How to Interpret Negative ΔG Without Assuming Complete Conversion

A reaction has a negative Gibbs free energy change, \(\Delta G<0\), but the reaction mixture still contains plenty of reactants after a long time. Has the thermodynamic prediction failed?

No. A negative \(\Delta G\) means that the reaction is thermodynamically favourable in the forward direction under the conditions being considered. It does not mean that every reactant particle will eventually become a product, that the reaction will happen quickly, or that the reaction will reach 100% conversion.

The key is to separate three ideas:

• Thermodynamic feasibility: Is the forward reaction favoured under the current conditions?
• Reaction rate: How quickly does the reaction proceed?
• Equilibrium: What mixture of reactants and products is present when there is no net change?

These are related, but they are not interchangeable. A negative \(\Delta G\) tells you about the direction in which a system can lower its Gibbs energy. It does not, by itself, tell you how far the reaction proceeds.

01What does a negative \(\Delta G\) actually predict?

Imagine placing hydrogen and iodine gases in a sealed container. They can react to form hydrogen iodide:

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

Suppose the mixture initially contains mostly hydrogen and iodine, with very little hydrogen iodide.

You might predict that, if the reaction is favourable, the reactants should keep disappearing until almost all of them are gone. That sounds reasonable, but it misses an important detail: the thermodynamic driving force changes as the composition changes.

At the beginning, forming hydrogen iodide may be strongly favoured. As more hydrogen iodide forms, the conditions change. Eventually, the forward and reverse reactions balance, and the system reaches equilibrium.

At equilibrium, both reactants and products can remain present in substantial amounts.

The important point is that \(\Delta G\) is not necessarily a fixed value throughout a reaction. It depends on the current conditions, including the relative amounts of reactants and products.

02The crucial distinction: \(\Delta G^\circ\) versus \(\Delta G\)

There’s a small symbol in the Gibbs equation that changes how you should interpret the result.

In the 2017 NSW Chemistry syllabus, you use:

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

The superscript \(\circ\) means standard-state conditions. This is the standard Gibbs free energy change, not necessarily the Gibbs free energy change for the mixture currently sitting in a reaction vessel.

That distinction is important.

• \(\Delta G^\circ<0\): The forward reaction is thermodynamically favoured from the standard state.
• \(\Delta G^\circ>0\): The reverse reaction is thermodynamically favoured from the standard state.
• \(\Delta G^\circ=0\): The standard state is also an equilibrium state.

For an actual reaction mixture, we use \(\Delta G\), without the superscript. Its value depends on the current composition.

For a reaction at constant temperature and pressure:

Notice that equilibrium does not mean the reaction has stopped. The forward and reverse reactions continue at equal rates, so the overall composition stays constant.

Why does \(\Delta G\) change during a reaction?

Think of Gibbs free energy as a landscape.

A reaction moves in the direction that lowers the system’s Gibbs energy. Initially, the forward reaction might lower it substantially. As the composition changes, the driving force becomes smaller.

Eventually, the system reaches a minimum. Moving further towards products would now increase Gibbs energy, so the reverse direction becomes favourable instead.

This is why a negative value at the beginning does not guarantee complete conversion.

The more precise relationship involves the reaction quotient, \(Q\):

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

Here:

• \(R\) is the gas constant, \(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
• \(T\) is the absolute temperature in kelvin.
• \(Q\) is the reaction quotient, calculated using the current composition.
• \(\ln\) is the natural logarithm.

The reaction quotient compares the current amounts of products and reactants. As products accumulate, \(Q\) changes, which changes the actual Gibbs energy change.

At equilibrium, \(Q=K\), where \(K\) is the equilibrium constant. Therefore:

\[ \Delta G=0 \]

Combining these relationships gives:

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

This equation connects the standard Gibbs energy change to the equilibrium constant.

Extension beyond the 2017 HSC calculation requirement

The \(Q\)-\(K\)-\(\Delta G\) relationships provide a deeper thermodynamic explanation of equilibrium. They are useful for understanding the question in this article, but should not be confused with the specific 2017 syllabus requirement to calculate \(\Delta G^\circ\) using enthalpy and entropy.

What does a negative \(\Delta G^\circ\) tell you about equilibrium?

The equation

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

shows that a negative standard Gibbs energy change corresponds to \(K>1\).

This means products are favoured at equilibrium, relative to the standard-state reference.

However, \(K>1\) does not mean that the equilibrium mixture contains only products.

The magnitude of \(K\) matters:

• A value of \(K\) slightly greater than 1 indicates a relatively modest product preference.
• A very large \(K\) indicates a strong product preference.
• Neither statement, by itself, establishes exactly 100% conversion.

The initial quantities, reaction stoichiometry, and equilibrium conditions also matter.

03What does the experiment actually show?

The distinction between thermodynamic favourability and complete conversion becomes easier to understand when you consider a real reversible reaction.

Consider nitrogen dioxide and dinitrogen tetroxide:

\[ \ce{2NO2(g) <=> N2O4(g)} \]

Nitrogen dioxide is a brown gas. Dinitrogen tetroxide is colourless or nearly colourless.

This gives us a way to monitor the changing composition of the reaction mixture.

The experimental reasoning

The unanswered question is straightforward: when nitrogen dioxide forms dinitrogen tetroxide, does the brown gas disappear completely?

A useful experimental design follows this sequence:

Step 1: Prepare a sealed gas cell.

The cell contains a mixture of \(\ce{NO2}\) and \(\ce{N2O4}\). It is sealed so that gas cannot escape or enter during the investigation.

Step 2: Control the temperature.

Place the cell in a temperature-controlled bath. Keep the temperature constant while the mixture approaches equilibrium.

Step 3: Measure the colour change.

Pass light through the cell and measure its absorbance at a wavelength absorbed by \(\ce{NO2}\).

At a fixed wavelength and path length, absorbance is proportional to the concentration of the absorbing species under the Beer-Lambert relationship.

Step 4: Monitor the reading over time.

If the forward reaction is occurring, the concentration of brown \(\ce{NO2}\) decreases, so its absorbance decreases.

What should happen if the reaction goes to completion?

The absorbance attributable to \(\ce{NO2}\) should approach zero.

Instead, the absorbance can reach a stable, non-zero value. This indicates that some nitrogen dioxide remains in the mixture.

The system has reached an equilibrium composition rather than converting every reactant molecule into product.

Why the original visual comparison is not enough

Simply comparing the apparent colour of two different tubes is a weak measurement.

Different tube widths, gas pressures, and light paths can change the apparent intensity, even when the chemical composition is similar.

A better design uses the same sealed cell, a fixed optical path, a controlled temperature, and a calibrated spectrophotometer.

A further improvement is to approach equilibrium from both directions. A nitrogen-dioxide-rich mixture and a dinitrogen-tetroxide-rich mixture should approach the same equilibrium composition when temperature and pressure are identical.

The experiment demonstrates an important limitation: measuring colour tells us about composition, not Gibbs energy directly. We need thermodynamic information and equilibrium data to establish the driving force.

The brown colour is evidence that reactant remains. It is not evidence that the reaction has failed.

Safety note: Nitrogen dioxide is toxic. This apparatus describes a controlled laboratory demonstration, not an experiment to prepare or handle the gas outside an appropriately equipped laboratory.

A brief historical connection

The theoretical framework behind this interpretation developed through the work of Josiah Willard Gibbs in the 1870s.

Gibbs investigated the conditions for equilibrium in thermodynamic systems, including systems containing different phases and chemical substances. His work provided a general mathematical way to describe equilibrium using thermodynamic properties.

This was not one decisive laboratory experiment in which Gibbs measured a reaction’s Gibbs energy. It was a theoretical development that could be tested against observations of real equilibrium systems.

You can read more about this development in the historical discussion of Gibbs’s work on heterogeneous equilibrium.

The connection to modern chemistry is straightforward: thermodynamics predicts the equilibrium condition, while experimental measurements reveal the composition that the system actually reaches.

04Keep feasibility, rate, and equilibrium separate

These three ideas are often mixed together in exam answers.

Consider rusting.

The formation of iron oxides is thermodynamically favourable under ordinary environmental conditions. However, the reaction can be slow because the reaction pathway has an activation energy barrier.

A negative Gibbs energy change does not remove that barrier.

What does a catalyst change?

A catalyst provides an alternative reaction pathway with a lower activation energy.

This increases the reaction rate, allowing equilibrium to be reached more quickly.

However, a catalyst does not change the initial or final thermodynamic states. It does not change:

• \(\Delta G^\circ\) at a given temperature;
• the equilibrium constant \(K\);
• the equilibrium composition.

For a reversible reaction, it accelerates both forward and reverse reactions.

The key exam distinction is that a catalyst changes kinetics, not thermodynamic feasibility.

If a question says a reaction has negative \(\Delta G^\circ\) but proceeds slowly, there is no contradiction.

For a fuller explanation of the composition of equilibrium mixtures, see HSC Chemistry Module 5 Equilibrium: Complete Topic Summary.

05Worked examples

06A reliable method for interpreting negative \(\Delta G\)

When you see a Gibbs energy question, work through these decisions in order.

1. Check the symbol. Is the question giving you \(\Delta G^\circ\) or the actual \(\Delta G\)?
2. Identify the conditions. If you have calculated \(\Delta G^\circ\) using \(\Delta H^\circ-T\Delta S^\circ\), your conclusion concerns standard-state conditions.
3. State the thermodynamic direction. A negative value favours the forward reaction as written; a positive value favours the reverse direction.
4. Do not predict the rate. You need kinetic information, such as activation energy, to discuss how quickly the reaction proceeds.
5. Do not predict complete conversion. Use the equilibrium constant and initial composition to determine the equilibrium mixture.
6. If current composition matters, the more advanced \(Q\)-versus-\(K\) comparison determines the actual direction of net change.

If you need to revise how the enthalpy and entropy terms produce \(\Delta G^\circ\), see How to Compare Enthalpy and Entropy in Gibbs Energy.

Syllabus note: 2017 versus 2025

The calculation \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\) belongs to the 2017 Chemistry Stage 6 syllabus, Year 11 Module 4: Drivers of Reactions, which includes entropy and Gibbs free energy.

NESA’s Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028. Students studying that later syllabus should check its specified content rather than assume that every 2017 calculation remains examinable.

You can check the 2017 Chemistry syllabus guide and NESA’s 2025 syllabus implementation information.

07Questions and solutions

Question 1

A reversible reaction has a standard Gibbs free energy change of

\[ \Delta G^\circ=-8.0\ \mathrm{kJ\,mol^{-1}} \]

at 298 K.

A student concludes that the reaction will convert all reactants into products.

Is this conclusion justified? Explain what the negative value actually predicts.

Solution 1

No. A negative standard Gibbs free energy change indicates that the forward reaction is thermodynamically favourable under standard-state conditions, but it does not establish complete conversion.

The reaction can proceed towards products until equilibrium is reached. At equilibrium, the forward and reverse reactions occur at equal rates, and the mixture can contain both reactants and products.

The equilibrium constant determines how strongly products are favoured. The sign of \(\Delta G^\circ\) alone does not give the exact percentage conversion.

The student has confused thermodynamic favourability with complete conversion.

Question 2

Consider the reversible reaction:

\[ \ce{2NO2(g) <=> N2O4(g)} \]

A sealed gas cell is maintained at a constant temperature of 298 K. A spectrophotometer measures the absorbance of nitrogen dioxide at a fixed wavelength and optical path length.

Two trials are conducted:

• Trial A begins with a nitrogen-dioxide-rich mixture.
• Trial B begins with a dinitrogen-tetroxide-rich mixture.

The following illustrative data are collected.

Absorbance is given in arbitrary units.

a. Explain the changes in absorbance in each trial.

b. What does the final absorbance suggest about the extent of the reaction?

c. Explain why the results provide evidence for dynamic equilibrium rather than complete conversion.

Solution 2

The two trials approach the same equilibrium composition, with nitrogen dioxide remaining in the mixture. The reaction does not achieve complete conversion.

Part a:

In Trial A, absorbance decreases from 0.80 to 0.34.

This indicates that the concentration of brown nitrogen dioxide decreases as it reacts to form dinitrogen tetroxide.

In Trial B, absorbance increases from 0.08 to 0.34.

This indicates that dinitrogen tetroxide decomposes to produce nitrogen dioxide through the reverse reaction.

Part b:

The final absorbance is 0.34, which is greater than zero.

Because absorbance is proportional to nitrogen dioxide concentration under the stated conditions, nitrogen dioxide remains present at equilibrium.

Therefore, the forward reaction has not converted all nitrogen dioxide into dinitrogen tetroxide.

Part c:

Both trials approach the same final absorbance despite starting with different compositions.

This supports the conclusion that both mixtures reach the same equilibrium state at the specified temperature.

At equilibrium, the forward and reverse reaction rates are equal. The concentrations remain constant even though both reactions continue.

The experiment demonstrates that a reversible reaction can reach equilibrium with substantial quantities of both reactants and products.

The non-zero absorbance is particularly important: equilibrium does not require the disappearance of either substance.

Question 3

Consider the reversible reaction:

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

At a particular temperature, the forward reaction has a negative standard Gibbs free energy change.

A student prepares a mixture close to standard-state conditions but observes very little hydrogen iodide formation over the first 10 minutes.

A catalyst is then introduced, and hydrogen iodide forms much more rapidly.

The student concludes:

“The catalyst has made the reaction more spontaneous, so the final equilibrium must contain a greater percentage of hydrogen iodide.”

Evaluate this explanation.

Solution 3

The student is incorrect. The catalyst increases the reaction rate but does not change the thermodynamic favourability or final equilibrium composition.

The negative standard Gibbs free energy change indicates that the forward reaction is thermodynamically favourable under standard-state conditions.

However, the reaction may still proceed slowly because it has an activation energy barrier.

The initial lack of observable change does not disprove the thermodynamic prediction.

When the catalyst is introduced, it provides an alternative reaction pathway with lower activation energy. More particles can react per unit time, increasing the reaction rate.

However, a catalyst accelerates both the forward and reverse reactions.

It does not change:

• The standard Gibbs free energy change at the specified temperature.
• The equilibrium constant.
• The equilibrium composition.

The system simply reaches its existing equilibrium more quickly.

The student has confused an increase in reaction rate with an increase in thermodynamic favourability.

Question 4

Consider the reversible reaction:

\[ \ce{A(aq) <=> B(aq)} \]

At 298 K, its equilibrium constant is:

\[ K_c=4.0 \]

A reaction mixture currently contains:

\[ [A]=0.10\ \mathrm{mol\,L^{-1}} \]

\[ [B]=0.90\ \mathrm{mol\,L^{-1}} \]

Using the extension relationship

\[ \Delta G=RT\ln\left(\frac{Q}{K}\right) \]

determine:

a. The reaction quotient, \(Q\).

b. The actual Gibbs free energy change at 298 K.

c. The favoured net reaction direction.

d. Explain why the negative standard Gibbs energy change does not determine the direction in this particular mixture.

Solution 4

The reverse reaction is thermodynamically favoured in the current mixture, even though the standard Gibbs free energy change is negative.

Part a: Calculate the reaction quotient.

For the reaction:

\[ \ce{A(aq) <=> B(aq)} \]

the reaction quotient is:

\[ Q=\frac{[B]}{[A]} \]

Substitute the current concentrations:

\[ Q=\frac{0.90}{0.10}=9.0 \]

Therefore:

\[ Q=9.0 \]

Part b: Calculate the actual Gibbs free energy change.

Use:

\[ \Delta G=RT\ln\left(\frac{Q}{K}\right) \]

Substitute:

\[ \begin{aligned} \Delta G &=8.314(298)\ln\left(\frac{9.0}{4.0}\right)\\ &=8.314(298)\ln(2.25)\\ &=2009\ \mathrm{J\,mol^{-1}}\\ &=+2.01\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Part c: Determine the favoured direction.

The actual Gibbs free energy change is positive.

Therefore, the forward reaction is not thermodynamically favoured under the current conditions. The reverse reaction, converting B into A, is favoured.

Part d: Explain the difference between the two Gibbs energy values.

The standard Gibbs free energy change describes the reaction under standard-state conditions.

The actual Gibbs free energy change also depends on the current composition.

Here, the mixture contains relatively more B than the equilibrium composition favours. The reverse reaction therefore lowers the Gibbs energy.

This demonstrates why a negative \(\Delta G^\circ\) does not guarantee that the forward reaction is favoured in every possible mixture.

Question 5

Consider the Haber equilibrium:

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

At 298 K, take the equilibrium constant to be:

\[ K_p=5.8\times10^5 \]

Assume ideal gas behaviour.

A reaction vessel initially contains an equilibrium mixture at 298 K.

Two separate experiments are conducted.

Experiment A: An inert gas is added while the vessel volume and temperature remain constant.

Experiment B: An inert gas is added while the total pressure and temperature remain constant. The vessel volume doubles.

After each experiment, a catalyst is introduced.

A student makes three claims:

1. Since the standard Gibbs free energy change is negative, the forward reaction must remain favoured after both experiments.
2. Adding an inert gas must shift the equilibrium because the total pressure changes.
3. Adding a catalyst after Experiment B will restore the original equilibrium composition and increase the final ammonia yield.

Evaluate all three claims.

Calculate the standard Gibbs free energy change and, for each experiment, determine the actual thermodynamic direction immediately after the disturbance.

Solution 5

All three claims are incorrect or incomplete. The standard Gibbs free energy change is negative, but the effect of inert gas depends on whether volume or pressure is held constant. A catalyst cannot restore the original equilibrium composition.

Step 1: Calculate the standard Gibbs free energy change.

Use the relationship:

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

Substitute:

\[ \begin{aligned} \Delta G^\circ &=-8.314(298)\ln(5.8\times10^5)\\ &\approx-32\,900\ \mathrm{J\,mol^{-1}}\\ &\approx-32.9\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The standard Gibbs free energy change is negative.

This indicates a strong thermodynamic preference for ammonia formation from the standard state.

However, the initial mixture is already at equilibrium. Its actual Gibbs free energy change is zero.

Step 2: Analyse Experiment A – constant volume.

Adding an inert gas at constant volume does not change the partial pressures of nitrogen, hydrogen, or ammonia.

For an ideal gas:

\[ p_i=\frac{n_iRT}{V} \]

The amount of each reacting gas, temperature, and vessel volume remain unchanged.

Therefore, the reaction quotient does not change.

Since the system was initially at equilibrium:

\[ Q_p=K_p \]

It remains at equilibrium immediately after the inert gas is added.

There is no net shift in either direction.

The total pressure increases, but that does not automatically mean the equilibrium shifts. The partial pressures of the reacting gases are what matter for this calculation.

Step 3: Analyse Experiment B – constant pressure.

This experiment produces a different result.

The vessel volume doubles while the temperature remains constant.

The amounts of the reacting gases have not yet changed, so each reactive gas’s partial pressure is halved.

The reaction quotient is:

\[ Q_p= \frac{p_{\ce{NH3}}^2} {p_{\ce{N2}}p_{\ce{H2}}^3} \]

After the volume doubles:

\[ Q'_p= \frac{(p_{\ce{NH3}}/2)^2} {(p_{\ce{N2}}/2)(p_{\ce{H2}}/2)^3} \]

Simplifying:

\[ Q'_p=4Q_p \]

Because the original mixture was at equilibrium:

\[ Q_p=K_p \]

Therefore:

\[ Q'_p=4K_p \]

This means:

\[ Q'_p>K_p \]

The mixture now contains too much ammonia relative to the new equilibrium condition.

The reverse reaction is favoured, so some ammonia decomposes to form nitrogen and hydrogen.

Step 4: Calculate the actual Gibbs free energy change after expansion.

Using:

\[ \Delta G=RT\ln\left(\frac{Q}{K}\right) \]

and \(Q=4K\):

\[ \begin{aligned} \Delta G &=8.314(298)\ln(4)\\ &=3434\ \mathrm{J\,mol^{-1}}\\ &=+3.43\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The positive value confirms that the reverse reaction is thermodynamically favoured immediately after expansion.

Notice the surprising result:

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

but

\[ \Delta G>0 \]

There is no contradiction. The first value describes the standard-state reference, while the second describes the actual mixture after expansion.

The system moves towards a new equilibrium. This does not mean that the final mixture must become reactant-rich; the equilibrium constant is still very large.

Step 5: Evaluate the catalyst claim.

The catalyst does not change the equilibrium constant or the Gibbs energy change associated with the initial and final thermodynamic states.

It increases the rates of both forward and reverse reactions.

Consequently, the system reaches its new equilibrium more quickly, but the catalyst cannot restore the original equilibrium composition.

The final judgement is:

This question combines standard Gibbs energy, actual reaction conditions, gas equilibrium, and reaction kinetics. The important reasoning step is recognising that the same reaction at the same temperature can favour different net directions when its composition changes.

07What to study next

Once you can interpret the sign of Gibbs free energy correctly, the next useful skill is calculating the equilibrium composition rather than simply predicting its direction.

Focus on the relationship between:

• The equilibrium constant, \(K_c\) or \(K_p\).
• Initial reactant and product concentrations.
• Changes in equilibrium conditions.
• The difference between equilibrium position and reaction rate.

For that, continue with HSC Chemistry Module 5 Equilibrium: Complete Topic Summary.

The central distinction to remember is simple: a negative Gibbs free energy change tells you which direction is thermodynamically favoured under the specified conditions. It does not promise a fast reaction or complete conversion.