VCE Chemistry Units 3&4 Volume 1: Topic Tests

1 DECODE: VCE CHEMISTRY

VCE Chemistry Units 3&4

Volume 1: Topic TestsFIRST EDITION

(VCE Study Design 2017 − 2022)

Solutions available as a digital download:http://www.decodeguides.com.au

Dr Thushan Hettige

Dr Maoyuan Liu

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Volume 3Solutions Manual

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DECODE: VCE CHEMISTRY 2

ISBN: 978-1-922445-00-1

First published in 2020, by:

Decode Publishing Pty Ltd

ABN 16 640 806 686

PO Box 1007

Ashwood, VIC 3147

E-mail: [email protected]

A note about copyingWe get it. We were there ourselves no more than a few years ago. It’s tough being a VCE student. You wantto do the best you can, and to do so you need to have the best materials. Everyone else seems to have all theresources that you don’t have. You can’t afford to buy them all, and you don’t want to put that pressure on yourparents to buy more books for you, when they already work hard enough to send you to school. If you havefriends who cannot afford this book but would benefit from its contents, nothing we can do will stop you fromletting them copy it without paying for it, and we are honestly happy that they are going to benefit from thiswonderful resource.

But if you can afford it, we ask that you kindly spare a thought for the people who wrote this book. We arerecent students who took considerable time and effort, some to the detriment of other commitments, to collateour wisdom and expertise in these subjects and provide them in an easily-accessible book form. Please be con-siderate of the talented authors who wrote this book, and do the right thing.

Legal jargonCopyright ©Decode Publishing Pty Ltd 2020

All rights reserved.

With the exception of that which is permitted by the Australian Copyright Act of 1968, no part of this book maybe reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanicalor otherwise, without prior written permission.

Volume 3Solutions ManualScan QR code™ to open!

2(QR code not available in sample)


Preface.This study guide covers the necessary material required for the VCE Chemistry 3/4 Study Design 2017 - 2022and is designed to build problem solving skills, to assist students in learning each topic and in preparing forSACs. Volume 2 (Trial Examination) is the continuation of this book, and will consist of three full-length trialexaminations. This can be purchased separately from retailers or from the decode: website.

This volume consists of 7 topic tests, each formatted to the specifics of the end-of-year examination. Due tothe variable amount of content in each topic, the topic tests vary in the number of marks and in time duration(from 50 to 90 minutes). The number of marks allotted to each test is such that the proportion of multiple choicequestions, short answer questions and the time duration is the same as that of the end-of-year examination.

The topic tests are arranged in approximate order to the study design, with the exception of Topic Test 3 (Cellsand Electrolysis); this was combined as a single test since, although they are two different topics spaced apartin the study design, the integrated nature of these topics necessitated that they be examined together.

This book, as well as all decode: VCE study guides, is unique among all study guides currently in themarket. This is due to the hyper-detailed solutions to all problems in the solution manual, which is availableas a digital download from www.decodeguides.com.au. The majority of current VCE Chemistry practice ma-terial is very much question-oriented without any meaningful focus on the solutions to the questions. There isnothing more frustrating than attempting a question, getting it wrong or being stuck, and not being able to makehead-or-tail of a skeleton solution set with little-to-no explanation. It is also equally frustrating when you un-derstand the solution, but you still do not know how you could have approached the problem to generate thesolution on your own. We have endeavoured to ensure that the detailed explanations in this book will give youa good understanding of how you can get the answer to any of our problems. They will show you how to thinkwhen you see a problem how to use logic to get to your answer. They will also show you how to methodicallystructure your answer so that it is easy for the examiner to follow. In some cases, there will even be coverage oftheory to fill in potential knowledge gaps.

Remember, Chemistry is a problem-solving based subject and the best way to study it is through completinglots of problems. This subject can be a richly rewarding study for those who work hard at it. Good luck, andhave fun!

Please do not hesitate to e-mail me directly about any questions you have about the book or its contents.

Dr Thushan Hettige

[email protected]

Acknowledgements.I would like to thank my co-author, Maoyuan Liu, for his assistance in editing a number of the questions withinthis text.

I would like to acknowledge the National Institute of Advanced Industial Science and Technology (AIST),Japan, for their provision of spectra from their Spectral Database of Organic Compounds (SDBS), as citedwithin the text.





Table of Contents.Unit 3Unit 3 Area of Study 1 − Topic Test 1: Fuels 7Unit 3 Area of Study 2 − Topic Test 2: Reaction Rates and Equilibria 23Unit 3 Area of Study 1 and 2 − Topic Test 3: Cells and Electrolysis 47

Unit 4Unit 4 Area of Study 1 − Topic Test 4: Organic Chemistry 69Unit 4 Area of Study 1 − Topic Test 5: Analysis of Organic Compounds 81Unit 4 Area of Study 2 − Topic Test 6: Biochemistry 101Unit 4 Area of Study 2 − Topic Test 7: Metabolism 113




Course Map.

Unit 3: How can chemical processes be designed to optimise efficiency?

Area of Study Topic Topic Test

Unit 3 AOS 1What are the options for energyproduction?

Obtaining energy from fuelsFuel choices

Topic Test 1:Fuels

Galvanic cells as a source ofenergyFuel cells as a source of energy

Topic Test 3:Cells and Electrolysis

Unit 3 AOS 2How can the yield of a chemicalproduct be optimised?

Rate of chemical reactionsExtent of chemical reactions

Topic Test 2:Reaction Rates and Equilibria

Production of chemicals byelectrolysisRechargeable batteries

Topic Test 3:Cells and Electrolysis

Unit 4: How are organic compounds categorised, analysed and used?

Area of Study Topic Topic Test

Unit 4 AOS 1How can the diversity of carboncompounds be explained andcategorised?

Structure and nomenclature oforganic compoundsCategories, properties andreactions of organic compounds

Topic Test 4:Organic Chemistry

Analysis of organic compounds Topic Test 5: Analysis ofOrganic Compounds

Unit 4 AOS 2What is the chemistry of food?

Key food molecules Topic Test 6: Biochemistry

Metabolism of food in thehuman body

Topic Test 7: Metabolism





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STUDENT NUMBER

Letter

CHEMISTRYTopic Test 1: Fuels

Day DateReading time: *.** to *.** (10 minutes)

Writing time: *.** to *.** (1 hour)

QUESTION AND ANSWER BOOK

Structure of book

SectionNumber ofquestions

Number of questionsto be answered

Number ofmarks

A 12 12 12B 4 4 38

Total 50

Note: Data Book is NOT supplied. You will need to use your own!• Students are permitted to bring into the examination room: pens, pencils, highlighters,

erasers, sharpeners, rulers and one scientific calculator.• Students are NOT permitted to bring into the examination room: blank sheets of paper and/or

white out liquid/tape.Materials supplied

• Question and answer book of 15 pages.Instructions

• Complete all multiple-choice questions by circling your choice on the book.• Complete all short answer questions in the spaces provided.

Students are NOT permitted to bring mobile phones and/or any other unauthorised electronicdevices into the examination room.





SECTION A - Multiple-choice questions

Instructions for Section AAnswer all questions in pencil on the answer sheet provided for multiple-choice questions.Choose the response that is correct or that best answers the question.A correct answer scores 1, an incorrect answer scores 0.Marks will not be deducted for incorrect answers.No marks will be given if more than one answer is completed for any question.

Question 1Which of the following fuels can be considered carbon-neutral?

A. biodiesel

B. petrodiesel

C. coal seam gas

D. coal

Question 2The volume, in L, of hydrogen gas that needs to be combusted to produce 1000 kJ of energy at SLC is

A. 87.9 L.

B. 44.0 L.

C. 39.7 L.

D. 79.4 L.

Question 3Bioethanol

A. is composed of hydrocarbon molecules.

B. can be formed from the transesterification of vegetable oil.

C. has a higher energy content in kJ g−1 compared to all fossil fuels.

D. is produced from the anaerobic fermentation of glucose by yeast.




Questions 4 to 6 refer to the following information.

A 20 mL syringe that is half-retracted contains 10 mL of air in the barrel at a pressure of 100 kPa. Thetemperature of the air inside the syringe and the surroundings is 25 °C, and the pressure in the surrounding airis also 100 kPa. The room temperature and pressure are kept constant throughout the experiment.

Experiment 1.The tip of the syringe is sealed with a finger and the plunger retracted so that the volume of air inside the barrelis 20 mL.

A diagram is shown below:

10 mL 20 mL

Experiment 2.The tip of the syringe is left open and the plunger retracted so that air is aspirated into the syringe, and the finalvolume of air inside the barrel is 20 mL.

10 mL 20 mL

Question 4In Experiment 1, the final pressure inside the syringe is

A. 25 kPa.

B. 50 kPa.

C. 100 kPa.

D. 200 kPa.

Question 5In Experiment 2, the final pressure inside the syringe is

A. 25 kPa.

B. 50 kPa.

C. 100 kPa.

D. 200 kPa.





Question 6It is given that air is approximately 78% nitrogen by mole (for example, 1 mole of gas molecules in room airconsists of 0.78 moles of N2 gas). Assume that the composition of gases in the barrel of the syringe is identicalto that of the surrounding air.

The amount, in mole, of N2 gas in the syringe after retraction of the syringe in Experiment 1 is closest to

A. 8.1×10−4.

B. 6.3×10−4.

C. 3.1×10−4.

D. 4.0×10−4.

Question 7The thermochemical equation for the combustion of propane is shown below:

C3H8 (g) + 5 O2(g)→ 3 CO2 (g) + 4 H2O (l) ∆H =−2220 kJ mol−1

What will be the ∆H value, in kJ mol−1, for the thermochemical equation shown below?

32

CO2 (g) + 2 H2O (l)→12

C3H8 (g) +52

O2(g)

A. +1110

B. +4440

C. −4440

D. −1110




Question 8Three thermochemical equations are shown below:

S (s) + O2 (g)→ SO2 (g) ∆H =−297 kJ mol−1

2 SO2 (g) + O2 (g)→ 2 SO3 (g) ∆H =−197 kJ mol−1

S (s) +32

O2 (g)→ SO3 (g) ∆H =−x kJ mol−1

The value of x is closest to

A. −594.

B. −396.

C. −100.

D. +100.

Question 9Propane (C3H8) is most likely to be a major component of which of the following fuels or distillates?

A. petroleum gas

B. biogas

C. natural gas

D. petrol (gasoline)

Question 10The gas pressure inside a vessel is

A. the force applied onto the walls of the vessel.

B. the total kinetic energy of the gas molecules in the vessel.

C. the concentration of the gas inside the vessel.

D. the force applied onto a unit surface area of the vessel wall.





Question 11The mass, in g, of CO2 produced upon the combustion of 100 g of pentane (C5H12) is closest to

A. 100 g.

B. 200 g.

C. 300 g.

D. 400 g

Question 12A rigid vessel at 20 °C contains a mixture of only propane and oxygen in their exact stoichiometric ratios withno other gases being present, with the pressure inside the vessel being 100 kPa. The mixture was subsequentlycompletely combusted to produce gaseous carbon dioxide and liquid water. The temperature of the vessel wasbrought back to 20 °C.

What is the final pressure inside the vessel?

A. 200 kPa

B. 300 kPa

C. 50 kPa

D. 33 kPa




SECTION B - Short answer questions

Instructions for Section BAnswer all questions in the spaces provided. Write using black or blue pen.To obtain full marks for your responses you should

• give simplified answers with an appropriate number of significant figures to allnumerical questions; unsimplified answers will not be given full marks.

• show all working in your answers to numerical questions. No credit will be given foran incorrect answer unless it is accompanied by details of the working.

• make sure chemical equations are balanced and that the formulas for individualsubstances include an indication of state; for example, H2(g); NaCl(s)

Question 1 (9 marks)Methane is a chemical that can be used as a fuel to generate electricity. The molar enthalpy ofcombustion of methane at 25 °C is −882 kJ mol−1.

a. Define the term ’fuel’. 1 mark

b. Write down a thermochemical equation, including states, for the complete combustionof methane.

1 mark

The molar enthalpy of combustion of carbon monoxide (CO) is −283 kJ mol−1.

c. Write down a balanced equation, including states, for the incomplete combustion ofmethane.

Assume CO is the only carbon-based product produced.

1 mark





d. Therefore, using the information above, determine the molar enthalpy for the reactionwritten in part c.

2 marks

e. Compare and contrast the different sources of methane with respect to their renewabilityand environmental impacts related to sourcing and combustion.

4 marks




STUDENT NUMBER

Letter

CHEMISTRYTopic Test 2: Reaction Rates and Equilibria


Writing time: *.** to *.** (1 hour 15 minutes)


Structure of book



Number ofmarks

A 15 15 15B 5 5 45

Total 60











STUDENT NUMBER

Letter

CHEMISTRYTopic Test 3: Cells and Electrolysis

Day Date

Reading time: *.** to *.** (15 minutes)Writing time: *.** to *.** (1 hour 30 minutes)


Structure of book



Number ofmarks

A 18 18 18B 10 10 54

Total 72











STUDENT NUMBER

Letter

CHEMISTRYTopic Test 4: Organic Chemistry

Day Date

Reading time: *.** to *.** (10 minutes)Writing time: *.** to *.** (1 hour)


Structure of book



Number ofmarks

A 12 12 12B 7 7 38

Total 50











STUDENT NUMBER

Letter

CHEMISTRYTopic Test 5: Analysis of Organic Compounds

Day Date

Reading time: *.** to *.** (15 minutes)Writing time: *.** to *.** (1 hour 30 minutes)


Structure of book



Number ofmarks

A 18 18 18B 6 6 54

Total 72











STUDENT NUMBER

Letter

CHEMISTRYTopic Test 6: Biochemistry

Day Date

Reading time: *.** to *.** (15 minutes)Writing time: *.** to *.** (50 minutes)


Structure of book



Number ofmarks

A 10 10 10B 7 7 30

Total 40











STUDENT NUMBER

Letter

CHEMISTRYTopic Test 7: Metabolism

Day Date

Reading time: *.** to *.** (15 minutes)Writing time: *.** to *.** (1 hour)


Structure of book



Number ofmarks

A 12 12 12B 6 6 38

Total 50












Volume 2: Trial Examinations

FIRST EDITION(VCE Study Design 2017 − 2021)

Solutions available as a digital download:http://www.decodeguides.com.au

Dr Thushan Hettige

Dr Maoyuan Liu





ISBN: 978-1-922445-01-8

First published in 2020, by:

Decode Publishing Pty Ltd

ABN 16 640 806 686

Burwood, VIC 3125

PO Box 1007

Ashwood, VIC 3147

E-mail: [email protected]

A note about copyingWe get it. We were there ourselves no more than a few years ago. It’s tough being a VCE student. You want todo the best you can, and to do so you need to have the best materials. Everyone else seems to have all theresources that you don’t have. You can’t afford to buy them all, and you don’t want to put that pressure on yourparents to buy more books for you, when they already work hard enough to send you to school. If you havefriends who cannot afford this book but would benefit from its contents, nothing we can do will stop you fromletting them copy it without paying for it, and we are honestly happy that they are going to benefit from thiswonderful resource.

But if you can afford it, we ask that you kindly spare a thought for the people who wrote this book. We arerecent students who took considerable time and effort, some to the detriment of other commitments, to collateour wisdom and expertise in these subjects and provide them in an easily-accessible book form. Please beconsiderate of the talented authors who wrote this book, and do the right thing.

Legal jargonCopyright ©Decode Publishing Pty Ltd 2020


With the exception of that which is permitted by the Australian Copyright Act of 1968, no part of this bookmay be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,mechanical or otherwise, without prior written permission.




Preface.This study guide covers the necessary material required for the VCE Chemistry 3/4 Study Design 2017 - 2021and is designed to build problem solving skills, to assist students in learning each topic and in preparing forSACs. Volume 2 (Trial Examinations) consists of three full-length trial examinations. Note that Volume 1(Topic Tests) can be purchased separately from retailers or the website www.decodeguides.com.

This book, as well as all decode: VCE study guides, is unique among all study guides currently in themarket. This is due to the hyper-detailed solutions to all problems in the solution manual, which is availableas a digital download from www.decodeguides.com.au. The majority of current VCE Chemistry practicematerial is very much question-oriented without any meaningful focus on the solutions to the questions. Thereis nothing more frustrating than attempting a question, getting it wrong or being stuck, and not being able tomake head-or-tail of a skeleton solution set with little-to-no explanation. It is also equally frustrating when youunderstand the solution, but you still do not know how you could have approached the problem to generatethe solution on your own. We have endeavoured to ensure that the detailed explanations in this book will giveyou a good understanding of how you can get the answer to any of our problems. They will show you how tothink when you see a problem how to use logic to get to your answer. They will also show you how tomethodically structure your answer so that it is easy for the examiner to follow. In some cases, there will evenbe coverage of theory to fill in potential knowledge gaps.

The three trial examinations are designed to match the specifications of the 2017-2021 syllabus. Importantly,the exams are arranged approximately in order of increasing difficulty,

• Trial Exam 1 aims to match the easier end of VCAA’s spectrum of difficulty and focuses more on thebasic concepts, with a high weighting towards simple application questions.

• Trial Exam 2 aims to match the medium-to-harder end of VCAA’s spectrum of difficulty. The conceptsexamined here are slightly more advanced, requiring a more detailed understanding of the course contentand more complex reasoning skills.

• Trial Exam 3 is probably slightly more difficult than VCAA, and aims to replicate the difficulty of thehardest questions inpast VCAA examinations (obviously via originally written questions). Complexreasoning skills are required, and the solution manual is your friend. This will show you how to approachsuch problems. Importantly, the difficulty is in the reasoning and critical thinking, and there is strictly nocontent in the exam that is out of the scope of the study design.





Below is the suggested method by the author by which this book should be used:

• When exams are approaching and you want to start working through trial examinations, attemptTrial Examination 1 first. This examination is weighted more towards basic concepts. This exam, as wellas the other two, is deliberately designed to expose knowledge gaps and misconceptions. Use the SolutionManual to address these knowledge gaps and misconceptions and use the teaching points from the SolutionManual to strengthen your problem solving skills. Use your textbook and course summaries to further plugthose knowledge gaps. Use question books (including relevant questions from the topic tests in this book)to practice exam-style questions in the areas in which you are most weak to cement your knowledge andunderstanding.

• After addressing your knowledge gaps and attempting trial examinations from other sources (whererelevant), attempt Trial Examination 2. This exam is slightly harder, and your knowledge and problemsolving skills will be tested more rigorously. Again, use the Solution Manual as well as your otherresources to plug knowledge gaps.

• Attempt Trial Examination 3 a couple of weeks before your real VCE examination, so that you haveenough time to derive as much learning as you can from this book before the VCE examination.

Remember, Chemistry is a problem-solving based subject and the best way to study it is through completinglots of problems. This subject can be a richly rewarding study for those who work hard at it. Good luck, andhave fun!

Please do not hesitate to e-mail me directly about any questions you have about the book or its contents.

Dr Thushan Hettige

[email protected]

Acknowledgements.I would like to thank my co-author, Maoyuan Liu, for his assistance in editing a number of the questions withinthis text.

I would like to acknowledge the National Institute of Advanced Industial Science and Technology (AIST),Japan, for their provision of spectra from their Spectral Database of Organic Compounds (SDBS), as citedwithin the text.

Table of Contents.Trial Examination 1 5Trial Examination 2 37Trial Examination 3 75




STUDENT NUMBER

Letter

CHEMISTRYWritten examination (Trial 1)


Writing time: *.** to *.** (2 hours 30 minutes)


Structure of book



Number ofmarks

A 30 30 30B 10 10 90

Total 120











STUDENT NUMBER

Letter





Structure of book



Number ofmarks

A 30 30 30B 10 10 90

Total 120











STUDENT NUMBER

Letter





Structure of book



Number ofmarks

A 30 30 30B 7 7 90

Total 120










Copyright © Decode Publishing Pty Ltd 2020


With the exception of that which is permitted by the Australian Copyright Act of 1968, no part of thisbook may be reproduced, stored in a retrieval system, or transmitted in any form or by any means,

electronic, mechanical or otherwise, without prior written permission.

A note about copying

This complimentary eBook is provided to you as a free digital download on our website, available to theowner of any of our study guides. This is to make it as easy as possible for our students who have

Volumes 1 and 2 of this publication to access this solution manual, without breaking their backs. If allthree volumes were printed in a single book, it would be over 800 pages and weigh more than 1 kg and

neither your back, schoolbag nor the trees would be particularly happy.

Whilst obviously we would prefer that you do not do this, for our authors’ sake, there is nothing that wecan do to physically stop you from distributing this eBook, for free, to anyone who you think maybenefit from what we believe is a wonderful resource; we obviously encourage (conflict of interestalert!) that anyone holding this eBook purchase Volumes 1 and 2 to gain the most benefit. The only

thing we do ask is not to use any part of this book for commercial reasons, especially given the trust wehave placed in you by publishing this online. That would be very poor form, and illegal. On anothernote, we really hope you find this publication valuable. Our authors have worked extremely hard to

develop this resource for you.

1


Volume 3: Solution Manual

FIRST EDITION

This manual is to be used in conjunction with Volumes 1 and 2 available for purchase on the decode:website. Click the below link to access the decode: website and view a sample of Volumes 1 and 2. Or

scan the above QR code.

www.decodeguides.com.au

Dr Thushan Hettige

Dr Maoyuan Liu

2

http://www.decodeguides.com.au/

First published in 2020, by:Decode Publishing Pty LtdABN 16 640 806 686PO Box 1007Ashwood, VIC 3147E-mail: [email protected]

Legal jargonCopyright ©Decode Publishing Pty Ltd 2020All rights reserved.With the exception of that which is permitted by the Australian Copyright Act of 1968, no part of this bookmay be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,mechanical or otherwise, without prior written permission.

3

Preface.This is the solution manual of the decode: VCE Chemistry 3/4 Study Guide (1st edition). This studyguide covers the necessary material required for the VCE Chemistry 3/4 Study Design 2017 - 2022 andis designed to build problem solving skills for students studying VCE Chemistry Units 3/4, to assist themin learning each topic and in preparing for SACs.The feature that makes this study guide package different is the hyper-detailed solutions to all problemsin this solution manual. The majority of current VCE Chemistry practice material is very much question-oriented without any meaningful focus on the solutions to the questions. There is nothing more frustratingthan attempting a question, getting it wrong or being stuck, and not being able to make head-or-tail ofa skeleton solution set with little-to-no explanation. We have endeavoured to ensure that the detailedexplanations in this book will give you a good understanding of how you can get the answer to any ofour problems. They will show you how to think when you see a problem how to use logic to get to youranswer. They will also show you how to methodically structure your answer so that it is easy for theexaminer to follow. In some cases, there will even be coverage of theory to fill in potential knowledgegaps.

Remember, Chemistry is a problem-solving based subject and the best way to study it is through com-pleting lots of problems. This subject can be a richly rewarding study for those who work hard at it.Good luck, and have fun!

Dr Thushan Hettige

[email protected]

Acknowledgements.I would like to thank my co-author, Maoyuan Liu, for his assistance in editing a number of the questionswithin this text.I would like to acknowledge the National Institute of Advanced Industial Science and Technology (AIST),Japan, for their provision of spectra from their Spectral Database of Organic Compounds (SDBS), as citedwithin the text.

4

Table of Contents.Instructions: Click below on each of the titles to go straight to the corresponding section on the document.

Section 1: Model Solutions and Marking Scheme

Volume 1 (Topic Tests)

Topic Test 1: Fuels 8

Topic Test 2: Reaction Rates and Equilibria 18

Topic Test 3: Cells and Electrolysis 37

Topic Test 4: Organic Chemistry 53

Topic Test 5: Analysis of Organic Compounds 61

Topic Test 6: Biochemistry 74

Topic Test 7: Metabolism 82

Volume 2 (Trial Examinations)

Trial Examination 1 90



Section 2: Detailed Solutions

Volume 1 (Topic Tests)

Topic Test 1: Fuels 167

Topic Test 2: Reaction Rates and Equilibria 192

Topic Test 3: Cells and Electrolysis 240

Topic Test 4: Organic Chemistry 294

Topic Test 5: Analysis of Organic Compounds 325

Topic Test 6: Biochemistry 361

Topic Test 7: Metabolism 380

Volume 2 (Trial Examinations)




5

6

Section 1: Model Solutions and Marking Scheme

7

Model solutions and marking scheme.Topic Test 1: Fuels

SECTION A - Multiple-choice questions [Click on question number to go to Detailed Solutions.]

Question 1 2 3 4 5 6 7 8 9 10 11 12

Answer A A D B C C A B A D C C

SECTION B - Short Answer Questions

Question 1Part a. [Go to Detailed Solutions]

Comment Mark Allocation

Valid definition. 1 mark

Solution:"A fuel is a substance that can be reacted with other substances (often oxygen), leading to the releaseof energy (usually chemical) that can be harnessed for a specific purpose."

Part b. [Go to Detailed Solutions]


ALL of the following:

• correctly balanced equation

• correct states

• correct ΔH value.

Note:Technically, when the molar enthalpy of combustion isgiven at 25°C, the state of each chemical have to be thestate that it would be in at 25°C, so water should be aliquid. However, in the 2010 VCAA Assessment report,water was given as a gas. Therefore, it is likely that bothwater as a liquid or as a gas would be accepted in the VCEcontext.

1 mark

Solution:CH4 (g) + 2 O2 (g)→ CO2 (g) + 2 H2O (l or g), ΔH = − 890 kJ mol−1

8

Part c. [Go to Detailed Solutions]


BOTH of:

• correctly balanced equation

• correct states

An equivalent reaction equation with coefficients that aremultiples of the model solution shown below is acceptable.

1 mark

Solution:

CH4 (g) + 32O2 (g)→ CO (g) + 2 H2O (l or g)

Part d. [Go to Detailed Solutions]


Writing down/identifying equation (2) with its associatedΔH value of −283 kJ mol−1.

1 mark

Determining the molar enthalpy for the equation writtendown in part (c).For the equation shown above, −599 kJ mol−1 is theappropriate answer. However, if a different set ofcoefficients were used in part (c), then the answer for part(d) needs to correspond to that equation.

1 mark

Solution:CH4 (g) + 2 O2 (g)→ CO2 (g) + 2 H2O (l), ΔH = − 890 kJ mol−1... (1)

CO (g) + 12 O2 (g)→ CO2 (g), ΔH = − 283 kJ mol−1... (2)

Rearranging equation (2),

CO2 (g)→ CO (g) + 12 O2 (g) , ΔH = + 283 kJ mol−1... (3)

Adding equations (1) and (3), we get:

CH4 (g) + 32O2 (g)→ CO (g) + 2 H2O (l), ΔH = −882 + 283 = −607 kJ mol−1

9

Part e. [Go to Detailed Solutions]


Correctly describing the renewability of natural gas, coalseam gas and biogas.

• natural gas is nonrenewable

• coal seam gas is nonrenewable

• biogas is renewable

1 mark

Identifying at least ONE similarity in environmentalimpacts for all three sources of methane (eg. landclearance/habitat destruction).

1 mark

Identifying differences in environmental impacts betweenthe sources of methane as described above.

0.5 marks for each difference1 mark is the maximum

explains why greenhouse gas emissions are lower forbiogas compared to the other two (either of the followingexplanations are suitable):

• Biogas is carbon-neutral as the CO2 emitted wasrecently absorbed from atmosphere during growingof crops.

• Sewage-derived biogas convertsatmosphere-destined CH4 to the less potentgreenhouse gas CO2 leading to a drop in impactfrom greenhouse gases.

1 mark

Solution:Biogas is a renewable source of methane, whereas both natural gas and coal seam gas are non-renewable sources of methane.Environmental impacts (sourcing): Land clearance and habitat destruction are impacts common to allthree sources of methane. However, with sourcing natural gas and coal seam gas, there is a significantrisk of gas leak, which could be harmful to the community and release significant amounts of methane,a highly potent greenhouse gas, into the atmosphere. Coal seam gas also has the issue related tofracking (the use of fluid to crack the coal seam), including contamination of water in aquifers.Environmental impacts (combustion): There is little net greenhouse gas emissions for biogas, giventhat in the growth of the plant material required to produce the biogas, CO2 is consumed by the plants.The amount of this CO2 is near the amount of CO2 released during combustion of biogas. The netgreenhouse gas emissions for methane and coal seam gas (both fossil fuels), are significantly higherthan that for biogas.

10

Detailed Solutions.Topic Test 1: Fuels

Section A - Multiple Choice Questions

Question 1 (A) [Go to MCQ Answers]

What is meant by the phrase carbon-neutral?

This phrase means that the net amount of CO2 released into the atmosphere is close to zero, with anemphasis on the word “net”. In other words, there is a significant amount of CO2 released (in the con-sumption of the fuel), but there is also a nearly equal amount of CO2 consumed in the production of thefuel.

Biofuels have this property. To understand why, let’s look at the carbon cycle, and the journey of thecarbon atom.

Take six C atoms in six CO2 molecules. Suppose now we are growing sugarcane so that we can producebioethanol. When sugarcane grows (as a plant), it becomes bigger as it “sucks” CO2 from the atmospherein a means to produce glucose and large organic molecules that will comprise its structure. This is donevia photosynthesis, which is literally the use of light (“photo”) for the “synthesis” of glucose. Thereaction equation is below:

6 CO2 + 6 H2O→ C6H12O6 + 6 O2

The carefree carbon atoms in the atmosphere (present as CO2) are now stuck in a glucose molecule withinthe sugarcane.The glucose produced from the sugar-cane is extracted when the sugarcane is harvested, and convertedto ethanol via fermentation:

C6H12O6 (aq)→ 2 C2H5OH (aq) + 2 CO2

Notice that two of the carbon atoms are already “released” back in the atmosphere. The other four arestill stuck, this time in ethanol.

The ethanol is transported, mixed with petrol, and is now stuck in a fuel tank of a car. The car then burnsthe ethanol to operate, via combustion:

C2H5OH (l) +72

O2 (g)→ 2 CO2(g) + 3 H2O (l)

The C atoms are liberated again, to be free in the atmosphere as CO2. They will remain free untilsomething “sucks” them out of the atmosphere. Like a tree. Or the sea dissolves it and attaches it tocalcium rocks.

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Mind you, keep in mind that we have to harvest the sugarcane and transport the ethanol. We need vehiclesand machines to do this. What do they typically run on? Petrol or electricity, the latter which is mostlysourced from our good old friend coal. This is one reason why even biofuels are not fully carbon-neutral,especially so long as we rely on our friendly black rock on an ongoing basis.

Therefore, the answer is biodiesel as it is the only biofuel among the options.

The other three options are fossil fuels. The C atoms were buried underneath the ground for millions ofyears until they were unceremoniously driven out of the ground and stuck into thermal power plants andvehicle engines, and then expelled into the atmosphere. That is not carbon-neutral at all.


We know that we can burn (combust) hydrogen gas in air, and release energy this way. The Hindenburglearnt this the hard way.

(Source: Wikimedia Commons)

Hydrogen gas combusts in oxygen to produce water as the sole product, in the following equation:

2 H2 (g) + O2 (g)→ 2 H2O (l)

The question stem asks us to find out how much H2 gas (the volume, strictly) it takes to produce/releasea certain amount of energy. We, therefore, need to know how much energy is released when hydrogenis combusted. The information we need is a ∆H value. This can be found in the Data Book, under“11. Heats of combustion of common fuels”. The accompanying table says that the “molar heat ofcombustion” is “−282 kJ mol−1”.

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Remember!The molar heat of combustion is the energy released (as heat) when 1 mole of the fuel is combusted.

When we combust 1 mole of H2, we release 282 kJ of energy.

In this question, we want to release 1,000 kJ of energy, much more than the 282 kJ of energy.

Also, we are looking for the volume of H2 (at SLC) here − but we know that there is an intricaterelationship between moles of gas and volume at SLC (see the Pro Tip).

Remember!The molar volume at SLC is the volume that 1 mole of any gas (does not matter what gas) occupies,when the pressure that the gas is exerting is 100 kPa and the gas is at a temperature of 25 °C.

It turns out that 1 mole of a gas, at SLC, occupies a volume of 24.8 L.

This is why we say that Vm = 24.8 L mol−1 [L per mole of gas].

Pro-tip!If you are ever asked to find a volume at SLC (or STP), think moles!This is because we know the relationship between the amount (in mole) of gas and the volume the gasoccupies at SLC (or STP).

Therefore, let us work out how much (in mol) H2 it takes to release 1000 kJ of energy:–

combust 1 mole of H2→ release 282 kJ of energycombust 2 moles of H2→ release 2 × 282 = 564 kJ of energy

combust � moles of H2→ release �×282 = 1000 kJ of energy

You would expect a number bigger than 1 to fit the square. By ratios, we can say:

n(H2) =1000282

= 3.55 mol

We now need to know the volume that 3.55 mol of H2 occupies at SLC. We know that:

1 mol of H2→ occupies... 24.8 L at SLC2 mol of H2→ occupies... 2×24.8 = 49.6 L at SLC

3.55 mol of H2→ occupies...3.55×24.8 L at SLC

Therefore, completing the above calculation:

Vm = 3.55×24.8= 87.9 L

The answer is therefore (A).

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Question 3 (D) [Go to MCQ Answers]

Let us go through each option. These sorts of questions you have to plug each option into the sentenceand see whether the sentence is true.

• (A) Bioethanol is composed of hydrocarbon molecules.

– From the name bioethanol, we can be sure that this statement is not true. Hydrocarbons are“hydro” “carbons” and have only C and H atoms in their structure. Ethanol, recall, is analcohol and therefore has an OH group, and therefore an O atom.

• (B) Bioethanol can be formed from the transesterification of vegetable oil.

– Recall that “vegetable oil” is associated with “biodiesel”, not bioethanol. Biodiesel is formedfrom the transesterification of vegetable oil. Bioethanol is produced from the fermentation ofglucose from sugarcane and other plant sources.

• (C) Bioethanol has a higher energy content in kJ g−1 compared to all fossil fuels.

– A cursory look at section 11 of your Data Book will refute this claim.

– Octane is found in petrol (don’t forget the octane rating in the petrol station). Its heat ofcombustion is 47.9 kJ g−1. Ethanol’s is only 29.6 kJ g−1.

• (D) Bioethanol is produced from the anaerobic fermentation of glucose by yeast.

– As discussed above, this is true. The phrase anaerobic refers to the fact that this is donewithout oxygen, hence the “an” and “aerobic” which means “without oxygen”.

Question 4 (B) [Go to MCQ Answers]

This question tests your understanding of the gas laws.

In this question, we are expanding the volume from 10 to 20 mL. In other words, we are doubling (×2)the volume.

This is a demonstration of Boyle’s Law.

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Boyle’s Law!Boyle’s Law states that:

• in a closed system (in other words, gas cannot enter or exit the system),

• where we keep temperature constant,

• the pressure is inversely proportional to the volume.

In maths language:

p =constant

V(1)

p1V1 = p2V2 (2)

The above are two different ways of expressing the same idea above.

For example:

• doubling the volume (×2) causes the pressure to halve (÷2)

• dividing the volume by 3 (÷3) by compressing the vessel causes the pressure to triple (×3)

The finger is over the tip of the syringe, so gas cannot enter or escape the syringe. This is a closed system.

Therefore, since the volume is being doubled (×2), the pressure will halve (÷2). This means the pressuregoes from 100 to 50 kPa.

If you don’t believe me, try it yourself. As you pull the plunger, you’ll feel a lot of resistance and yourfinger being sucked in, as the lower pressure inside the vessel is trying to “suck” the plunger back to itsoriginal position.

Question 5 (C) [Go to MCQ Answers]

This situation is different as we have an open system; air can enter the syringe.

On this occasion, since air can enter the syringe, pulling the plunger back will not cause a change inpressure (making the answer (C)). It will cause an increase in the amount of gas in the vessel instead.In fact, by Avogadro’s Law, the amount of gas in the vessel would double. Gas will continue to enter thevessel until the pressure inside the syringe is equal to the pressure of the surrounding air (which is 100kPa). If you don’t believe me, think about what happens when you pop your ears in an aeroplane.

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Avogadro’s Law!Avogadro’s Law states that:

• in an open system (in other words, gas can enter or exit the system) with constant pressure,

• where we keep temperature constant,

• the amount of gas is proportional to the volume of the vessel.

In maths language:

n =V × constant (1)n1

V1=

n2

V2(2)

The above are two different ways of expressing the same idea above.

For example:

• if the volume is doubled (by pulling back on the plunger) gas will rush in until the total amountof gas is double what it used to be

• if the volume is divided by 3 by pushing air out of the plunger, the amount of gas will be 13 of

the original amount.


What happens to the amount of gas after retraction of the syringe in Experiment 1 (where you put a fingerover the tip)? Since air cannot go in or out, it does not change!

Therefore, the question is actually asking you in disguise to determine the amount of nitrogen gas in the10 mL syringe in the first place.

Remember that we know the intricate relationship between volume and moles in SLC and STP (seehere for a reminder of what molar volume is). Also, from the question stem, the pressure is 1 bar andtemperature 25 °C, which is just SLC!

Remember that 1 mole of gas occupies 24.8 L at SLC.

However, we have a mixture of gases that occupies only 10 mL (or 0.010 L). So, how much gas is in thismuch smaller volume? If we figure this out, we know that 78% of this gas is N2 and the problem willbecome easy. Using ratios:

1 mole of gas→ occupies 24.8 L0.5 mole of gas→ occupies 0.5×24.8 = 12.4 L� mol of gas→ occupies �×24.8 = 0.01 L

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You would expect the square to be smaller than 1. Therefore, you would expect the square to be equal to0.0124.8

rather than24.80.01

.

Alternatively, you can use the formula n =VVm

with V = 0.01 and Vm = 24.8.

Hence

n(gas) =0.01024.8

= 4.0×10−4 mol

Given that 78% of this is nitrogen, by mole, we can say:

n(N2) = 0.78×n(gas)

= 0.78×4.0×10−4

= 3.1×10−4 mol

The answer is therefore (C).


Remember!When manipulating thermochemical equations:

• multiplying the coefficients by x means you need to multiply ∆H by x

• changing the arrow direction means you need to change the sign of ∆H

• if you add thermochemical equations together, you add the ∆H values.

Let’s examine the thermochemical equations:

C3H8 (g) + 5 O2(g)→ 3 CO2 (g) + 4 H2O (l) ∆H =−2220 kJ mol−1(1)

32

CO2 (g) + 2 H2O (l)→12

C3H8 (g) +52

O2(g) ∆H =?(2)

Firstly, we can see that the chemicals are on opposite sides in reactions (1) and (2). We will have tochange the direction of the arrow.Secondly, the coefficients seem to all be halved. Therefore, we will also need to multiply the coefficientsby 1

2 in equation (1).

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Step 1.Change the direction of the arrow.

3 CO2 (g) + 4 H2O (l)→ C3H8 (g) + 5 O2(g) ∆H =+2220 kJ mol−1(3)

Note here that we have changed the sign of ∆H.

Step 2.

Multiply the coefficients by12

.

32

CO2 (g) + 2 H2O (l)→ 12

C3H8 (g) +52

O2(g) ∆H =+1110 kJ mol−1(2)

Note here that we have halved ∆H.

We arrived at our second equation. The answer is therefore (A).

Question 8 (B) [Go to MCQ Answers]

This is the same idea as Question 7.

Remember!When manipulating thermochemical equations:

• multiplying the coefficients by x means you need to multiply ∆H by x

• changing the arrow direction means you need to change the sign of ∆H

• if you add thermochemical equations together, you add the ∆H values.

This is sort of like simultaneous equations. Let’s examine them.

S (s) + O2 (g)→ SO2 (g) ∆H =−297 kJ mol−1 (1)

2 SO2 (g) + O2 (g)→ 2 SO3 (g) ∆H =−197 kJ mol−1 (2)

S (s) +32

O2 (g)→ SO3 (g) ∆H =−x kJ mol−1 (3)

We need to manipulate and summate equations (1) and (2) to derive equation (3).

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Observations:

• S (s) is on the LHS in equation (3). It is also on the LHS in equation (1). The coefficients are thesame.

– Therefore, to get equation (3), we probably need to add equation (1) to some permutation ofequation (2).

– We will not need to multiply equation (1) by anything.

– We will not need to switch the sides around in equation (1).

• SO3 (g) is on the RHS in equations (2) and (3). However, the coefficient of SO3 in equation (2) is2, whereas in equation 3 it is only 1.

– Therefore, we will not need to switch the sides around in equation (2).

– However, we will need to multiply equation (2) by12

to make the SO3 coefficient 1, like inequation (3).

– Then, we would be able to add (1) up with12× (2) to form equation (3), with luck.

• SO2 (g) is on equations (1) and (2), but not in (3). This means that we will have to cancel it out.Luckily, doing the above steps will lead to an equal amount of SO2 molecules on each side anyway.

Let’s do this.

(1) :S (s) + O2 (g)→ SO2 (g) ∆H =−297 kJ mol−1(1)

(2)× 12

:SO2 (g) +12

O2 (g)→ SO3 (g) ∆H =−98.5 kJ mol−1(4)

The SO2 molecules will cancel out when we add the equations together. The S and SO3 molecules areon the correct sides. Let’s now add the equations:

(1)+(4) :S (s) + O2 (g) +��SO2 (g) +12

O2 (g)→��SO2 (g) + SO3 (g) ∆H =−395.5 kJ mol−1

Simplifying:

(1)+(4) :S (s) +32

O2 (g)→SO3 (g) ∆H =−395.5 kJ mol−1

The answer is therefore (B).

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This question is largely recall.

Going through each option:

• Petroleum gas is...gas from petroleum. Petroleum is the other name for crude oil. Crude oil is ablack mixture of liquids, known as black gold due to the amount of money you can make from it.There are gases dissolved in it, which bubble out when you do fractional distillation. These gasesare collectively known as petroleum gas. They consist mainly of propane and butane.

• Biogas is gas derived from the anaerobic decomposition of recently dead plant/animal matter bybacteria; a product of this decomposition is methane. Biogas is a mixture mainly of CH4 and CO2.

• Natural gas is gas derived from “giant bubbles” under the ground. It is a fossil fuel. It is primarilycomposed of methane.

• Petrol is a distillate of crude oil; it is used in cars as fuel. It is a mixture of hydrocarbons ofapproximately 8 carbon atoms. It definitely does not contain propane.

The answer is therefore (A).

Question 10 (D) [Go to MCQ Answers]

This is a definition question. You are expected to know the definition of pressure (as per the VCAA studydesign).

The pressure of a gas in a vessel is the force the gas applies on the walls of the vessel, per unit surfacearea of the vessel.

To use a crude example, if we had a 1× 1× 1 m vessel (which, incidentally, is a 1000 L vessel) whichhas surface area 6 m2, and the gas inside the vessel is exerting a total force of 600,000 N (newtons) onthe walls of the vessel, the pressure the gas is exerting on the vessel is:

P =600,000 N

6 m2

= 100,000 N m−2

= 100,000 Pa= 100 kPa

Fun fact!The unit for pressure is the Pascal (Pa), which is equal to 1 N of force over an area of 1 m2 (N m−2 andPa are equivalent units).

However, in real life, we are dealing with pressures of thousands of Pascals, which is why you see theunit kilopascal (kPa); 1 kPa is 1,000 Pascals.

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This is a stoichiometry question.

We have an amount (in this case, mass) of a reactant (C5H10).We want to work out the amount of product (CO2) produced.

Pro-tip!If you know the amount of one chemical in a reaction, and you want to know the amount of anotherchemical, you want to find out the mole ratio between the chemicals. In other words, a reactionequation would be helpful.

Let us write the reaction equation for the combustion of pentane, in steps:

C5H12+ O2→ CO2 + H2O

There’s a quick way I like to balance these equations:

1. Balance the C atoms and H atoms first.

Here, there are 5 C atoms on the LHS, so we can stick a 5 in front of the CO2 to get 5 C atoms on theRHS.Also, there are 12 H atoms on the LHS. Since H2O has 2 H atoms each, we need 6 H2O molecules tomake up 10 H atoms on the RHS.

C5H12+ 8 O2→ 5 CO2 + 6 H2O

The mole ratio of pentane to CO2 is 1 : 5. Combusting 1 pentane molecule gives you 5 CO2 molecules.

Since we have a mass of pentane, but we have a mole ratio, let’s work out the amount of pentane (molarmass would be 72 g mol−1):

n(C5H12) =10072.0

= 1.39 mol

Since we know that 1 pentane molecule gives you 5 CO2 molecules:

n(CO2) = 5×n(C5H12)

= 5×1.39= 6.94 mol

Finally, working out the mass of CO2 (molar mass 44 g mol−1):

m(CO2) = 6.94×44.0= 306 g

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The answer is therefore (C).


This is a pretty tough question.

Let’s collect the information we know.

• We have a rigid vessel. It cannot change its volume. The volume is constant.

• The temperature of the vessel was originally 20 °C. At the end, it is still 20 °C. The temperatureis, effectively, constant.

• The initial pressure inside the vessel is 100 kPa. We want to find out the final pressure.

• The only gases in the vessel are propane and oxygen. They are in their stoichiometric ratios. Theyare combusted.

• We are left with CO2 (a gas) and water (a liquid).

For the purposes of this question, we have a vessel with a constant volume and a constant temperature.What is it in this vessel that changes? We had propane and oxygen as the original gases. When theyreact, there is no excess as the gases were in their stoichiometric ratios. We end up with only CO2 asthe product. You would imagine that the amount (in mol) of gas in the vessel changes. This would bethe reason why the pressure changes.

However, we have no numbers to play with here. We have a vessel at constant temperature and volume,and an initial pressure of 100 kPa, but no other data.

This is a question VCAA could spring on you. I am going to show you how to do it to get the edge overyour peers. Cue the devil horns.

The first key is writing the reaction equation down. You’ll see why I am doing this.

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

I will make a point here. For every 1 (one) C3H8 molecule and 5 (five) O2 molecules we react, we endup producing 3 (three) CO2 molecules and 4 water molecules. However, the water molecules are liquid.Hence, for every 1+5 = 6 gas reactant molecules we consume, we only end up producing 3 product gasmolecules. In other words, we end up halving the total number of gas molecules in the vessel.

Importantly, there has to be a relationship between the amount (n) of gas in the vessel and the pressure(P) inside the vessel. An easy cheat sheet way of remembering what that relationship is is to just look atthe ideal gas equation which is in your Data Book:

pV = nRT

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See how n and p are on opposite sides of the equation? Well, if we keep two of the variables aboveconstant apart from R (remembering that R is always constant anyway)

• if the other two variables are on the SAME side, then they are inversely proportional (eg p andV , Boyle’s law)

• if the other two variables are on OPPOSITE sides, then they are proportional.

This means that n and p are proportional, if we are keeping V and T constant (which is what we aredoing). In other words, if we double the amount of gas in the vessel, we double the pressure inside thatvessel. If we halve the amount of gas in the vessel, we halve the pressure inside that vessel.

Since we are halving the total number of gas molecules in the vessel, we end up halving the pressure.Since the original pressure was 100 kPa, the final pressure must be 50 kPa.

Imagine pumping gas into a rigid box that cannot expand. The pressure inside the box would increaseright? Similarly, if we sucked out gas, the pressure inside the box would decrease. On this occasion, thereaction removes gas molecules, effectively “sucking” them out.

Section B - Short Answer Questions

Question 1

Part a. [Go to Model Solutions and Marking Scheme]

This is essentially a definition question, which is largely recall.

"A fuel is a substance that can be reacted with other substances (often oxygen), leading to the release ofenergy (usually chemical) that can be harnessed for a specific purpose."

You need to know the definition of a fuel; it would be good if you could find a reliable definition from atextbook, for example.

Part b. [Go to Model Solutions and Marking Scheme]

A thermochemical equation is basically a balanced equation that includes theΔH value. Therefore, thefirst step here is to write a balanced chemical equation for this reaction.

So, this is just an example of balancing a combustion eqauation. We have this so far:

CH4 + O2→ CO2 + H2O

The products of combustion of a hydrocarbon or an organic compound composed of only carbon, hydro-gen and oxygen are carbon dioxide and water.

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To balance combustion equations, use the following steps:

1. Balance the C atoms first by writing the appropriate coefficient next to the CO2.

2. Balance the H atoms then by writing the appropriate coefficient next to the H2O.

3. Balance the O atoms by writing the appropriate coefficient next to the O2.

Let’s start with Step 1 (C atoms).

CH4 + O2→ CO2 + H2O

There is one C atom in CH4. Therefore, there would be only one C atom in CO2.

Now, for Step 2 (H atoms).

CH4 + O2→ CO2 + 2 H2O

There are 4 H atoms in CH4. Therefore, you need to have 2 H2O molecules (each H2O molecule has 2 Hatoms).

Let’s go to Step 3 (O atoms).

CH4 + 2 O2→ CO2 + 2 H2O

There are 4 O atoms on the RHS (1 CO2 and 2 H2O). Therefore, you will have 2 O2 molecules on LHS.

Now, add states. Methane is a gas at room temperature - oxygen and carbon dioxide are also gases. Wateris a tricky one. There are two ways of looking at it:

• the combustion reaction usually releases so much energy that water is evaporated as a gas - by thislogic H2O would be a gas

• the products are cooled back down to 25°C after the reaction, where water would be a liquid

Conventionally, when we write thermochemical equations for enthalpies of combustion and the quotedenthalpies are given at 25°C, the states of all species will be the states that they exist at 25°C. This isbecause changing the state of the product changes theΔH value (there is an enthalpy change in the inter-conversion of water from liquid to gas). The ΔH value given assumes water to be a liquid. However,this distinction is usually not discussed in VCE, and VCAA Assessment reports have given water as agas. Therefore, whilst liquid is the technically correct state, either liquid or gas should be accepted atVCE level.

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The final step is to assign the correctΔH value. This can be found in the data book, which contains the∆H value of many different fuels. As per the data book, the ∆H value (the molar heat of combustion)of methane is −890 kJ mol−1, meaning this would be the total enthalpy change if you combusted 1 moleof CH4. Now, if we assign a ΔH value to the above equation, this would be the enthalpy change if youreact 1 mole of CH4 with 2 moles of O2. Hence, the thermochemical equation is:

CH4 (g) + 2 O2 (g)→ CO2 (g) + 2 H2O (l), ΔH = − 890 kJ mol−1

Part c. [Go to Model Solutions and Marking Scheme]

This is similar to part b, except you need to know that when writing equations of incomplete combustion,the products are conventionally carbon MONOxide (CO) and water (anyhow, the question asks you toassume that CO is the carbon-based product that is produced). In reality, when you perform incompletecombustion, you produce a variety of products in varying quantities, such as a tiny amount of CO2, CO,graphite (elemental carbon, C), and when you incompletely combust large hydrocarbons, you can alsoproduce smaller unburnt hydrocarbons by pyrolytic cracking. Anyway, balancing incomplete combustionequations - you use the exact same steps as for part b.

So we have the following equation:

CH4 + O2→ CO + H2O

Balancing the C atoms - they are already balanced (1 on each side).

Balancing the H atoms:

CH4 + O2→ CO + 2 H2O

Balancing the O atoms: there are 3 O atoms on the RHS, except the only oxygen-containing species onthe LHS is O2 - therefore one could say that you could only add an even number of O atoms on the LHS.We can get around this issue by using fractions as coefficients. This is a perfectly acceptable conventionto use. Hence, balancing the O atoms and including states, we get -

CH4 (g) + 32 O2 (g)→ CO (g) + 2 H2O (l)

Alternatively, if you do not like fractions, you can double the coefficients:

2 CH4 (g) + 3 O2 (g)→ 2 CO (g) + 4 H2O (l)

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Part d. [Go to Model Solutions and Marking Scheme]

This is a rather tricky question. Let’s put out the facts on the table. What do we know? We know thethermochemical equation for the complete combustion of methane:

CH4 (g) + 2 O2 (g)→ CO2 (g) + 2 H2O (l), ΔH = − 890 kJ mol−1... (1)

We are also given that the molar enthalpy of combustion of CO is −283 kJ mol−1. What does thisequation look like? Since there are no H atoms in CO, the only possible product in the combustion of COwould be CO2. Hence, we can write this thermochemical equation:

CO (g) + 12 O2 (g)→ CO2 (g), ΔH = − 283 kJ mol−1... (2)

Now, we need to use the above two equations to derive a ΔH value for the following equation:

CH4 (g) + 32 O2 (g)→ CO (g) + 2 H2O (l)... (3)

Now, equation (3) most closely resembles equation (1), except we have to somehow get rid of the CO2and replace it with CO. Fortunately, we are given a thermochemical equation that converts between COand CO2 - equation (2). However, equation (1) and (2) have the CO2 on the same side. Hence, we needto reverse equation (2) to get the CO2 onto the opposite side so we can cancel out the CO2 by combiningequations (1) and (2). Reversing equation (2):


Remember that reversing a reaction equation means you change the sign of the ΔH value. Lets putequations (1) and (4) side by side for comparison:

CH4 (g) + 2 O2 (g)→ CO2 (g) + 2 H2O (l), ΔH = − 890 kJ mol−1... (1)


You can now just add equations (1) and (4) to, in effect, replace the CO2 with CO. Adding equationsmeans adding ΔH values:

CH4 (g) + 2 O2 (g) + CO2 (g)→ CO2 (g) + 2 H2O (l) + CO (g) + 12 O2 (g), ΔH = − 607 kJ mol−1

Simplifying this equation by cancelling out the CO2 and 12 the O2:

CH4 (g) + 32 O2 (g)→ CO (g) + 2 H2O (l), ΔH = − 607 kJ mol−1

There’s your answer!

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Part e. [Go to Model Solutions and Marking Scheme]

This question is difficult and requires you to think before you write. With these kinds of questions, ensurethat you have a good structure and you plan your response.

Let’s read the question again:“Compare and contrast the different sources of methane with respect to their renewability andenvironmental impacts related to sourcing and combustion.”

The first part of this question is - what are the different sources of methane? This requires recall!

Remember that methane can be sourced from:

• fossil fuels; methane is a hydrocarbon that will be a product of the fossilisation of plant materialthat has been deceased millions of years ago, and subjected to extremely high pressures via burial.The methane can be present -

– in gaseous deposits under the ground as part of natural gas– in liquid deposits, dissolved in crude oil (the methane comes out as association gas after

drilling into an oil well - this is very peripheral knowledge and is probably not examinable)

– in solid deposits, mixed in with coal, as coal seam gas

• biofuels; the methane can be sourced from recently deceased plant matter or sewage (where bacte-ria produce methane as a metabolite) as biogas

Now, what are the renewabilities of these different sources of methane? Remember that fossil fuels arenonrenewable, as the process of fossilisation takes millions of years and is an extremely slow process,whereas biofuels are renewable as the plants that provide the biomass for the generation of biogas canconstantly be grown. Therefore, the fossil-fuel derived sources of methane, being natural gas and coalseam gas, are nonrenewable whereas biogas is renewable.

Let us now focus on the environmental impact related to sourcing the fuel. With respect to coal seam gasand natural gas, both require drilling into the ground. To set this up, it is necessary to clear some land,which necessitates destruction of natural habitat for animals. This could affect ecosystems in negativeways. Even with biogas, biogas derived from biomass may involve clearing up land for the perpetualgrowth and harvesting of plants for biogas production. This leads to manipulation of natural habitats,which can also have negative ecological effects.

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Additional issues with respect to sourcing of fuels include:

• Gas leaks in the drilling of natural gas. This leads to significant amounts of methane being releasedinto the atmosphere; methane is a potent greenhouse gas

• Fracking. Fracking is a practice that is sometimes used to extract methane from coal seams. Itinvolves the pumping of fluid into the coal seam under very high pressures (alongside the drill),leading to the splitting of the coal seam and the subsequent release of methane gas. The issuehere is that the fracking fluid, that contains a mixture of chemicals, can seep through the rock andcontaminate water from groundwater aquifers that are sometimes used as sources of drinking water.

• Loss of water from aquifers. Drilling of methane from coal seams leads to the extraction of alot of water (called produced water) that is mixed in the coal seam. This depressurises the coalseam, which can cause water from nearby aquifers to seep into the coal seam, potentially removinga source of drinking water for the locals.

How about combustion?

The first thing to talk about are greenhouse gas emissions. Carbon dioxide is the typical greenhousegas. It goes without saying that burning methane from coal seams and from natural gas leads to theevolution of carbon dioxide into the atmosphere. Biogas is a little different. Yes, the methane frombiogas produces carbon dioxide when combusted, whih evolves into the atmosphere. However, keep inmind that we source biogas, at least partially, from biomass (recently deceased plant material). Now,these plants would have recently absorbed carbon dioxide from the atmosphere via photosynthesis, in thesynthesis of its structure and large organic molecules, in the course of its life. Therefore, it is believed thatthe CO2 evolved from the combustion of biogas is comparable to the CO2 absorbed from the atmosphereduring the production of one of the sources of biogas (biomass).

The burning of natural gas and coal seam gas is associated with some issues. In particular, there is somesulfur and nitrogen which, when combusted, lead to the evolution of sulfur and nitrogen oxides. Thesecan both lead to the production of acid rain, and nitrogen oxides in particular can lead to photochemicalsmog. That said, this particular issue is significantly less prominent in the combustion of gas comparedto the combustion of coal or crude oil.

Question 2

Part a. [Go to Model Solutions and Marking Scheme]

For an energy content to be able to be expressed in kJ mol−1, your substance has to have a known molarmass. In other words, your substance must be pure with a known formula. However, fuel oil is a mixtureof different hydrocarbons, whose exact composition cannot be perfectly ascertained.

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VCE Chemistry Units 3&4 Volume 1: Topic Tests