Why Endothermic Reactions Can Occur at Room Temperature

Mix aqueous citric acid with sodium hydrogencarbonate at room temperature. The mixture fizzes, carbon dioxide escapes, and the temperature drops. If the reaction is absorbing heat, why does it happen without being heated?

An endothermic reaction can occur at room temperature because absorbing heat does not mean a reaction is impossible or thermodynamically unfavourable. Particles already have energy at room temperature, so some can overcome the activation energy barrier. The overall reaction can also be spontaneous if its increase in entropy is large enough to outweigh its positive enthalpy change.

There are three different questions hiding inside this observation:

• Enthalpy: Does the reaction absorb or release heat?
• Activation energy: What energy barrier must reacting particles overcome?
• Spontaneity: Is the forward reaction thermodynamically favoured under these conditions?

You need all three ideas to explain why a cold, fizzing reaction can happen in a warm room.

01The experiment: a reaction that cools itself

Consider the reaction between citric acid and sodium hydrogencarbonate, commonly called bicarbonate of soda.

When their aqueous solutions are mixed, carbon dioxide gas forms. Under suitable experimental conditions, the overall process is endothermic.

The balanced equation is:

\[ \ce{C6H8O7(aq) + 3NaHCO3(aq)} \]

\[ \ce{-> Na3C6H5O7(aq) + 3CO2(g) + 3H2O(l)} \]

Citric acid reacts with sodium hydrogencarbonate to produce sodium citrate, carbon dioxide, and water.

The bubbles are evidence that a chemical change is taking place. The falling temperature gives us information about the energy transferred during that change.

How the first experiment works

Picture a simple laboratory arrangement.

A glass beaker contains sodium hydrogencarbonate solution. A thermometer sits in the liquid, with its tip below the surface but not touching the glass.

Citric acid solution, initially at the same temperature, is added from a second beaker. The mixture is stirred gently.

The thermometer records the temperature while carbon dioxide bubbles rise and escape from the open beaker.

Suppose someone argues that a chemical reaction cannot happen at room temperature unless it releases heat.

What would that model predict?

If the reaction occurs, the mixture should become warmer.

Now compare that prediction with an illustrative set of observations.

These values are invented for teaching and are not measurements from a published experiment.

The solutions react even though the temperature falls.

That challenges the idea that only exothermic reactions can occur without continuous external heating.

It does not, by itself, prove the precise enthalpy change. Cooling can also be influenced by evaporation, initial temperature differences, heat exchange with the room, and the escape of gas.

A better experiment must account for those effects.

Improving the apparatus

The first design has a problem: an open glass beaker exchanges heat with the air and bench.

If the reaction mixture becomes colder than the room, heat can enter from its surroundings. The measured temperature drop may therefore be smaller than it would be in a better-insulated container.

A revised arrangement uses a polystyrene cup, an insulating lid, and a temperature probe.

The lid must allow carbon dioxide to escape. A gas-producing reaction should not be sealed in a rigid container without appropriate pressure controls.

A careful investigation would:

• Bring both solutions to the same initial temperature.
• Measure the masses and concentrations used.
• Record temperature continuously before, during, and after mixing.
• Use a suitable non-reacting control to investigate temperature drift caused by the apparatus and surroundings.
• Repeat the measurements and account, where possible, for heat absorbed by the cup and heat carried away by escaping gas.

The important observation is not just that the mixture gets cold. It is that the corrected temperature change supports a net absorption of heat by the chemical process.

For an endothermic reaction, the sign convention is:

\[ \Delta H>0 \]

Here, \(\Delta H\) is the enthalpy change of the reaction. A positive value means the chemical system absorbs heat at constant pressure.

If you need a refresher on that sign convention, see Endothermic and Exothermic Reactions for HSC Chemistry.

02How much heat does the reaction absorb?

A temperature drop tells us the direction of energy transfer. Calorimetry lets us estimate its magnitude.

For a solution, we use:

\[ q_{\text{solution}}=mc\Delta T \]

where:

• \(q_{\text{solution}}\) is heat transferred to the solution, in joules (J).
• \(m\) is the mass of solution, in grams (g).
• \(c\) is its specific heat capacity, in \(\text{J g}^{-1}\text{K}^{-1}\).
• \(\Delta T=T_f-T_i\), where \(T_f\) and \(T_i\) are final and initial temperatures.

For a dilute aqueous solution, we often approximate its specific heat capacity using water’s value:

\[ c=4.18\ \text{J g}^{-1}\text{K}^{-1} \]

This is an approximation, not an exact property of every solution.

If heat exchange with the container and external environment is negligible, the chemical process gains approximately the heat lost by the liquid:

\[ q_{\text{reaction}}\approx-q_{\text{solution}} \]

03Reaction enthalpy is not the activation energy

We now know that the forward reaction can absorb heat.

But a second question remains: how do the particles get the reaction started?

Imagine a reaction occurring along a pathway with a temporary high-energy arrangement of atoms.

Reactant particles must reach that arrangement before they can form products. This requires an activation energy, represented by \(E_a\).

A reaction-energy profile separates two different vertical energy differences:

• The gap from reactants to products represents the overall enthalpy change, \(\Delta H\).
• The gap from reactants to the top of the pathway represents the forward activation energy, \(E_{a,\text{forward}}\).

For this illustrative pathway:

\[ \Delta H=+25\text{ kJ mol}^{-1} \]

but:

\[ E_{a,\text{forward}}=60\text{ kJ mol}^{-1} \]

Those are not interchangeable quantities.

The reaction needs to access a pathway with a barrier of \(60\text{ kJ mol}^{-1}\), while its products end up \(25\text{ kJ mol}^{-1}\) higher in energy than its reactants.

Why room temperature can be enough

At room temperature, particles are moving, rotating, vibrating, and exchanging energy.

They do not all have the same energy.

Some collisions can access a reaction pathway with sufficient energy and suitable molecular orientation to overcome its activation barrier.

It is tempting to imagine activation energy as an amount of heat that must first be supplied to the entire beaker.

That is not how it works.

A reaction does not require every particle to possess the activation energy simultaneously. Individual successful reaction events can occur as particles exchange energy.

The activation energy helps determine the reaction rate. Temperature, concentration, molecular orientation, and the reaction mechanism also matter.

A reaction with a very large activation barrier may proceed so slowly that nothing noticeable happens in an ordinary experiment, even when the forward reaction is thermodynamically favourable.

What happens in the reverse direction?

This is where an energy diagram becomes especially useful.

If the forward reaction is endothermic, its products lie above its reactants in enthalpy.

The reverse reaction therefore goes from a higher-energy state to a lower-energy state.

So:

\[ \Delta H_{\text{reverse}} =-\Delta H_{\text{forward}} \]

For the same simple pathway, the HSC energy-profile relationship is:

\[ E_{a,\text{reverse}} =E_{a,\text{forward}}-\Delta H \]

If the forward reaction absorbs \(25\text{ kJ mol}^{-1}\), the reverse reaction releases \(25\text{ kJ mol}^{-1}\).

And if both directions use the same peak, the reverse activation barrier is smaller.

This is a pathway relationship, not a prediction that the reverse reaction must dominate every mixture. Actual forward and reverse rates also depend on the amounts of reactants and products present.

04Why the forward reaction can be thermodynamically favourable

We have explained why an activation barrier does not automatically prevent a room-temperature reaction.

But overcoming a barrier is not enough to tell us whether the forward reaction is thermodynamically favoured.

For that, enthalpy alone is insufficient.

Consider the carbon dioxide produced in the acid-hydrogencarbonate reaction.

Before the reaction, the atoms that will form carbon dioxide are part of dissolved chemical species. Afterwards, carbon dioxide molecules can occupy a gas phase and escape into the surrounding air.

The gas has many possible positions and arrangements. That gives an intuitive reason why changes involving gas formation and dispersal can have important entropy effects.

The formal quantity is entropy, \(S\). It describes the number of microscopic ways matter and energy can be distributed while producing the same macroscopic state.

Greater dispersal often indicates increased entropy, but counting gas molecules or looking at bubbles is not sufficient to calculate the exact entropy change of a reaction.

For thermodynamic predictions at constant temperature and pressure, we use Gibbs free energy:

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

where:

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

For a forward change under the stated conditions:

For an endothermic reaction, \(\Delta H\) is positive.

However, if \(\Delta S\) is also positive, the term \(T\Delta S\) can be larger than \(\Delta H\).

Then \(\Delta G\) is negative.

That is how an entropy increase can make an endothermic process thermodynamically favourable.

05The contrasting case: releasing heat does not guarantee an observable reaction

Think about a piece of dry timber sitting in air at room temperature.

Its combustion is strongly exothermic. Yet the timber does not normally burst into flames while lying on a table.

Why not?

Combustion has an activation barrier. Under ordinary conditions, the reaction rate is extremely low without an ignition source. Once burning begins, heat released by the reaction can help sustain it.

Now compare that with the aqueous acid-hydrogencarbonate reaction.

The fizzing reaction can proceed noticeably at room temperature while absorbing heat.

The contrast is not that endothermic reactions are faster than exothermic reactions. They are not necessarily.

It is that enthalpy and activation energy answer different questions.

There is one more subtle point.

An endothermic reaction can proceed while the reaction mixture cools, but that cooling may change the conditions controlling its spontaneity and rate.

If the reaction is reversible, the mixture may eventually reach equilibrium. The forward reaction does not necessarily continue until every reactant has disappeared.

This becomes particularly important when temperature, gas pressure, or product concentration changes during an experiment.

06Questions and solutions

All numerical data in the following questions are illustrative. Unless stated otherwise, activation energies refer to the same simple reaction pathway, and enthalpy changes are for the forward direction as written.

Question 1

A hypothetical reaction has a forward activation energy of \(50\text{ kJ mol}^{-1}\) and an enthalpy change of \(+18\text{ kJ mol}^{-1}\).

Calculate the reverse activation energy, state whether the reverse reaction is endothermic or exothermic, and explain whether these values alone prove that the forward reaction is spontaneous at room temperature.

Solution 1

The reverse activation energy is \(32\text{ kJ mol}^{-1}\), and the reverse reaction is exothermic. The information is insufficient to determine forward spontaneity.

For the reverse activation energy:

\[ \begin{aligned} E_{a,\text{reverse}} &=E_{a,\text{forward}}-\Delta H\\ &=50-18\\ &=32\text{ kJ mol}^{-1} \end{aligned} \]

For the reverse enthalpy change:

\[ \Delta H_{\text{reverse}} =-18\text{ kJ mol}^{-1} \]

The reverse reaction releases heat because it converts higher-enthalpy products back into lower-enthalpy reactants.

However, the signs of enthalpy change and the activation energies do not determine spontaneity.

We would need additional thermodynamic information, such as Gibbs free energy change under the stated conditions.

The tempting mistake is to conclude that the lower reverse activation energy automatically makes the reverse reaction thermodynamically favoured. Activation energy concerns the kinetic pathway; it does not establish the preferred overall direction.

Question 2

A student investigates the same endothermic reaction in two open beakers.

Both trials begin with \(40.0\text{ g}\) of mixed solution at \(23.0^\circ\text{C}\). In each trial, \(0.0120\text{ mol}\) of a limiting reactant is consumed.

One trial is conducted in a warm room, and the other in a cool room.

The student also performs a suitable non-reacting control in each room.

The student claims that the reaction must absorb more heat in the cool room because the observed temperature drop is greater.

Assume the control temperature changes accurately estimate non-reaction temperature drift, the effects are additive, and both mixtures have a specific heat capacity of \(4.18\ \text{J g}^{-1}\text{K}^{-1}\).

(a) Calculate the corrected temperature change for each trial.

(b) Estimate the molar enthalpy change per mole of limiting reactant consumed.

(c) Evaluate the student’s conclusion and identify an important limitation of the correction.

Solution 2

Both trials give the same corrected temperature change, \(-2.3^\circ\text{C}\), and an approximate enthalpy change of \(+32.0\text{ kJ mol}^{-1}\). The larger uncorrected temperature drop in the cool room does not establish a larger reaction enthalpy.

(a) Correct the measured temperature changes.

For the warm room:

\[ \begin{aligned} \Delta T_{\text{reaction}}&=21.0-23.0\\ &=-2.0^\circ\text{C} \end{aligned} \]

The control warms by \(+0.3^\circ\text{C}\).

Subtracting that drift:

\[ \Delta T_{\text{corrected}} =-2.0-(+0.3) =-2.3^\circ\text{C} \]

For the cool room:

\[ \begin{aligned} \Delta T_{\text{reaction}}&=20.1-23.0\\ &=-2.9^\circ\text{C} \end{aligned} \]

The control cools by \(0.6^\circ\text{C}\).

\[ \Delta T_{\text{corrected}} =-2.9-(-0.6) =-2.3^\circ\text{C} \]

The apparent disagreement disappears after the assumed environmental correction.

(b) Calculate the approximate enthalpy change.

Using either corrected result:

\[ \begin{aligned} q_{\text{solution}} &=mc\Delta T\\ &=(40.0)(4.18)(-2.3)\\ &=-384.6\text{ J}\\ &=-0.3846\text{ kJ} \end{aligned} \]

Therefore:

\[ q_{\text{reaction}}\approx+0.3846\text{ kJ} \]

Dividing by the amount of limiting reactant consumed:

\[ \begin{aligned} \Delta H &\approx\frac{0.3846}{0.0120}\\ &=+32.0\text{ kJ mol}^{-1} \end{aligned} \]

Both trials support the same approximate positive reaction enthalpy.

(c) Evaluate the conclusion.

The cool room removes additional heat from the reaction mixture, making the uncorrected temperature drop greater.

The warm room can transfer heat into the colder reaction mixture, reducing the observed drop.

The student has confused heat exchange with the environment with heat absorbed by the chemical reaction.

An important limitation is that a non-reacting control may not reproduce everything occurring in the real reaction. Gas escape, evaporation, and changing solution composition can affect the energy balance differently.

The corrected value is therefore a model-based estimate, not an exact measurement of reaction enthalpy.

Question 3

A hypothetical reversible reaction follows:

\[ \ce{R(aq) <=> P(aq)} \]

The reactants have a relative energy of \(0\text{ kJ mol}^{-1}\), and the products have a relative energy of \(+24\text{ kJ mol}^{-1}\).

An uncatalysed pathway has a peak at \(+86\text{ kJ mol}^{-1}\). A catalysed pathway between the same reactants and products has a peak at \(+52\text{ kJ mol}^{-1}\).

For a particular product-rich mixture at \(298\text{ K}\), the Gibbs free energy change for the forward reaction is \(+4.0\text{ kJ mol}^{-1}\).

A student claims:

“The catalyst lowers the forward activation energy but leaves the reverse activation energy unchanged. Since the forward reaction becomes easier, the equilibrium mixture will contain more product.”

Determine the forward and reverse activation energies for both pathways. Then evaluate the student’s claims and predict the initially favoured net direction in the stated mixture.

Solution 3

The catalyst lowers both activation energies, but it does not change the reaction enthalpy, Gibbs free energy change, or equilibrium position. The stated product-rich mixture initially favours the reverse reaction.

For the uncatalysed pathway:

\[ \begin{aligned} E_{a,\text{forward}}&=86-0\\ &=86\text{ kJ mol}^{-1}\\ E_{a,\text{reverse}}&=86-24\\ &=62\text{ kJ mol}^{-1} \end{aligned} \]

For the catalysed pathway:

\[ \begin{aligned} E_{a,\text{forward}}&=52-0\\ &=52\text{ kJ mol}^{-1}\\ E_{a,\text{reverse}}&=52-24\\ &=28\text{ kJ mol}^{-1} \end{aligned} \]

The results are:

The catalyst lowers both activation energies by \(34\text{ kJ mol}^{-1}\).

The forward reaction remains endothermic:

\[ \Delta H=+24\text{ kJ mol}^{-1} \]

The reverse reaction remains exothermic:

\[ \Delta H_{\text{reverse}} =-24\text{ kJ mol}^{-1} \]

Now consider the thermodynamic information.

For the actual mixture:

\[ \Delta G_{\text{forward}} =+4.0\text{ kJ mol}^{-1} \]

Therefore, the reverse direction is thermodynamically favoured at that instant.

Adding a catalyst does not change that sign. It provides a faster pathway for both directions, so the system can approach equilibrium more rapidly.

The first claim fails because both directions use the same catalysed pathway between the same endpoints. Lowering its peak also lowers the reverse barrier.

The second claim fails because equilibrium is determined by thermodynamics, not by which pathway has the smaller activation energy.

Question 4

A hypothetical endothermic reaction has the following standard-state quantities, assumed approximately constant over the temperature range considered:

\[ \begin{aligned} \Delta H^\circ&=+30.0\text{ kJ mol}^{-1}\\ \Delta S^\circ&=+102\text{ J mol}^{-1}\text{K}^{-1}\\ E_{a,\text{forward}}&=95.0\text{ kJ mol}^{-1} \end{aligned} \]

A researcher examines the reaction at \(288\text{ K}\) and \(308\text{ K}\).

(a) Determine the standard Gibbs free energy change at each temperature.

(b) Estimate the temperature at which the standard Gibbs free energy change becomes zero.

(c) At \(308\text{ K}\), no visible product forms during a short observation period. Explain whether this contradicts the Gibbs free energy prediction.

(d) Explain whether adding a catalyst at \(288\text{ K}\) would make the forward reaction thermodynamically favourable under standard-state conditions.

Solution 4

The standard forward reaction is thermodynamically unfavourable at \(288\text{ K}\) but favourable at \(308\text{ K}\). Its crossover temperature is approximately \(294\text{ K}\). A slow observed rate does not contradict the thermodynamic prediction.

(a) Calculate Gibbs free energy at each temperature.

First convert the entropy change:

\[ \Delta S^\circ =0.102\text{ kJ mol}^{-1}\text{K}^{-1} \]

At \(288\text{ K}\):

\[ \begin{aligned} \Delta G^\circ &=30.0-(288)(0.102)\\ &=30.0-29.376\\ &=+0.624\text{ kJ mol}^{-1} \end{aligned} \]

The positive result means the standard forward change is not thermodynamically favoured at \(288\text{ K}\).

At \(308\text{ K}\):

\[ \begin{aligned} \Delta G^\circ &=30.0-(308)(0.102)\\ &=30.0-31.416\\ &=-1.416\text{ kJ mol}^{-1} \end{aligned} \]

The negative result means the standard forward change is favoured at \(308\text{ K}\).

(b) Find the crossover temperature.

Set standard Gibbs free energy change equal to zero:

\[ 0=\Delta H^\circ-T\Delta S^\circ \]

Rearrange:

\[ \begin{aligned} T&=\frac{\Delta H^\circ}{\Delta S^\circ}\\ &=\frac{30.0}{0.102}\\ &=294.1\text{ K} \end{aligned} \]

This is approximately \(21.0^\circ\text{C}\).

With the assumed constant enthalpy and entropy values, temperatures above this point favour the standard forward change, and temperatures below it do not.

(c) Explain the lack of visible reaction.

The forward activation energy is relatively large at \(95.0\text{ kJ mol}^{-1}\).

Although the standard Gibbs free energy change at \(308\text{ K}\) is negative, the actual reaction may still be too slow to produce a visible change during the observation period.

We cannot calculate a specific rate from the activation energy alone. We would need further kinetic information.

The tempting inference is that a negative Gibbs free energy change guarantees an immediately observable reaction. It does not.

(d) Evaluate the catalyst.

A catalyst can increase the forward and reverse reaction rates by providing alternative pathways with lower activation energies.

It cannot change \(\Delta H^\circ\), \(\Delta S^\circ\), or \(\Delta G^\circ\) at a fixed temperature.

Therefore, at \(288\text{ K}\), the standard forward reaction remains thermodynamically unfavourable even after adding a catalyst.

There is a further limitation: these calculations concern standard-state changes. An actual mixture with different concentrations or gas pressures could have a different instantaneous thermodynamic driving force.

Question 5

Consider the hypothetical reversible reaction:

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

The forward reaction is endothermic.

The system contains a liquid solution and a gas-filled headspace inside a rigid, sealed vessel.

For this idealised model, its reaction quotient is:

\[ Q=\frac{[B]}{[A]}\frac{p_G}{p^\circ} \]

where \([A]\) and \([B]\) are the aqueous concentrations, \(p_G\) is the partial pressure of gaseous product G, and \(p^\circ=1.00\text{ bar}\) is the reference pressure.

The equilibrium constants are:

Initially, the vessel is at equilibrium at \(298\text{ K}\), with:

\[ [A]=[B]=0.100\text{ mol L}^{-1} \]

and:

\[ p_G=4.00\text{ bar} \]

The vessel is rapidly cooled to \(288\text{ K}\) before appreciable chemical reaction occurs. It is then held at \(288\text{ K}\) by a thermostat.

Assume ideal-gas behaviour, constant vessel and liquid volumes, negligible gas dissolution, and no immediate concentration changes during cooling or venting.

At any time after cooling, but before or upon reaching the new equilibrium, a valve can remove a specified fraction of the gaseous product. Venting is fast enough that the liquid composition is unchanged at that instant.

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

(b) Show whether removing 50% of the gaseous product will always cause an initial forward net reaction, regardless of how long the system has been held at \(288\text{ K}\) before venting.

(c) Would removing only 10% of the gaseous product always produce the same direction? Establish the condition that separates forward and reverse behaviour.

(d) Explain the heat transfer associated with the initial response to cooling and the response after 50% venting. Would the thermometer necessarily show the solution cooling during the endothermic forward reaction?

Solution 5

Cooling initially favours the reverse reaction. Removing 50% of the gas then guarantees a forward response, regardless of when the gas is removed during the approach to equilibrium. Removing only 10% does not guarantee either direction.

The crucial point is that cooling changes both the gas pressure and the equilibrium constant. Venting changes the reaction quotient again, without changing the equilibrium constant at that temperature.

(a) Determine the direction immediately after cooling.

The initial vessel is at \(298\text{ K}\) and equilibrium:

\[ Q=K=4.00 \]

Immediately after rapid cooling, the amounts of gas and dissolved substances have not changed.

For an ideal gas at fixed volume:

\[ \frac{p_2}{p_1}=\frac{T_2}{T_1} \]

Therefore:

\[ \begin{aligned} p_2 &=4.00\times\frac{288}{298}\\ &=3.87\text{ bar} \end{aligned} \]

The solution concentrations are still equal.

Thus:

\[ \begin{aligned} Q &=\frac{0.100}{0.100} \times\frac{3.87}{1.00}\\ &=3.87 \end{aligned} \]

But at \(288\text{ K}\):

\[ K=2.50 \]

Since \(Q>K\), the mixture has too much product relative to the equilibrium condition.

The initial net reaction is reverse:

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

This reverse process is exothermic.

Notice the trap: cooling lowered the gas pressure, which alone tends to favour the gas-producing forward reaction. However, cooling also lowered \(K\) by a larger factor. The combined result favours the reverse reaction.

(b) Investigate removing 50% of the gas.

Immediately after cooling:

\[ Q=3.87 \]

As the reverse reaction proceeds towards equilibrium at \(288\text{ K}\), its reaction quotient decreases towards:

\[ Q=K=2.50 \]

Therefore, at any permitted venting time:

\[ 2.50\leq Q_{\text{before}}\leq3.87 \]

Removing half the gaseous product halves its pressure instantly, while the dissolved concentrations remain unchanged.

Consequently:

\[ Q_{\text{after}}=0.50Q_{\text{before}} \]

The possible values lie between:

\[ \begin{aligned} Q_{\text{after,min}}&=1.25\\ Q_{\text{after,max}}&=1.94 \end{aligned} \]

Every one of these values is below the equilibrium constant:

\[ Q_{\text{after}}<2.50 \]

Therefore, the net reaction must move forwards immediately after the 50% vent.

This conclusion does not depend on whether the vessel was vented just after cooling or after it had already reached the new equilibrium.

The result comes from tracking the entire possible range of reaction quotients, not from assuming the system has already equilibrated.

(c) Investigate removing only 10% of the gas.

Removing 10% of G leaves 90% of its previous partial pressure.

Therefore:

\[ Q_{\text{after}}=0.90Q_{\text{before}} \]

If venting happens immediately after cooling:

\[ \begin{aligned} Q_{\text{after}} &=0.90(3.87)\\ &=3.48 \end{aligned} \]

Since \(3.48>2.50\), the reverse reaction remains favoured.

If venting happens after the system reaches equilibrium at \(288\text{ K}\):

\[ \begin{aligned} Q_{\text{after}} &=0.90(2.50)\\ &=2.25 \end{aligned} \]

Now \(2.25<2.50\), so the forward reaction is favoured.

The boundary occurs when:

\[ 0.90Q_{\text{before}}=2.50 \]

Therefore:

\[ Q_{\text{before}} =\frac{2.50}{0.90} =2.78 \]

So:

• If \(Q_{\text{before}}>2.78\), a 10% vent initially leaves the reverse reaction favoured.
• If \(Q_{\text{before}}<2.78\), the forward reaction becomes favoured.
• If \(Q_{\text{before}}=2.78\), the vent leaves the system at equilibrium at that instant.

This is the surprising result. The same intervention can produce opposite net reaction directions depending on how far the system has already progressed towards equilibrium.

(d) Explain the heat transfer.

Immediately after cooling, the net reaction is reverse.

Because the forward reaction is endothermic, the reverse reaction is exothermic and releases heat to the surroundings.

The thermostat removes this released heat as necessary to maintain \(288\text{ K}\).

After removing 50% of the gaseous product, the forward reaction becomes thermodynamically favoured.

The forward reaction absorbs heat.

However, the thermometer does not necessarily show a further temperature drop.

The vessel is maintained at \(288\text{ K}\) by a thermostat. As the endothermic reaction proceeds, the thermostat supplies the required heat, keeping the temperature approximately constant.

That distinction is the conceptual centre of the problem:

An endothermic reaction absorbs heat, but the reaction mixture does not have to become colder if its surroundings continuously replace the absorbed energy.

The actual direction of a reversible reaction must be determined from its thermodynamic state, not simply from whether its forward reaction absorbs or releases heat.

07What to take into the next topic

When you encounter an endothermic reaction at room temperature, don’t use its positive enthalpy change to predict that nothing will happen.

Separate the problem into three decisions.

First, use \(\Delta H\) to establish the direction of heat transfer.

Second, use the activation energy and reaction conditions to reason about how readily particles can react.

Third, use Gibbs free energy, or equilibrium information when appropriate, to decide which direction is thermodynamically favoured.

The next useful question is what happens when the temperature changes while a reversible reaction is already at equilibrium. Heating usually favours the endothermic direction, but that affects the equilibrium position as well as the reaction rates.

That distinction is developed further in How Temperature Changes \(K_{\mathrm{eq}}\) in Endothermic and Exothermic Reactions.