What Does ΔG = 0 Mean? Gibbs Free Energy and Equilibrium

At approximately \(0^\circ\mathrm{C}\), ice and liquid water can exist together without either phase steadily taking over. Yet molecules at the ice surface are still leaving the solid and joining the liquid, while other molecules are freezing.

For the melting process at this temperature and pressure, \(\Delta G=0\).

This means the system is at thermodynamic equilibrium: there is no net Gibbs free energy driving force favouring either melting or freezing. It does not mean the water has zero energy, or that molecular activity has stopped.

That distinction is central to understanding Gibbs free energy. A zero value describes the balance between two possible directions of change, not the absence of energy or activity.

Predict what would happen if you raised the temperature slightly above \(0^\circ\mathrm{C}\). Melting becomes thermodynamically favourable. Lower the temperature slightly below \(0^\circ\mathrm{C}\), and freezing becomes favourable.

The zero marks the boundary between those two conditions.

01An experiment that makes \(\Delta G=0\) visible

Consider a simple experimental question:

At what temperature are solid water and liquid water equally stable at atmospheric pressure?

A useful experimental arrangement would contain:

• A transparent vessel holding a mixture of ice and liquid water.
• A temperature probe positioned inside the mixture.
• A temperature-controlled water bath surrounding the vessel.
• A camera or suitable balance to monitor changes in the amount of ice.
• A controlled pressure of approximately \(1\ \mathrm{atm}\).

The water bath allows the temperature to be adjusted while the probe records the actual sample temperature. Monitoring the ice fraction reveals which phase is becoming more stable.

What should the experiment show?

If solid and liquid water have different thermodynamic stabilities, one phase should become favoured.

The predictions are:

• Below the melting point, freezing is favoured. The amount of ice increases.
• Above the melting point, melting is favoured. The amount of ice decreases.
• At the melting point, neither phase has a net thermodynamic advantage. Both can coexist at equilibrium.

The following table shows the expected pattern. These are illustrative observations, not experimental measurements.

Here, \(\Delta G_{\mathrm{fus}}\) is the Gibbs free energy change when solid water melts into liquid water.

There is an important experimental limitation.

Suppose you simply heat an ice-water mixture and notice that its temperature stays near \(0^\circ\mathrm{C}\) while the ice disappears. That temperature plateau does not prove equilibrium.

The heater is supplying energy. That energy is being absorbed as latent heat of fusion, allowing ice to melt without a corresponding rise in temperature.

A better design holds the sample at a fixed temperature without continuously supplying net heat, then monitors the phase amounts over time. Repeating the measurement slightly above and below the melting point provides a useful comparison.

Even this experiment does not measure Gibbs free energy directly. It measures temperature and changes in phase composition. Thermodynamics provides the explanation for the observed behaviour.

02Why does Gibbs free energy become zero?

The relationship between Gibbs free energy, enthalpy, and entropy is:

\[ \Delta G=\Delta H-T\Delta S \]

Where:

• \(\Delta G\) is the Gibbs free energy change, measured in \(\mathrm{kJ\,mol^{-1}}\).
• \(\Delta H\) is the enthalpy change, measured in \(\mathrm{kJ\,mol^{-1}}\).
• \(T\) is the absolute temperature, measured in kelvin.
• \(\Delta S\) is the entropy change, measured in \(\mathrm{kJ\,mol^{-1}\,K^{-1}}\) when enthalpy is expressed in kilojoules.

For melting, two things happen:

1. Energy is absorbed to overcome the organised structure of the solid. Therefore, \(\Delta H_{\mathrm{fus}}>0\).
2. The liquid has more accessible molecular arrangements than the solid. Therefore, \(\Delta S_{\mathrm{fus}}>0\).

These contributions compete.

At lower temperatures, the enthalpy cost of melting is greater than the entropy contribution. Freezing is favoured.

As temperature increases, the \(T\Delta S\) contribution becomes larger. Eventually, the two contributions balance.

At the melting point:

\[ \Delta G_{\mathrm{fus}}=0 \]

Therefore:

\[ \Delta H_{\mathrm{fus}} = T_{\mathrm{m}}\Delta S_{\mathrm{fus}} \]

The melting point is the temperature at which the solid and liquid phases have equal Gibbs free energy per mole.

03What does \(\Delta G=0\) mean for a chemical reaction?

The same thermodynamic principle applies to reversible chemical reactions.

Consider:

\[ \ce{A <=> B} \]

Initially, imagine a mixture containing mostly A.

If the forward reaction is thermodynamically favourable, converting some A into B lowers the Gibbs free energy of the system.

As the reaction proceeds, the composition changes. Eventually, the system reaches equilibrium.

At this point:

\[ \boxed{\Delta G=0} \]

There is no net thermodynamic driving force for the forward or reverse reaction.

However, the system has not stopped reacting.

The forward reaction continues converting A into B, while the reverse reaction converts B into A. At equilibrium, these two reaction rates are equal.

Consequently, the concentrations of A and B remain constant over time.

This is called dynamic equilibrium.

Think about Gibbs free energy as a landscape

Imagine plotting the total Gibbs free energy of a reacting mixture against the extent of the forward reaction.

The mixture moves towards lower Gibbs free energy when the forward reaction is favoured.

At equilibrium, the graph reaches a minimum. Its slope is zero.

The slope tells you whether a small amount of additional forward reaction would increase or decrease Gibbs free energy.

At the minimum, neither a small forward change nor a small reverse change can lower the Gibbs free energy.

This gives us an important distinction:

\(\Delta G=0\) does not mean \(G=0\).

• \(G\) represents the Gibbs free energy of the system.
• \(\Delta G\) represents the Gibbs free energy change associated with a process.

The system can have a substantial amount of Gibbs free energy while the change associated with a small forward reaction is zero.

The three possible Gibbs free energy values

For a reaction written in the forward direction:

These signs refer to the actual reaction conditions.

There is another distinction worth keeping clear: thermodynamic feasibility is not reaction rate.

A negative \(\Delta G\) tells you that a reaction direction is thermodynamically favourable. It does not tell you how quickly the reaction occurs.

For example, a reaction can have a negative Gibbs free energy change but proceed extremely slowly because it has a large activation energy.

A catalyst can increase the rates of both forward and reverse reactions, but it does not change the equilibrium Gibbs energy condition.

For a closer comparison of thermodynamic favourability, reaction rate, and conversion, see How to Interpret Negative ΔG Without Assuming Complete Conversion.

04The important distinction: \(\Delta G=0\) versus \(\Delta G^\circ=0\)

This is where many students lose marks.

The two expressions look almost identical, but they describe different things.

The small circle, \(^{\circ}\), indicates standard-state quantities.

In the 2017 NSW HSC Chemistry syllabus, the Gibbs free energy equation is generally written:

\[ \Delta G^\circ = \Delta H^\circ-T\Delta S^\circ \]

This equation uses standard enthalpy and entropy changes.

If a question asks you to find the temperature at which standard Gibbs free energy becomes zero, you set:

\[ \Delta G^\circ=0 \]

Rearranging gives:

\[ T=\frac{\Delta H^\circ}{\Delta S^\circ} \]

This is the standard-state crossover temperature.

It does not automatically mean that any actual reaction mixture is at equilibrium at that temperature.

Why not?

Because the actual Gibbs free energy change also depends on the current composition of the mixture.

An extension: How composition changes actual Gibbs free energy

A more advanced thermodynamic relationship is:

\[ \Delta G=\Delta G^\circ+RT\ln Q \]

Where:

• \(\Delta G\) is the actual Gibbs free energy change.
• \(\Delta G^\circ\) is the standard Gibbs free energy change.
• \(R\) is the gas constant, \(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
• \(T\) is the absolute temperature in kelvin.
• \(Q\) is the dimensionless reaction quotient, calculated using the current composition.

This relationship is an extension beyond the standard 2017 HSC Gibbs-energy calculation.

It explains why standard Gibbs free energy and actual Gibbs free energy can have different values.

At equilibrium:

\[ \Delta G=0 \]

The reaction quotient then equals the equilibrium constant:

\[ Q=K \]

Combining the equations gives:

\[ \Delta G^\circ=-RT\ln K \]

Therefore, if:

\[ \Delta G^\circ=0 \]

then:

\[ K=1 \]

But the current reaction quotient does not have to equal one.

A mixture can have \(\Delta G^\circ=0\) while its actual \(\Delta G\) is positive or negative.

05Which syllabus does this belong to?

2017 Chemistry Stage 6 syllabus: Gibbs free energy, entropy, and reaction spontaneity are included in Year 11 Module 4, Drivers of Reactions. The standard Gibbs free energy equation and its application to predicting spontaneity are part of this syllabus.

2025 Chemistry 11-12 syllabus: This new syllabus begins implementation for Year 11 in Term 1 of 2028 and Year 12 in Term 4 of 2028. The first HSC examination under the new syllabus is scheduled for 2029.

Gibbs free-energy calculations are not explicitly named in the published 2025 syllabus outcomes and course overview. Do not automatically treat the 2017 Gibbs calculation requirements as mandatory 2025 syllabus content.

For the official syllabus information, see:

• NESA: Chemistry Stage 6 Syllabus (2017)
• NESA: Chemistry 11-12 Syllabus (2025)

The remaining questions include some extension reasoning to help you understand what the zero actually means, rather than simply memorising the equation.

06Questions and solutions

Question 1

A reversible reaction occurs in a sealed container at constant temperature and pressure. After some time, the concentrations of reactants and products remain constant.

The reaction has reached equilibrium.

a) What is the actual Gibbs free energy change, \(\Delta G\), at equilibrium?

b) Are the forward and reverse reactions still occurring?

c) Explain why constant concentrations do not mean that the reaction has stopped.

Solution 1

a) The actual Gibbs free energy change is zero.

At equilibrium:

\[ \Delta G=0 \]

There is no net thermodynamic driving force favouring either reaction direction.

b) Yes. Both reactions continue.

The forward and reverse reactions are still occurring.

c) The concentrations remain constant because the forward and reverse reaction rates are equal.

The forward reaction converts reactants into products, while the reverse reaction converts products back into reactants.

When these rates are equal, there is no net change in concentration.

This is dynamic equilibrium, not a situation in which all molecular activity has stopped.

The common mistake is to interpret \(\Delta G=0\) as meaning that no reaction can occur. It actually means that neither direction has a net thermodynamic advantage.

Question 2

The melting of a particular substance has the following thermodynamic data:

\[ \Delta H_{\mathrm{fus}} =+8.40\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S_{\mathrm{fus}} =+30.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Calculate the Gibbs free energy change for melting at \(268\ \mathrm{K}\).

Determine whether melting or freezing is thermodynamically favoured.

Solution 2

Melting has a positive Gibbs free energy change, so freezing is thermodynamically favoured at \(268\ \mathrm{K}\).

First, convert entropy into kilojoules:

\[ \Delta S_{\mathrm{fus}} =\frac{30.0}{1000} \]

\[ =0.0300\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Now use:

\[ \Delta G=\Delta H-T\Delta S \]

Substitute the values:

\[ \begin{aligned} \Delta G &=8.40-(268\times0.0300)\\ &=8.40-8.04\\ &=+0.36\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The result is positive.

This means melting is not thermodynamically favourable under the specified conditions. The reverse process, freezing, is favoured.

Notice that a positive \(\Delta G\) does not mean melting is impossible. It means the reverse direction is favoured under these conditions.

Question 3

A reversible reaction has the following standard Gibbs free energy change at \(310\ \mathrm{K}\):

\[ \Delta G^\circ=0 \]

The reaction quotient for the current mixture is:

\[ Q=0.25 \]

Assume ideal behaviour.

Using the extension relationship:

\[ \Delta G=\Delta G^\circ+RT\ln Q \]

a) Calculate the actual Gibbs free energy change.

b) Determine the thermodynamically favoured direction.

c) Explain why the reaction is not necessarily at equilibrium.

Solution 3

The actual Gibbs free energy change is approximately \(-3.58\ \mathrm{kJ\,mol^{-1}}\), so the forward reaction is favoured.

Part a: Calculate the actual Gibbs free energy change.

Use:

\[ \Delta G=\Delta G^\circ+RT\ln Q \]

Substitute:

\[ \Delta G=0+8.314(310)\ln(0.25) \]

Since:

\[ \ln(0.25)=-1.3863 \]

we obtain:

\[ \begin{aligned} \Delta G &=8.314(310)(-1.3863)\\ &=-3573\ \mathrm{J\,mol^{-1}}\\ &=-3.57\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

To three significant figures:

\[ \boxed{\Delta G=-3.57\ \mathrm{kJ\,mol^{-1}}} \]

Part b: Determine the favoured direction.

The actual Gibbs free energy change is negative.

Therefore, the forward reaction is thermodynamically favoured.

Part c: Explain why the system is not necessarily at equilibrium.

At equilibrium:

\[ \Delta G=0 \]

and:

\[ Q=K \]

Here, the reaction quotient is \(0.25\), not one.

Because the standard Gibbs free energy change is zero, the equilibrium constant is one for this reaction at this temperature.

Therefore, the current mixture does not satisfy the equilibrium condition.

The forward reaction proceeds until the reaction quotient reaches its equilibrium value.

The tempting mistake is to see \(\Delta G^\circ=0\) and immediately conclude that the reaction is at equilibrium. The actual composition determines whether that conclusion is valid.

Question 4

A student investigates the melting of ice using a beaker containing both ice and liquid water.

The beaker is heated continuously. The thermometer records a temperature close to \(0^\circ\mathrm{C}\) for several minutes, while the amount of ice steadily decreases.

The student concludes:

“The temperature is constant, so \(\Delta G=0\). This means the sample must be at equilibrium, even though the ice is melting.”

Evaluate this conclusion.

Suggest a modification to the experimental method that would provide stronger evidence for equilibrium.

Solution 4

The student’s conclusion is not justified. A constant temperature during continuous heating does not establish equilibrium.

The temperature plateau can occur because the energy supplied by the heater is being absorbed as latent heat of fusion.

This energy allows ice to melt without a corresponding temperature increase.

The amount of ice decreases because energy is continuously entering the sample.

The student has confused a constant temperature with a system experiencing no net change.

There is an important distinction:

• A mixture of pure ice and water at its melting point can have \(\Delta G_{\mathrm{fus}}\approx0\).
• A sample receiving continuous heat can still undergo net melting because energy is being supplied externally.

A better experimental design would use a temperature-controlled bath.

The student should:

1. Place a mixture of ice and water in the controlled bath.
2. Set the bath to approximately \(0^\circ\mathrm{C}\).
3. Allow the sample to reach thermal equilibrium.
4. Stop any net heating or cooling of the sample.
5. Monitor the temperature and ice mass over time.

The student should then repeat the experiment slightly above and below the melting point.

At approximately \(0^\circ\mathrm{C}\), the phase amounts should remain stable on average when there is no net heat transfer.

Above the melting point, melting should be favoured. Below it, freezing should be favoured.

This revised method provides stronger evidence that the two phases are in equilibrium.

However, even this method does not directly measure \(\Delta G\). It uses observable changes in phase composition to test the thermodynamic prediction.

Question 5

Two separate samples of pure water are held at \(273.15\ \mathrm{K}\) and approximately \(1\ \mathrm{atm}\).

Both samples contain ice and liquid water.

Their compositions are:

Assume both samples are initially at phase equilibrium.

A student makes the following argument:

“Only one sample can be at equilibrium. If \(\Delta G=0\), the system must have one particular equilibrium composition. The other sample must undergo a net phase change until it reaches that composition.”

The student then makes a second claim:

“If 1.00 kJ of heat is supplied to sample X while its temperature is maintained at \(273.15\ \mathrm{K}\), no ice can melt because \(\Delta G=0\).”

Evaluate both claims.

Calculate the mass of ice that melts when the heat is supplied.

Assume:

• The samples contain pure water only.
• Pressure remains constant.
• No water evaporates or escapes.
• All supplied heat is used to melt ice.
• Surface effects are negligible.
• The molar enthalpy of fusion of water is \(6.01\ \mathrm{kJ\,mol^{-1}}\).
• The molar mass of water is \(18.0\ \mathrm{g\,mol^{-1}}\).

Solution 5

Both samples can be at equilibrium, and supplying 1.00 kJ of heat can melt approximately 3.00 g of ice in sample X. The resulting mixture does not have to return to its original composition.

This question exposes an important limitation in the idea that equilibrium always means one unique composition.

Part a: Can both samples be at equilibrium?

Yes.

At the melting point, the molar Gibbs free energies of solid and liquid water are equal:

\[ G_{\mathrm{m,ice}} = G_{\mathrm{m,liquid}} \]

Therefore:

\[ \Delta G_{\mathrm{fus}}=0 \]

Neither phase has a thermodynamic advantage over the other.

This means sample X and sample Y can both be at equilibrium, even though their phase proportions are very different.

For a fixed amount of pure water at the melting point, the total Gibbs free energy is unchanged by redistributing material between the solid and liquid phases, provided both phases coexist and surface effects are negligible.

There is no requirement for a 50:50 mixture or any other unique ice-to-water ratio.

Part b: Calculate the amount of ice melted.

The supplied heat is:

\[ q=1.00\ \mathrm{kJ} \]

The molar enthalpy of fusion is:

\[ \Delta H_{\mathrm{fus}} =6.01\ \mathrm{kJ\,mol^{-1}} \]

Use:

\[ q=n\Delta H_{\mathrm{fus}} \]

Rearrange:

\[ n=\frac{q}{\Delta H_{\mathrm{fus}}} \]

Substitute:

\[ n=\frac{1.00}{6.01} \]

\[ n=0.166\ \mathrm{mol} \]

Convert the number of moles into mass:

\[ m=nM \]

\[ m=0.166(18.0) \]

\[ \boxed{m=2.99\ \mathrm{g}} \]

Approximately \(3.00\ \mathrm{g}\) of ice melts.

The new composition of sample X is therefore approximately:

• Ice: \(87.0\ \mathrm{g}\).
• Liquid water: \(13.0\ \mathrm{g}\).

The temperature remains at the melting point because the supplied energy is being used for the phase change rather than raising the temperature.

Part c: Must the sample return to its original composition?

No.

Once the heat supply stops, the sample can settle into a new equilibrium containing approximately \(87.0\ \mathrm{g}\) of ice and \(13.0\ \mathrm{g}\) of liquid water.

At \(273.15\ \mathrm{K}\), the two phases still have equal molar Gibbs free energies.

There is no thermodynamic driving force requiring the additional liquid water to freeze and restore the original 90.0 g to 10.0 g ratio.

This is the subtle point: \(\Delta G_{\mathrm{fus}}=0\) does not identify one unique phase proportion.

For a chemical reaction, equilibrium generally corresponds to a particular composition where the Gibbs free energy of the reacting mixture reaches a minimum.

For pure water at its melting point, the Gibbs free energy is instead independent of the proportion of solid and liquid phases, under the assumptions given.

The student’s first claim confuses phase equilibrium with the particular composition usually associated with chemical reaction equilibrium.

The second claim ignores the external energy supplied by the apparatus. Zero Gibbs free energy change means there is no net thermodynamic preference for melting or freezing when the system is left at equilibrium. It does not prevent an externally supplied energy transfer from changing the phase proportions.

07What should you work out next?

When you encounter a zero Gibbs free energy value, first identify which quantity the question gives you.

• If \(\Delta G=0\), the actual reaction or phase change is at equilibrium under the specified conditions.
• If \(\Delta G^\circ=0\), the standard-state Gibbs free energy change is zero. You may be looking for a temperature at which its sign changes.
• If a reaction mixture is given, its actual composition matters. Standard-state values alone may not determine the favoured direction.

Once you understand this distinction, the next useful step is connecting Gibbs free energy to the equilibrium constant and calculating the composition of an equilibrium mixture.

That is where thermodynamic feasibility becomes a prediction about how much reactant and product will actually be present.