3. Module 5 Trends
● Module 5 made up ~¼ of the exam in the 2021 HSC.
○ Accounted for 19/80 SAQ marks in 2021 (Q22, 26, 31, 33)
○ Compared to previous years: 2020 had 20/80 marks (although 4 marks wasn’t really M5, it
was just maths), and 2019 had only 12/80 marks (this seems to be an outlier year.)
● I anticipate that M5 will continue to make up ~¼ of the exam.
● M5 also has a high potential for difficult/curveball-style questions with
hidden traps. This makes M5 a grounded conceptual understanding of M5
especially important.
● Furthermore, although M5 has a large number of dotpoints, it has received
less attention compared to, say, M6 (which has less dotpoints!).
Consequently, some M5 dotpoints are under-assessed in NESA exams.
4.
5.
6. Crystal Ball Gazing
Disclaimer: these slides are my own opinion, and the past does not necessarily predict the future. Trust these slides at your own risk!
Very high likelihood:
● Keq calculation question (2019-2021)
● Qualitative equilibrium shift question (using LCP or collision theory) (2019-2021)
● Ksp calculation question (2021, did not appear in 2020 or 2019 SAQ, but
present every year in MCQ). Notably, questions involving the common ion effect
has also never been asked…
These three topics would, in my opinion, likely make up 10+ marks of M5 already!
Hence, if you are short on time, it may be best to study these question types, as you
would likely hit 50+% of the M5 questions asked on the exam. These topics are also
relatively straightforward.
7. Crystal Ball Gazing
Possible questions(even more speculation now! Again, trust at your own risk.)
● Static vs dynamic equilibrium (has not been properly assessed)
● Using ΔG to determine spontaneity of reaction/equilibrium (assessed in 2021
and arguably 2019 Q30)
● Prac to determine the value of Keq (not assessed, but I believe this dotpoint
would be difficult to assess; the majority of the state would likely be unfamiliar
with the prac).
● Process of dissolution of ionic compounds in water (questions have not
specifically assessed the process of dissolution in terms of bonds/forces
broken/formed).
● Cycad Fruit (not formally assessed).
9. Keq calculations
● No difficult Keq calculation has ever been asked by NESA.
○ However, this could change…
● A typical calculation involves:
○ Writing/balancing the equation (if required)
○ Writing Keq expression
○ Drawing ICE/RICE table
○ Making the small x assumption
○ Solving for x
○ Using x to find whatever value the question is asking for
● Let’s go through the common mistakes that I have seen students make in
each step!
● We’ll then go through a few weirder calculation types that could
conceivably be asked as they do fall within syllabus (although how
likely/unlikely remains to be seen…)
10. Common errors
● Writing/balancing the equation
○ This will likely be done for you already, but always double check equations before writing
your Keq/ICE table. Schools have sneakily given students unbalanced equations in the past.
● Writing Keq expression
○ Unlikely to make mistakes here :)
● Drawing ICE/RICE table
○ Not putting in concentrations!
■ Many questions (esp. involving gases) will give a volume and moles of each species; it
is up to you to convert moles to concentration.
■ I dislike fractions in ICE tables, it’s far cleaner to quickly write up a few c = n/v
expressions right before your ICE table.
■ Beware the 1.00L questions: I have seen students lose marks in school assessment
tasks because they subbed in the moles (which equals concentrations due to volume of
1.00L), but lose marks because they did not explicitly state they were subbing in
concentrations…
11. Common errors
● Drawing ICE/RICE table (cont.)
○ You are allowed to assume the concentration of 0 for certain species if they are in the
equation, but the question stem does not specify the amount of that species. However, best
to write down that you are making this assumption.
○ Not taking stoichiometric ratios into account for the C row.
○ Not knowing which way equilibrium is supposed to shift:
■ As a rule, equilibrium will always shift towards the side with a 0 as its concentration.
■ If you have 0’s on both sides of the table, you’ve done the question wrong!
■ If all species have a non-zero concentration, use Q to determine which way
equilibrium will shift.
12. Common errors
● Making the small x assumption
○ Many students do not actually understand why they are making this assumption: let’s try an
example question to understand how this assumption works.
13. Common errors
● Making the small x assumption
○ We can see that the small x assumption only holds when Keq is quite small. This is because
when Keq is small, we can say that equilibrium will likely not shift very far right. Luckily, our
initial conditions of 0.15M H2 and 0.25M I2 is also not very far to the right!
○ Always substitute x back into your Keq expression to check that the assumption is valid.
■ Try subbing our answer just then back in Keq - what do you find?
○ As a rough rule of thumb, if our calculated x is more than 5% of the smallest value we
subtract x from, the assumption is a bit dodgy.
When writing out the assumption, I assume everyone already has their own ways
of writing it out. I encourage you to write too much as opposed to too little.
E.g. We assume x is very small as Keq = 3.44 x 10-3 << 1. This indicates that equilibrium will lies far to
the left, which is similar to our initial conditions. ∴ 0.15 - x ≈ 0.15, and 0.25 - x ≈ 0.25.
14. Common errors
● Solving for x
○ Generally only see silly mistakes/calculator entry errors here!
● Using x to find whatever value the question is asking for.
○ Read the question!!!
○ Chemistry is not maths, finding x does not mean the question is finished.
○ Even if x does correspond to the answer you want, you are still likely missing a unit.
○ Always answer the question in your very last line. X = ____ should never be the last line
in a question.
○ E.g. ∴ [HI] at equilibrium = 6.9 x 10-3 (2 s.f.)
○ Also, remember your sig figs!
16. Same question, different flavour
● Changing reaction conditions:
○ Since Keq is very big indicating equilibrium lies far to the right, we can still make the small x
assumption if we change our initial conditions to also be far to the right!
○ This means that once again, our initial conditions are very close to where equilibrium lies,
making our small x assumption once again valid.
○ Explain your working out! An extra row in the ICE table may not be enough.
○ The same guidelines apply: check your assumption, and explain your assumption!
18. One more Keq question
● In this question, the small x assumption cannot be made since x is neither
big nor small.
● Luckily, the design of the question allows us to solve this question purely
algebraically! No assumptions required.
I personally believe this question type likely won’t be asked, but it doesn’t hurt to
know just in case! I believe this question type would be in syllabus as it does not
require anything beyond basic algebra, but tricks students who tend to blindly
use the small x assumption.
Try using the small x assumption in this question - what happens?
20. Equilibrium Shifts
● There are three methods for justifying the direction of an equilibrium shift:
○ LCP
○ Collision Theory
○ Reaction Quotient
● These methods are generally used in different contexts:
○ LCP: bread-and-butter method, use whenever the other two methods don’t apply. Generally
worth 2 marks.
○ Collision Theory: questions will generally specify that collision theory be used as the
justification in the question stem. Will generally be worth more marks. We will talk about
C.T. more today!
○ Reaction Quotient: used as a quantitative method; you will be given numbers to sub into Q
and asked to compare with Keq - generally it is pretty obvious when you need to use Q!
21. LCP
Generally, students don’t make too many errors with LCP.
Some pointers:
● Don’t forget to define LCP before using it just in case defining it is in the
marking criteria.
● Don’t forget to link the direction of equilibrium shift to the concentrations of
reactants/products. Students have lost marks in the past for correctly
predicting the direction of equilibrium shift; however the question asked for
“what would happen to the concentration of ___”.
22. Collision Theory
Essentially, two statements that you should state if a question asks you to use it:
1. For a collision between reactant particles to be successful, the particles must
collide with both sufficient energy and the correct orientation.
2. The rate of reaction is proportional to the frequency of successful collisions.
● Consequently, there are three fundamental “probabilities” involved in
determining the rate of reaction:
○ 1. The probability for a collision to occur in the first place.
○ 2. The probability for said collision to have sufficient energy.
○ 3. The probability for said collision to have the correct orientation.
● A successful collision occurs if all of 1, 2, and 3 are true!
● Consequently, increasing the probability of one or more of 1, 2, or 3 will
increase the rate of reaction (through pressure, volume, temperature…).
23. How do we answer Collision Theory Qs?
● All questions asking you to explain the impact of a disturbance using
collision theory have a similar answer structure:
○ First, quote collision theory.
○ Then, mention which probability(s) is being changed (out of the three listed before:
probability of collision occurring, of sufficient energy, or of correct orientation), and whether it
is increasing or decreasing.
○ Consequently, predict what would happen to the rate of forward reaction and rate of
reverse reaction*, and from this, which way equilibrium will shift.
○ It’s good to mention how the rate of forward and reverse reactions will move to reach
equilibrium again, although this may not be required.
● Some questions only require you to justify what happens to the rate of
forward/reverse reaction and not the shift in equilibrium - if this happens,
your job is easier! Use the same structure, but stop at the *.
24. How do we answer Collision Theory Qs?
● E.g. Use collision theory to explain why increasing [X] in X ⇌ Y shifts
equilibrium to the right:
○ Collision theory states that ________ (quote C.T. here)
○ As a result, the number of collisions for the forward reaction is being increased, whilst the
number of collisions for the reverse reaction is constant.
○ Consequently, the rate of forward reaction will increase whilst the rate of reverse reaction
remains constant.
○ This causes equilibrium to shift to the right as there is a net forward reaction. Eventually, the
rate of forward reaction will decrease and the rate of reverse reaction would increase,
causing the two to become equal again, thus restoring equilibrium.
25. CO2 (g) + H2 (g) ⇌ CO (g) + H2O (g)
What happens to equilibrium if [CO] is decreased? Explain making reference to collision theory.
26. Collision Theory
● Fundamental probabilities affecting rate of reaction:
○ 1. The probability for a collision to occur in the first place.
○ 2. The probability for said collision to have sufficient energy.
○ 3. The probability for said collision to have the correct orientation.
● Changing pressures/concentrations:
○ Affects probability 1. If a reactant is added/removed, affects the rate of forward reaction only.
If a product is added/removed, affects the rate of reverse reaction only.
● Changing volumes:
○ Essentially a more complex change in pressures
○ Hard to justify changes using collision theory alone: we need to bring in rate expressions to
explain this: see speaker notes.
■ As a result, I doubt a HSC question will ask you to explain this. I have not seen
questions even in school papers asking students to explain a volume shift using C.T.!
27. Collision Theory
● Fundamental probabilities affecting rate of reaction:
○ 1. The probability for a collision to occur in the first place.
○ 2. The probability for said collision to have sufficient energy.
○ 3. The probability for said collision to have the correct orientation.
● Changing Temperatures:
○ Affects both probability 1 and 2.
○ Will affect both the rate of forward and reverse reaction! However, will affect one more than
the other.
○ To determine which rate increases more, we need to use the Maxwell-Boltzmann
distribution to really explain it.
○ However, the HSC seems to accept more simplistic answers as we will shortly see.
○ Let’s quickly explore the Maxwell-Boltzmann distribution:
30. Collision Theory
● Fundamental probabilities affecting rate of reaction:
○ 1. The probability for a collision to occur in the first place.
○ 2. The probability for said collision to have sufficient energy.
○ 3. The probability for said collision to have the correct orientation.
● Adding a catalyst
○ Affects probability 2 and possibly 3.
○ However, affects both forward and reverse rate of reaction equally - and so there is no net
disturbance to the equilibrium!
■ The mathematical reason why this is the case is beyond the syllabus.
● Adding an inert gas
○ Does not affect any of the probabilities above, therefore does not affect rate of reaction!
31. Some questions
A sealed syringe of gas containing a mixture of NO2 and N2O4 is used to investigate the properties of an
equilibrium system. NO2 and N2O4 form an equilibrium as follows:
2NO2 (g) ⇌ N2O4 (g)
Brown colourless
The sealed syringe was compressed such that its volume was halved. Describe the colour change of the
sealed syringe from the moment of compression until equilibrium is re-established. Make reference to
collision theory in your response.
33. Predicting precipitate formation
A 45 mL sample of 0.015 M calcium chloride is added to 55 mL of 0.030 M
sodium sulfate. Predict if a precipitate will form.
Extension: Calculate the volume of calcium chloride needed for a precipitate to
begin forming. (Should be out of syllabus, I believe it requires quadratic formula
unless there’s an easy way to do it which I haven’t found)
34. Finding Solubility using Ksp
● At the HSC level, you should expect questions to involve the common ion effect;
finding solubility just from Ksp in pure water would be a MCQ question.
○ 0.054g of Silver hydroxide was added to 250mL of water, and it is known that
some amount of Silver hydroxide dissolves. 1.680g of sodium hydroxide was
then added to the same solution. Calculate the mass of dissolved Silver
hydroxide which precipitated out of solution when sodium hydroxide was added.
You may assume that the volume of the solution remains constant throughout.
37. Rate-time & Concentration-time Graphs
Common Mistakes:
● Events not lining up at the same point in time:
○ Reaching equilibrium
○ Onset of changes
● Rate-time graph lines not meeting at equilibrium (Ksp exception)
● Temperature change:
○ Instantaneous rate-time jump, gradual concentration-time change
● Volume change:
○ Concentrations will change as a ratio of their initial amount - e.g. if volume halves, all
concentrations will double - THIS IS REGARDLESS OF STOICHIOMETRIC RATIO.
38. Rate-time & Concentration-time Graphs
Common Mistakes:
● Concentration-graph lines changing in correct stoichiometric ratios
○ Stoichiometric ratios apply for pretty much every change after the disturbance, when going
from the position after disturbance but before equilibrium, to final equilibrium.
● For concentration-time graphs after return to equilibrium after a disturbance,
the concentration cannot return all the way back to baseline.
○ This is because LCP also states that an equilibrium shift in response to a disturbance will
only partially counteract the initial disturbance!
39.
40. Identify the mistakes in this figure below, and draw a rate-time graph illustrating
the difference between a catalysed and uncatalysed reaction.
41. The rate-time graph below (with a rather unique shape) represents the rate of
the forward and the reverse reaction for a particular reaction when a particular
disturbance occurred. You may assume that the disturbance occurred
instantaneously. Give an example of a equilibrium reaction and two disturbances
which could lead to the graph below.
42.
43. From Ruse Trials 2022*
At room temperature, it is known that solid iodine forms an equilibrium with purple iodine
vapor:
I2 (s) ⇌ I2 (g)
Some iodine solid was placed in a sealed flask and allowed to reach equilibrium. Which of
the following disturbances would cause the purple colour in the flask to be more intense
when equilibrium is re-established?
A. Decreasing the volume of the flask
B. Increasing temperature
C. Adding more I2 (s)
*I don’t have a copy of the paper, hopefully this question captures the essence of the question!
44. Funny troll question (won’t come up in the HSC)
A cylindrical syringe containing H2, Cl2, and HCl gas is at equilibrium:
H2 (g) + Cl2 (g) ⇌ 2HCl (g)
Cl2 is green-yellow, whilst the other two gases are colourless. The syringe is
compressed so that its volume is halved, and the gas mixture allowed to reach
equilibrium.
Anna says: “the colour of the syringe got darker yellow”. However, Brendan says:
“the colour of the syringe did not change”.
A: Only Anna could be correct
B: Only Brendan could be correct
C: Both Anna and Brendan could be correct
D: Neither Anna or Brendan could be correct.
Editor's Notes
X = 3.44 x 10-3 (2sf)
X = 3.4369…e-3
Subbing back in gives us 1.3075…x10-3, which is already a bit different from our original Keq value.
Subbing back in gives us 1.3578…x10-2, which is already a bit different from our original Keq value.
X = 2.6 x 10-4 (2sf)
X = 0.32 (2sf)
Using the small x assumption gives us a negative denominator in the Keq expression - not possible!
All rates of reactions can be written in the form k[A]^a[B]^b, where k is a constant, [A] and [B] are reactant concentrations, and a and b are stoichiometric coefficient. This is called a rate expression.
When volume is changed, the rate expression also changes as [A] and [B] change.
However, due to the fact that stoichiometric coefficients are involved, the side with the bigger sum of a+b will see a bigger magnitude in change:
So, if volume is increased, all concentrations would increase. However, the side with a bigger a+b will increase more, and so would end up with a faster rate of reaction.
Consequently, equilibrium will shift towards the side with a smaller a+b sum!
This logic will also work for volume decreases.
Overall, the fact that you need to do all of this work to explain volume changes properly means that I really don’t think this will be asked in the HSC. However for the sake of completion I have included the explanation here!
Less brown than
After mixing, [ca] = 6.75e-3, [so4] = 1.65e-2
Q = 1.11e-4, which is greater than ksp of 4.93e-5, so precipitate forms.
Volume required is 7.87mL
Remember to use ICE table and change initial conditions: assume NaOH was added first.
1.766466 x 10^-2g/L in pure water
1.487x 10^-5g/L in 1.68g of NaOH
Let [ca] = x, [sr] = y
F = 2(x+y)
Using ksp: 3.9e-11 = x(2(x+y))^2, 2.9e-9 = y(2(x+y))^2
Divide two ksp: 3.9e-11/2.9e-9 = [ca]/[sr], and so [ca] = 0.0134…[sr]
Sub back into equation, we get [sr] = 8.9e-4, [ca] = 1.2e-5, [f] = 1.8e-3
A
Catalyst addition (obvious answer)
Volume decrease for reaction with same stoichiometric ratio left and right summed
Example: water-gas shift reaction (CO + H2O ⇌ CO2 + H2), catalyst is Fe2O3
C: depends on angle of looking at the gas (front on vs side view) XD