Why a Catalyst Does Not Change Gibbs Energy: HSC Chemistry

Two sealed reaction vessels contain the same starting mixture. A catalyst is added to one. Three minutes later, the catalysed vessel contains nearly four times as much product. Has the catalyst made the reaction more thermodynamically favourable?

No. A catalyst can increase the reaction rate, but it does not change the Gibbs free energy change, \(\Delta G\), for the same overall chemical reaction under the same conditions. Gibbs energy depends on the chemical states being compared, not on the pathway taken between them.

The distinction is between two questions:

• Feasibility: Is a reaction thermodynamically favoured in a particular direction?
• Rate: How quickly does the reaction proceed?

A catalyst changes the second. It doesn’t change the first.

Before going further, predict what would happen if both vessels were left under identical conditions for long enough.

For a simple reversible reaction that can reach equilibrium, both would approach the same equilibrium composition. The catalyst changes how quickly that composition is reached, not the thermodynamic conditions that determine it.

That prediction gives us a way to investigate the claim.

01An experiment that separates reaction rate from Gibbs energy

Consider a hypothetical reversible reaction:

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

Suppose A is colourless and B absorbs blue-green light. Their concentrations can therefore be determined using a calibrated optical measurement.

This is a model chemical system, not a claim about a particular real substance. The measurements below are illustrative data constructed to demonstrate the reasoning.

The apparatus

Imagine two identical sealed glass reaction cells placed side by side in a temperature-controlled water bath at \(300\ \mathrm{K}\).

Each contains:

• An initial concentration of A of \(1.00\ \mathrm{mol\,L^{-1}}\).
• No B initially.
• The same solution volume and solvent.
• The same pressure and temperature throughout the comparison.

One cell contains a small amount of catalyst. The other is the control.

A light source sits on one side of each cell. Light travels horizontally through the solution and reaches a detector on the opposite side.

The detectors are connected to a recording system that tracks the optical signal over time. Calibration converts this signal into the concentration of B.

The catalyst is assumed not to absorb significantly at the measurement wavelength or interfere with the calibration.

What should the first experiment show?

There are two competing predictions.

If the catalyst changes the thermodynamic favourability of forming B, the catalysed system might approach a different equilibrium composition.

If it changes only the reaction pathway, B should form faster in the catalysed cell, but both cells should eventually approach the same equilibrium composition.

The distinction is important because an early measurement cannot tell us which prediction is correct.

The misleading three-minute result

Suppose the experiment produces these illustrative measurements.

All concentrations are rounded. The values at 60 minutes represent mixtures that are effectively at equilibrium within the measurement precision.

At three minutes, the catalysed vessel contains \(0.75\ \mathrm{mol\,L^{-1}}\) of B, compared with only \(0.20\ \mathrm{mol\,L^{-1}}\) in the control.

That is a substantial difference.

But by 60 minutes, both contain approximately \(0.77\ \mathrm{mol\,L^{-1}}\) of B.

The early difference was evidence of a faster reaction, not a different equilibrium.

In fact, comparing product concentrations after three minutes would have produced the wrong conclusion about the catalyst’s thermodynamic effect.

Improving the experimental design

The initial experiment has an important weakness: it assumes that product concentration measured at a chosen time tells us something about equilibrium.

A better investigation would:

1. Continue recording until concentrations become stable within experimental uncertainty.
2. Confirm that both cells remain at the same temperature.
3. Use catalyst-only blanks to check whether the catalyst affects the optical measurement.
4. Repeat the experiment using different initial mixtures, including mixtures containing mostly B.
5. Check for side reactions or permanent binding of A or B to the catalyst.

Starting from mostly B is particularly useful.

If the proposed equilibrium is genuine, a system starting with excess B should approach the equilibrium composition from the opposite direction. The catalyst should accelerate that approach as well.

There is another limitation. An experiment like this measures concentrations and rates, not Gibbs energy directly.

Agreement between the final compositions supports the prediction that the catalyst has not changed the equilibrium. The deeper explanation comes from what Gibbs energy actually represents.

02Why the Gibbs energy change stays the same

Imagine that A can become B through a sequence of molecular rearrangements.

Without a catalyst, the rearrangement might require passing through a high-energy molecular configuration. Relatively few molecular encounters successfully complete that pathway.

A catalyst provides an alternative mechanism. It might temporarily bind A, rearrange its bonds, and release B.

A simplified two-step mechanism could be written as:

\[ \ce{A + Cat -> I} \]

\[ \ce{I -> B + Cat} \]

Here, \(\ce{Cat}\) represents the catalyst, and \(\ce{I}\) is a temporary intermediate.

The catalyst appears at the beginning and is regenerated at the end. Adding the steps gives the net reaction:

\[ \ce{A -> B} \]

The intermediate disappears from the overall equation, as does the unchanged catalyst.

Gibbs energy is a state function

Gibbs free energy, \(G\), is a thermodynamic property of a system.

A state function depends on the system’s state, such as its temperature, pressure, and chemical composition. It does not depend on the sequence of steps used to reach that state.

For a process between specified initial and final states:

\[ \Delta G=G_{\mathrm{final}}-G_{\mathrm{initial}} \]

Here:

• \(G_{\mathrm{initial}}\) is the Gibbs energy of the starting state.
• \(G_{\mathrm{final}}\) is the Gibbs energy of the finishing state.
• \(\Delta G\) is the difference between them.

The quantities can be expressed in joules or kilojoules. For a reaction expressed per mole of reaction as written, we commonly use \(\mathrm{kJ\,mol^{-1}}\).

Now apply this to the catalyst.

The catalysed and uncatalysed pathways begin with the same overall reactants and end with the same overall products. Under the stated assumptions, the catalyst is regenerated and does not alter the chemical states being compared.

Therefore, the Gibbs energy difference between those states is unchanged.

This remains true even if the catalysed mechanism involves several intermediates.

A lower energy barrier is not a lower Gibbs energy change

A reaction pathway can have a high barrier while still leading to a thermodynamically favourable final state.

The barrier affects how readily the reaction takes place. The difference between the initial and final states determines its overall Gibbs energy change.

There is one technical distinction worth keeping straight. A Gibbs-energy landscape uses an activation free energy, commonly written \(\Delta G^\ddagger\), to describe the barrier. The activation energy \(E_a\) commonly taught using enthalpy or potential-energy profiles is not necessarily numerically identical to \(\Delta G^\ddagger\).

Both descriptions separate the barrier from the overall thermodynamic change.

For more practice interpreting activation-energy diagrams, see Catalysts and Energy Profiles: HSC Chemistry Explained.

The important result here is that reducing a barrier does not require changing the starting or finishing Gibbs energy.

03Gibbs energy predicts feasibility, not speed

At constant temperature and pressure, the sign of the reaction Gibbs energy change tells us the thermodynamically favoured direction.

For the current composition of a reacting mixture, the precise quantity is the reaction Gibbs energy, \(\Delta_\mathrm{r}G\).

These signs describe thermodynamic tendency, not reaction speed.

A reaction can be thermodynamically favourable but extremely slow. A catalyst can make it faster without changing that underlying tendency.

A reaction can also be thermodynamically unfavourable in the forward direction while still experiencing individual forward reaction events. What matters is the direction of the net change.

At equilibrium, forward and reverse reactions continue, but their rates are equal. The mixture’s composition therefore remains constant.

Standard Gibbs energy is not always the Gibbs energy of the current mixture

You will often see \(\Delta G^\circ\) rather than \(\Delta G\).

The superscript circle indicates standard-state conditions. For reaction calculations, \(\Delta G^\circ\) is the standard Gibbs energy change per mole of reaction as written.

The actual thermodynamic driving force also depends on the current composition.

For example, changing the amounts of A and B can change the direction in which the mixture tends to react, even though \(\Delta G^\circ\) remains the same at a fixed temperature.

This distinction becomes important when interpreting reactions that start with only reactants, or when a system is disturbed after reaching equilibrium.

04Calculating Gibbs energy with and without a catalyst

For standard-state thermodynamic data, we can use:

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

Where:

• \(\Delta G^\circ\) is standard Gibbs energy change, usually in \(\mathrm{kJ\,mol^{-1}}\).
• \(\Delta H^\circ\) is standard enthalpy change, usually in \(\mathrm{kJ\,mol^{-1}}\).
• \(T\) is absolute temperature, measured in kelvin (\(\mathrm{K}\)).
• \(\Delta S^\circ\) is standard entropy change, often supplied in \(\mathrm{J\,mol^{-1}\,K^{-1}}\).

Entropy measures the dispersal of energy among available microscopic arrangements. It is not simply a count of how messy the system looks.

Enthalpy accounts for the heat-related energy change at constant pressure. Entropy and temperature together contribute to the reaction’s Gibbs energy change.

For a given reaction under the same standard-state conditions, adding a catalyst does not change its \(\Delta H^\circ\) or \(\Delta S^\circ\). Consequently, it cannot change \(\Delta G^\circ\).

Syllabus note: Chemistry Stage 6 (2017)

The numerical use of \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\), including interpreting reaction spontaneity and temperature effects, is part of Year 11 Module 4, Drivers of Reactions, in the NSW Chemistry Stage 6 Syllabus (2017).

As at November 2026, this remains the syllabus being taught. NESA’s Chemistry 11-12 Syllabus (2025) is scheduled for Year 11 implementation in 2028, with its first HSC examination in 2029.

The discussion here of standard versus actual reaction Gibbs energy, and the later light-driven reaction challenge, adds conceptual depth. Do not assume every extension in this article is an explicit assessment requirement of either syllabus edition.

05What can change Gibbs energy instead?

A useful way to recognise the catalyst misconception is to compare it with conditions that genuinely can change the thermodynamic situation.

The table assumes the same chemical reaction is being considered, with no unexpected side reactions or permanent changes to the catalyst.

A contrasting case: heating a catalysed reaction

Suppose a reaction has:

\[ \Delta H^\circ=+28.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+90.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

At \(298\ \mathrm{K}\), the entropy contribution is:

\[ T\Delta S^\circ=(298)(0.0900) \]

That gives \(26.82\ \mathrm{kJ\,mol^{-1}}\), so:

\[ \Delta G^\circ=+1.18\ \mathrm{kJ\,mol^{-1}} \]

At \(330\ \mathrm{K}\):

\[ \Delta G^\circ=28.0-(330)(0.0900) \]

\[ \Delta G^\circ=-1.70\ \mathrm{kJ\,mol^{-1}} \]

We have assumed that the supplied standard enthalpy and entropy changes remain approximately constant over this temperature range.

The sign has changed because the temperature changed, not because a catalyst was added.

If a poorly controlled catalysed experiment warmed from 298 K to 330 K, we could not simply attribute its changed reaction behaviour to catalysis.

This is why controlling temperature is essential when testing the effect of a catalyst.

06The diagnostic test for catalyst questions

When a question asks whether a catalyst has changed Gibbs energy, work through these distinctions:

1. Is the overall reaction unchanged? A different net reaction, including a coupled reaction, can have a different Gibbs energy change.
2. Are temperature and other thermodynamic conditions unchanged? Changing them can change the thermodynamic result.
3. Are the starting compositions the same? This matters when interpreting the direction of net reaction in a mixture.
4. Is the evidence about rate or final equilibrium? More product after a short time is not automatically evidence of a different equilibrium.
5. Is there an external energy supply? A system maintained away from equilibrium by light or electricity is not simply an ordinary closed equilibrium reaction with a catalyst.

A final subtlety: adding a catalyst introduces material into a vessel, so the total Gibbs energy of a vessel containing catalyst need not be identical to that of a vessel without it. The claim is that the Gibbs energy change of the specified net chemical reaction is unchanged when the catalyst is regenerated and the relevant conditions are held fixed.

This also explains why the related statement about equilibrium yield is true. For that separate issue, see Why a Catalyst Does Not Change Equilibrium Yield.

07Questions and solutions

These five questions build from a standard Gibbs energy calculation to harder problems involving measurement errors, changes in equilibrium, and a reaction maintained away from equilibrium by an external energy source.

All numerical data are original and illustrative unless otherwise stated.

Question 1

A hypothetical reaction has the following standard thermodynamic data:

\[ \Delta H^\circ=-24.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=-40.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

At \(300\ \mathrm{K}\), calculate \(\Delta G^\circ\).

A catalyst is then introduced without changing the temperature, pressure, or overall reaction.

State the new value of \(\Delta G^\circ\), and explain whether the catalyst makes the reaction more thermodynamically favourable.

Solution 1

The standard Gibbs energy change is \(-12.0\ \mathrm{kJ\,mol^{-1}}\), both before and after adding the catalyst. The forward reaction is thermodynamically favoured under the standard-state conditions, but the catalyst does not increase that favourability.

First, convert the entropy change:

\[ \Delta S^\circ=-0.0400\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Then calculate:

\[ \begin{aligned} \Delta G^\circ &=\Delta H^\circ-T\Delta S^\circ\\ &=-24.0-(300)(-0.0400)\\ &=-24.0+12.0\\ &=-12.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative result means the forward reaction is favoured under standard-state conditions at 300 K.

The catalyst does not change \(\Delta H^\circ\), \(\Delta S^\circ\), or the thermodynamic states of the overall reaction.

Therefore:

\[ \Delta G^\circ_{\mathrm{after}} =-12.0\ \mathrm{kJ\,mol^{-1}} \]

The tempting error is to assume that a faster reaction must have a more negative Gibbs energy change. Rate and thermodynamic favourability are different properties.

Question 2

A hypothetical reversible reaction is:

\[ \ce{D(aq) <=> E(aq)} \]

Its standard thermodynamic data are:

\[ \Delta H^\circ=+21.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+60.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Two sealed vessels contain identical initial solutions of D, with no E initially.

• Vessel A is held at \(300\ \mathrm{K}\) and contains a catalyst.
• Vessel B is held at \(400\ \mathrm{K}\) without a catalyst.

Both vessels eventually produce some E.

A student makes two claims:

1. Because \(\Delta G^\circ\) is positive at 300 K, the catalyst must have changed Gibbs energy to allow E to form.
2. Because \(\Delta G^\circ\) is negative at 400 K, vessel B must contain more E after exactly two minutes.

Evaluate both claims.

Also predict which vessel will have the greater equilibrium proportion of E, assuming ideal solution behaviour, unchanged reaction enthalpy and entropy over the temperature range, and no side reactions.

Solution 2

Both claims are incorrect. Vessel B will have the greater equilibrium proportion of E, but the information is insufficient to decide which vessel contains more E after two minutes.

First, convert entropy:

\[ \Delta S^\circ=0.0600\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

At 300 K:

\[ \begin{aligned} \Delta G^\circ &=21.0-(300)(0.0600)\\ &=+3.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

At 400 K:

\[ \begin{aligned} \Delta G^\circ &=21.0-(400)(0.0600)\\ &=-3.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The standard Gibbs energy changes are different because the temperatures are different.

Why Claim 1 fails

A positive \(\Delta G^\circ\) does not mean that no E molecules can ever form.

The starting mixture contains only D. It is not the standard-state mixture used to define \(\Delta G^\circ\).

For an ideal reversible system, some E can form from an initially E-free mixture, even when \(\Delta G^\circ>0\). The system proceeds towards an equilibrium containing both substances.

The catalyst changes the rate at which this happens. It does not change the standard Gibbs energy change at 300 K.

Why Claim 2 fails

A negative \(\Delta G^\circ\) indicates that products are favoured under standard-state conditions at 400 K.

It does not tell us how quickly the uncatalysed reaction takes place.

At two minutes, the catalysed vessel at 300 K might contain more E than the uncatalysed vessel at 400 K. The opposite is also possible.

We would need kinetic data to determine that comparison.

Which vessel has more E at equilibrium?

Vessel B, at 400 K.

The reaction is endothermic, so increasing temperature favours E at equilibrium. For this ideal one-to-one reaction, the change from positive to negative \(\Delta G^\circ\) also indicates a change from an equilibrium favouring D to one favouring E.

The higher equilibrium proportion of E in vessel B results from temperature, not the presence or absence of catalyst.

This problem exposes two separate mistakes: confusing standard Gibbs energy with the behaviour of an actual mixture, and using thermodynamic favourability to predict reaction speed.

Question 3

A researcher investigates the reversible reaction:

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

Two sealed vessels are maintained at \(300\ \mathrm{K}\). They begin with the same amount of A and no B.

The researcher uses an optical instrument calibrated to report the concentration of B in \(\mathrm{mol\,L^{-1}}\).

One vessel contains a catalyst.

The instrument gives these readings:

An independent calibration establishes that the catalyst produces an additional optical signal equivalent to \(0.07\ \mathrm{mol\,L^{-1}}\) of B, even when no B is present.

This interference is additive and constant across the concentration range.

Assume the reaction is the only chemical transformation, and all other experimental conditions are identical.

(a) Determine the corrected B concentration in the catalysed vessel at three and 120 minutes.

(b) Calculate the average rate of B formation over the first three minutes in each vessel.

(c) A researcher concludes that the catalyst changes Gibbs energy because the original 120-minute readings differ. Evaluate the conclusion.

(d) Suggest one further measurement that would strengthen the thermodynamic interpretation.

Solution 3

After correcting for optical interference, both vessels approach the same B concentration of \(0.61\ \mathrm{mol\,L^{-1}}\). The catalyst increases the measured early reaction rate, but the evidence does not support a catalyst-induced change in Gibbs energy.

(a) Correct the readings

The catalyst adds an apparent concentration of:

\[ 0.07\ \mathrm{mol\,L^{-1}} \]

At three minutes:

\[ [B]_{\mathrm{corrected}}=0.52-0.07 \]

\[ [B]_{\mathrm{corrected}}=0.45\ \mathrm{mol\,L^{-1}} \]

At 120 minutes:

\[ [B]_{\mathrm{corrected}}=0.68-0.07 \]

\[ [B]_{\mathrm{corrected}}=0.61\ \mathrm{mol\,L^{-1}} \]

Both vessels therefore give the same late-time B concentration within the supplied precision.

(b) Calculate average rates

For the uncatalysed vessel:

\[ \begin{aligned} \bar v_{\mathrm{uncat}} &=\frac{0.17-0.00}{3}\\ &=0.0567 \end{aligned} \]

For the catalysed vessel, use the corrected readings:

\[ \begin{aligned} \bar v_{\mathrm{cat}} &=\frac{0.45-0.00}{3}\\ &=0.150 \end{aligned} \]

Both results are in \(\mathrm{mol\,L^{-1}\,min^{-1}}\).

The catalysed vessel’s average rate is approximately:

\[ \frac{0.150}{0.0567}\approx2.65 \]

times the uncatalysed rate over this interval.

(c) Evaluate the Gibbs energy claim

The conclusion is unsupported.

The apparent difference at 120 minutes is entirely explained by the stated optical interference.

Once corrected, both vessels show the same final concentration of B.

Their repeated stable readings are consistent with the two mixtures approaching the same equilibrium.

The catalyst has not been shown to change the thermodynamic relationship between A and B.

The important experimental lesson is that an instrument reading is not always the same as the quantity it is intended to measure. A measurement bias can imitate a chemical effect.

(d) Strengthen the investigation

Repeat the experiment using a mixture initially rich in B, and follow its approach towards equilibrium.

If it approaches the same equilibrium composition from the opposite direction, that would strengthen the conclusion that the observed late-time composition represents genuine equilibrium.

An independent analytical method for measuring A and B would also help establish that the catalyst’s optical interference has been corrected properly.

Neither measurement is a direct measurement of Gibbs energy, but both improve the validity of the thermodynamic interpretation.

Question 4

A reversible gas reaction is represented by:

\[ \ce{A(g) + B(g) <=> C(g)} \]

At a fixed temperature of \(300\ \mathrm{K}\):

\[ K_c=2.0\ \mathrm{L\,mol^{-1}} \]

Two identical vessels, each initially of volume \(1.00\ \mathrm{L}\), contain the same equilibrium gas mixture:

\[ [A]=1.00\ \mathrm{mol\,L^{-1}} \]

\[ [B]=1.00\ \mathrm{mol\,L^{-1}} \]

\[ [C]=2.00\ \mathrm{mol\,L^{-1}} \]

Vessel 1 has a solid catalyst that does not occupy significant gas volume. Vessel 2 has no catalyst.

Both gases are compressed isothermally to a volume of \(0.500\ \mathrm{L}\). The compression is sufficiently rapid that negligible reaction occurs during the volume change.

The vessels are then allowed to establish equilibrium.

Finally, both are expanded isothermally back to \(1.00\ \mathrm{L}\), again sufficiently rapidly that negligible reaction occurs during the expansion.

(a) Determine the direction of net reaction immediately after compression.

(b) Calculate the equilibrium concentrations in the compressed vessels.

(c) Determine the initial direction of net reaction immediately after re-expansion.

(d) Explain why these changes do not contradict the statement that a catalyst leaves Gibbs energy unchanged.

Solution 4

Compression initially drives the net reaction forwards, and both vessels reach the same compressed equilibrium concentrations: \(1.50\ \mathrm{mol\,L^{-1}}\) each of A and B, and \(4.50\ \mathrm{mol\,L^{-1}}\) of C. Re-expansion then drives the reaction backwards. None of these changes requires the catalyst to alter Gibbs energy.

(a) Direction immediately after compression

The reaction quotient is:

\[ Q_c=\frac{[C]}{[A][B]} \]

Immediately after the volume is halved, each concentration doubles.

Therefore:

\[ [A]=2.00\ \mathrm{mol\,L^{-1}} \]

\[ [B]=2.00\ \mathrm{mol\,L^{-1}} \]

\[ [C]=4.00\ \mathrm{mol\,L^{-1}} \]

Substitute:

\[ \begin{aligned} Q_c &=\frac{4.00}{(2.00)(2.00)}\\ &=1.0\ \mathrm{L\,mol^{-1}} \end{aligned} \]

Since \(Q_c<K_c\), the net reaction proceeds forwards.

The compression has altered the current mixture’s thermodynamic driving force, even though the temperature and standard Gibbs energy change are unchanged.

(b) Equilibrium concentrations after compression

Let \(x\) be the concentration of C formed after compression, measured in \(\mathrm{mol\,L^{-1}}\).

The new equilibrium concentrations are:

\[ [A]=2.00-x \]

\[ [B]=2.00-x \]

\[ [C]=4.00+x \]

Using the numerical concentration values in \(\mathrm{mol\,L^{-1}}\):

\[ 2.0=\frac{4.00+x}{(2.00-x)^2} \]

Rearranging:

\[ 2(2-x)^2=4+x \]

\[ 2x^2-9x+4=0 \]

Factorise:

\[ (2x-1)(x-4)=0 \]

The mathematical possibilities are \(x=0.50\) and \(x=4.00\).

But \(x=4.00\) would require more A and B to react than are present. It would produce negative reactant concentrations.

The physically possible solution is:

\[ x=0.50\ \mathrm{mol\,L^{-1}} \]

Hence:

\[ [A]=[B]=1.50\ \mathrm{mol\,L^{-1}} \]

\[ [C]=4.50\ \mathrm{mol\,L^{-1}} \]

Both vessels reach this same equilibrium composition.

The catalyst in Vessel 1 affects the time required to get there.

(c) Direction after re-expansion

When the volume doubles, all concentrations immediately halve:

\[ [A]=[B]=0.750\ \mathrm{mol\,L^{-1}} \]

\[ [C]=2.25\ \mathrm{mol\,L^{-1}} \]

Calculate the reaction quotient:

\[ \begin{aligned} Q_c &=\frac{2.25}{(0.750)(0.750)}\\ &=4.0\ \mathrm{L\,mol^{-1}} \end{aligned} \]

Now \(Q_c>K_c\).

The reverse reaction is favoured, so C is converted back into A and B.

After sufficient time, each vessel returns to its original equilibrium composition.

(d) Why Gibbs energy has not been changed by the catalyst

The catalyst has not altered the standard Gibbs energy change of the chemical reaction.

Compression and expansion have changed the current composition and pressure, and therefore the reaction Gibbs energy of those mixtures.

Those are changes to the thermodynamic state, not the catalytic pathway.

The tempting argument is that because a catalyst cannot change Gibbs energy, it cannot change anything about how the system responds to compression.

That is incorrect. The catalyst can accelerate the response to a genuine thermodynamic disturbance without causing the disturbance.

The return to the original equilibrium composition also shows why the chemical system’s initial and final states matter. Over the complete return to the original state, the net Gibbs energy change of the system is zero, even though individual stages involve reactions, heat exchange, and mechanical work.

Question 5

A hypothetical reversible reaction occurs in solution:

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

Its standard thermodynamic data at \(300\ \mathrm{K}\) are:

\[ \Delta H^\circ=+15.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+40.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Three sealed vessels contain the same solution, with:

\[ [A]+[B]=1.00\ \mathrm{mol\,L^{-1}} \]

All three contain the same reusable catalyst, and a temperature-control system maintains them at exactly \(300\ \mathrm{K}\).

The catalyst can absorb light of a particular wavelength and use that absorbed energy to drive conversion of A into B. It is regenerated after this light-driven process.

The vessels are treated differently:

• Vessel 1 is kept in darkness.
• Vessel 2 is illuminated continuously with light of the wavelength absorbed by the catalyst.
• Vessel 3 receives light of the same total intensity, but a filter removes the wavelength absorbed by the catalyst.

After a long time, the measured B concentrations become approximately constant:

The dark mixture is known to be at chemical equilibrium. Assume ideal dilute-solution behaviour, no side reactions or catalyst degradation, and no loss of A or B. All values are illustrative.

A student argues:

“Vessel 2 has reached a new equilibrium because its concentration stopped changing. The catalyst must therefore have changed Gibbs energy, even though the temperature stayed constant.”

(a) Calculate the standard Gibbs energy change at 300 K.

(b) Use the concentrations to determine whether the stable mixture in Vessel 2 is at ordinary chemical equilibrium.

(c) Explain what is maintaining the unusually high concentration of B.

(d) Predict what happens if the lamp illuminating Vessel 2 is switched off while temperature, catalyst concentration, and all other conditions remain unchanged.

(e) Explain why the entire observation does not violate the principle that a catalyst cannot change the Gibbs energy of a chemical reaction.

Solution 5

The catalyst has not changed the chemical reaction’s standard Gibbs energy. Vessel 2 is in a light-driven steady state, not ordinary chemical equilibrium. Its high B concentration is maintained by continuous energy input, and B will decrease towards the dark equilibrium concentration when the lamp is switched off.

(a) Calculate the standard Gibbs energy change

First, convert the entropy change:

\[ \Delta S^\circ=0.0400\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Then:

\[ \begin{aligned} \Delta G^\circ &=15.0-(300)(0.0400)\\ &=15.0-12.0\\ &=+3.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Under standard-state conditions, the forward reaction is thermodynamically unfavourable.

This value is unchanged by the catalyst in all three vessels.

(b) Determine whether Vessel 2 is at equilibrium

Vessel 2 is not at ordinary chemical equilibrium.

We can show this by comparing its reaction quotient with the equilibrium constant inferred from Vessel 1.

For this ideal one-to-one reaction:

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

At equilibrium in Vessel 1:

\[ [B]=0.23\ \mathrm{mol\,L^{-1}} \]

\[ [A]=1.00-0.23=0.77\ \mathrm{mol\,L^{-1}} \]

Therefore, to the precision of the illustrative measurements:

\[ K_c\approx\frac{0.23}{0.77}\approx0.30 \]

For Vessel 2:

\[ [B]=0.70\ \mathrm{mol\,L^{-1}} \]

\[ [A]=1.00-0.70=0.30\ \mathrm{mol\,L^{-1}} \]

Hence:

\[ Q_c=\frac{0.70}{0.30}\approx2.33 \]

Since \(Q_c>K_c\), the chemical mixture has a thermodynamic tendency to react backwards, from B to A.

Yet its B concentration remains constant while the lamp is on.

That tells us the system is not simply resting at chemical equilibrium.

The values are consistent with the Gibbs energy calculation: a positive standard Gibbs energy change favours A under standard conditions, and the dark equilibrium contains more A than B.

(c) Explain the stable B concentration

The lamp continuously transfers energy into the system through photons absorbed by the catalyst.

That energy supports conversion of A into B, even though the composition of the illuminated mixture favours the reverse chemical reaction thermodynamically.

At the stable observed composition, light-driven production of B balances its net removal through the other reaction processes.

There is no overall accumulation of B, but energy is still entering and being dissipated.

This is a non-equilibrium steady state.

The crucial distinction is that constant concentration does not, by itself, prove thermodynamic equilibrium.

In ordinary chemical equilibrium, there is no net thermodynamic driving force for a small advance of the reaction.

In the illuminated vessel, the light-driven process continually opposes the mixture’s tendency to return towards dark equilibrium.

(d) Predict what happens when the lamp is switched off

B will decrease from approximately \(0.70\ \mathrm{mol\,L^{-1}}\) towards \(0.23\ \mathrm{mol\,L^{-1}}\).

When illumination stops, the external energy supply driving additional A-to-B conversion disappears.

The remaining mixture contains more B than its dark equilibrium composition.

The net reaction therefore proceeds backwards:

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

until the ordinary equilibrium composition is approached.

The catalyst is still present, so it can continue to accelerate the thermal reactions. But without the light input, it cannot maintain the previous B concentration.

(e) Resolve the apparent contradiction

There is no contradiction because the illuminated vessel is not simply undergoing the same undriven reaction under otherwise identical conditions.

The chemical reaction’s standard Gibbs energy change remains:

\[ \Delta G^\circ=+3.0\ \mathrm{kJ\,mol^{-1}} \]

The catalyst has not altered that thermodynamic quantity.

Instead, the apparatus continuously supplies energy through absorbed light. That energy can drive a process which would not proceed in the same net direction unaided at the observed composition.

The temperature remaining constant does not mean that no energy is entering the system. The temperature-control system can remove energy as heat while the lamp continues to supply it.

This is the hidden assumption in the student’s argument: treating a vessel that is closed to matter as though it were also isolated from energy transfer.

The two are not equivalent.

A catalyst can speed up an energetically permitted process. A light-powered catalytic system can additionally use supplied energy to maintain a chemical state away from equilibrium.

In neither case has the catalyst itself changed the Gibbs energy difference of the specified chemical reaction.

This leads to the next useful distinction in chemical thermodynamics: the difference between a reaction’s standard Gibbs energy, its actual driving force at a particular composition, and the energy supplied when a process is coupled to another process. Those three ideas explain how reactions can be fast, slow, thermodynamically favoured, or externally driven without contradicting one another.