How to Calculate the Temperature When a Reaction Becomes Spontaneous

A sealed glass bulb contains a pale brown gas. Heat it, and the gas becomes much darker. Cool it, and the colour fades. The chemical reaction is reversible, so what makes one direction more favourable at a particular temperature?

To find the temperature at which a reaction becomes spontaneous under standard-state conditions, use the Gibbs free energy equation:

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

Set \(\Delta G^\circ=0\) and rearrange:

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

Here, \(T_*\) is the temperature where the standard Gibbs energy changes sign, provided the enthalpy and entropy changes can be treated as constant. Convert your units first, calculate \(T_*\) in kelvin, and then determine whether the reaction is spontaneous above or below that temperature.

The important catch: reaching this temperature doesn’t suddenly switch the reaction on. It changes which direction is thermodynamically favoured under standard-state conditions. How quickly the reaction happens is a separate question.

NSW syllabus note (November 2026): Gibbs free energy and reaction spontaneity are explicitly covered in Year 11 Module 4, Drivers of Reactions, of the Chemistry Stage 6 Syllabus (2017). The replacement Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. Gibbs free-energy calculations are not listed as required content in that new edition, so treat this calculation method as extension material if you’re following the 2025 syllabus.

01Why heating can make a reaction more favourable

Consider the reversible reaction responsible for the changing colour:

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

Dinitrogen tetroxide, \(\ce{N2O4}\), is colourless. Nitrogen dioxide, \(\ce{NO2}\), is brown.

When one \(\ce{N2O4}\) molecule breaks apart, it forms two \(\ce{NO2}\) molecules.

Before reading further, predict what happens to the freedom of movement of the particles. Does splitting one gas molecule into two tend to increase or decrease entropy?

It tends to increase entropy. Two independently moving gas molecules have more possible arrangements and distributions of energy than one molecule under comparable conditions.

The forward reaction also absorbs heat, making it endothermic.

That creates two competing effects:

• Enthalpy: Breaking \(\ce{N2O4}\) into \(\ce{NO2}\) requires a net input of energy. This opposes the forward reaction.
• Entropy: Forming two separate gas molecules increases the number of accessible particle arrangements. This favours the forward reaction.

At lower temperatures, the enthalpy cost dominates the standard Gibbs energy calculation. At higher temperatures, the entropy contribution becomes more important.

The temperature we’re looking for is where those contributions balance.

02What an experiment can actually tell us

The \(\ce{NO2}/\ce{N2O4}\) equilibrium is a well-established demonstration of temperature-dependent equilibrium.

Imagine trying to use it to locate the exact temperature where the forward reaction becomes spontaneous.

The first design: watch the colour

Place a sealed bulb containing the equilibrium gas mixture in a temperature-controlled bath.

As the temperature increases, the mixture becomes darker. Cooling it favours the formation of colourless \(\ce{N2O4}\), so the colour becomes paler.

These observations support the conclusion that dissociation into \(\ce{NO2}\) is endothermic.

But suppose you predicted that the gas would remain colourless until one particular temperature, then suddenly turn brown.

That prediction fails. The mixture can contain both gases below and above the calculated crossover temperature. The equilibrium composition changes continuously rather than switching between pure reactants and pure products.

A colour change also cannot tell us whether the system has reached equilibrium, or exactly how much of each gas is present, without further measurements.

The revised design: measure the equilibrium composition

A more informative experimental arrangement would contain:

1. A sealed gas cell inside a temperature-controlled chamber.
2. A temperature probe to measure the gas temperature.
3. A light source directed through one transparent window of the cell.
4. A calibrated optical detector on the opposite side to estimate the concentration of brown \(\ce{NO2}\).
5. A pressure sensor connected to the gas volume to measure total pressure.

The calibrated optical signal, temperature, and total pressure can be used to estimate the partial pressures of both gases, assuming ideal-gas behaviour.

At each temperature, the readings must be allowed to stabilise before estimating the equilibrium composition.

The difference between the two designs is important.

Watching the colour shows that temperature changes the equilibrium composition. Measuring the equilibrium composition allows us to calculate an equilibrium constant and locate its relationship to the Gibbs energy boundary.

Neither experiment directly measures the rate at which the reaction becomes thermodynamically favourable.

Nitrogen dioxide is toxic, so this is an explanation of an established demonstration and an illustrative measurement design, not a home experiment.

03The Gibbs energy method, explained

The standard Gibbs energy change combines the enthalpy and entropy changes:

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

Each quantity has a particular meaning.

The degree symbol indicates standard-state quantities. The values refer to the balanced reaction as written, and the standard reference pressure for gases is conventionally \(1\ \mathrm{bar}\).

For a reaction with positive \(\Delta H^\circ\) and positive \(\Delta S^\circ\), increasing temperature makes the negative term \(-T\Delta S^\circ\) larger in magnitude. Eventually, it can outweigh the positive enthalpy term.

What does the sign of Gibbs energy mean?

At \(\Delta G^\circ=0\), the standard-state driving force changes sign. This does not mean every mixture at that temperature is at equilibrium.

We’ll return to that distinction when interpreting the gas experiment.

Finding the temperature where the sign changes

At the crossover temperature:

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

Therefore:

\[ \begin{aligned} 0&=\Delta H^\circ-T_*\Delta S^\circ\\ T_*\Delta S^\circ&=\Delta H^\circ\\ T_*&=\frac{\Delta H^\circ}{\Delta S^\circ} \end{aligned} \]

This calculation only gives a physically meaningful positive temperature if the signs of \(\Delta H^\circ\) and \(\Delta S^\circ\) allow a positive result.

And even then, we must check which side of the temperature is spontaneous.

05Does a reaction always become spontaneous when heated?

No. The answer depends on both signs.

These conclusions assume the enthalpy and entropy changes remain constant and the phases and standard states remain appropriate.

The two middle cases do not have a positive temperature where the forward Gibbs energy changes sign. Blindly calculating \(\Delta H^\circ/\Delta S^\circ\) can produce a negative temperature, which is not a meaningful crossover here.

If \(\Delta S^\circ=0\), there is no temperature-dependent entropy term in this approximation, so don’t divide by zero.

Solution

Step 1: Interpret the energy and entropy changes.

The negative enthalpy means association releases heat.

The negative entropy means two separate gas particles become one gas molecule, reducing their freedom of movement.

Low temperatures favour the enthalpy contribution. Higher temperatures increase the penalty from decreasing entropy.

Step 2: Convert the entropy units.

\[ \Delta S^\circ=-0.210\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Step 3: Calculate the temperature.

\[ \begin{aligned} T_*&=\frac{-84.0}{-0.210}\\ &=400\ \mathrm{K} \end{aligned} \]

Both signs are negative, so the result is a positive temperature.

Step 4: Test the two temperatures.

At \(300\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ_{300} &=-84.0-300(-0.210)\\ &=-21.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Association is favoured.

At \(450\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ_{450} &=-84.0-450(-0.210)\\ &=+10.5\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Association is not favoured under standard-state conditions.

Step 5: Explain the apparent contradiction.

Heating increases particle kinetic energy and generally increases collision frequency and reaction rates.

But this is a kinetic effect. The Gibbs energy calculation describes thermodynamic favourability, not speed.

Above \(400\ \mathrm{K}\), the entropy penalty outweighs the energy released when the particles combine. The reverse reaction is favoured under standard-state conditions.

That means a reaction can become faster while its forward direction becomes less thermodynamically favourable.

06How does this connect to actual equilibrium measurements?

Return to the nitrogen dioxide experiment.

We calculated that \(\Delta G^\circ=0\) at approximately \(326\ \mathrm{K}\).

What would the improved experiment measure at this temperature?

For an ideal-gas equilibrium:

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

we can write the dimensionless pressure-based equilibrium constant:

\[ K_p= \frac{(p_{\ce{NO2}}/p^\circ)^2} {p_{\ce{N2O4}}/p^\circ} \]

Here, \(p_{\ce{NO2}}\) and \(p_{\ce{N2O4}}\) are the equilibrium partial pressures of the two gases, and \(p^\circ=1\ \mathrm{bar}\) is the standard reference pressure.

Thermodynamics predicts that \(\Delta G^\circ=0\) corresponds to \(K_p=1\).

For example, consider these illustrative equilibrium pressures at the approximate crossover temperature:

• \(p_{\ce{NO2}}=0.20\ \mathrm{bar}\)
• \(p_{\ce{N2O4}}=0.040\ \mathrm{bar}\)

Then:

\[ \begin{aligned} K_p&=\frac{(0.20/1.00)^2}{0.040/1.00}\\ &=1.00 \end{aligned} \]

The mixture contains both substances.

In fact, the partial pressure of \(\ce{NO2}\) is five times that of \(\ce{N2O4}\). Therefore, \(K_p=1\) certainly doesn’t mean equal amounts of reactants and products.

What if the experiment is repeated at other temperatures?

The following values are illustrative model predictions consistent with the rounded thermodynamic data, not recorded measurements.

The increasing equilibrium constant is consistent with the gas becoming browner when heated.

However, a mixture’s immediate direction of change also depends on its current composition.

Extension: the difference between \(Q\) and \(K\)

For a mixture that has not necessarily reached equilibrium, calculate the reaction quotient \(Q_p\) using the same pressure expression:

\[ Q_p= \frac{(p_{\ce{NO2}}/p^\circ)^2} {p_{\ce{N2O4}}/p^\circ} \]

The difference is that \(Q_p\) uses the current partial pressures, while \(K_p\) describes the equilibrium condition at that temperature.

• If \(Q_p<K_p\), the net reaction is favoured forwards.
• If \(Q_p>K_p\), the net reaction is favoured backwards.
• If \(Q_p=K_p\), the mixture is at equilibrium.

This is why a positive standard Gibbs energy does not mean a reaction is impossible. A particular mixture can still move forwards if its composition creates a favourable driving force.

For a fuller explanation of how heating changes the equilibrium constant, see How Temperature Changes Keq in Endothermic and Exothermic Reactions.

The extra \(Q_p\) calculations and their Gibbs energy interpretation are useful extensions beyond the basic 2017 syllabus temperature-threshold method.

07Limits of the temperature calculation

The equation \(T_*=\Delta H^\circ/\Delta S^\circ\) is useful, but the calculation depends on several assumptions.

1. Enthalpy and entropy are treated as constant.

In reality, both can change with temperature because substances have different heat capacities. Over a wide temperature range, the simple crossover calculation may only be an estimate.

2. The substances stay in the assumed physical states.

If a reactant melts or a product condenses, the thermodynamic data for the original reaction may no longer apply.

3. Standard-state favourability is not the same as the direction in every mixture.

A real reaction mixture may have concentrations or partial pressures very different from standard conditions.

4. Spontaneity does not measure reaction speed.

A negative Gibbs energy does not guarantee a visible reaction within a practical time. Activation energy can still create a large kinetic barrier.

A catalyst can lower that barrier and speed up the approach to equilibrium. It does not change the reaction’s Gibbs energy or the crossover temperature.

These limitations become especially important when interpreting calorimetry, phase changes, and gas equilibrium experiments.

08Questions and solutions

The questions below use original, illustrative thermodynamic data unless stated otherwise. Assume standard-state conditions and constant enthalpy and entropy changes within the stated temperature ranges, except where the question introduces a change of physical state.

Question 1

A hypothetical decomposition reaction is:

\[ \ce{R(s) <=> S(s) + T(g)} \]

Its standard thermodynamic data are:

\[ \Delta H^\circ=+48.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+120\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Calculate:

a) The crossover temperature in kelvin and degrees Celsius.

b) Whether the forward reaction is spontaneous under standard-state conditions at \(300\ \mathrm{K}\) and \(500\ \mathrm{K}\).

c) Why the reaction becomes more favourable when heated.

Solution 1

The crossover temperature is \(400\ \mathrm{K}\), or approximately \(127^\circ\mathrm{C}\). The forward reaction is not thermodynamically favoured at \(300\ \mathrm{K}\), but is favoured at \(500\ \mathrm{K}\).

First, convert the entropy change:

\[ \Delta S^\circ=0.120\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Calculate the crossover temperature:

\[ \begin{aligned} T_*&=\frac{48.0}{0.120}\\ &=400\ \mathrm{K} \end{aligned} \]

Converting to Celsius:

\[ 400-273.15=126.85^\circ\mathrm{C} \]

At \(300\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ &=48.0-300(0.120)\\ &=+12.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The positive result means the forward reaction is not favoured.

At \(500\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ &=48.0-500(0.120)\\ &=-12.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative result means the forward reaction is favoured.

The decomposition increases entropy because it produces a gas. At higher temperatures, the entropy contribution outweighs the positive enthalpy change.

The tempting mistake is to say that the reaction becomes spontaneous simply because it is endothermic. Endothermicity alone is not enough. The positive entropy change is what allows the forward reaction to become favourable at high temperature.

Question 2

Two hypothetical reaction steps can be coupled:

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

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

Their standard thermodynamic data are:

Assume one mole of each step occurs together, so that intermediate \(B\) is consumed as it forms.

a) At \(500\ \mathrm{K}\), is Step 1 favoured under standard-state conditions? Is the overall transformation favoured?

b) Determine the temperature at which the overall transformation changes thermodynamic favourability.

c) At \(800\ \mathrm{K}\), determine whether Step 1 and the overall transformation are favourable. Explain the apparent contradiction.

Solution 2

The overall transformation is favoured at \(500\ \mathrm{K}\), even though Step 1 is not. The overall crossover occurs at approximately \(767\ \mathrm{K}\). At \(800\ \mathrm{K}\), Step 1 is favoured but the overall transformation is not.

The key is that the Gibbs energy of coupled steps must be considered together.

Adding the chemical equations cancels intermediate \(B\):

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

For the overall reaction, enthalpy and entropy changes add.

\[ \begin{aligned} \Delta H^\circ_{\mathrm{net}} &=54.0-100\\ &=-46.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

\[ \begin{aligned} \Delta S^\circ_{\mathrm{net}} &=90-150\\ &=-60\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

Therefore:

\[ \Delta S^\circ_{\mathrm{net}} =-0.060\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

At \(500\ \mathrm{K}\), Step 1 has:

\[ \begin{aligned} \Delta G^\circ_1 &=54.0-500(0.090)\\ &=+9.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 1 alone is not favoured.

Step 2 has:

\[ \begin{aligned} \Delta G^\circ_2 &=-100-500(-0.150)\\ &=-25.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The overall Gibbs energy is:

\[ \begin{aligned} \Delta G^\circ_{\mathrm{net}} &=+9.0-25.0\\ &=-16.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The overall transformation is favoured. The strongly favourable second step more than compensates for the first step’s unfavourable standard Gibbs energy, provided the processes are genuinely coupled.

For the overall crossover:

\[ \begin{aligned} T_*&=\frac{-46.0}{-0.060}\\ &=766.7\ \mathrm{K} \end{aligned} \]

Because the net reaction has negative enthalpy and negative entropy changes, it is favoured below this temperature.

At \(800\ \mathrm{K}\), Step 1 has:

\[ \begin{aligned} \Delta G^\circ_1 &=54.0-800(0.090)\\ &=-18.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Step 1 is now favoured.

But Step 2 has:

\[ \begin{aligned} \Delta G^\circ_2 &=-100-800(-0.150)\\ &=+20.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta G^\circ_{\mathrm{net}} &=-18.0+20.0\\ &=+2.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The overall transformation is not favoured under standard-state conditions.

The trap is treating the spontaneity of an intermediate step as the spontaneity of the complete process. Heating helps Step 1, but the entropy penalty of Step 2 eventually dominates the combined reaction.

A favourable overall Gibbs energy establishes thermodynamic feasibility, not a guarantee that the coupled reaction pathway will proceed quickly.

Question 3

A hypothetical reaction is investigated in an insulated-cup calorimeter:

\[ \ce{P(aq) -> Q(aq)} \]

A \(0.0500\ \mathrm{mol}\) quantity of \(P\) reacts completely.

The following illustrative measurements are recorded:

A separate reference-data calculation gives:

\[ \Delta S^\circ=-140\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Assume the cup’s heat capacity is negligible, the solution has the given specific heat capacity, and heat loss is initially negligible.

a) Estimate the reaction’s molar enthalpy change.

b) Calculate the approximate temperature above which the forward reaction is no longer thermodynamically favoured under standard-state conditions.

c) Determine its favourability at \(350\ \mathrm{K}\).

d) Suppose the calorimeter actually loses some heat to the surroundings. Explain whether the calculated crossover temperature would be too high or too low.

Solution 3

The reaction’s estimated molar enthalpy change is \(-50.2\ \mathrm{kJ\,mol^{-1}}\). Its crossover temperature is approximately \(358\ \mathrm{K}\), and it is thermodynamically favoured at \(350\ \mathrm{K}\). Heat loss would make the calculated crossover temperature too low.

First, determine the temperature rise:

\[ \Delta T=28.0-22.0=6.0\ \mathrm{K} \]

The solution gains heat:

\[ \begin{aligned} q_{\mathrm{solution}} &=mc\Delta T\\ &=100.0(4.18)(6.0)\\ &=2508\ \mathrm{J} \end{aligned} \]

Here, \(m\) is solution mass, \(c\) is specific heat capacity, and \(q\) is heat transferred.

Since the solution warms, the reacting system releases heat:

\[ q_{\mathrm{reaction}}=-2508\ \mathrm{J} \]

The molar enthalpy change is:

\[ \begin{aligned} \Delta H^\circ &\approx\frac{-2.508}{0.0500}\\ &=-50.16\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

This approximation treats the measured reaction enthalpy as representative of the standard value.

Now convert entropy:

\[ \Delta S^\circ=-0.140\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

The crossover temperature is:

\[ \begin{aligned} T_*&=\frac{-50.16}{-0.140}\\ &=358.3\ \mathrm{K} \end{aligned} \]

Since both enthalpy and entropy changes are negative, the reaction is favourable below the crossover.

At \(350\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ &=-50.16-350(-0.140)\\ &=-1.16\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The negative value confirms that the forward reaction is favoured.

Finally, consider heat loss.

If some released heat escapes into the environment, the observed temperature increase is smaller than the true insulated temperature increase.

That makes the calculated heat release too small in magnitude. The estimated reaction enthalpy becomes less negative than its true value.

Because the magnitude of \(\Delta H^\circ\) is underestimated while \(\Delta S^\circ\) is held fixed, the calculated crossover temperature is too low.

The tempting conclusion is that the thermometer reading is sufficient to determine the spontaneous temperature. It isn’t. The entropy data must also be known, and systematic heat loss affects the result.

Question 4

A hypothetical reaction is written as:

\[ \ce{X(s) -> Y(g)} \]

At temperatures where \(X\) is solid, its standard thermodynamic data are:

\[ \Delta H^\circ=+60.0\ \mathrm{kJ\,mol^{-1}} \]

\[ \Delta S^\circ=+150\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

A student calculates \(T_*=400\ \mathrm{K}\) and concludes that the forward reaction becomes favourable just above \(400\ \mathrm{K}\).

However, pure \(X\) melts reversibly at \(350\ \mathrm{K}\), with a molar enthalpy of fusion of \(+7.00\ \mathrm{kJ\,mol^{-1}}\).

Assume:

• The liquid is the stable form of \(X\) above \(350\ \mathrm{K}\).
• \(Y\) remains gaseous over the temperatures considered.
• Enthalpy and entropy changes can be treated as constant within each phase range.
• The fusion transition occurs at equilibrium at \(350\ \mathrm{K}\).

a) Determine the entropy change of fusion.

b) Find the corrected enthalpy and entropy changes for \(\ce{X(l) -> Y(g)}\).

c) Calculate the corrected crossover temperature and classify the reaction at \(400\ \mathrm{K}\) and \(420\ \mathrm{K}\).

d) Show that the Gibbs energy change for the reaction is continuous at the melting temperature.

Solution 4

The corrected crossover temperature is approximately \(408\ \mathrm{K}\), not \(400\ \mathrm{K}\). The forward reaction is still not favoured at \(400\ \mathrm{K}\), but becomes favourable by \(420\ \mathrm{K}\).

The original calculation used thermodynamic data for a reactant that is no longer in its stable physical state.

At the reversible melting temperature, the Gibbs energy change of fusion is zero:

\[ \Delta G^\circ_{\mathrm{fus}}=0 \]

Therefore:

\[ \begin{aligned} \Delta S^\circ_{\mathrm{fus}} &=\frac{\Delta H^\circ_{\mathrm{fus}}}{T_m}\\ &=\frac{7.00}{350}\\ &=0.0200\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \end{aligned} \]

Here, \(T_m\) is the melting temperature.

Converting entropy to joules:

\[ \Delta S^\circ_{\mathrm{fus}} =20.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

The thermodynamic data for the solid reaction include starting from solid \(X\).

Starting from liquid \(X\) instead removes the energy and entropy changes associated with melting.

Thus:

\[ \begin{aligned} \Delta H^\circ_{\mathrm{liquid}} &=60.0-7.00\\ &=53.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Similarly:

\[ \begin{aligned} \Delta S^\circ_{\mathrm{liquid}} &=150-20.0\\ &=130\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

The corrected crossover is:

\[ \begin{aligned} T_*&=\frac{53.0}{0.130}\\ &=407.7\ \mathrm{K} \end{aligned} \]

At \(400\ \mathrm{K}\), using the correct liquid reactant:

\[ \begin{aligned} \Delta G^\circ &=53.0-400(0.130)\\ &=+1.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The forward reaction is not favoured.

At \(420\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ &=53.0-420(0.130)\\ &=-1.6\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The forward reaction is now favoured.

To check continuity, calculate the reaction Gibbs energy at the melting temperature.

Using solid \(X\):

\[ \begin{aligned} \Delta G^\circ_{\mathrm{solid}} &=60.0-350(0.150)\\ &=+7.50\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Using liquid \(X\):

\[ \begin{aligned} \Delta G^\circ_{\mathrm{liquid}} &=53.0-350(0.130)\\ &=+7.50\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Both descriptions agree at the phase transition.

The change of physical state creates a change in the slope of the Gibbs energy relationship, not a sudden jump in Gibbs energy at equilibrium melting.

This exposes a limitation of the usual threshold formula: a mathematically correct calculation can give the wrong physical conclusion when its assumed reactant state is no longer valid.

Question 5

Consider the equilibrium:

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

At \(326\ \mathrm{K}\), an ideal-gas mixture is initially at equilibrium in a sealed, temperature-controlled cylinder fitted with a movable piston.

Its partial pressures are:

\[ p_{\ce{NO2}}=0.20\ \mathrm{bar} \]

\[ p_{\ce{N2O4}}=0.040\ \mathrm{bar} \]

At this temperature, take \(K_p=1.00\).

The cylinder initially has volume \(V_0\).

An operator then performs the following sequence:

1. The gas is rapidly compressed to half its original volume while simultaneously being heated to \(360\ \mathrm{K}\). The changes are fast enough that no significant reaction occurs during the adjustment.
2. The gas is held at this new temperature and volume until equilibrium is reached. At \(360\ \mathrm{K}\), take \(K_p=8.00\).
3. The gas is then rapidly cooled back to \(326\ \mathrm{K}\) while being expanded back to \(V_0\), again without significant reaction during the adjustment.
4. The mixture is finally allowed to reach equilibrium.

Assume ideal-gas behaviour, no leakage, no other reactions, unchanged total amounts of nitrogen and oxygen, and the stated equilibrium constants.

a) Immediately after the first rapid compression and heating, which direction is thermodynamically favoured?

b) Immediately after the rapid return to \(326\ \mathrm{K}\) and \(V_0\), which direction is favoured?

c) What are the final equilibrium partial pressures?

d) Explain why considering only Le Chatelier’s principle for compression would give an incomplete prediction, and why the final result does not depend on the path taken through the temperature and volume changes.

Solution 5

The first adjustment initially favours the forward reaction, despite compression. Returning to the original temperature and volume initially favours the reverse reaction. After the final equilibration, the partial pressures return to \(0.20\ \mathrm{bar}\) for \(\ce{NO2}\) and \(0.040\ \mathrm{bar}\) for \(\ce{N2O4}\).

The crucial idea is that temperature changes the equilibrium constant, while rapid changes in temperature and volume also change the current reaction quotient.

Initially:

\[ \begin{aligned} Q_p&=\frac{(0.20)^2}{0.040}\\ &=1.00=K_p \end{aligned} \]

The standard pressure of \(1\ \mathrm{bar}\) is understood in this numerical substitution, making \(Q_p\) dimensionless.

The gas is initially at equilibrium.

Part a: Compression and heating

For a fixed number of ideal-gas molecules:

\[ p\propto\frac{T}{V} \]

Here, \(p\) is partial pressure, \(T\) is absolute temperature, and \(V\) is volume.

During the rapid adjustment, the number of molecules of each species is initially unchanged.

Each partial pressure is therefore multiplied by the factor:

\[ \begin{aligned} r&=\frac{360}{326}\times2\\ &=2.21 \end{aligned} \]

For this reaction, the pressure quotient contains two powers of \(\ce{NO2}\) pressure but only one power of \(\ce{N2O4}\) pressure.

Multiplying every partial pressure by \(r\) therefore multiplies \(Q_p\) by \(r\).

Immediately after the adjustment:

\[ Q_p=2.21(1.00)=2.21 \]

But at \(360\ \mathrm{K}\):

\[ K_p=8.00 \]

Since:

\[ Q_p<K_p \]

the forward reaction is favoured.

This is the first surprising result.

Compression on its own would favour the side with fewer gas molecules, which is \(\ce{N2O4}\).

However, the simultaneous temperature increase raises \(K_p\) enough that the actual compressed and heated mixture still favours producing more \(\ce{NO2}\).

The temperature effect wins for these particular conditions.

Part b: Returning to the original temperature and volume

The hot, compressed mixture is allowed to reach equilibrium before the return adjustment.

At that point:

\[ Q_p=K_p=8.00 \]

Now the operator rapidly cools and expands the cylinder.

Because the return adjustment is fast, the amounts of each species are initially unchanged.

Each partial pressure is divided by the same factor \(r=2.21\).

The reaction quotient is also divided by this factor:

\[ \begin{aligned} Q_{p,\mathrm{return}} &=\frac{8.00}{2.21}\\ &=3.62 \end{aligned} \]

At the restored temperature of \(326\ \mathrm{K}\):

\[ K_p=1.00 \]

Therefore:

\[ Q_p>K_p \]

The reverse reaction is now favoured.

Two \(\ce{NO2}\) molecules tend to combine to form \(\ce{N2O4}\).

The tempting shortcut is to say that expanding a gas always favours the side with more gas molecules.

But expansion is only one part of the state change. The mixture has also been cooled, and its composition was altered during the previous equilibration.

The immediate reaction direction must be determined from the actual reaction quotient and the equilibrium constant at the restored temperature.

Part c: The final equilibrium state

Once the original temperature and volume are restored, the system has the same total amounts of nitrogen and oxygen as it had initially.

The chemical reaction has not created or removed any atoms, and no gas has escaped.

At the original temperature and volume, the equilibrium constant and the total chemical inventory are also unchanged.

Therefore, the final equilibrium composition must be the original equilibrium composition:

\[ p_{\ce{NO2}}=0.20\ \mathrm{bar} \]

\[ p_{\ce{N2O4}}=0.040\ \mathrm{bar} \]

The mixture does not necessarily return to equilibrium instantly when the piston and temperature return to their original settings. It first undergoes the reverse reaction until equilibrium is restored.

Part d: Why the path matters temporarily but not finally

During the sequence, temperature, pressure, equilibrium composition, and reaction quotient all change.

The forward direction can be favoured during a compression if the temperature-dependent increase in \(K_p\) is sufficiently strong.

The reverse direction can then be favoured during expansion if the mixture’s current composition creates \(Q_p>K_p\).

These temporary changes depend on the path taken.

But the final equilibrium state depends on the final temperature, volume, and conserved amount of material, assuming a unique equilibrium and the absence of other reactions or phase changes.

The deeper lesson is that a reaction does not have one universal temperature at which it suddenly starts proceeding forwards.

The Gibbs threshold describes standard-state thermodynamic favourability. The actual direction of change in a particular reacting mixture depends on both temperature and composition.

09What to decide before using the crossover formula

When a question asks for the temperature at which a reaction becomes spontaneous, first identify exactly what it means.

If it supplies \(\Delta H^\circ\) and \(\Delta S^\circ\) and asks about standard-state spontaneity, calculate \(T_*=\Delta H^\circ/\Delta S^\circ\), after matching units and checking signs.

If the question involves an actual equilibrium mixture, pressure changes, or concentration changes, the standard-state crossover alone may not decide the reaction direction.

And if the temperature range crosses a phase transition, check whether the thermodynamic data still describe the correct reaction.

That distinction leads directly into the next useful problem: predicting the equilibrium composition at a chosen temperature, rather than only deciding whether the standard forward reaction is thermodynamically favoured.