Guide to HSC Chemistry Module 5: Equilibrium and Acid Reactions

Are you staring at the HSC Chemistry Module 5 syllabus wondering where to start and what equilibrium even means? 🤔

I was there too – until I discovered the game-changing strategies that turned Module 5 Chemistry from being my most difficult topic into one which I aced with confidence and ease!

Having helped countless students go from struggling with the module to achieving Band 6 responses, I’m sharing my BEST tips and techniques that will transform YOUR understanding of equilibrium and help you master Module 5 Chemistry in no time!

So what are you waiting for?

Let’s get started!

General Overview of HSC Chemistry Module 5: Equilibrium and Acid Reactions 

Topic 1: Static and Dynamic Equilibrium

Topic 2: Factors that Affect Equilibrium

Topic 3: Calculating the Equilibrium Constant

Topic 4: Solution Equilibria

How Do I Get a Band 6 for this Module? 

General Overview of HSC Chemistry Module 5: Equilibrium and Acid Reactions

As the name suggests, the entire Module 5 Chemistry syllabus focuses on equilibrium, making it a vital concept to understand in order to do well in HSC Chemistry. 

But what is equilibrium? 🤔

We can think of equilibrium as a balancing of scales. 

Put simply, equilibrium occurs when no visible change in a reaction can be perceived. 

Visible changes refer to changes in colour, precipitation or state, such as the formation of gas.

Can you think of any other visible changes?

At the core of Module 5 Chemistry is the idea that Equilibrium can be either static or dynamic.

Then we’ll look at the factors that can change equilibrium, and then apply collision theory to learn how to determine the direction in which equilibrium shifts.

So, the topics under this module are:

• Topic 1: Static and Dynamic Equilibrium
• Topic 2: Factors that Affect Equilibrium
• Topic 3: Calculating the Equilibrium Constant
• Topic 4: Solution Equilibria

Looking for some extra help with HSC Chemistry? We have an incredible team of HSC Chemistry coaches and mentors here to share their expert tips in personalised lessons to help you succeed!

WHEN and HOW will Module 5 Chemistry be ASSESSED?🧐 

But first, you might be wondering how you will actually be tested on the knowledge you’ve learnt throughout Module 5 Chemistry.

Module 5 is usually assessed as a sit-down exam, with a focus on applying concepts (such as Le Chatelier’s Principle, which we’ll talk about soon) and doing calculations. 

Practicals are usually reserved for Module 6, but it’s important to mention that I’ve witnessed schools teach these modules out of order and examined Module 5 as a research task!

The image above is taken from the NESA Sample Assessment Schedules (at the bottom of this page), which can be useful for determining how YOUR school might structure the HSC Chemistry Modules and Assessments!

Topic 1: Static and Dynamic Equilibrium

Inquiry Q: What happens when chemical reactions do not go through to completion?

Under this topic, ‘Static and dynamic have the same definitions as in any other sense. 

For example, when athletes warm up before sports games, some stretches will be static, like holding a toe touch, while others are dynamic, such as walking lunges. 

Now here’s the common misconception that might get you confused. 

If I just referred to ‘equilibrium’ as a state of no visible change, doesn’t that mean the reaction is at a standstill? How can something that’s a dynamic, moving process be at equilibrium ⁉️

The key here is the specific wording.

There are no physical changes in a reaction at equilibrium, but despite this, reactions continue to occur. 

For example, the ozone layer is composed of O3 molecules, but when these absorb radiation it turns into our usual diatomic O2 molecules:

But that leaves all these single and lonely oxygen molecules floating around.

We know oxygen likes to be in a pair (diatomic) which means all the lonely oxygen atoms form 2O pairs, which then become O₂ molecules.

This is an example of an equilibrium reaction occurring without any physical changes. 

The key is that In dynamic equilibrium, the forward and reverse reactions occur at constant rates!

Let’s now explore this concept using the Haber Process:

The Haber Process

The equation for the Haber Process is: 

\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)

Here, the double tipped arrow shows that this reaction can go in both the forward and the reverse direction. 

As we just mentioned, in dynamic equilibrium these reactions occur at constant rates.

Here’s an analogy that helped me understand this idea and one that I use often when teaching:

Imagine a friend sends you lots of TikToks. For every one TikTok you share with your friend, they share two back with you.

The number of TikToks you are exchanging remains the same (constant). So, the “rate” of TikToks being sent back and forth will be constant.

Static equilibrium is easier to identify than dynamic equilibrium, as static equilibrium reaches completion so there cannot be a reverse reaction!

When a reaction proceeds to completion, it means no further reactions are possible.

The example I like to think of is burning wood for a campfire.

It’s impossible to ‘unburn’ wood because once it’s burnt, it’s just a husk of carbon.

We are limited by the amount of wood available and the reaction will be completed once the wood is used up.

This is an example of static equilibrium, as the reaction is balanced and has reached completion.

Need to review more concepts from Topic 1? Be sure to check out HSC Together which has FREE video resources explaining concepts within each syllabus dot point!

Topic 2: Factors that Affect Equilibrium

Inquiry Q: What factors affect equilibrium and how?

Now that we’re familiar with dynamic equilibrium, let’s dive a little deeper. 

Le Chatelier’s Principle

Le Chatelier’s Principle (LCP) is a concept that comes up a lot in HSC Chemistry Module 5. 

It can be quite confusing, so here’s a scenario that helped me wrap my head around it (and hopefully helps you too):

Imagine two people sitting on either end of a seesaw, if they weigh the same amount, the perfectly balanced seesaw will be parallel to the ground. 

Adding a third person to the left side will cause it to become much heavier and the right side will rise. 

As we all know, for the seesaw to become balanced again, a fourth person with the same weight as the third must be added to the right side.

This essentially describes Le Chatelier’s Principle. 

The seesaw represents dynamic equilibrium when it is parallel to the ground. The people getting on (or off) the seesaw represent the “changes” affecting the equilibrium of a reaction.

The difference is that in chemical reactions, the changes that affect equilibrium are not people moving from one side to another, but rather, the ‘factors’ that manipulate a reaction. 

Factors that Affect Equilibrium:

These factors include changes in:

1. Concentration,
2. Temperature and
3. Volume (or Pressure). 

Once one of these conditions is changed, LCP essentially states that the “system” (chemical reaction) fights to overcome or “counteract” this change by either 

1. accelerating the forward reaction or,
2. shifting to the reverse reaction,

      …with the goal of reestablishing equilibrium. 

I know that was a bit of a wordy sentence, so let’s solidify our understanding with an example.

When talking about shifting equilibrium, I used to find it extremely confusing when referring to ‘left’ and ‘right’.

You can use the seesaw or balancing of scales metaphor to visualise the effect of these changes on the equilibrium of a system.

We can identify how various changes to the system will affect the position of equilibrium using the chemical reaction of nitrogen dioxide to dinitrogen tetroxide.

At equilibrium, the equation for this reaction is:

2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \quad \Delta H = -57.2 \, \text{kJ/mol} ,

where NO₂ gas is reddish brown in colour, and  N₂O₄ is colourless.

#1: Changing Concentration

If we were to increase the concentration of our NO₂ in the solution, the increase in molecules would result in a “shift to the right” in equilibrium, as the forward reaction is favoured, producing more N₂O₄.

As you’ve probably guessed,

I’m sure you’ve already guessed, the same is also true in reverse.

If we increase the concentration of N₂O₄, more N₂O₄ molecules are available to react (or collide) to produce NO₂ molecules, shifting the equilibrium to the left.

Note, this reaction has distinctive colours depending on whether the forward or reverse reaction is being favoured.

We can (and you will often be asked to) identify these changes visually as well.

#2: Changing Temperature

Now, let’s take a look at how temperature affects this reaction and its equilibrium position.

As you can see in the equation above, the forward reaction is exothermic with a change in enthalpy of -57.2 kJ/mol.

We are given the change in enthalpy for the forward reaction, and we can use this information to find the enthalpy of the reverse reaction.

The reverse reaction is a “flipped” version of the forward reaction. This means the change in enthalpy will have the same magnitude (number value) but opposite sign, making it +57.2 kJ/mol.

If we increase the temperature, this change shifts equilibrium. The endothermic reaction would now be preferred since the extra energy (heat) being introduced to the system needs to be countered.

The effect of this change results in the reverse reaction being favoured, producing more NO₂.

As we know from Year 11 chemistry, exothermic reactions release heat whereas endothermic reactions absorb heat. 

If we follow this thinking, we can see that when we decrease the temperature the forward reaction will continue to be preferred, producing more N₂O₄ molecules and releasing heat.

#3: Changing Volume or Pressure

The last shifts to equilibrium that is important to us in HSC Chemistry Module 5 are changes in volume or pressure.

When we have equilibrium reactions where we have only gaseous components, volume and pressure are interchangeable as they have the same effect on equilibrium as each other.

The easiest way that helped me understand this is by imagining our equilibrium system occurring in a stoppered container.

The syringe is half-filled with our solution.

By pushing the plunger in and halving the volume we are also increasing the pressure. This is because the same number of particles in our system is now contained in a smaller space.

If you don’t like crowded spaces then you can relate to our NO₂ and N₂O₄ molecules – they are feeling pretty squashed and claustrophobic!

The system wants to combat this change, doing so by favouring the reaction with a lower number of molecules. 

Have a look at our equation again

2NO_2(g) \rightleftharpoons N_2O_4(g) \quad \Delta H = -57.2 \, \text{kJ/mol}

The forward reaction produces one molecule of N₂O₄ for every two molecules of NO₂ whereas the reverse reaction produces two molecules of NO₂ for every molecule of N₂O₄. 

We can see that the forward reaction results in fewer molecules compared to the reverse reaction.

In a crowded space you would want to decrease the amount of people around you right? 

So does the system!

This is why the forward reaction is preferred when we decrease the volume (or increase the pressure).

If we now pull the plunger back out, giving the system more space to occupy in the syringe as before, there is now more space to fill, and with more space come more molecules (which for this reaction is the NO₂ molecules) meaning the reverse reaction is favoured in order to re-establish equilibrium.

Now that you’ve learnt the specific changes that can affect equilibrium, HSC Chemistry Module 5 requires us to look at two other concepts that give us more answers for why these changes affect our reactions – Collision Theory, and Activation Energy.

Collision Theory

As you may recall from Year 11, collision theory states that reactions occur when molecules “collide” with each other in set conditions, allowing them to react with one another. 

Let’s apply the collision theory to each of the changes that can affect equilibrium.

#1: Increasing Concentration

When we increase the concentration of NO₂ molecules in our system, there are more of them available to successfully collide and react with one another.

This simply means that more forward reactions occur due to a greater number of available molecules. 

Both the forward and reverse reactions are still occurring, however, the forward reaction is favoured due to the increased presence of NO₂ .

This results in the production of more N₂O₄ molecules to balance and re-establish the equilibrium.

#2: Decreasing Volume

The effect of collision theory on our system, when we decrease the volume, is very similar.

Less space for the molecules causes them to collide more frequently, favouring the forward reaction and producing more N₂O₄ molecules.

#3: Changing Temperature

The change in temperature can also be explained using collision theory.

The concept of activation energy is crucial to fully understanding this concept.

The activation energy for a reaction is the amount of energy required for a reaction to occur.

We know that exothermic reactions have a lower activation energy compared to endothermic reactions.

Now, if we increase the temperature, there is more energy in our system.

Like kids on a sugar high, our molecules are moving around much faster creating greater possibilities for collisions to occur.

However, adding energy to a system that wants to overall release energy (as our forward reaction is exothermic) is like someone telling you to put a jumper on when you’re already sweaty and hot. 

You feel like doing the opposite, and so does the system.

While more reactions occur because of faster molecules and more collisions, the system also has the energy for more reverse reactions. 

As the reverse reaction is endothermic, it absorbs the added energy (increase in temperature), resulting in the reverse reaction being favoured and more NO₂ molecules being produced to re-establish equilibrium.

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Topic 3: Calculating the Equilibrium Constant

Inquiry Q: How can the position of equilibrium be described and what does the equilibrium constant represent?

Now that we are comfortable with equilibrium in both its static and dynamic form, we can delve into its applications. 

In science, we love constants and formulas, as well as calculating values.

This is where the equilibrium constant (K_eq) is born.

The equilibrium constant is the ratio of concentrations of the products to the reactants at the specific temperature of the system.

Okay, that was a lot of words, have a look at the diagram below to see what I mean.

This is the general example for a K_eq equation.

However this can still be pretty confusing (at least it initially was for me), so let’s use it on our previous equilibrium reaction.

And now we can calculate the K_eq if we have equilibrium concentrations.

So if the concentration of NO₂ is 0.3M and the concentration of N2O4 is 0.52M (at equilibrium) and we’ll assume the reaction is taking place at SLC (25C or 298K).

The important thing here is to ensure we write the temperature of the system after our calculated K_eq.

We’ve already learnt that temperature is one of the changes that can shift the equilibrium of a reaction and because of this, it is crucial we indicate that the temperature has not changed.

Magnitude of K_eq

A large K_eq indicates that the forward reaction is favoured to establish equilibrium. 

Meanwhile, a smaller K_eq means the reverse reaction is favoured.

I initially found this a really hard concept to grasp because I didn’t understand why!

It wasn’t until I compared it to fractions that I could fully understand what a smaller or larger K_eq meant.

We’ve now calculated the K_eq for a system that is already at equilibrium, but as budding scientists, there is much more to explore.

For example, let’s take this example where we initially have 0.1M of NO₂. When the system reaches equilibrium we have 0.01M of NO₂. What is the K_eq for this reaction?

So, the K_eq is 450 at 298K.

The Reaction Quotient, ‘Q’

In both examples above, we were told that the system was at equilibrium.

But what about when we don’t know? 

Often, you will be given the K_eq and concentrations of a system and will be asked to determine whether it is at equilibrium.

This is where Q, the reaction quotient, comes in.

Let’s say we have a closed system that has reached equilibrium, and the K_eq is 27 at 298K. We have a second system with concentrations of 0.8M and 0.3M for NO₂ and N₂O₄ respectively.

Now, calculate both the Q and K_eq for this reaction:

Since our Q is smaller than the K_eq for this reaction, we know that our second system is not at equilibrium and that for it to reach equilibrium it will need to shift to the right.

(Remember a shift to the right means that more products will need to be formed to establish equilibrium. We can confirm this using the fraction metaphor from above)

Check out our Top 5 Study Tips for HSC Chemistry to make your life 100x easier and score that Band 6!

Topic 4: Solution Equilibria

Inquiry Q: How can the position of equilibrium be described and what does the equilibrium constant represent?

When we add salt to a cup of cold water, we often have to stir it with a teaspoon to get it to dissolve.

But when we add salt to a pot of boiling pasta the salt disappears straight away. 

The one-word answer for this is “solubility”.

Solubility is the ability of a substance to dissolve in an aqueous solution, forming different products.

But some elements and compounds are more soluble than others.

Okay, but now what?

The solubility of a substance can be determined using a K_sp value, the sister equation to our K_eq. 

K_sp values are the equilibrium constants for the solubility of a substance. 

Essentially this value indicates “how soluble a substance is”.

The K_sp can be found using the dissolution equation (or net ionic equation for the relevant compound). 

When a substance is less soluble, a precipitate is formed. 

This is the powdery, gritty or chunky solid you see when insoluble components are formed in solution. 

Solubility Rules

Now, the Module 5 Chemistry course requires us to know whether a substance will be soluble or form a precipitate.

Chemists love their rules, and once again, Module 5 Chemistry has its own set of solubility rules that you should commit to memory:

If you’re overwhelmed by how you’re going to memorise it all (I know I was), I promise that with practice, it gets easier! 

Think of it this way, if you think about going to the gym for the first time, you wouldn’t expect to be able to bench press or squat 100kg- you have to work up to it.

This is the same for memorising the solubility table!

The more practice questions I did, the more I familiarised myself with and actively remembered these rules.

This is a definitive relationship I’ve observed across my experience working with hundreds of HSC Chemistry students!

Don’t wait until it’s too late to master this module! Let our experienced HSC Chemistry Tutors support your studies at our Campuses in Hornsby, Chatswood or the Hills, at your home or online.

How to Score a Band 6 in HSC Chemistry

So, how do I score a Band 6 for this Module?

Tip #1: Map out equilibrium 

Just like the module, your mind map will encompass all things to do with our central focus, Equilibrium. 

Grab the largest piece of paper from your nearest news agency, and start mapping out everything you know about Equilibrium! 

Design your mind map so that each branch explores its respective inquiry questions. 

This forms a big-picture summary that relates one part of the module to another and is useful for future revision.

Feel free to add sticky notes containing solutions to harder questions onto the mind map, so you can easily reflect on what you need to master.

Needing to organise your HSC Chemistry notes? I recommend all my students to use our FREE HSC Dashboards that will help you stay on top of your HSC Chemistry Module 5 revision and give you the added benefit of accessing tons of useful resources!

Tip #2: Identify holes in your understanding 

As mentioned, practice (and reviewing your mistakes!) is your gateway to scoring well in Module 5 Chemistry.

We’ve compiled an ultimate list of all HSC Chemistry Past Papers along with marking guidelines, which is the #1 resource that I get my students to practise from.

It’s always best to do as many of these papers as possible to get a feel of the exam question styles and the level of thinking and understanding required to ace the higher band 6 questions. 

But identifying what you don’t understand is not enough! Practice, practice and more practice will enable you to master the topics you are struggling with.

So what are you waiting for? Get grinding!

Tip #3: Apply theory in your pracs

As you are writing your practical report, ensure you are explaining what you see. 

This means applying the concepts you’ve learnt in class to what you observe in your pracs. 

If you have any trouble remembering what has happened in class, it is good to revise your practical experiments at home online. 

While revising each practical, ask yourself these questions: 

1. What are the molecular interactions that have caused the equilibrium position to shift?
2. How have outside factors affected the molecular interactions within the reaction?

By asking yourself this question every time you see a change in the reaction, you’re essentially mastering your understanding of chemistry on a molecular level!

And that wraps up our guide to Module 5 Chemistry: Equilibrium and Acid Reactions!

Good luck 💪

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