HSC Chemistry Module 5 Equilibrium: Complete Topic Summary

A reversible reaction creates a strange problem. If reactants can become products and products can turn back into reactants, why doesn’t the reaction keep changing forever? And if the concentrations eventually stop changing, does that mean the molecules themselves have stopped reacting?

Predict it first: imagine sealing a reversible reaction inside a flask and leaving it for long enough. Do you expect the forward reaction to stop completely?

It doesn’t. The forward and reverse reactions continue, but eventually they occur at the same rate. That idea, dynamic equilibrium, is the thread connecting almost everything in HSC Chemistry Module 5.

For the 2026 HSC, Year 12 Chemistry still follows the Chemistry Stage 6 Syllabus introduced in 2017. This summary covers the Module 5 equilibrium content: reversible reactions, open and closed systems, Le Chatelier’s principle, \(K_{eq}\), reaction quotient \(Q\), dissolution, solubility rules, \(K_{sp}\), and precipitation. We will only touch on acid and base dissociation as an equilibrium application. Detailed acid-base theory, pH, and titration belong in Module 6.

01Why reactions do not always go to completion

Consider a general reversible reaction:

\[
\ce{A + B <=> C + D}
\]

At the beginning, there may be lots of A and B but almost no C or D. Collisions between A and B are therefore common, so the forward reaction is fast.

As C and D accumulate, something changes. They can now collide and undergo the reverse reaction.

Eventually:

\[
\text{rate of forward reaction} = \text{rate of reverse reaction}
\]

This is dynamic equilibrium.

The concentrations of A, B, C, and D become constant, but molecules continue reacting in both directions.

A useful picture is two rooms at a party. People keep moving between the rooms, but suppose 10 people move left each minute and 10 move right each minute. The number of people in each room stays constant even though individuals are still moving.

The rooms represent reactants and products. People crossing between them represent reactions.

The analogy breaks because molecules are not making regrettable social decisions. Their behaviour comes from collisions, energy, and probability.

Equilibrium does not mean equal concentrations

This is one of the easiest traps in the module.

At equilibrium:

\[
\text{forward rate} = \text{reverse rate}
\]

It does not follow that:

\[
[\text{reactants}] = [\text{products}]
\]

One equilibrium might contain mostly products. Another might contain mostly reactants.

We describe this as the position of equilibrium.

Why a closed system matters

For equilibrium to become established, the reacting substances normally need to remain in the system.

A closed system can exchange energy with its surroundings, but not matter. A sealed flask is the obvious example.

An open system can exchange both energy and matter.

Imagine:

\[
\ce{H2O(l) <=> H2O(g)}
\]

In a sealed bottle, water evaporates and water vapour condenses. Eventually the two rates can become equal.

In an open dish, water vapour continually escapes into the room. The reverse process cannot keep up, so the system does not establish the same equilibrium.

Static and dynamic equilibrium

A static equilibrium has no overall change and no continuing opposing microscopic processes.

A dynamic equilibrium also has no observable overall change, but opposing processes continue at equal rates.

Chemical equilibria are generally dynamic.

What do enthalpy, entropy, and collision theory contribute?

Module 5 builds on ideas from Year 11.

Enthalpy concerns energy changes. Entropy concerns the dispersal of energy and matter. Together, they help explain why some changes are thermodynamically favourable while others require continuing energy input.

But thermodynamic favourability is not the same as speed.

A reaction might be favourable but extremely slow because particles rarely collide with enough energy to overcome the activation energy.

Collision theory therefore explains the rate at which equilibrium is approached. Thermodynamics helps explain which direction is favoured.

Quick check: A sealed equilibrium mixture looks completely unchanged for ten minutes. Have the reactions stopped?

02Le Chatelier’s principle: predicting how equilibrium shifts

Once a system has reached equilibrium, disturb it.

Add more reactant. Compress the container. Heat it.

What happens?

Le Chatelier’s principle says that when an equilibrium system is disturbed, the system shifts in the direction that partially opposes that disturbance.

Be careful with the wording. The disturbance is not magically cancelled. The equilibrium simply adjusts.

Changing concentration

Consider:

\[
\ce{Fe^{3+}(aq) + SCN^-(aq) <=> FeSCN^{2+}(aq)}
\]

If more \(\ce{Fe^{3+}}\) is added, the system now has excess reactant.

The forward reaction is favoured, consuming some of the added \(\ce{Fe^{3+}}\) and producing more \(\ce{FeSCN^{2+}}\).

So the equilibrium shifts right.

Remove a product, and the same logic applies. The system shifts right to replace some of what was removed.

A good decision rule is:

• add something -> shift away from it
• remove something -> shift towards it

That is a shortcut, not the underlying cause. Collision frequencies change first, which changes the forward or reverse reaction rate.

Changing pressure or volume

Pressure matters mainly for gaseous equilibria.

Consider:

\[
\ce{N2(g) + 3H2(g) <=> 2NH3(g)}
\]

There are four moles of gas on the left and two on the right.

Now decrease the volume.

All gas particles become more crowded, so pressure increases. The equilibrium shifts towards the side with fewer moles of gas, which is the right.

Therefore more \(\ce{NH3}\) is favoured.

Increasing the volume does the opposite.

Common trap: pressure does not automatically make equilibrium shift right. Count gaseous coefficients.

If both sides contain the same total number of moles of gas, changing pressure does not favour either side.

Changing temperature

Temperature is different because heat participates in the energy balance of the reaction.

For an exothermic forward reaction, imagine heat as a product:

\[
\ce{Reactants <=> Products + heat}
\]

Heating the system effectively adds a product, so equilibrium shifts left.

Cooling removes heat, so equilibrium shifts right.

For an endothermic forward reaction:

\[
\ce{Reactants + heat <=> Products}
\]

Heating shifts the equilibrium right.

This leads to a rule worth knowing rather than memorising as four disconnected cases:

Increasing temperature favours the endothermic direction. Decreasing temperature favours the exothermic direction.

Catalysts are the awkward exception

Suppose you add a catalyst.

Would the equilibrium shift?

No.

A catalyst provides an alternative pathway with a lower activation energy. It increases both forward and reverse reaction rates.

The system reaches equilibrium faster, but its equilibrium position is unchanged.

This distinction matters:

• kinetics asks how fast
• equilibrium asks where the system settles

A catalyst changes the first, not the second.

Quick check: For an exothermic forward reaction, the temperature is increased. What happens to the equilibrium position?

03\(K_{eq}\) and \(Q\): putting a number on equilibrium

Le Chatelier’s principle predicts shifts qualitatively. Sometimes we need a number.

For:

\[
\ce{aA(aq) + bB(aq) <=> cC(aq) + dD(aq)}
\]

the equilibrium constant expression is:

\[
K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}
\]

The square brackets mean equilibrium concentration in \(\mathrm{mol\,L^{-1}}\). The lowercase letters \(a\), \(b\), \(c\), and \(d\) are the coefficients from the balanced equation.

The coefficients become powers.

For example:

\[
\ce{H2(g) + I2(g) <=> 2HI(g)}
\]

gives:

\[
K_{eq}=\frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}
\]

What does the size of \(K_{eq}\) tell you?

If \(K_{eq}\) is much greater than 1, products are favoured at equilibrium.

If \(K_{eq}\) is much less than 1, reactants are favoured.

If \(K_{eq}\) is around 1, substantial amounts of both are normally present.

It does not tell you how fast the reaction occurs.

A reaction can have an enormous \(K_{eq}\) and still take ages to reach equilibrium.

Pure solids and liquids are omitted

For a heterogeneous equilibrium such as:

\[
\ce{CaCO3(s) <=> CaO(s) + CO2(g)}
\]

the pure solids do not appear in the equilibrium expression.

The amount of solid may change, but its concentration in the thermodynamic sense used here remains effectively constant.

Only temperature changes \(K_{eq}\)

Changing concentration, pressure, or volume may shift the equilibrium position, but after the system settles again at the same temperature, the value of \(K_{eq}\) is unchanged.

Changing temperature changes \(K_{eq}\).

This is why temperature is fundamentally different from the other disturbances.

\(Q\): where is the system right now?

The reaction quotient, \(Q\), uses the same expression as \(K_{eq}\), but the concentrations do not need to be equilibrium concentrations.

Compare the two:

For ordinary equilibrium:

• \(Q < K_{eq}\): reaction proceeds forward
• \(Q = K_{eq}\): system is at equilibrium
• \(Q > K_{eq}\): reaction proceeds in reverse

Think of \(K_{eq}\) as the destination and \(Q\) as your current location.

Unsaturated, saturated, and supersaturated

An unsaturated solution can dissolve more solute under the current conditions.

A saturated solution is in equilibrium with undissolved solute.

A supersaturated solution contains more dissolved solute than would normally be stable at that temperature. It is unstable and may crystallise when disturbed.

Solubility rules are useful, but qualitative

Suppose you mix two ionic solutions.

You can first use solubility rules to predict whether an insoluble product is likely.

For example:

\[
\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}
\]

The important net ionic equation is:

\[
\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}
\]

Solubility rules are excellent for identifying likely precipitates. \(K_{sp}\) allows us to make the prediction quantitatively.

Solubility and food processing

The syllabus also asks students to investigate how solubility equilibria can be involved in traditional Aboriginal and Torres Strait Islander food-processing practices, including examples involving cycad foods. Repeated washing or leaching can transfer water-soluble toxic substances from plant material into the surrounding water, which is then removed.

The chemical idea is familiar: continually removing dissolved material can allow more of that material to enter solution.

Specific preparation practices depend on the food and community, so this chemical model should not be treated as one universal processing method.

05\(K_{sp}\): measuring the solubility equilibrium

For a sparingly soluble salt:

\[
\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}
\]

the solubility product is:

\[
K_{sp}=[\ce{Ag+}][\ce{Cl-}]
\]

The solid is omitted.

Now consider:

\[
\ce{CaF2(s) <=> Ca^{2+}(aq) + 2F-(aq)}
\]

The coefficient of 2 matters:

\[
K_{sp}=[\ce{Ca^{2+}}][\ce{F-}]^2
\]

Suppose the molar solubility of \(\ce{CaF2}\) is \(s\ \mathrm{mol\,L^{-1}}\).

For every mole of \(\ce{CaF2}\) that dissolves:

\[
[\ce{Ca^{2+}}]=s
\]

and:

\[
[\ce{F-}]=2s
\]

Therefore:

\[
K_{sp}=s(2s)^2=4s^3
\]

This stoichiometric step is where many \(K_{sp}\) calculations go wrong.

Quick check: If the molar solubility of \(\ce{AgCl}\) in pure water is \(s\), what is its \(K_{sp}\) expression in terms of \(s\)?

Do not compare \(K_{sp}\) values blindly

A larger \(K_{sp}\) often suggests greater solubility when salts have the same ion ratio.

But comparing \(K_{sp}\) values for something like \(\ce{AgCl}\) and \(\ce{CaF2}\) directly can mislead you because their equilibrium expressions have different powers.

Calculate molar solubility if you need a fair comparison.

The common ion effect

Return to:

\[
\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}
\]

Now add a soluble chloride such as \(\ce{NaCl}\).

The extra \(\ce{Cl-}\) shifts the equilibrium left. More \(\ce{AgCl}\) is favoured as a solid, so the solubility of \(\ce{AgCl}\) decreases.

This is the common ion effect.

Importantly, \(K_{sp}\) has not changed because the temperature has not changed.

06Predicting precipitation with \(Q_{sp}\)

Suppose two clear solutions are mixed. You know their ion concentrations and a \(K_{sp}\) value.

Will a precipitate actually appear?

Use \(Q_{sp}\).

For:

\[
\ce{PbI2(s) <=> Pb^{2+}(aq) + 2I-(aq)}
\]

\[
Q_{sp}=[\ce{Pb^{2+}}][\ce{I-}]^2
\]

Then compare:

• \(Q_{sp}K_{sp}\): precipitation is favoured

There is one nasty trap: when solutions are mixed, both are diluted.

Do not use their original concentrations directly.

07How the whole module fits together

Module 5 becomes much easier when you stop treating it as a collection of unrelated rules.

Nearly every problem is asking some version of the same question:

What direction will this chemical system move, and where will it eventually settle?

Start with the physical picture. Reversible reactions can proceed in both directions. In a suitable closed system, their rates can become equal, producing dynamic equilibrium.

Then choose the right tool.

If the question changes concentration, pressure, volume, or temperature, think about Le Chatelier’s principle.

If it asks about reaction direction from numerical concentrations, calculate \(Q\) and compare it with \(K_{eq}\).

If it asks where a reaction lies at equilibrium, interpret \(K_{eq}\).

If an ionic solid is dissolving, think of dissolution and precipitation as opposing reactions.

If the question asks about the maximum dissolved amount of a sparingly soluble salt, use \(K_{sp}\).

If two solutions are mixed and you need to decide whether a precipitate forms, first account for dilution, calculate \(Q_{sp}\), and compare it with \(K_{sp}\).

And whenever you see a catalyst, resist the temptation to shift anything. A catalyst changes how quickly equilibrium is reached, not the position of equilibrium.

That framework is the useful bridge into Module 6. Acid ionisation is not a completely new kind of chemistry. It is another reversible system, governed by the same equilibrium logic you have just built.