the cell. Cell 1 is a dry cell (b)

(c)

(1 mark) State ONE problem associated with using the cell. (1 mark) The voltage fluctuates because the ions in the electrolyte do not easily migrate between the half-cells; the battery becomes polarised (charge build-up) and the current drops. Describe, using half-equations, the chemistry of the chosen cell. (3 marks) 2+ Anode: Zn (s) Zn (aq) + 2e Cathode: NH4+ (aq) + MnO2 (s) + H2O (l) + eMn(OH)3 (s) + NH3 (aq) The cathode is a carbon rod in contact with manganese dioxide and the electrolytic paste containing ammonium ions and zinc chloride. The anode is zinc.

53

6.

Students performed the following first-hand investigation.

The beaker initially contained 500.0 mL of 0.025 mol L-1 copper sulfate solution. After several hours the colour of the solution had become lighter and a reddish deposit had formed on the surface of the zinc. The reddish deposit was removed from the zinc and dried. The deposit was found to weigh 0.125 g. (a)

Explain the observations and write a balanced overall equation for the reaction which occurred. (3 marks) The solution became lighter as the concentration of the blue copper sulfate decreased. Cu2+ ions were reduced to form copper (a reddish solid), which deposited on the surface of the zinc. The overall equation is: Zn (s) + Cu 2+ (aq)

(b)

Zn2+ (aq) + Cu (s)

E = +1.10 V

Calculate the concentration of the copper sulfate solution remaining after the reaction. (3 marks) Initial moles of copper sulfate in solution = c x V = 0.025 x 0.5000 mol = 0.0125 mol Moles of copper formed = 0.125 g/63.55 g = 0.0020 moles Since 1 mole of copper is formed from 1 mole of copper ions, Moles of copper ions remaining in solution = 0.0125 – 0.0020 = 0.0105 mol Volume of solution = 500.0 mL Hence concentration of final solution = 0.00105/0.5000 = 0.021 mol/L (2 s. f.)

54

Section 5 NUCLEAR ENERGY AS A SOURCE OF MATERIALS 1.

What is radioactivity? The spontaneous emission of particles and/or energy from the nucleus of an atom.

2.

What is the difference between a stable and a radioactive isotope? Do radioactive isotopes have the same chemical and physical properties as stable isotopes of the same element? Explain your answer using examples of stable and radioactive isotopes of the same element. A stable isotope does not emit particles from its nucleus, whereas a radioactive nucleus is unstable and, emitting particles or energy, moves to a more stable arrangement of particles within the nucleus. Unstable nuclei are either too big (like uranium, with 92 protons and a mass number ≥ 235) or have a neutron/proton ratio which is too large. Isotopes with a high n/p ratio tend to emit beta particles (which reduces the ratio by a neutron changing into a proton plus a high energy electron, the beta particle). Atoms which have a big nucleus tend to emit alpha particles, which reduces the total number of protons and neutrons. Further changes then emit beta particles and these reduce the neutron/proton ratio. Radioisotopes and stable isotopes have identical chemical and physical properties (apart from a slight difference in atomic mass). Carbon-12 is stable, whereas carbon-14 is radioactive and emits beta and gamma particles.

3.

When is a nucleus unstable? Describe conditions for the formation of alpha and for beta particles. Unstable nuclei are either too big (like uranium, with 92 protons and a mass number ≥ 235) or have a neutron/proton ratio which is too large. Isotopes with a high n/p ratio tend to emit beta particles (which reduces the ratio by a neutron changing into a proton plus a high energy electron, the beta particle). Atoms which have a big nucleus tend to emit alpha particles, which reduces the total number of protons and neutrons. Further changes then emit beta particles and these reduce the neutron/proton ratio.

55

4.

Draw up a table to compare alpha, beta and gamma radiation in terms of particles, mass, charge, ionising ability and penetrating ability. Write a suitable nuclear equation to show the formation of each type of radiation.

Type of radiation Particle or not

Alpha

Beta

Gamma

Particle, 2 protons, 2 neutrons (equivalent to a helium nucleus)

Particle, high energy electron formed from the break-up of a neutron into a proton plus electron (the beta particle)

Energy only, no particle

Relative Mass

4 (mass of 2 protons plus 2 neutrons)

1/2000 0 (no mass) (as an electron has only 1/2000 mass of a proton or neutron)

Charge

+2

-1

0 (no charge)

Ionising ability

High (high charge and mass)

High (charged and high speed)

Low (no mass or charge so no momentum)

Penetrating ability

Very low; stopped by tissue paper and 1 cm of air Does not penetrate skin

Moderate; 10 cm of air and will pass through Al foil but stopped by Pb Penetrates skin

High; stopped by 5 cm of lead or 15 cm of concrete Penetrates body tissues

Equation for formation

222 86

Rn 

218 84

Po +

4 2

He

228 88

Ra 

228 89

Ac +

0 1

e

No equation as no change in particles

56

5.

Write an equation for the alpha decay of radon-222. Rn 

222 86

6.

Po +

4 2

He

Write an equation for the beta decay of radium-228. 228 88

7.

218 84

Ra 

228 89

Ac +

0 1

e

What is meant by the half-life of a radioactive isotope and what is its relevance? The half-life is a measure of the rate of disintegration of radioisotopes. Each isotope has its own half-life. The half-life of an isotope is the time required for half the atoms in a given sample to undergo radioactive decay. For any particular radioisotope the half-life is independent of the initial amount of the isotope present. Radioisotopes which have a short half-life will disintegrate rapidly. The level of radiation from a sample drops rapidly. Isotopes, such as technetium-99m which has a half-life of 6 hours, lose their radioactivity very quickly, and are useful for medical diagnosis. Iodine-131 has a half-life of 8 days, long enough for treatment, but short enough that the patient does not need to be isolated for too long. Cobalt-60, used in gamma ray treatment of cancer cells from an instrument outside the body, has a half-life of 5.3 years. The isotope which releases the gamma rays does not need to be replaced in the instrument too often.

8.

Consider the following graph representing the decay of strontium-90.

(a)

What is the approximate half-life of this isotope?

27 years

(b)

A sample which contained 4 g of this isotope was analysed. How much strontium-90 would remain in the sample after approximately 80 years had elapsed? 0.5 g (2 g left after 27 years, 1 g left after 54 years, 0.5 g after 81 years) 57

9.

What is a transuranic element? An element with atomic number >93 i.e. after uranium, in the Periodic Table. The majority of these elements are artificially made (by bombardment with neutrons in a nuclear reactor or positively charged particles in an accelerator). However, there are traces of plutonium and neptunium in natural samples of uranium.

10.

Describe how transuranic elements are produced in a nuclear reactor. Give 2 examples. Transuranic elements (e.g. neptunium) are made in a reactor by bombardment of other nuclei by neutrons 238 239 239 0 1 92 U + 0 n  92 U  1 e + 93 Np Plutonium is bombarded with neutrons to produce an isotope of americium, which is an alpha emitter and is used in domestic smoke detectors. 239 241 0 1 94 Pu + 2( 0 n)  95 Am + 1 e

11.

Describe how artificial isotopes are produced in a cyclotron. Give 2 examples. Artificial isotopes are produced by bombardment of nuclei by high speed positive particles such as helium (alpha particles), protons or nuclei of other atoms, such as carbon, in linear accelerators or in cyclotrons. Californium-245 is made by bombardment of curium-242 with alpha particles. 242 245 1 4 96 Cm + 2 He  98 Cf + 0 n Phosphorus-30 is formed by bombardment of aluminium with alpha particles. 27 30 1 4 13 Al + 2 He  15 P + 0 n

12.

How is radiation detected? Describe 3 different methods of detection and the principle which allows this method of detection. Geiger counter. This method depends on the ionising ability of radiation, especially alpha and beta emission. The charged particles enter the detecting tube and the noble gas is ionised. An electric current flows and is detected. Photographic film. Gamma rays are a form of electromagnetic radiation and can change the silver compounds on a photographic film. Scintillation counter. This method involves the use of certain chemicals which give out a flash of light when hit by radiation such as alpha, beta or gamma rays.

58

13.

Complete the table to summarise the uses of at least 2 radioactive isotopes in industry and 2 in medicine. Choose the 2 examples in the medicine category so that 1 isotope is used for diagnosis and the other for therapy.

Use in :

Radioisotope used

Type of radiation

Properties of this type of radiation

Explanation of how it is used

Problems associated with its use

INDUSTRY Co-60

Beta and gamma

Gamma Destroys large molecules

Sterilisation of surgical instruments Bacteria are killed by the gamma radiation

Workers must be protected from gamma radiation

Am-241

Alpha

Ionises air Long halflife (>400 years)

In smoke detectors. Current flows when air ionised. Smoke blocks flow of ions, no current. Alarm goes off.

Special precautions for disposal of smoke detectors.

Diagnosis

I-123

Gamma

Gamma rays penetrate out through skin to be detected outside body. Short halflife (13 hours)

Ingested, goes to thyroid. Concentrates in thyroid, allows photo image of thyroid.

Shielding of workers, others for short period.

Treatment

Co-60

Gamma (beta)

Penetrating, kills living tissues, including cancer cells Moderate half-life (5 years)

Irradiation of affected area over a period of weeks kills tumours. Isotope lasts in instrument.

Shielding of workers.

MEDICINE

59

14.

The table below provides a summary of transuranic element synthesis. ELEMENT AND SYMBOL

ATOMIC NUMBER

MASS NUMBER

YEAR DISCOVERED

SYTHESISED FROM

SYNTHESIS METHOD

Neptunium (Np)

93

239

1940

U-238

Neutron in, electron out

Plutonium (Pu)

94

239

1940

Np-239

Electron out

Americium (Am)

95

241

1944

Pu-239

2 neutrons in, electron out

Curium (Cm)

96

242

1944

Pu-239

 particle in, neutron out

Berkelium (Bk)

97

243

1949

Am-241

Californium (Cf)

98

245

1952

Cm-242

 particle in, 2 neutrons out  particle in, neutron out

Einsteinium (Es)

99

253

1953

U-238

Fermium (Fm)

100

255

1955

U-238

Mendelevium (Md)

101

256

1957

Es-253

15 neutrons in, 7 electrons out 17 neutrons in, 8 electrons out  particle in, neutron out

Nobelium (No)

102

254

1961

Cm-246

C-12 in, 4 neutrons out

Lawrencium (Lw)

103

257

1969

Cf-252

B-10 in, 5 neutrons out

Choose 3 transuranic elements from the above list and write nuclear equations to show how they are synthesised. Ensure that the 3 examples involve bombardment by different particles. Americium: Formed by neutron bombardment of plutonium-239 239 241 0 1 94 Pu + 2( 0 n)  95 Am + 1 e Californium-245 is made by bombardment of curium-242 with alpha particles. 242 245 1 4 96 Cm + 2 He  98 Cf + 0 n Nobellium-254 is made by bombardment of curium-246 with a carbon nucleus. 246 12 254 1 96 Cm + 6 C  102 No + 4 0 n

60

15.

Research two recent (since 1969) discoveries of elements. Identify the:  atomic number of the new element.  year of discovery.  process used to discover the new element. Give a brief description of the nuclear chemistry involved for each element researched. (Note that the discovery of element 118 has since been disregarded by the original “discovery” team on the basis that the experimental results have not been able to be repeated.) Information which might be useful for this research task (extracted from an article on transuranium elements at www.infoplease.com): Transuranium elements are radioactive elements with atomic numbers greater than that of uranium (at. no. 92). All the transuranium elements of the actinide series were discovered as synthetic radioactive isotopes at the Univ. of California at Berkeley or at Argonne National Laboratory. In order of increasing atomic number they are neptunium, plutonium, americium, curium, berkelium, californium, einsteinium, fermium, mendelevium, nobelium, and lawrencium. Of these only neptunium and plutonium occur in nature; they are produced in minute amounts in the radioactive decay of uranium. Much of the study of the transuranium elements has taken place at the Lawrence Berkeley National Laboratory (at Berkeley, Calif.) and at the Joint Institute for Nuclear Research in Dubna, Russia; workers at both locations share credit for the independent discovery of rutherfordium, dubnium, and seaborgium (at. no. 104, 105, and 106 respectively), which are the first three transactinide elements. A German team at the Institute for Heavy Ion Research at Darmstadt discovered bohrium, hassium, meitnerium, darmstadtium, roentgenium, and ununbium (at. no. 107 through 112). The Dubna laboratory, with assistance from Berkeley, claims to have synthesized ununquadium (at. no. 114), and working jointly with the Lawrence Livermore National Laboratory (at Livermore, Calif.) claims to have produced ununtrium (at. no. 113) and ununpentium (at. no. 115). The Berkeley team claimed to have produced ununhexium (at. no. 116) and ununoctium (at. no. 118), but later retracted the claim for ununoctium after other laboratories failed to reproduce Berkeley's results and a reanalysis of their data did not show the production of the element. Other research teams have since synthesized ununhexium directly. Up to and including fermium (at. no. 100), the transuranium elements are produced by the capture of neutrons; the transfermium elements are synthesized by the bombardment of transuranium targets with light particles or, more recently, by projecting medium-weight elements at targets of other medium-weight elements. Isotopes of the transuranium elements are radioactive because their large nuclei are unstable, and the superheavy elements in particular have very short half-lives. In initial experiments, the formation of elements 118 and 116 was attempted by accelerating a beam of krypton-86 (8636Kr) ions to an energy of 449 million electron volts and directing the beam onto targets of lead-208 (20882Pb). 208

82Pb

+ 8636Kr



293

118Uuo

+ 1n

It was hypothesised that element 118 nucleus would decay less than a millisecond after its formation by emitting an -particle. This would result in an isotope of element 116 (mass number 289, containing 116 protons and 173 neutrons). This isotope of element 116, would also undergo further -decay processes to an isotope of element 114 and so on down to at least element 106.

61

Other Questions relating to Section 5 (Note: these questions are not necessarily of HSC standard but will be useful to test your knowledge and understanding of this topic) 1.

A radioactive isotope is (A) only found after uranium on the periodic table (B) an atom that has an unstable nucleus (C) a version of an element that has different chemical properties (D) an element that has a stable nucleus and only emits radiation irregularly

2.

12

3.

An example of a transuranic element is (A) 207Pb 235 (B) U 239 (C) Np 14 (D) C

4.

A transuranic element (A) generally has a stable nucleus (B) is made when a heavy nucleus is bombarded by neutrons or high speed particles (C) will have a molar mass less than 235 g/mole (D) is formed when a fissionable element like uranium 235 undergoes nuclear breakdown

5.

An example of a commercially produced radioisotope used in medicine is (A) 235U 14 (B) C 206 (C) Pb (D) 131I

6.

Technetium-99m is commonly used in medicine for the diagnosis of certain diseases. It is produced from the radioactive decay of molybdenum-99. Molybdenum-99 (rather than technetium-99m) is transported to the hospitals around Australia. A possible reason for this would be (A) technetium is too poisonous to be transported 99 (B) Mo is less dangerous 99m (C) Tc is too heavy to transport because of its density (D) 99Mo has a longer half-life than 99mTc

C is a stable isotope but 14C is radioactive. The reason for this is (A) 14C is exactly the same as 14N but with fewer electrons 12 (B) C was discovered before 14C (C) the ratio of neutrons to protons is too high in 14C (D) there can only be one isotope of the element inside the “zone of stability” for protonneutron ratios

62

7.

Most commercially produced isotopes have short half-lives. For instance, 131I is used in the diagnosis and treatment of thyroid gland diseases. It has a half-life of 8 days. If 1 gram of 131I was supplied to a hospital, how many grams of 131I would be left after 24 days? (A) 0.5 g (B) 0.25 g (C) 1.0 g (D) 0.125 g

8.

The earliest means of detecting radiation was (A) AAS (B) the Geiger-Muller counter (C) the scintillation counter (D) a photographic film

9.

A scintillation counter (A) measures the potential difference across a gas that has been ionized by radiation (B) displays a streak of water or alcohol droplets in the path of some radioactive particle (C) detects the flash given off when certain nuclei break down (D) detects the dark spots produced on photographic film that has been affected by radiation

10.

Which of the following radioisotopes is not likely to be used in medicine for the purpose of diagnosis? (A) 238U 131 (B) I 18 (C) F (D) 99mTc

11.

Identify the transuranic element. (A) calcium (B) cerium (C) chromium (D) curium

12.

Which instrument is used to detect radiation from radioactive isotopes? (A) pH meter (B) Geiger counter (C) burette (D) atomic absorption spectrophotometer (AAS)

13.

Consider the following nuclear reaction 238 234 92 U  90 Th + X Identify X (A) -radiation (B) a beta particle (C) an alpha particle (D) a neutron 63

Questions relating to Section 9.2.5 (similar to past HSC questions) 1.

2.

A radioactive isotope is likely to demonstrate the following property: (A) too many protons for the number of electrons in the atom (B) too many protons and neutrons in the atom (C) too many electrons in the outer shell of the atom (D) too many electrons for the number of neutrons in the atom

Describe the conditions for a nucleus to be unstable.

(2 marks)

The nucleus is too big (too many protons and neutrons) or the balance of neutrons and protons is outside the “zone of stability”. 3.

The following diagram shows the decay series for uranium-238.

Use the diagram to identify and write the nuclear equation for: (a)

an alpha decay process Radon-222 decays to polonium-218 222 86

(b)

Rn 

218 84

Po +

4 2

He

a beta decay process Thorium-234 decays to protactinium-234 234 234 0 90 Th  91 Pa + 1 e

(2 marks)

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4.

Describe the different ways both commercial isotopes and transuranic elements are produced. (4 marks) Commercial isotopes and transuranic elements are produced by bombardment of nuclei by either neutrons (in a reactor) or charged particles (in an accelerator, such as a cyclotron). Artificial isotopes are produced by bombardment of nuclei by high speed positive particles such as helium (alpha particles), protons or nuclei of other atoms, such as carbon, in linear accelerators or in cyclotrons. The charged particles are accelerated to enormous speeds by moving through a magnetic field. When the optimum speed is reached, the charged particle crashes into the target nucleus. Phosphorus-30 is formed by bombardment of aluminium with alpha particles. 27 30 1 4 13 Al + 2 He  15 P + 0 n Transuranic elements (e.g. neptunium) are made in a reactor by bombardment of other nuclei by neutrons 238 239 239 0 1 92 U + 0 n  92 U  1 e + 93 Np Plutonium is bombarded with neutrons to produce an isotope of americium, which is an alpha emitter and is used in domestic smoke detectors. 239 241 0 1 94 Pu + 2( 0 n)  95 Am + 1 e Some transuranic elements are produced by bombardment with charged particles. Californium-245 is made by bombardment of curium-242 with alpha particles. 242 245 1 4 96 Cm + 2 He  98 Cf + 0 n

5.

Discuss the benefits and problems associated with the use of ONE named radioactive isotope in industry. (4 marks) Cobalt-60. Benefits Used for sterilising medical supplies, as gamma rays are emitted. These destroy large molecules such as DNA and hence kill bacteria on instruments, bandages, etc. The isotope has a relatively long half-life (5 years), so the isotope does not need to be replaced for many years, yet the source is able to produce a reasonable level of intensity. Problems Gamma radiation is very penetrating (gamma penetrates 5 cm of lead and 15 cm of concrete) and causes cell damage and/or death of cells. Workers in industry (here the medical staff) need to be protected from the radiation source by thick lead shields.

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6.

Discuss the criteria considered when choosing radioisotopes for use in medicine. (4 marks) 3 criteria considered when choosing an appropriate isotope are:  the type of radiation emitted  the half-life of the isotope  the chemical specificity of the isotope (which organs it targets). If the isotope is used for diagnosis, it must be a gamma emitter, as the radiation must be detected outside the body. Only gamma rays readily penetrate through tissues to reach the external detector (film). The isotope should have a short half-life (hours), so that the radiation level drops rapidly and the patient is only exposed to radiation for a short time. The isotope must target particular organs or tissue types in the body. Technetium-99m targets tissues where cells are dividing rapidly. As a result, tumours, infection or inflammation can be detected. If the isotope is used as treatment (to destroy cells in the body), the rays emitted are normally gamma, often in conjunction with beta. It is the gamma radiation that kills the target tissues. The half-life of the isotope should be longer than for diagnosis (8 days in the case of iodine123, used to treat thyroid problems) as days, rather than hours, are needed for successful treatment. As with diagnosis, the isotope should be chemically specific in that it targets the normal biochemistry of the organ to be treated (calcium or strontium isotopes are used to treat bone cancers). If a beam of gamma rays, from an instrument outside the body, is used to destroy tumour cells (as in the case of cobalt-60), then the isotope must be a gamma emitter as it must penetrate the body tissues to the target organ and destroy the target cells. The half-life is relatively long (5 years for Co-60), so that the isotope does not need to be replaced regularly but the intensity of the beam is still strong enough to destroy the cells targeted. Since the isotope used is external to the body, it does not need to be matched to the chemical metabolism of the target cells.

7.

Assess the impact of the production of radioisotopes for society and the environment. (5 marks) The answer must include an overall assessment statement. (Deduct 1 mark if not included) Benefits of the use of isotopes for society should be identified, using specific examples. Generalisations should include medical diagnosis AND treatment AND use in industry and at least ONE specific example from medicine and ONE from industry should be used. (3 marks) Problems for society and the environment. Answer could include discussion of problems associated with the long half-life of some isotopes, the problems for storage of isotopes or radioactive wastes and the need for shielding of researchers and workers when using isotopes. Issues associated with the safety of nuclear reactors could also be included. (2 marks)

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Answers to Multiple Choice Questions Question No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Section 1 Page 14 C B C D A A D B A B A C

Section 2 Page 21 C A C C C C

Section 3 Page 28 C D C B C C C A C D C D

Section 3 Page 32 D A B A B

Section 4 Page 48 B D D A A D C D A D B B

Section 5 Page 62 B C C B D D D D C A D B C

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HSC CHEMISTRY STUDENT WORKBOOKS About the author:

Bronwen Hegarty is an experienced teacher of HSC Chemistry and has been the convenor of Examination Committees and Chemistry writing teams. She continues to teach Intensive Chemistry Revision courses and is working to assist Chemistry teachers and their students in a number of schools. She has recognised the need for these workbooks in schools, not only to provide revision for students throughout their Year 11 and HSC courses, but also to reduce the workload on teachers in preparing course materials and sequencing lesson plans. The workbooks are designed to be a low cost alternative to creating and photocopying teacher-designed student worksheets, assignments and revision questions. The format will facilitate student-centred teaching and learning and provide a framework for students to monitor their progress through the HSC syllabus. Teacher Editions of the workbooks include a CD and are available only to teachers, to provide them with fully worked solutions for each workbook.

Student Workbooks available: Year 12: Core Module 1: Production of Materials Core Module 2: The Acidic Environment Core Module 3: Chemical Monitoring and Management Option: Industrial Chemistry Option: Shipwrecks, Corrosion and Conservation Option: Forensic Chemistry Year 11: Preliminary Chemistry Note: Teacher Editions of the workbooks (including worked solutions to questions on CD) are available only to teachers.

To obtain copies of these workbooks, contact Bronwen Hegarty: Mobile 0402 890 724 or [email protected] 70