Why Spontaneous Reactions Can Be Slow: HSC Chemistry

A bottle of hydrogen peroxide can sit on a shelf for months without producing obvious bubbles. Add a suitable catalyst, and oxygen can be released rapidly. If the decomposition is already thermodynamically favourable, why doesn’t it happen quickly on its own?

Because spontaneity and reaction rate describe two different things. Spontaneity tells us whether a chemical change is thermodynamically favoured under particular conditions. Reaction rate tells us how quickly that change occurs. A reaction can be spontaneous but extremely slow because its molecules must overcome a large activation-energy barrier.

Here’s the prediction worth testing: if a catalyst makes a spontaneous reaction faster, does it also make the reaction more spontaneous?

No. A catalyst provides a faster reaction pathway without changing the thermodynamic difference between the initial and final states. Understanding why requires looking at both the molecules and the experiment used to measure their reaction.

01Why favourable reactions don’t always happen quickly

Hydrogen peroxide decomposes to form water and oxygen:

\[ \ce{2H2O2(aq) -> 2H2O(l) + O2(g)} \]

Under ordinary conditions, this decomposition is thermodynamically favourable. Yet hydrogen peroxide solutions can remain largely undecomposed for considerable periods, especially when properly stored and protected from substances that catalyse decomposition.

What is preventing the molecules from reacting rapidly?

The molecules must rearrange their bonds

Each hydrogen peroxide molecule contains an oxygen-oxygen single bond:

\[ \ce{H-O-O-H} \]

During decomposition, bonds are rearranged to form water molecules and oxygen molecules.

The overall reaction releases energy, but that does not mean the molecules can move directly from reactants to products without overcoming a barrier.

The reaction must proceed through a suitable molecular pathway. Reaching the necessary reactive arrangements requires energy.

Think about riding a bicycle down a hill with a locked gate across the road.

The downhill slope represents a thermodynamically favourable change. The gate represents a barrier preventing an easy transition.

Unlocking the gate doesn’t make the bottom of the hill any lower. It simply makes the journey possible without overcoming such a difficult obstacle.

A catalyst plays a similar role by providing a lower-barrier chemical pathway.

The analogy has a limit: molecules aren’t bicycles waiting at a gate. Their energies and arrangements constantly change, and some can react even without a catalyst. A high barrier makes successful reactions rare, not necessarily impossible.

Activation energy controls how readily a reaction starts

The activation energy, \(E_a\), is the energy barrier associated with a reaction pathway.

For a reaction to occur through that pathway, molecules must reach a suitable high-energy arrangement. At a particular temperature, only a proportion of molecular encounters have the necessary energy and orientation.

If the activation energy is large, successful reactions may be uncommon.

A catalyst changes the mechanism so that the reaction can proceed through an alternative pathway with a lower activation energy.

There are two separate energy comparisons here:

• The height from the reactants to the pathway’s peak relates to the activation energy and therefore the reaction rate.
• The vertical difference between reactants and products on an enthalpy profile represents the enthalpy change, \(\Delta H\).

An exothermic reaction has \(\Delta H<0\), but that alone does not establish spontaneity. Thermodynamic feasibility depends on Gibbs free energy, which also accounts for entropy.

The important point is that lowering an activation barrier does not change the initial and final thermodynamic states.

02An experiment that separates reaction rate from feasibility

Suppose you’re investigating hydrogen peroxide decomposition in a school laboratory.

The problem is straightforward: hydrogen peroxide appears to produce little oxygen on its own, but oxygen appears much faster when manganese dioxide is introduced.

Is manganese dioxide making an otherwise impossible reaction possible, or merely making an already favourable reaction faster?

We can use oxygen production to investigate.

The first experimental design

Imagine preparing two identical conical flasks.

Each contains \(10.0\ \mathrm{mL}\) of \(0.400\ \mathrm{mol\,L^{-1}}\) hydrogen peroxide solution.

• Flask A contains hydrogen peroxide without an added catalyst.
• Flask B contains the same solution but receives a small amount of manganese dioxide, \(\ce{MnO2(s)}\).

Each flask is fitted with an airtight bung. A delivery tube runs from the bung to a horizontal gas syringe.

As oxygen forms, it travels through the tube and pushes the gas syringe’s plunger outwards.

The syringe provides a direct measurement of gas volume against time.

What should this experiment show?

If manganese dioxide increases the reaction rate, Flask B should produce oxygen more rapidly.

Flask A might still produce oxygen. However, if its reaction is sufficiently slow, the gas collected during a short experiment could be too small to measure reliably.

The first design has a problem

Suppose you add the manganese dioxide before inserting the bung.

Oxygen may begin escaping immediately.

By the time the bung is fitted, some gas has already been lost. The measured volume will underestimate the oxygen actually produced.

This problem is especially important when studying a fast catalysed reaction.

The design also needs better temperature control because hydrogen peroxide decomposition releases heat. If Flask B becomes warmer, temperature and catalyst addition could both affect its measured rate.

Revising the experiment

A stronger experimental design would use an internal holder to keep the catalyst separate from the hydrogen peroxide until the apparatus is sealed.

Both flasks would be tilted in the same way at the start of timing, releasing the catalyst in Flask B. Flask A would undergo the same movement without catalyst addition.

Both would be maintained at the same temperature using a controlled water bath.

Keep the initial hydrogen peroxide volume, concentration, apparatus, temperature, and timing consistent. Repeat the comparison to assess reliability.

This isolates catalyst addition much more effectively.

The apparatus should be used only under appropriate laboratory supervision and chemical risk controls.

What would the measurements mean?

Consider this illustrative dataset.

Illustrative data only: These are invented values to demonstrate interpretation, not observations from a performed experiment.

Both flasks begin with \(10.0\ \mathrm{mL}\) of \(0.400\ \mathrm{mol\,L^{-1}}\) hydrogen peroxide. Temperature is maintained at \(25^\circ\mathrm{C}\), and the gas syringe can reliably distinguish changes of approximately \(0.5\ \mathrm{mL}\) or greater.

In the catalysed flask, oxygen production is initially rapid and then slows as the hydrogen peroxide concentration falls.

In the control flask, so little gas is produced during the first two minutes that the instrument cannot resolve a reliable volume.

But after ten minutes, a small amount of oxygen is detectable.

The important observation is not that the uncatalysed reaction was impossible. It was simply much slower under these illustrative conditions.

03Why a catalyst changes speed without changing the final equilibrium

Manganese dioxide provides an alternative pathway for hydrogen peroxide decomposition.

It participates in the reaction mechanism but is regenerated overall, rather than being consumed as a stoichiometric reactant.

That explains why adding a relatively small amount can substantially increase the rate.

But consider a different prediction.

If a catalyst makes a reaction faster, should it also allow a greater proportion of reactants to turn into products at equilibrium?

Not at a fixed temperature when the catalyst acts solely as a catalyst.

For a reversible reaction, the catalyst increases the rates of both forward and reverse processes. It does not alter the equilibrium constant or equilibrium composition.

This gives us a useful distinction between three observations:

For hydrogen peroxide, the theoretical amount of oxygen also depends on how much peroxide was initially present.

A catalyst cannot create additional oxygen atoms.

If a gas collection experiment produces more oxygen than allowed by the balanced equation and starting amount, the problem is not insufficient catalytic power. Some measurement or chemical assumption needs checking.

A contrasting example: Why diamonds don’t turn into graphite

Hydrogen peroxide is not the only case where thermodynamics and kinetics disagree.

At ordinary room temperature and atmospheric pressure, graphite is thermodynamically more stable than diamond.

The transformation can be represented as:

\[ \ce{C(s, diamond) -> C(s, graphite)} \]

Yet a diamond doesn’t visibly turn into graphite while sitting in a drawer.

Diamond has a rigid, three-dimensional covalent network. Converting it into graphite requires extensive rearrangement of carbon bonding and structure.

The transformation has a substantial kinetic barrier.

So diamond is an example of a metastable material: it persists for extremely long periods despite another form being thermodynamically more stable under those conditions.

This exposes the misconception particularly well.

A material doesn’t have to be the most thermodynamically stable possible form to persist.

And a reaction doesn’t have to be non-spontaneous to be too slow for us to notice.

04The thermodynamic test: Gibbs free energy

Syllabus note (November 2026): The NSW Chemistry Stage 6 Syllabus (2017) remains in use for current HSC students. Its Year 11 Module 4, Drivers of Reactions, explicitly includes entropy, Gibbs free energy, and spontaneity calculations. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. The Gibbs-energy calculation below is labelled for the 2017 syllabus and should not be assumed to be assessed content under the 2025 edition.

Under constant temperature and pressure, Gibbs free energy provides the thermodynamic test for whether a specified forward change is favourable.

A negative Gibbs free energy change means that the stated forward change is thermodynamically favourable.

A positive value means the forward change is thermodynamically unfavourable under the stated conditions.

Neither sign tells us how quickly the reaction occurs.

The 2017 syllabus equation

For standard-state thermodynamic data:

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

Where:

• \(\Delta G^\circ\) is the standard Gibbs free energy change, usually in \(\mathrm{kJ\,mol^{-1}}\) of reaction.
• \(\Delta H^\circ\) is the standard enthalpy change, in \(\mathrm{kJ\,mol^{-1}}\) of reaction.
• \(T\) is absolute temperature in kelvin, \(\mathrm{K}\).
• \(\Delta S^\circ\) is the standard entropy change, converted to \(\mathrm{kJ\,K^{-1}\,mol^{-1}}\) of reaction.

The superscript \(\circ\) indicates standard-state quantities.

The sign convention for the forward reaction is:

The distinction between \(\Delta G^\circ\) and the actual Gibbs free energy change matters.

A negative \(\Delta G^\circ\) does not mean every possible mixture will keep reacting forwards. As concentrations change, the actual thermodynamic driving force changes.

For a more detailed treatment of this limitation, see How to Interpret Negative ΔG Without Assuming Complete Conversion.

05How to diagnose a reaction that appears to have stopped

Imagine monitoring oxygen production in a flask.

The gas syringe stops moving.

There are at least three possible explanations:

1. The reaction is occurring so slowly that the change is not measurable over the observation period.
2. The system has reached equilibrium, so there is no net change.
3. A reactant has been exhausted, or another practical limitation is preventing further product collection.

These explanations are not interchangeable.

For a reversible reaction at equilibrium, forward and reverse reactions continue at equal rates. Adding a catalyst speeds both directions without producing a sustained net shift.

For a mixture that is away from equilibrium but kinetically trapped, adding a suitable catalyst may allow a measurable change towards equilibrium.

And for a reaction whose limiting reactant has already been completely consumed, a catalyst cannot generate more product from reactant that is no longer present.

That gives you a useful exam method:

The strongest answers identify which question the evidence actually addresses.

06Questions and solutions

Question 1

A hydrogen peroxide decomposition experiment produces \(24.0\ \mathrm{mL}\) of oxygen in the first \(40.0\ \mathrm{s}\) when manganese dioxide is present.

An otherwise identical flask without added catalyst produces less than the gas syringe’s detectable limit of \(0.5\ \mathrm{mL}\) over the same period.

a. Calculate the average oxygen collection rate for the catalysed reaction in \(\mathrm{mL\,s^{-1}}\).

b. A student concludes that the uncatalysed reaction is non-spontaneous. Explain whether the observation supports this conclusion.

Solution 1

The catalysed flask has an average oxygen collection rate of \(0.600\ \mathrm{mL\,s^{-1}}\). The uncatalysed result does not establish that the reaction is non-spontaneous.

For part a, use:

\[ \begin{aligned} r_V&=\frac{\Delta V}{\Delta t}\\ &=\frac{24.0\ \mathrm{mL}}{40.0\ \mathrm{s}}\\ &=0.600\ \mathrm{mL\,s^{-1}} \end{aligned} \]

This is the average volume of oxygen collected each second during the first 40 seconds.

For part b, the absence of detectable gas does not prove the absence of reaction.

The uncatalysed decomposition may have a sufficiently high activation-energy barrier that very little oxygen forms during the observation period.

The gas syringe also has a detection limit. A small amount of reaction could occur without registering a reliable volume.

The tempting mistake is treating “too slow to observe” as equivalent to “thermodynamically impossible”.

Question 2

Two experiments use \(10.0\ \mathrm{mL}\) of \(0.400\ \mathrm{mol\,L^{-1}}\) hydrogen peroxide.

In Experiment A, manganese dioxide is added and the gas syringe records \(47.0\ \mathrm{mL}\) of oxygen after two minutes.

In Experiment B, a different group reports \(55.0\ \mathrm{mL}\) of oxygen using the same initial amount of hydrogen peroxide and catalyst.

Assume the molar gas volume is \(24.8\ \mathrm{L\,mol^{-1}}\) at the stated temperature and pressure.

a. Calculate the theoretical maximum volume of oxygen from the initial hydrogen peroxide.

b. Assuming all oxygen is collected without loss and the syringe contains only oxygen at the stated conditions, calculate the percentage of hydrogen peroxide decomposed in Experiment A.

c. Explain why the result from Experiment B demands further investigation. Suggest two specific factors that should be checked, and explain why adding more catalyst cannot account for the excess volume.

Solution 2

The theoretical maximum oxygen volume is \(49.6\ \mathrm{mL}\). Experiment A corresponds to approximately \(94.8\%\) decomposition under the ideal collection assumptions, while Experiment B exceeds the amount permitted by the stated starting material.

Part a: Theoretical maximum

First calculate the initial moles of hydrogen peroxide.

Using \(n=cV\), where \(c\) is concentration and \(V\) is solution volume:

\[ \begin{aligned} n(\ce{H2O2}) &=0.400(0.0100)\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

From the balanced equation:

\[ \ce{2H2O2 -> 2H2O + O2} \]

The maximum oxygen amount is:

\[ \begin{aligned} n(\ce{O2}) &=\frac{0.00400}{2}\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

Therefore:

\[ \begin{aligned} V(\ce{O2}) &=nV_m\\ &=0.00200(24.8)\\ &=0.0496\ \mathrm{L} \end{aligned} \]

Converting to millilitres:

\[ V_{\max}=49.6\ \mathrm{mL} \]

Part b: Percentage decomposed

Under the stated ideal assumptions, the fraction of theoretical gas collected equals the fraction of peroxide decomposed.

\[ \begin{aligned} \%\text{ decomposed} &=\frac{47.0}{49.6}\times100\\ &=94.8\% \end{aligned} \]

The result indicates that most, but not all, of the initial hydrogen peroxide has decomposed.

Without the ideal collection assumptions, the gas volume alone would not establish an exact percentage conversion.

Part c: Investigating the apparent excess

Experiment B reports:

\[ 55.0\ \mathrm{mL}>49.6\ \mathrm{mL} \]

This exceeds the maximum volume of dry oxygen permitted by the stated starting amount under the specified conditions.

At least one assumption or measurement must be wrong.

Two useful checks are:

• Gas identity and measurement conditions: Is the collected gas entirely oxygen? Could water vapour, another gas, or a temperature or pressure difference be affecting the volume reading?
• Starting amount and apparatus calibration: Was the hydrogen peroxide concentration measured correctly, and was the gas syringe correctly zeroed and read?

A catalyst can change the rate at which oxygen is produced, but it cannot increase the number of oxygen atoms present in the original reactants.

Increasing catalyst mass does not remove the stoichiometric constraint.

The deeper lesson is that reaction kinetics cannot override conservation of matter.

Question 3

A hypothetical exothermic reaction is:

\[ \ce{P(g) <=> Q(g)} \]

Its forward enthalpy change is:

\[ \Delta H=-35.0\ \mathrm{kJ\,mol^{-1}} \]

An energy-profile model gives these forward activation energies:

Assume each pathway can be represented by a single energy maximum and that the same catalyst is available to both reaction directions.

a. Calculate the reverse activation energy for each pathway.

b. A student argues that because the catalyst lowers the forward activation energy so much, it must increase the equilibrium proportion of \(\ce{Q}\). Explain why this argument is incorrect.

c. Can the information in the question establish the sign of \(\Delta G\) under arbitrary reaction conditions? Explain.

Solution 3

The uncatalysed reverse activation energy is \(125\ \mathrm{kJ\,mol^{-1}}\), and the catalysed reverse activation energy is \(80.0\ \mathrm{kJ\,mol^{-1}}\). The catalyst lowers both barriers without changing the reaction’s equilibrium position.

Part a: Reverse activation energies

The reaction is exothermic, so the products lie below the reactants on the enthalpy profile.

For a single-peak energy model:

\[ E_{a,r}=E_{a,f}-\Delta H \]

Here, \(E_{a,r}\) and \(E_{a,f}\) are the reverse and forward activation energies.

For the uncatalysed pathway:

\[ \begin{aligned} E_{a,r} &=90.0-(-35.0)\\ &=125\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For the catalysed pathway:

\[ \begin{aligned} E_{a,r} &=45.0-(-35.0)\\ &=80.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The catalyst reduces the modelled activation energy by \(45.0\ \mathrm{kJ\,mol^{-1}}\) in each direction.

Part b: Why equilibrium does not shift

The catalyst changes the available reaction pathway, not the enthalpy or Gibbs free energy of the initial and final states.

At a fixed temperature, it increases the rates of both the forward and reverse reactions.

The equilibrium constant remains unchanged, so the equilibrium composition remains unchanged.

The student has confused a lower activation barrier with a larger thermodynamic driving force.

Part c: What the energy profile cannot establish

The enthalpy change is negative, so the reaction is exothermic.

However, spontaneity depends on Gibbs free energy, which includes the contribution from entropy and depends on the conditions and mixture composition.

A negative \(\Delta H\) alone is insufficient to establish the sign of \(\Delta G\) for every possible set of conditions.

The energy profile is useful for understanding reaction barriers. It is not, by itself, a complete thermodynamic spontaneity test.

Question 4

2017 syllabus Gibbs-energy connection

A hypothetical reversible reaction occurs in a rigid vessel:

\[ \ce{X(g) <=> Y(g)} \]

At \(450\ \mathrm{K}\), the following values apply:

\[ \Delta G^\circ=+5.2\ \mathrm{kJ\,mol^{-1}} \]

\[ K_c=0.25 \]

These are consistent rounded thermodynamic and equilibrium values.

The initial concentrations are:

\[ [\ce{X}]=0.400\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{Y}]=0.0400\ \mathrm{mol\,L^{-1}} \]

A suitable catalyst is present.

a. Determine the initial spontaneous net reaction direction. Explain why this is possible despite the positive standard Gibbs free energy change.

b. Calculate the equilibrium concentrations in the closed vessel, assuming no side reactions.

c. Once equilibrium has been reached, additional \(\ce{Y}\) is injected so that its concentration increases immediately by \(0.0800\ \mathrm{mol\,L^{-1}}\). The vessel volume and temperature remain constant. Determine the new equilibrium concentrations and explain why the final concentration of \(\ce{Y}\) can be greater than before injection even though the net reaction after injection proceeds in reverse.

Solution 4

The reaction initially proceeds towards \(\ce{Y}\), despite \(\Delta G^\circ>0\). After the injection, it proceeds in reverse, but the new equilibrium concentration of \(\ce{Y}\) is still greater than its previous equilibrium concentration.

Part a: Initial direction

The reaction quotient is:

\[ Q_c=\frac{[\ce{Y}]}{[\ce{X}]} \]

Initially:

\[ \begin{aligned} Q_c&=\frac{0.0400}{0.400}\\ &=0.100 \end{aligned} \]

Since:

\[ Q_c<K_c \]

the net reaction proceeds forwards towards \(\ce{Y}\).

The positive \(\Delta G^\circ\) describes the standard-state thermodynamic preference.

It does not mean every possible mixture must react backwards.

The actual mixture contains relatively little \(\ce{Y}\), so forming more \(\ce{Y}\) is thermodynamically favourable until equilibrium is reached.

Part b: Original equilibrium

Because this is a one-to-one conversion in a closed, fixed-volume vessel, the total concentration remains:

\[ 0.400+0.0400=0.440\ \mathrm{mol\,L^{-1}} \]

At equilibrium:

\[ \frac{[\ce{Y}]}{[\ce{X}]}=0.25 \]

Let the equilibrium concentration of \(\ce{X}\) be \(x\).

Then the concentration of \(\ce{Y}\) is \(0.25x\).

\[ \begin{aligned} x+0.25x&=0.440\\ 1.25x&=0.440\\ x&=0.352 \end{aligned} \]

Therefore:

\[ [\ce{X}]_{\mathrm{eq}}=0.352\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{Y}]_{\mathrm{eq}}=0.0880\ \mathrm{mol\,L^{-1}} \]

Part c: After injecting additional product

Immediately after adding \(\ce{Y}\):

\[ [\ce{Y}]=0.0880+0.0800 \]

\[ [\ce{Y}]=0.168\ \mathrm{mol\,L^{-1}} \]

The concentration of \(\ce{X}\) remains \(0.352\ \mathrm{mol\,L^{-1}}\) at that instant.

The new reaction quotient is:

\[ \begin{aligned} Q_c&=\frac{0.168}{0.352}\\ &=0.477 \end{aligned} \]

Since \(Q_c>K_c\), the system now reacts in reverse.

However, the total concentration has increased because additional \(\ce{Y}\) was injected:

\[ 0.352+0.168=0.520\ \mathrm{mol\,L^{-1}} \]

At the new equilibrium:

\[ \begin{aligned} 1.25[\ce{X}]&=0.520\\ [\ce{X}]&=0.416 \end{aligned} \]

Hence:

\[ [\ce{Y}]=0.25(0.416) \]

\[ [\ce{Y}]=0.104\ \mathrm{mol\,L^{-1}} \]

The new equilibrium concentrations are therefore:

• \([\ce{X}]=0.416\ \mathrm{mol\,L^{-1}}\)
• \([\ce{Y}]=0.104\ \mathrm{mol\,L^{-1}}\)

The net reverse reaction reduces \(\ce{Y}\) from \(0.168\) to \(0.104\ \mathrm{mol\,L^{-1}}\).

But the final \(0.104\ \mathrm{mol\,L^{-1}}\) remains greater than the original equilibrium value of \(0.0880\ \mathrm{mol\,L^{-1}}\).

This is the crucial distinction: the direction of reaction after a disturbance does not tell you whether the final amount of product must be below its value before the disturbance.

The catalyst simply allows the system to approach the appropriate equilibrium faster.

Question 5

A reversible gas-phase reaction is represented by:

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

At two temperatures, its equilibrium constants are:

Assume ideal-gas behaviour and a rigid, sealed vessel. The interconversion is one-to-one, and the combined concentration of \(\ce{A}\) and \(\ce{B}\) remains \(1.00\ \mathrm{mol\,L^{-1}}\) throughout.

Initially, at \(300\ \mathrm{K}\), the vessel contains only \(\ce{A}\).

An effective catalyst is introduced, and the system is allowed to reach equilibrium.

The vessel is then heated to \(360\ \mathrm{K}\), with the catalyst still present, and allowed to reach equilibrium again.

Next, the catalyst is completely isolated from the reacting mixture without changing its composition. The vessel is cooled rapidly to \(300\ \mathrm{K}\).

Without the catalyst, the reaction rate at \(300\ \mathrm{K}\) is so low that no measurable concentration change occurs during the following several days.

Two identical vessels are prepared using this entire sequence. At the end, a catalyst is reintroduced into one vessel but not the other.

a. Determine the equilibrium concentrations after the first stage at \(300\ \mathrm{K}\).

b. Determine the concentrations after equilibration at \(360\ \mathrm{K}\). Explain the net reaction direction during this stage, despite the higher temperature.

c. Immediately after the uncatalysed vessel has cooled to \(300\ \mathrm{K}\), is its mixture at equilibrium? Is the forward reaction thermodynamically favourable? Explain why the lack of visible change does not settle either question.

d. Predict the concentration of \(\ce{B}\) in each vessel after sufficient time for the catalysed vessel to equilibrate, but before the uncatalysed vessel has changed measurably.

e. A student claims that the catalyst must have increased the equilibrium yield of \(\ce{B}\), because the two vessels eventually contain different amounts of \(\ce{B}\). Identify the exact flaw in this argument.

Solution 5

The mixture reaches \(80\%\) \(\ce{B}\) at its first \(300\ \mathrm{K}\) equilibrium, falls to \(50\%\) \(\ce{B}\) at \(360\ \mathrm{K}\), and then remains kinetically trapped at approximately \(50\%\) after cooling without a catalyst. Reintroducing the catalyst at \(300\ \mathrm{K}\) allows that vessel to return to \(80\%\) \(\ce{B}\). The equilibrium constant has never been changed by the catalyst.

Part a: First equilibrium at 300 K

For the one-to-one reaction:

\[ K_c=\frac{[\ce{B}]}{[\ce{A}]}=4.0 \]

The total concentration is:

\[ [\ce{A}]+[\ce{B}]=1.00 \]

Let \([\ce{A}]=x\). Then \([\ce{B}]=4x\).

\[ \begin{aligned} x+4x&=1.00\\ 5x&=1.00\\ x&=0.200 \end{aligned} \]

Therefore:

\[ [\ce{A}]=0.200\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{B}]=0.800\ \mathrm{mol\,L^{-1}} \]

The system initially moves forwards because it starts with only reactant \(\ce{A}\).

Part b: Equilibrium at 360 K

At \(360\ \mathrm{K}\):

\[ K_c=1.0 \]

Therefore:

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

Because their combined concentration remains \(1.00\ \mathrm{mol\,L^{-1}}\):

\[ [\ce{A}]=[\ce{B}] =0.500\ \mathrm{mol\,L^{-1}} \]

Heating causes the equilibrium proportion of \(\ce{B}\) to decrease from \(0.800\) to \(0.500\).

Immediately after the temperature change, before substantial reaction occurs, the mixture still has:

\[ Q_c=\frac{0.800}{0.200}=4.0 \]

But the new equilibrium constant is only \(1.0\).

Since \(Q_c>K_c\), the net reaction proceeds backwards towards \(\ce{A}\).

The tempting prediction is that heating should produce more \(\ce{B}\) because reactions generally become faster.

That confuses reaction speed with the equilibrium position.

Heating can increase microscopic reaction rates while causing the net reaction to proceed towards reactants.

Part c: Cooling without the catalyst

Immediately after cooling, assume the concentrations remain:

\[ [\ce{A}]=[\ce{B}] =0.500\ \mathrm{mol\,L^{-1}} \]

At \(300\ \mathrm{K}\), the equilibrium constant returns to \(4.0\).

But the reaction quotient is still:

\[ Q_c=\frac{0.500}{0.500}=1.0 \]

Therefore:

\[ Q_c<K_c \]

The mixture is not at equilibrium at \(300\ \mathrm{K}\).

The forward reaction is thermodynamically favourable because the current mixture contains less \(\ce{B}\) than its equilibrium composition at that temperature.

Nevertheless, the reaction can remain almost unchanged for several days because its uncatalysed rate is extremely low.

This is the key result: the mixture has a thermodynamic tendency to react forwards, yet it shows almost no measurable progress.

Part d: Comparing the two final vessels

The vessel receiving the catalyst approaches the \(300\ \mathrm{K}\) equilibrium again:

\[ [\ce{B}]=0.800\ \mathrm{mol\,L^{-1}} \]

The vessel without the catalyst remains approximately:

\[ [\ce{B}]=0.500\ \mathrm{mol\,L^{-1}} \]

during the stated observation period.

Both vessels have the same temperature, the same equilibrium constant, and the same overall thermodynamic relationship between \(\ce{A}\) and \(\ce{B}\).

Their observed compositions differ because one has had time to reach equilibrium through an accessible pathway, while the other remains kinetically trapped.

Part e: Resolving the apparent contradiction

The student’s argument fails because it compares an equilibrium mixture with a non-equilibrium mixture.

The catalysed vessel reaches the equilibrium composition appropriate to \(300\ \mathrm{K}\).

The uncatalysed vessel has not reached equilibrium, even though its composition appears constant over several days.

If its reaction continues over a sufficiently long period, with no side reactions or other changes, it will also approach the same \(300\ \mathrm{K}\) equilibrium composition.

The catalyst has not increased the equilibrium yield.

It has increased the rate at which the system approaches that yield.

The strongest part of this problem is recognising that an apparently stationary mixture can be away from equilibrium. A reaction can be thermodynamically driven forwards without changing enough to be detected on the timescale of the experiment.

07Where this leads next

When a reaction appears to stop, the next decision is whether it has reached equilibrium or whether its progress is simply too slow to observe.

Activation energy and catalysts help explain the rate. Gibbs free energy helps explain thermodynamic feasibility under specified conditions. The reaction quotient and equilibrium constant help determine which direction a reversible system should move from its current composition.

That final distinction is developed in Keq Calculations and Reaction Direction | HSC Chemistry.

The useful habit is to separate three questions whenever you analyse a reaction: is the change favourable, how quickly can it occur, and has the system reached equilibrium? The same experiment can give very different answers to each.