Each year, the VCAA subtly upgrades the VCE Chemistry data book. Each year, I print it and annotate it to show students the wealth of useful information hidden within it (most of which, is in plain sight).
This year, the VCAA has changed some “constants” and added some interesting functional groups to the spectroscopy tables. Smaller things are changed, too. All the protons in the 1H NMR table are now in bold; not just the ambiguous ones.
Start using this annotated version of the data book for your year 11 and year 12 chemistry homework exercises. While you can’t take this annotated version into the final examination (or into most SACs), seeing the annotations frequently throughout the two years will help you find things faster in the final examination.
Do you have feedback? Any comments? Do you require 1-to-1 chemistry tutoring? Email me at email@example.com and I’ll get back to you personally.
Inspired by the formula booklets used by VCE Physics and VCE Maths Methods, here’s an 8-page Chemistry formula booklet you can use for your Year 11 and 12 Chemistry assignments. This custom-made booklet is a collection of reliable formulae that I have been using to answer VCE Chemistry questions while teaching and tutoring around Melbourne.
There are 76 formulae on 8 pages. At least 10 of these formulae aren’t in the three main chemistry textbooks. Orders are shipped in A4-sized booklet that resembles the VCAA Data Booklet.
Orders from schools, students and tutors are all welcome. Price includes free international delivery and a 10% voucher for the T-shirt store.
James Kennedy achieved outstanding A-level results in 2006 in Maths, Chemistry, Physics and Biology. Those excellent grades (which equate to an ATAR of 99+) earned him a BA (Hons) degree and a Masters degree in Natural Sciences from the University of Cambridge.
Shortcut formulae were just one of the techniques James used to pass his A-level exams and get into Cambridge. Along with structured revision, revision guides, practice papers and study notes on wall-cards, James used shortcut formulae to save precious time in the examination hall. You can get your own copy of these original shortcut formulae – revised and updated for the 2017-2021 VCE Chemistry course – for just $55 including free international shipping. Click here to get your copy.
This post concludes the Periodic Table Smoothie experiment.
Recall that we’ve just finished adding one mole of nitrogen gas and created a bizarre boron polymer at the bottom of our vessel. The temperature was 350 °C and the pressure in our vessel was 891 kPa.
Today, we’re going to add 1.00 mole of oxygen gas, stand back and observe.
This is disappointing news.
Many of the substances in our vessel react (more accurately, explode) in the presence of oxygen but the ignition temperature for all of those explosions to take place is at least 500 °C. The temperature of our vessel is set at just 350 °C. At this temperature, nothing would actually happen.
There’s not enough activation energy to break bonds in the reactant particles in order to get the reaction started. We call this activation energy (EA) in chemistry. If we were to add a source of excessive heat (e.g. a matchstick), the vessel would explode.
Should we heat up the vessel to 500 °C and blow up the experiment right here?
If we did, the following reactions would happen:
Enough of these reactions – particularly the first three – are sufficiently exothermic to trigger a chain reaction – at least up to the reaction of oxygen with beryllium carbide. The vessel would bang, explode, and shatter. The helium would float away, dangerous lithium amide would fly out sideways, and polyborazine powder, whatever that is, would land on the floor.
Let’s not ignite our experiment – not yet.
Conclusion after adding 1.00 mole of oxygen gas
Amount in mol
Pressure: 891 kPa (higher than before due to the addition of nitrogen gas) Temperature: 350 °C (vessel is still being maintained at constant temperature)
Oxygen was relatively uneventful. Let’s add fluorine and see what happens.
Let’s add fluorine gas
The following three reactions would all occur as 1.00 mole of fluorine gas is added:
These two products are quite interesting:
HF, hydrogen fluoride, an aqueous solution of which was used by Breaking Bad’s Walter White to dissolve evidence (his victims)
NF3, nitrogen trifluoride, is used as an etching agent when making printed circuit boards (PCBs)
Let’s add neon gas
When 1.00 mole of neon gas is added, the total pressure inside the vessel increases but no reaction occurs. The concentrations of all the other gases present are unaffected.
That concludes our Periodic Table Smoothie experiment. The most interesting conclusion was the discovery of polyborazine, the bizarre solid that collected at the bottom of the vessel.
Also of interest was how easily we created ammonia, one of the simplest of biological compounds, just by mixing elements together. Could the compounds necessary for life be so easy to create that their existence is an inevitable consequence of the Big Bang? Is life inevitable? If the Big Bang were to happen all over again, would life occur? And would it look any different?
This book contains 50 lies taught in the VCE Chemistry course.
These lies include well-meaning simplifications of the truth, mistakes in the textbook, and, in a few extreme cases, blatant falsehoods.
This book isn’t a criticism of the VCE Chemistry course at all. In fact, I just want to highlight the sheer complexity of Chemistry and the need to make sweeping generalisations at every level so it can be comprehensible to our students. This is a legitimate practice called constructivism in pedagogical circles. (Look that up.)
Many of these ‘lies’ taught at VCE level will be debunked by your first-year chemistry lecturers at university.
Here’s a preview of some of the lies mentioned in the book. Check out all 50 by clicking the download link at the bottom of the page.
The content you’re learning now is probably not as true as it seems. Chemistry is a set of models that explain the macro level sometimes at the expense of detail. The more you study Chemistry, the more precise these models become, and they’ll gradually enlighten you with a newfound clarity about the inner workings of our universe. It’s profound.
Rules taught as ‘true’ usually work 90% of the time in this subject. Chemistry has rules, exceptions, exceptions to exceptions, and exceptions to those – you’ll need to peel pack these layers of rules and exceptions like an onion until you reach the core, where you’ll find Physics and Specialist Maths.
Enjoy this book. I hope it emboldens you to question everything you’re told, and encourages you to read beyond the courses you’re taught in school.
Today, we’re going to answer the following question:
When 200 grams of ice is added to a bucket containing 1.00 litre of hot water, what’s the final temperature of the water?
To answer the question, we’re going to need to make some assumptions. We’ll take 1.000 litre of pure water at 80.00°C and add 200.0 g of ice (at -10.00°C) to it. What’s the final temperature of the water?
Part 1: Heat transfer method
The following equation can calculate the temperature at thermal equilibrium of any number of objects in thermal contact.
I love this equation because it’s several lines of maths shorter than the version taught in school. With this equation, you don’t even need to convert the temperatures into kelvin. Celsius works just fine.
Let’s set up the equation so that the addition series contains the variables in the question.
Now, let’s substitute the gives values into the equation. The specific heat capacity of water is 4200 J kg-1 K-1, and that of ice is 2100 J kg-1 K-1.
Great! Adding 200.0 g of ice to 1.000 L of water decreases the temperature from 80.00°C to 71.80°C.
But we’ve forgotten something. The ice will melt as soon as it hits the hot water. Since melting is an endothermic process, heat energy from the water will actually be absorbed, thus reducing the final temperature even further.
Part 2: Let’s take into account the fact that the ice melts!
Remember our formula from part 1.
The amount of energy required to melt ice can be calculated using the latent heat equation:
Removing that amount of heat energy from the system results in the following equation:
Great! Now, we’ve calculated that the final temperature of the water would be 57.36°C after the addition of the ice. That’s equal to 330.5 kelvin, which will be useful later.
However, we’ve forgotten to take something else into account: how much heat will be lost as radiation from the surface of the bucket?
Part 3: What’s the rate of heat loss from the bucket by radiation?
The rate of heat lost by radiation can be calculated by using the Stefan-Boltzmann equation, below.
P is the rate at which heat energy is radiated from the surface of the bucket in watts. Emissivity, e, of water is 0.95, and the surface area, A, should be around 0.0707 m2 for a one-litre bucket. Calculation of A is shown below. Assuming that the radius of the surface of the bucket is 6cm:
Plugging that value into the equation, we can find P. We’ll assume that the experiment is being conducted at room temperature and the temperature of the surroundings is 20.00°C (29.03 K).
This means that 2.928 joules of energy are emitted from the surface of the bucket every second. Ten minutes later, the bucket would have lost 1756.8 joules of energy due to radiation from the surface. But what about emission of radiation from the sides of the bucket?
Let’s say that our bucket is made from highly polished aluminium (which has emissivity 0.035) and it holds exactly 1.2 litres of water. We need to calculate the dimensions of the bucket.
Assuming it has straight sides (i.e. it’s a cylinder), the bucket had volume equal to the following formula:
The surface area of our bucket (excluding the open surface at the top) is:
The rate of energy radiation from the sides would therefore be:
It’s interesting to note how very little radiation is emitted from the shiny aluminium bucket, while lots more radiation is emitted from the surface of the water. This is because relatively ‘dark’ water has a much higher emissivity than shiny aluminium. Total emission from the bucket is therefore:
After ten minutes, the bucket would have lost the following amount of energy:
Let’s factor this amount of energy loss into our final temperature equation.
Not much energy is lost via radiation! Finally, let’s find the peak wavelength of the radiation emitted by the object using Wien’s law.
Part 4: What’s the wavelength of the radiation being emitted by the bucket?
Here’s Wien’s law from Unit 1 Physics…
The radiation emitted from the resulting bucket of water lies firmly in the infra-red part of the electromagnetic spectrum. The bucket would be clearly visible on an infra-red camera!
Next week, we’ll begin a new a Chemistry-themed project called Periodic Table Smoothie. More next week.
Demonstrate electrolysis with an electrolytic cell in a petri dish.
1 × Large petri dish
1 × DC Power pack
~50 mL Distilled water dH2O(l)
~3 g potassium nitrate powder KNO3(s)
2 × Graphite electrodes
2 × Wires with crocodile clips
1 × Clamp and stand
1 × Very strong static magnet
1 × Roll of sticky tape (any type)
~10 drops of universal indicator
~50 mL dilute HNO3(aq)
~50 mL dilute KOH(aq)
1 × Spatula
Place petri dish on clean, light-coloured bench and add distilled water until it is two thirds full
Add ~10 drops of universal indicator and observe the colour. Q: What pH is the distilled water? (You’ll be surprised!) Q: Why is/isn’t the colour green?
Add ~3 g of potassium nitrate to the petri dish and stir using a spatula until completely dissolved
Adjust the pH of the distilled water carefully using the nitric acid and potassium hydroxide as required. Try to make the universal indicator colour green (as pictured) ~pH 7
Attach one electrode to each of two wires using crocodile clips
Dip each graphite electrode into the green solution at opposite sides of the petri dish. Hold these electrodes (and wires) in position by in position by sticky-taping each wire to the surface of the workbench
Demonstrate the strength of the magnet by attaching it to the clamp. Carefully, clamp the magnet into the clamp and position the magnet 2 mm above the surface of the green solution
Ensuring the power is turned off, very carefully, attach the wires to the DC power pack according to the manufacturer’s instructions
Turn the voltage to zero (or very low) and turn on the power pack
Turn the voltage up slowly (12 volts worked well) and observe any changes you might see in the Kennedy Rainbow Cell
Turn off the power pack and stir the solution. Explain why the colour goes back to being green. (If it’s not green, explain that, too!)
Turn the magnet upside-down (TURN OFF THE POWER FIRST)
Reverse the polarity of the wires
Use AC current instead of DC
Use different indicators
Why would using NaCl(aq) be dangerous in this cell?
Make your own risk assessment before carrying out this experiment
The strong magnet is capable of attracting both wires to itself. Don’t be touching the exposed parts of the crocodile clips when this happens. If this does happen, immediately turn off the power pack and fix the problem. Secure the wires with more tape. Don’t touch the electrodes while the Cell is operating.
Don’t use chloride salts or hydrochloric acid in this experiment. The voltages involved can cause the production of toxic chlorine gas if sodium chloride is used. Use nitric acid and potassium nitrate instead.
Make sure the wires don’t touch each other.
Again, make your own risk assessment before carrying out this experiment
This cell is potentially dangerous. I accept no responsibility for and loss, damage or injury caused by the operation of a Kennedy Rainbow Cell. If you’re under 18, always get adult permission before you make this type of cell.
The groups credited for creating them – in Japan, Russia and the US – have spent several years gathering enough evidence to convince experts from Iupac and its physics equivalent, the International Union of Pure and Applied Physics, of the elements’ existence. All four are highly unstable superheavy metals that exist for only a fraction of a second. They are made by bombarding heavy metal targets with beams of ions, and can usually only be detected by measuring the radiation and other nuclides produced as they decay.
Only positively-charged fragments from mass spectrometers produce a peak on the spectrum. Uncharged free radical fragments are not detected because they lack a positive charge.
Weak acids with a lower Ka value are the weakest… this means that they ionise to a lesser extent when in aqueous solution, giving rise to a lower concentration of available H3O+(aq) and a higher pH.
The conversion of triglycerides (a type of ester) into biodiesel (another type of ester) is called transesterification.
The covalent bonds between deoxyribose and phosphate groups in DNA form a group of atoms called a phosphodiester group.
Ether bonds and glycosidc bonds are not the same. Ether bonds are C-O-C. Glycosidic bonds are a type of covalent bond that joins a carbohydrate (sugar) molecule to another group, which may or may not be another carbohydrate.
Amide groups and peptide groups are not the same, either. Amide groups are CONH. Peptide groups are CONH between amino acid residues in a polypeptide chain. Nylon, for example, has amide groups (CONH) which aren’t called peptide groups.
Ether: C-O-C Ester: COO Amine: NH2 Amide: CONH
The molar mass of any amino acid without its Z-group is 74 gmol-1.
The molar mass of glucose, fructose and galactose (all monosaccharides) is 180 gmol-1. By coincidence, aspirin is also 180 gmol-1.
The molar mass of sucrose is 342 gmol-1 because (180*2)-18=342.
In general, energy is required to break bonds. Energy is released when bonds are formed.
Use the formula C-(H/2) to find how many C=C are present in a fatty acid (only works for fatty acids).
Use the shortcut formula (Ka/[acid])^0.5 to find % ionisation of a weak acid.
Use -log(Ka) to find the exact pH at the end point of an indicator.
Use the quick titration formula for rapid multi-choice titration questions: c1v1/ratio1 = c2v2/ratio2
A hydrogen bond is an intermolecular bond that forms between O-H groups. The covalent bond between the O and the H is not a hydrogen bond.
Can you write the half-equation for the reaction occurring at the anode in an ethanol-oxygen fuel cell with an alkaline electrolyte? Tip: start by writing the known reactants and products then use KOHES(OH) to balance your equation.
The products of a titration determine the pH at the equivalence point. For example, the the pH at the equivalence point in a titration between CH3COOH(aq) and NaOH(aq) is around 8.5 because at equivalence point, only products are present: Na+(aq) and CH3COO–(aq). The ethanoate ion (CH3COO–(aq)) is a weak base, which makes the solution produced slightly basic.
If you have absolutely no clue in the multiple choice sections, pick C. In the last 4 years of VCE Chemistry examinations, C has been correct 50% more of the time than B.
The multiple choice questions really do get harder towards the end. I’ve done the statistics.
Use your reading time wisely. During reading time, read all the questions with the following idea in mind: “how would I do this question?” without actually doing the question.
Bring sharp pencils.
Sleep early tonight (before 9pm). At this stage, getting enough sleep is far more important than revising those tiny details that may or may not come up in the examination.
The VCE Chemistry examination is only 22 days away. As you complete at least one practice paper each dayand correct them ccording to your revision timetable, you’ll be finding that you’ve already mastered certain topics while others remain difficult.
Patterns emerge in student readiness: each year, electrolysis is the worst-studied topic on the course. Because VCAA has a reputation for asking questions on topics that students repeatedly got wrong in previous years; I decided to test this hypothesis by getting real data from recent examination reports and displaying it on a scatterplot of:
how difficult each topic is (% of marks lost) on the x-axis
how often the topic is asked (marks per paper) on the y-axis
The results were fascinating. While it’s impossible to say with any certainty which topics will be on the examination this year, previous years’ examination papers have placed more emphasis on the difficult topics (electrolysis, Ka, redox and biofuels). Focus your revision on these topics again this year.
Conclusion: Focus your Chemistry revision this week on your least favourite topics… those topics will probably be worth more marks in the examination!
Calorimetry can be a confusing topic. Avoid common errors by following these essential tips:
Always label the units of E (kJ or J) above the E. This is the most common source of error in calorimetry calculations. Try this quick way to remember the required units of E: If there’s ΔH in the equation, the units are kJ; otherwise, the units are J.
In E=mcΔT, all the variables refer to the mass of water being heated. A common error among students is to use the mass of limiting reactant instead of the mass of water. Generally, m in this equation is 100 g or a similar round number.
Never convert ΔT to kelvin. Temperature changes are the same in kelvin and celcius… never add 273 when finding ΔT.
No calibration step? Use m×c instead. Because E=mcΔT and E=CfΔT, it therefore follows that Cf ≡ m×c. For example, if we’re heating a 100.0 g of water without a Cf, we should use Cf = 100×4.18 = 418 J K-1 instead.
In ΔH = E/n, n denotes the number of moles of limiting reactant. Never add up the number of moles of reactants: use the number of moles of limiting reagent only.
Calculate twice. Students most often make mistakes when converting hours or days into seconds. Many answers are therefore wrong by a factor of 60. Do your calculations twice: once while doing the question and again when you check over your answers at the end of the SAC or examination.
Know a ballpark figure. Neutralisation and solubility reactions tend to have 2-digit ΔH values; combustion reactions tend to have a 3-digit ΔH and explosive reactions tend to have a 4-digit ΔH. If you get a 5-digit ΔH value, you’ve probably forgotten to convert your answer into kilojoules!
Remember the ‘+’ or ‘-‘ sign! The calculator doesn’t know whether the answer should be positive or negative. Think about it yourself instead: endothermic reactions need a ‘+’ sign and exothermic need a ‘-‘ sign. VCAA awards a whole mark for getting the ‘+’ or ‘-‘ sign correct! It’s possibly the easiest mark in the whole paper.
Consider getting a home tutor who can answer your questions and explain difficult concepts to you. Students learn much faster with a tutor than on their own.
Redox can be a confusing topic for VCE Chemistry students. It’s also taught right at the end of the year, when students are tired and some teachers are rushing their lessons so they can finish the course before the end of Term 3. Student motivation levels are at their lowest time of the year, which means that students often finish the course with an incomplete understanding of Redox.
Fortunately, there are six universal principles that are always true in Redox no matter what type of cell is being studied.
First, here’s a reminder of the types of cells you need to have studied in this course.
Primary (can’t be recharged)
Secondary (can be recharged)
Fuel Cells (reactants are supplied continuously)
Electroplating Cells (no overall reaction)
Electrolytic Cells (non-spontaneous reaction)
Commercial Cells (usually molten electrolytes)
Recharge reaction of a secondary cell (non-spontaneous)
Now, here are the six universal Redox principles.
1. The strongest oxidant at the cathode reacts with the strongest reductant at the anode (SOC SRA)
To predict which species will react with each other, circle all the species present at the cathode on the electrochemical series. The highest species on the left will always react. Now, circle all the species present at the anode… the lowest species on the right will react.
2. The half-reaction with the highest E° value is always positive
In all cells, the half-equation with the highest electrode potential (also called ‘reduction potential’ or E° value) always occurs at the positive electrode. Similarly, the half-equation with the lowest electrode potential (E°) will always occurs at the negative electrode.
3. OIL RIG
Oxidation is loss of electrons. Reduction is gain of electrons.
4. ←AN OIL RIG CAT→
Anode reaction (oxidation reaction) is whichever reaction is happening to the left in the electrochemical series.
Cathode reaction (reduction reaction) is whichever reaction is happening to the right in the electrochemical series.
5. Electrons always flow in this order (RACO)
Reductant → anode → cathode → oxidant
6. In the internal circuit, cations always flow to the cathode, and anions always flow to the anode.
The internal circuit might be an electrolyte or a salt bridge that contains soluble weak oxidants and reductants such as KNO3(aq) (potassium nitrate). Either way:
cations always flow to the cathode; and
anions always flow to the anode.
Keep practicing redox questions by completing past papers, Checkpoints and Lisachem questions. If you need more help, contact me via the Get a Tutor button in the site’s menu bar. Students learn much faster with a tutor than on their own.
Track your progress in VCE Chemistry with this A3 size progress tracker. Cross out or colour in each box as you complete it, and write your scores in . Start at the bottom (highlighted) and work your way upwards.
A ‘minimum expected level of examination preparation’ of 26 examination papers is labelled on the chart. Write your percentage scores in each of the boxes as you mark each paper. When you’re achieving past/practice examination scores concordantly above 90%, you’re ready to sit the VCE Chemistry examination.
1. Develop excellent study skills. Cultivate ideal study habits such as waking up early, reading your notes before school, doing all homework on time and studying even when there’s no homework set.
2. Stay committed and know what you want and WHY. People who know why they do what they do are far more likely to persist and put in the huge number of hours required to excel at that particular skill. All successful people were driven by a higher. Find your why and you’ll feel more motivated to study VCE.
3. Keep motivation levels high and consistent throughout the year. Remind yourself constantly why you’re studying the VCE subjcets you’ve chosen.
4. Do not “over-indulge” in VCE tutoring. Your tutors and teachers can only take you so far. The highest-achieving students are those who are self-motivated: they push themselves and study even when nobody asked them to. Become self-motivated and use your tutoring time wisely to maximise your performance in VCE exams.
2. There are two things you need to do: make great notes and do practice questions.
3. Build on your notes from external sources (other people’s notes and the textbook)
4. Mark your questions – or get them marked! Akhil says that while it’s an excellent learning exercise to practice marking questions by yourself, it’s also necessary to get your practice papers and Checkpoints questions marked by a teacher or tutor because they’ll be more vigilant with sticking to the marking scheme and can pick up slight errors in wording that are easy to miss if you mark your own work.