Why an Endothermic Reaction Can Be Spontaneous at High Temperature

A piece of limestone sits in a hot kiln, absorbs heat, and breaks down into calcium oxide and carbon dioxide. The reaction is endothermic: it needs energy from its surroundings. So why can it happen spontaneously at high temperature?

Before going further, make a prediction. Does heating eventually turn the reaction exothermic, or does something else make it favourable?

It stays endothermic. At sufficiently high temperature, an increase in entropy can outweigh the energy absorbed by the reaction. The Gibbs free energy change then becomes negative, making the forward reaction thermodynamically favourable under the specified conditions.

That’s the explanation, but there’s a catch. A reaction can be favourable without being fast, and a hot reaction vessel doesn’t guarantee complete conversion.

01The experiment: why does heated limestone release a gas?

Limestone is mainly calcium carbonate, \(\ce{CaCO3}\). When it decomposes, it forms calcium oxide (quicklime) and carbon dioxide:

\[ \ce{CaCO3(s) <=> CaO(s) + CO2(g)} \]

Breaking down calcium carbonate absorbs heat, so the forward reaction has a positive enthalpy change, \(\Delta H>0\).

If releasing heat were the only thing that made reactions favourable, you would predict that this decomposition could never be spontaneous.

Yet limestone is converted into quicklime in industrial kilns. The challenge is explaining how an energy-absorbing reaction can become favourable.

What would the first experiment measure?

Imagine a laboratory investigation using a thermogravimetric analyser. This instrument measures a sample’s mass as its temperature changes.

The apparatus works in this order:

1. A small calcium carbonate sample sits in a heat-resistant pan suspended from a sensitive balance.
2. A furnace surrounds the sample, while a temperature sensor measures the temperature near it.
3. A controlled gas stream passes through the heated chamber.
4. Gas leaving the chamber passes through an outlet to a detector that can identify carbon dioxide.
5. The balance records any change in the mass of the solid sample.

What would you expect if calcium carbonate decomposes?

The sample should lose mass as gaseous \(\ce{CO2}\) leaves. If the reaction proceeds to completion, one mole of calcium carbonate produces one mole of carbon dioxide.

The known behaviour of calcium carbonate supports that prediction: when heated under suitable conditions, it decomposes to calcium oxide and carbon dioxide.

But here’s the weakness in the first experimental design.

Mass loss alone does not prove that the reaction became spontaneous under standard-state conditions.

Other substances in an impure sample might evaporate. More importantly, the flowing gas continuously removes carbon dioxide, changing the chemical conditions around the solid.

The temperature is not the only variable.

How would you improve the experiment?

A stronger investigation would use a pure, suitably prepared sample, identify the gas leaving the furnace, and compare decomposition under controlled carbon dioxide pressures.

You could also use a separate closed, temperature-controlled vessel containing both calcium carbonate and calcium oxide. A pressure sensor would measure the carbon dioxide building up inside.

The two arrangements test different things:

• In the flowing-gas apparatus, carbon dioxide is carried away, helping maintain a low concentration of gaseous product near the sample.
• In the closed vessel, carbon dioxide accumulates. The increasing gas pressure can eventually oppose further decomposition.

The crucial comparison is what happens at the same temperature when the carbon dioxide pressure differs.

This is a model investigation, not a report of a particular experiment performed by tutorgum. Thermogravimetric and gas-pressure methods have been used in published investigations of calcium carbonate decomposition.

02The missing idea: entropy can favour the products

Think about the products of limestone decomposition.

On the reactant side, calcium carbonate is a solid. Its particles occupy positions within a crystal structure.

On the product side, calcium oxide remains solid, but carbon dioxide is a gas. The carbon dioxide molecules can move throughout the space available to them.

Imagine students initially confined to allocated seats in an assembly hall. Once they are free to spread out across the school oval, there are far more possible arrangements.

The seated students represent particles constrained within a solid. The students moving around the oval represent gas molecules with many accessible positions.

The analogy has a limit: real entropy also depends on the ways particles can distribute energy, not just where they can move.

Entropy, represented by \(S\), measures how many microscopic arrangements of matter and energy are accessible to a system.

For calcium carbonate decomposition, producing gaseous carbon dioxide creates a substantial increase in these possibilities.

Therefore:

\[ \Delta S>0 \]

Here, \(\Delta S\) is the entropy change of the reaction, calculated as the entropy of the products minus that of the reactants.

We now have two competing effects.

The reaction has an energy cost, but it also produces an entropy gain.

Higher temperature makes that entropy gain more important when deciding which direction is thermodynamically favourable.

03How Gibbs free energy combines both effects

The two influences are brought together using Gibbs free energy.

For a reaction under standard-state conditions:

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

where:

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

The superscript \(\circ\) means that the thermodynamic quantities refer to standard states at the specified temperature. For the gaseous substances in this article, the standard pressure is 1 bar.

The equation gives three possibilities.

For an endothermic reaction with positive entropy change, the equation becomes:

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

At a low temperature, the positive enthalpy contribution may be larger than the entropy contribution.

At a high temperature, \(T\Delta S^\circ\) becomes larger. Because we subtract it, \(\Delta G^\circ\) decreases and can become negative.

Notice what hasn’t happened: \(\Delta H^\circ\) hasn’t suddenly become negative.

Heating makes the positive entropy change more influential. It doesn’t need to reverse the reaction’s enthalpy sign.

Syllabus scope: NSW Chemistry (2017 edition)

The Gibbs free energy equation, standard-state calculations, and predictions about temperature and spontaneity are part of Year 11 Module 4, Drivers of Reactions, in the NSW Chemistry Stage 6 Syllabus (2017).

That syllabus applies to Chemistry courses taught in 2026. The Chemistry 11-12 Syllabus (2025) begins implementation for Year 11 in 2028, with its first HSC examination in 2029. This article follows the 2017 syllabus content; it does not assume that the same material is assessed in the 2025 edition.

05The experimental catch: carbon dioxide pressure changes the answer

Return to our two experimental arrangements.

Suppose calcium carbonate and calcium oxide are both present in a sealed vessel. Decomposition releases carbon dioxide into the available space.

As more carbon dioxide forms, its partial pressure increases.

Eventually, the system can reach a position where the forward and reverse reactions occur at equal rates. There is then no net change in composition.

For this solid-gas equilibrium, the relevant gas variable is the partial pressure of carbon dioxide, \(p_{\mathrm{CO_2}}\).

At a fixed temperature, there is an equilibrium carbon dioxide pressure, \(p_{\mathrm{eq}}\), at which calcium carbonate and calcium oxide can coexist.

Compare the actual carbon dioxide pressure with that equilibrium pressure:

This explains why two samples at the same high temperature can behave differently.

A stream of gas that removes carbon dioxide can favour decomposition. A carbon dioxide-rich atmosphere can favour the reverse reaction.

Using the same simplified model as the worked example, the following approximate equilibrium pressures are consistent with its thermodynamic assumptions.

Illustrative model values, not independently measured data. They assume ideal-gas behaviour and both solids are available.

Look at \(1000\ \mathrm{K}\).

The standard Gibbs energy is positive, but if the actual carbon dioxide pressure is only \(0.010\ \mathrm{bar}\), decomposition is favourable because that pressure is below the equilibrium value of \(0.050\ \mathrm{bar}\).

This is not a contradiction. Standard Gibbs energy describes standard-state conditions. Actual reaction favourability also depends on the current composition and gas pressure.

The corresponding HSC equilibrium idea is explored in How Temperature Changes Keq in Endothermic and Exothermic Reactions.

07Why spontaneous doesn’t mean fast or complete

There’s one more misconception worth removing.

Suppose the calculation gives \(\Delta G^\circ<0\). Would you necessarily observe a large amount of product immediately?

No.

Gibbs free energy predicts thermodynamic favourability. Reaction rate depends on kinetic factors, including the activation energy.

Even a thermodynamically favourable decomposition may take time because heat must move through the solid, bonds must rearrange, and carbon dioxide must escape from the reacting material.

A useful distinction is:

In a heated limestone particle, decomposition can start near the surface while material deeper inside remains unreacted. How quickly the interior reacts depends partly on heat transfer and the escape of carbon dioxide.

The thermodynamic prediction and the rate prediction are answering different questions.

A contrasting case: heating cannot rescue every endothermic reaction

Imagine a different reaction with:

\[ \Delta H^\circ>0,\qquad \Delta S^\circ<0 \]

The reaction absorbs heat and decreases entropy.

Because entropy change is negative, the term \(-T\Delta S^\circ\) is positive.

Therefore:

\[ \Delta G^\circ=(+\Delta H^\circ)+T|\Delta S^\circ| \]

Here, \(|\Delta S^\circ|\) means the positive magnitude of the entropy change.

Both contributions make \(\Delta G^\circ\) positive. Increasing temperature makes the entropy penalty larger.

Under the constant-sign assumptions, this reaction cannot become thermodynamically favourable under standard-state conditions merely by heating it.

So the useful rule isn’t “endothermic reactions become spontaneous when heated”.

It is: An endothermic reaction can become spontaneous at high temperature when it also has a sufficiently positive entropy change.

08Questions and solutions

The first question checks the Gibbs energy calculation. The later questions combine experimental evidence, gas pressure, equilibrium, and reaction kinetics.

All thermodynamic and equilibrium values supplied in these questions are illustrative unless stated otherwise.

Question 1

An endothermic reaction has:

\[ \begin{aligned} \Delta H^\circ&=+48.0\ \mathrm{kJ\,mol^{-1}}\\ \Delta S^\circ&=+80.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

Assume these values remain constant.

Calculate \(\Delta G^\circ\) at \(500\ \mathrm{K}\) and \(750\ \mathrm{K}\). State whether the reaction is spontaneous under standard-state conditions at each temperature, and explain whether increasing temperature changes its enthalpy sign.

Solution 1

The reaction is nonspontaneous under standard-state conditions at \(500\ \mathrm{K}\), but spontaneous at \(750\ \mathrm{K}\). It remains endothermic at both temperatures.

First, convert entropy into kilojoules:

\[ \Delta S^\circ = +0.0800\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

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

\[ \begin{aligned} \Delta G^\circ &=48.0-(500)(0.0800)\\ &=+8.00\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Since \(\Delta G^\circ>0\), the forward reaction is not thermodynamically favourable under standard-state conditions.

At \(750\ \mathrm{K}\):

\[ \begin{aligned} \Delta G^\circ &=48.0-(750)(0.0800)\\ &=-12.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Now \(\Delta G^\circ<0\), so the forward reaction is favourable.

The tempting mistake is to say the reaction must have become exothermic. It hasn’t. The supplied enthalpy change is still positive.

What changed is the size of the entropy contribution, \(T\Delta S^\circ\).

Question 2

A researcher heats a \(5.00\ \mathrm{g}\) sample of limestone containing calcium carbonate and an unknown hydrated impurity.

During heating at \(1000\ \mathrm{K}\):

• The sample pan loses \(1.50\ \mathrm{g}\).
• A gas analysis system measures \(1.10\ \mathrm{g}\) of released carbon dioxide.
• The remaining \(0.40\ \mathrm{g}\) of mass loss is identified as water vapour.
• Carbon dioxide is continuously removed from the heated chamber.

Assume all detected carbon dioxide came from calcium carbonate decomposition, and no solid material escaped.

A student claims:

“The full \(1.50\ \mathrm{g}\) loss came from calcium carbonate decomposition. Also, because decomposition occurred at \(1000\ \mathrm{K}\), its standard Gibbs energy must have been negative.”

Evaluate both claims. Calculate the mass of calcium carbonate that decomposed, and propose a revised investigation that would isolate the influence of carbon dioxide pressure.

Solution 2

Both claims are incorrect. Only \(1.10\ \mathrm{g}\) of the mass loss was carbon dioxide, corresponding to approximately \(2.50\ \mathrm{g}\) of calcium carbonate decomposed. Observing decomposition also does not establish that \(\Delta G^\circ<0\).

The first error is a measurement error.

The balance records every substance that leaves the pan, not only the products of calcium carbonate decomposition.

The amount of calcium carbonate decomposed is determined from the measured carbon dioxide:

\[ \begin{aligned} n(\ce{CO2}) &=\frac{1.10}{44.0}\\ &=0.0250\ \mathrm{mol} \end{aligned} \]

Using the \(1:1\) reaction ratio:

\[ n(\ce{CaCO3})=0.0250\ \mathrm{mol} \]

Therefore:

\[ \begin{aligned} m(\ce{CaCO3}) &=(0.0250)(100.1)\\ &\approx2.50\ \mathrm{g} \end{aligned} \]

Treating the full \(1.50\ \mathrm{g}\) loss as carbon dioxide would overestimate decomposition because \(0.40\ \mathrm{g}\) was water vapour.

The second error confuses actual reaction conditions with standard-state conditions.

Using the earlier illustrative calcium carbonate model:

\[ \Delta G^\circ_{1000} =+25.0\ \mathrm{kJ\,mol^{-1}} \]

The standard Gibbs energy is positive. However, continuous removal of carbon dioxide can keep its actual partial pressure below the equilibrium pressure, allowing the forward reaction to be favourable.

To isolate the pressure effect, the researcher should use pure, pre-characterised calcium carbonate and compare trials at the same temperature under different controlled carbon dioxide partial pressures.

The mass sensor and gas detector should both remain in use. Their combined readings distinguish actual decomposition from unrelated mass loss.

The deeper experimental point is that observing a reaction is not the same as controlling the conditions needed to explain why it occurs.

Question 3

Consider the calcium carbonate equilibrium at \(1000\ \mathrm{K}\).

The illustrative equilibrium carbon dioxide pressure is \(0.050\ \mathrm{bar}\), and the standard Gibbs energy for decomposition is \(+25.0\ \mathrm{kJ\,mol^{-1}}\).

Two sealed vessels contain both calcium carbonate and calcium oxide, with enough of each solid to reach the predicted equilibrium.

• Vessel A initially has \(p_{\mathrm{CO_2}}=0.010\ \mathrm{bar}\).
• Vessel B initially has \(p_{\mathrm{CO_2}}=0.200\ \mathrm{bar}\).

Both vessels remain at \(1000\ \mathrm{K}\), and carbon dioxide behaves as an ideal gas.

Predict the initial net reaction direction and final carbon dioxide pressure in each vessel. Explain why the outcomes are compatible with the positive standard Gibbs energy.

Solution 3

Vessel A initially favours decomposition, while vessel B initially favours the reverse reaction. Both approach a carbon dioxide pressure of \(0.050\ \mathrm{bar}\), provided the specified solids remain available.

In Vessel A:

\[ 0.010\ \mathrm{bar}<0.050\ \mathrm{bar} \]

The carbon dioxide pressure is below equilibrium.

Therefore, the forward reaction is favoured:

\[ \ce{CaCO3(s) -> CaO(s) + CO2(g)} \]

As decomposition produces carbon dioxide, its pressure increases towards \(0.050\ \mathrm{bar}\).

In Vessel B:

\[ 0.200\ \mathrm{bar}>0.050\ \mathrm{bar} \]

The carbon dioxide pressure exceeds equilibrium.

The reverse reaction is favoured:

\[ \ce{CaO(s) + CO2(g) -> CaCO3(s)} \]

As carbon dioxide is consumed, its pressure decreases towards \(0.050\ \mathrm{bar}\).

The tempting conclusion is that a positive \(\Delta G^\circ\) rules out forward decomposition at this temperature.

But \(\Delta G^\circ\) refers to standard states. Neither vessel initially has carbon dioxide at its standard pressure of 1 bar.

The actual reaction direction depends on the gas composition and pressure.

The most important lesson is that the sign of standard Gibbs energy and the direction of change in a non-standard mixture are related, but they are not identical questions.

Question 4

A researcher studies an unknown reaction and obtains the following standard Gibbs energy values:

Assume \(\Delta H^\circ\) and \(\Delta S^\circ\) remain constant between these temperatures.

At \(700\ \mathrm{K}\), the researcher heats a fresh reactant-rich mixture for two minutes but detects almost no product.

A student argues that the Gibbs energy calculations must be wrong because a spontaneous reaction should happen immediately.

Determine \(\Delta H^\circ\), \(\Delta S^\circ\), and the temperature at which \(\Delta G^\circ=0\). Then evaluate the student’s argument and describe two experimental checks that would help distinguish slow kinetics from an equilibrium limitation.

Solution 4

The reaction has \(\Delta H^\circ=+62.0\ \mathrm{kJ\,mol^{-1}}\), \(\Delta S^\circ=+100\ \mathrm{J\,mol^{-1}\,K^{-1}}\), and a standard Gibbs-energy crossover temperature of \(620\ \mathrm{K}\). The lack of detectable product after two minutes does not invalidate these results.

Start with:

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

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

\[ 12.0=\Delta H^\circ-500\Delta S^\circ \]

At \(700\ \mathrm{K}\):

\[ -8.0=\Delta H^\circ-700\Delta S^\circ \]

In these equations, Gibbs energy and enthalpy are in \(\mathrm{kJ\,mol^{-1}}\), so entropy must be expressed in \(\mathrm{kJ\,mol^{-1}\,K^{-1}}\).

Subtract the second equation from the first:

\[ \begin{aligned} 20.0&=200\Delta S^\circ\\ \Delta S^\circ &=0.100\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \end{aligned} \]

Therefore:

\[ \Delta S^\circ =+100\ \mathrm{J\,mol^{-1}\,K^{-1}} \]

Substitute into the \(500\ \mathrm{K}\) equation:

\[ \begin{aligned} 12.0&=\Delta H^\circ-50.0\\ \Delta H^\circ&=+62.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

The crossover occurs when:

\[ \begin{aligned} T&=\frac{\Delta H^\circ}{\Delta S^\circ}\\ &=\frac{62.0}{0.100}\\ &=620\ \mathrm{K} \end{aligned} \]

The reaction is endothermic with positive entropy change, so its standard Gibbs energy becomes negative above \(620\ \mathrm{K}\) under the stated assumptions.

Why, then, was little product observed at \(700\ \mathrm{K}\)?

A negative standard Gibbs energy does not tell us the reaction rate. There may be a substantial activation-energy barrier, making product formation slow on the two-minute timescale.

Alternatively, the mixture may have reached a product-limited equilibrium under its actual non-standard conditions. The experiment hasn’t yet supplied enough information to distinguish those possibilities.

Two useful checks are:

1. Monitor the composition over a much longer period, with and without a suitable catalyst where one is available. If a catalyst increases the rate but leaves the final equilibrium composition unchanged, that supports a kinetic explanation for the initial delay.

2. Approach equilibrium from different starting compositions at the same temperature and pressure. If reactant-rich and product-rich mixtures approach the same final composition, this supports an equilibrium interpretation. The researcher must also allow sufficient time and check for competing reactions.

The tempting mistake was to treat spontaneity as a stopwatch. Gibbs energy describes a thermodynamic driving force, not how quickly the system responds.

Question 5

A sealed piston-cylinder contains calcium carbonate, calcium oxide, and carbon dioxide.

The following model data apply:

Initially, the system is at equilibrium.

The vessel is then heated rapidly to \(1200\ \mathrm{K}\) while its gas volume is reduced to \(0.100\ \mathrm{L}\).

Assume:

• The temperature and volume change occur too quickly for the reaction to proceed during the change itself.
• The gas is pure, ideal carbon dioxide, with \(R=0.08314\ \mathrm{L\,bar\,mol^{-1}\,K^{-1}}\).
• The apparatus subsequently remains at \(1200\ \mathrm{K}\) and \(0.100\ \mathrm{L}\), with enough heat supplied to maintain that temperature.
• The solids occupy negligible volume, and no gas leaks unless deliberately vented.

(a) Calculate the carbon dioxide pressure immediately after heating and compression, before any chemical reaction.

(b) Predict the initial net reaction direction, then determine the final pressure after reaction has proceeded as far as the available solids permit.

(c) The vessel is then vented until the carbon dioxide pressure is \(0.650\ \mathrm{bar}\). The valve is closed, with temperature and volume unchanged. Predict the new net reaction direction and final pressure.

(d) Explain why a student using only the statement “heating favours an endothermic reaction” would get the first reaction direction wrong.

Solution 5

The rapid heating and compression raise the carbon dioxide pressure to \(3.00\ \mathrm{bar}\). The reverse reaction is initially favoured, even though the reaction is endothermic in the forward direction and the temperature has increased. The available calcium oxide is exhausted before the gas can fall to its equilibrium pressure. Subsequent venting makes forward decomposition favourable again.

The key is that temperature and volume change together.

Part (a): Pressure immediately after heating and compression

Before chemical reaction occurs, the number of moles of gas is unchanged.

Using the combined ideal gas law:

\[ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \]

Here, \(p_1\) and \(p_2\) are initial and final gas pressures, \(V_1\) and \(V_2\) are gas volumes, and \(T_1\) and \(T_2\) are absolute temperatures.

Rearrange:

\[ p_2 = p_1\frac{V_1T_2}{V_2T_1} \]

Substitute:

\[ \begin{aligned} p_2 &=0.050\frac{(5.00)(1200)}{(0.100)(1000)}\\ &=3.00\ \mathrm{bar} \end{aligned} \]

The rapid compression outweighs the effect of the higher temperature on the equilibrium preference.

Part (b): Initial direction and endpoint

At \(1200\ \mathrm{K}\), the supplied equilibrium pressure is \(1.65\ \mathrm{bar}\).

But immediately after the change:

\[ 3.00\ \mathrm{bar}>1.65\ \mathrm{bar} \]

Therefore, the reverse reaction is initially favoured:

\[ \ce{CaO(s) + CO2(g) -> CaCO3(s)} \]

Carbon dioxide is consumed, so its pressure decreases.

It is tempting to assume the pressure must fall all the way to \(1.65\ \mathrm{bar}\). However, the reaction needs calcium oxide, and the vessel contains only \(0.00100\ \mathrm{mol}\).

Calculate the moles of carbon dioxide that would have to be consumed to reach \(1.65\ \mathrm{bar}\).

At constant temperature and volume:

\[ \Delta n=\frac{\Delta pV}{RT} \]

where \(\Delta n\) is the change in gas amount and \(\Delta p\) is the required pressure change.

\[ \begin{aligned} \Delta n &=\frac{(3.00-1.65)(0.100)} {(0.08314)(1200)}\\ &=0.00135\ \mathrm{mol} \end{aligned} \]

The reverse reaction would need \(0.00135\ \mathrm{mol}\) of calcium oxide, but only \(0.00100\ \mathrm{mol}\) exists.

Therefore, all the calcium oxide is consumed first.

The actual pressure decrease is:

\[ \begin{aligned} \Delta p &=\frac{\Delta nRT}{V}\\ &=\frac{(0.00100)(0.08314)(1200)} {0.100}\\ &=0.998\ \mathrm{bar} \end{aligned} \]

So the final pressure is:

\[ \begin{aligned} p_{\mathrm{final}} &=3.00-0.998\\ &=2.00\ \mathrm{bar} \end{aligned} \]

The pressure stops at approximately \(2.00\ \mathrm{bar}\), rather than \(1.65\ \mathrm{bar}\), because no calcium oxide remains for the reverse reaction.

The endpoint is consistent with thermodynamics. The equilibrium-pressure condition for coexistence of both solids cannot be enforced once one solid has been exhausted.

Part (c): What happens after venting?

After venting, the carbon dioxide pressure is \(0.650\ \mathrm{bar}\).

This is below the equilibrium pressure:

\[ 0.650\ \mathrm{bar}<1.65\ \mathrm{bar} \]

Therefore, calcium carbonate decomposition is now favoured.

To reach equilibrium again, the gas pressure must increase by:

\[ \Delta p=1.65-0.650=1.00\ \mathrm{bar} \]

The amount of carbon dioxide required is:

\[ \begin{aligned} \Delta n &=\frac{(1.00)(0.100)} {(0.08314)(1200)}\\ &=0.00100\ \mathrm{mol} \end{aligned} \]

Approximately \(0.00100\ \mathrm{mol}\) of calcium carbonate must decompose.

There is enough calcium carbonate available. In fact, the earlier reverse reaction increased its amount from \(0.00500\ \mathrm{mol}\) to \(0.00600\ \mathrm{mol}\).

Thus, decomposition can proceed until the pressure reaches:

\[ p_{\mathrm{final}}=1.65\ \mathrm{bar} \]

Both solids are then present again, and the system can reach equilibrium.

Part (d): Why the temperature-only prediction fails

For an endothermic reaction, increasing temperature increases the equilibrium tendency towards products. That principle is correct.

But the student overlooked two additional constraints.

First, the simultaneous compression increases carbon dioxide pressure enormously. The initial pressure rises from \(0.050\) to \(3.00\ \mathrm{bar}\), exceeding the new equilibrium pressure. The actual reaction therefore initially runs in reverse.

Second, reaction direction alone does not determine how far the change can proceed. The finite amount of calcium oxide prevents the system from reaching the two-solid equilibrium pressure during the first stage.

This is the conceptual hinge: the effect of heating on the equilibrium constant is not the same as the net reaction direction after several conditions change together.

A hotter system can initially react backwards, and a thermodynamically favoured direction can stop because a required reactant is exhausted.

The next useful connection is between Gibbs energy and equilibrium. Gibbs energy tells us which direction is favourable under specified conditions, while equilibrium calculations help explain where the reaction stops. When solving a new problem, first identify whether it asks about standard-state spontaneity, the actual mixture, or the rate of reaction. Those are three different questions, even when they describe the same chemistry.