How to Predict the Sign of Entropy Change from a Reaction Equation

A gas syringe attached to a flask of marble chips and hydrochloric acid begins to fill with carbon dioxide. Before looking at the syringe, could you predict whether the reacting system’s entropy increases or decreases?

Yes. The reaction produces a gas from reactants that contain no gaseous substances:

\[ \ce{CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)} \]

The formation of \(\ce{CO2(g)}\) gives particles much greater freedom to move and strongly indicates a positive entropy change, \(\Delta S > 0\).

To predict the sign of entropy change from a balanced equation, count the moles of gas on each side first. More gas usually means increasing entropy; less gas usually means decreasing entropy. Then check for changes of state and effects involving dissolved particles. If these clues compete, you may need standard molar entropy data.

Syllabus note (November 2026): Predicting entropy changes from balanced chemical equations is explicitly included in Year 11 Module 4, Drivers of Reactions, of the NSW Chemistry Stage 6 Syllabus (2017). That syllabus remains in use in 2026 and 2027. The Chemistry 11-12 Syllabus (2025) begins implementation in Year 11 in 2028, with its first HSC examination in 2029. References to prescribed Module 4 content in this article apply specifically to the 2017 syllabus, not automatically to the new course.

01Why gaseous particles are the strongest clue

Imagine releasing a collection of ping-pong balls from a small container into an entire sports hall.

The number of balls hasn’t changed, but there are far more ways to arrange them throughout the larger space.

Gas particles behave somewhat like this. They can move throughout their available volume, giving them many possible positions and ways of distributing energy.

A solid is different. Its particles generally occupy relatively restricted positions, although they still vibrate.

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

The analogy has a limit: molecules don’t just change position. They rotate, vibrate, interact, and exchange energy. Entropy includes these possibilities, so counting positions alone isn’t an exact calculation.

For a reaction, the entropy change is:

\[ \Delta S=S_{\mathrm{products}}-S_{\mathrm{reactants}} \]

Here, \(\Delta S\) is the change in entropy of the reacting system.

• \(\Delta S>0\): the system’s entropy increases.
• \(\Delta S<0\): the system’s entropy decreases.
• \(\Delta S=0\): there is no net entropy change.

We usually make these predictions at a specified temperature, often under standard conditions.

A reliable method for predicting the sign

Start with the balanced equation, including its state symbols.

Step 1: Identify the gaseous substances.

Look for \((g)\). Don’t count solids, liquids, or aqueous substances as gases.

Step 2: Add the coefficients of the gaseous substances.

For example, in:

\[ \ce{2SO2(g) + O2(g) -> 2SO3(g)} \]

there are three stoichiometric moles of gas on the reactant side and two on the product side.

Step 3: Calculate the change in gaseous moles.

\[ \Delta n_{\mathrm{gas}} =n_{\mathrm{gas,products}} -n_{\mathrm{gas,reactants}} \]

The symbol \(n_{\mathrm{gas}}\) represents the sum of the gaseous coefficients for the reaction as written. It is not the volume of gas measured in an experiment.

For sulfur dioxide reacting with oxygen:

\[ \Delta n_{\mathrm{gas}}=2-(2+1)=-1 \]

Gas particles are being combined into fewer gas molecules, so we predict \(\Delta S<0\).

Step 4: Check what else changes.

A substantial gas change usually provides a strong prediction. But if gaseous amounts are unchanged, or there are significant competing effects involving dissolved ions and phase changes, you may need entropy data.

These are physical clues, not mathematical guarantees.

Also, you must balance the equation before counting gaseous coefficients. If that’s the difficult part, revisit how to predict products and balance chemical equations.

02What a gas-collection experiment can actually tell us

Return to calcium carbonate reacting with hydrochloric acid.

We want to test the prediction that gas forms. More importantly, we want to understand what an observation can establish and what it cannot.

The first design: collecting gas over water

Picture a conical flask sitting on the bench.

Calcium carbonate chips rest at the bottom. Dilute hydrochloric acid reacts with their surfaces, producing carbon dioxide.

A rubber bung closes the flask. A delivery tube passes through the bung and leads into a water trough.

At the other end, the tube opens beneath an inverted measuring cylinder that was initially filled with water.

As carbon dioxide enters the cylinder, it displaces water. The volume of collected gas can be estimated from the cylinder markings.

What would you predict before the reaction starts?

If carbon dioxide is produced and collected successfully, bubbles should enter the measuring cylinder, and the water level inside it should fall.

That observation would support the prediction that the reaction generates a gas.

But there is a problem with this design.

Carbon dioxide dissolves in water. Some of the gas may enter the surrounding water instead of reaching the top of the measuring cylinder.

Gas can also escape while the acid is being added and the flask is being sealed.

Consequently, the measured volume could be lower than the amount predicted by the balanced equation.

The revised design: using a gas syringe

A more suitable collection method connects the reaction flask directly to an airtight gas syringe.

The revised arrangement uses a sealed flask with a way to introduce the acid after sealing, such as an appropriate stoppered addition system. A delivery tube leads from the flask to the gas syringe.

As carbon dioxide forms, it travels into the syringe and pushes the plunger outwards.

This removes the need to collect the gas over water, reducing one important source of carbon dioxide loss.

The temperature and pressure should also be recorded because they affect the gas volume.

Even the improved apparatus has limitations. Carbon dioxide can still remain dissolved in the reaction liquid, gas may leak, and heating can expand the gas already present.

The important distinction is that the apparatus measures gas production, not entropy directly.

To quantify reaction entropy, we use thermodynamic data. The gas-collection experiment gives evidence that helps us understand why our initial entropy prediction is reasonable.

03When equal gaseous moles do not mean zero entropy change

Suppose you count the gaseous coefficients and find exactly the same total on both sides.

A tempting conclusion is:

“If the number of gas molecules doesn’t change, entropy must stay constant.”

That’s not necessarily true.

Different gas molecules have different structures, masses, rotational motions, and vibrational motions. Their accessible microscopic energy arrangements aren’t identical.

So even when gas amounts remain equal, the entropy can change.

04Why dissolved ions can overturn your prediction

Gas formation is usually the strongest clue, but aqueous reactions introduce another effect.

Imagine dissolving an ionic solid.

The ions leave their crystal lattice and move into the surrounding water. That gives the ions more freedom to move.

You might immediately predict an increase in entropy.

But water molecules interact strongly with dissolved ions.

Around a positively charged ion, nearby water molecules tend to orient with their partially negative oxygen ends facing the ion. This is part of the formation of a hydration shell.

The dissolved ions become more dispersed, while nearby water molecules experience constraints on their orientations.

Those two effects compete.

This is why you shouldn’t use “more dissolved particles means positive entropy” as an absolute rule.

The reliable hierarchy is:

1. Use the balanced equation and state symbols to identify major changes, especially gases.
2. Make a qualitative prediction only when one effect clearly dominates.
3. When effects compete, calculate \(\Delta_{\mathrm r}S^\circ\) from supplied data.

There is one more distinction worth keeping clear.

A negative entropy change does not automatically mean a reaction is thermodynamically unfavourable. Enthalpy and entropy both contribute to feasibility.

Under the 2017 syllabus, their relationship is expressed by:

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

Here, \(\Delta G\) is Gibbs free energy change, \(\Delta H\) is enthalpy change, and \(T\) is absolute temperature in kelvin.

At constant temperature and pressure, a negative \(\Delta G\) indicates a thermodynamically favourable direction under the stated conditions. It does not guarantee a fast reaction.

This broader distinction is explained in Entropy and Enthalpy Changes: HSC Chemistry Explained.

05Questions and solutions

The first question checks the basic gas rule. The remaining questions test what happens when measurements, state symbols, ionic hydration, and competing processes make that rule less reliable.

Question 1

Consider the electrolysis of liquid water:

\[ \ce{2H2O(l) -> 2H2(g) + O2(g)} \]

Predict the sign of the system’s entropy change and explain your reasoning using the balanced equation.

Solution 1

The entropy change is predicted to be positive, \(\Delta S>0\), because liquid water is converted into gaseous hydrogen and oxygen.

The reactants contain no gaseous substances.

The products contain two stoichiometric moles of hydrogen gas and one of oxygen gas.

Therefore:

\[ \Delta n_{\mathrm{gas}}=(2+1)-0=+3 \]

The products have much greater freedom of movement and access to more microscopic arrangements.

The electrical energy required for electrolysis does not contradict this positive entropy prediction. Energy requirements and entropy changes are different properties, and a positive \(\Delta S\) alone does not establish spontaneity.

Question 2

A sealed cylinder with a freely moving piston initially contains 40.0 mL of air at 298 K and 100 kPa.

Inside the cylinder, an aqueous hydrogen peroxide sample decomposes completely:

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

After the reaction, the gas temperature has increased to 323 K, and the piston indicates a total gas volume of 56.0 mL. The pressure remains at 100 kPa.

Assume ideal gas behaviour, no leaks, no air consumption, and negligible water vapour and gas dissolution.

A student subtracts \(56.0-40.0\) and concludes that the reaction produced 16.0 mL of oxygen.

Determine whether the student is correct, calculate the moles of oxygen produced, and predict the sign of the reaction entropy change.

Solution 2

The student has overestimated the oxygen volume because some of the expansion is caused by heating. Approximately \(4.71\times10^{-4}\) mol of oxygen forms, and the reaction entropy is predicted to be positive.

Step 1: Correct the final gas volume to the initial temperature.

Because pressure is constant, the ideal gas law gives \(V/T\) proportional to the amount of gas.

Correcting the final volume from 323 K to 298 K:

\[ \begin{aligned} V_{f,298}&=56.0\left(\frac{298}{323}\right)\\ &=51.7\ \mathrm{mL} \end{aligned} \]

Here, \(V_{f,298}\) means the final amount of gas expressed as the volume it would occupy at 298 K and the same pressure.

Step 2: Find the additional gas volume at 298 K.

\[ \begin{aligned} V_{\mathrm{added}} &=51.7-40.0\\ &=11.7\ \mathrm{mL} \end{aligned} \]

The actual oxygen production corresponds to approximately 11.7 mL at the initial temperature and pressure, not 16.0 mL.

Step 3: Calculate the oxygen amount.

Convert 11.7 mL to 0.0117 L and use \(pV=nRT\).

\[ \begin{aligned} n&=\frac{pV}{RT}\\ &=\frac{100(0.0117)}{8.314(298)}\\ &\approx4.71\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Step 4: Predict the entropy sign independently of the volume change.

The balanced equation produces one mole of oxygen gas per mole of reaction, from substances initially described as aqueous and liquid.

\[ \Delta n_{\mathrm{gas}}=1-0=+1 \]

So we predict \(\Delta S>0\).

The key distinction is that the observed volume change includes thermal expansion. The standard reaction entropy concerns the chemical transformation at specified conditions, not every simultaneous change occurring inside the cylinder.

Question 3

Two ionic solids dissolve in separate samples of water:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

\[ \ce{CaCl2(s) -> Ca^2+(aq) + 2Cl-(aq)} \]

A student argues that calcium chloride must have the larger positive entropy change because one formula unit produces three dissolved ions rather than two.

Use the following rounded standard molar entropy values at 298 K to assess the argument.

Calculate the entropy change for each dissolution and explain why the number of dissolved ions does not determine the sign.

Solution 3

Sodium chloride dissolution has a positive standard entropy change of \(+43.4\ \mathrm{J\,mol^{-1}\,K^{-1}}\), while calcium chloride dissolution has a negative change of \(-44.7\ \mathrm{J\,mol^{-1}\,K^{-1}}\).

The student’s argument is therefore incorrect.

Step 1: Calculate the sodium chloride entropy change.

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=[59.0+56.5]-72.1\\ &=+43.4\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

For sodium chloride, the net entropy increases.

Step 2: Calculate the calcium chloride entropy change.

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=[-53.1+2(56.5)]-104.6\\ &=59.9-104.6\\ &=-44.7\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

For calcium chloride, the net entropy decreases under the standard-state calculation.

Step 3: Explain the difference.

Both dissolutions free ions from a crystal lattice, increasing the ions’ possible positions.

However, the surrounding water molecules interact with the dissolved ions. Calcium ions have a \(2+\) charge and can strongly constrain nearby water molecules through hydration.

For calcium chloride, the net effect represented by the tabulated thermodynamic values is a decrease in entropy.

The negative tabulated value for \(\ce{Ca^2+(aq)}\) is not a negative absolute entropy. Individual aqueous-ion entropies use a reference convention in which \(S^\circ(\ce{H+(aq)})=0\).

The result does not mean calcium chloride cannot dissolve. Whether a process is thermodynamically favourable depends on Gibbs free energy, not the sign of the system’s entropy change alone.

Question 4

Two reactions both produce a white silver chloride precipitate.

In Reaction A, silver ions react with chloride ions that are already dissolved:

\[ \ce{Ag+(aq) + Cl-(aq) -> AgCl(s)} \]

In Reaction B, solid sodium chloride is added to a solution containing silver ions:

\[ \ce{NaCl(s) + Ag+(aq) -> AgCl(s) + Na+(aq)} \]

A student predicts that both reactions must decrease entropy because both form the same solid precipitate.

Use the standard molar entropy values below, at 298 K, to determine the sign of each reaction.

Explain how Reaction B can be understood as two linked processes, and why observing the same precipitate does not settle the entropy question.

Solution 4

Reaction A decreases standard reaction entropy by \(32.9\ \mathrm{J\,mol^{-1}\,K^{-1}}\), while Reaction B increases it by \(10.5\ \mathrm{J\,mol^{-1}\,K^{-1}}\).

Although both form silver chloride, they start from different physical states.

Step 1: Calculate Reaction A.

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=96.3-(72.7+56.5)\\ &=-32.9\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

Two separate aqueous ions combine into a solid. The negative result agrees with the qualitative prediction that forming a solid from dissolved ions can reduce entropy.

Step 2: Calculate Reaction B.

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=(96.3+59.0)\\ &\quad -(72.1+72.7)\\ &=+10.5\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

This time, the overall entropy increases.

Step 3: Identify the extra process.

Reaction B includes the dissolution of sodium chloride:

\[ \ce{NaCl(s) -> Na+(aq) + Cl-(aq)} \]

Its standard entropy change is:

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=(59.0+56.5)-72.1\\ &=+43.4\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

The chloride ions then react with silver ions to form silver chloride.

Combining the two steps:

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=43.4+(-32.9)\\ &=+10.5\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

The chloride ions produced during dissolution are consumed during precipitation, so they cancel from the overall chemical equation.

This illustrates an important property of entropy: it is a state function. The overall entropy change depends on the initial and final states, not on the route used to describe the transformation.

The student’s mistake was focusing only on the visible precipitation and ignoring the simultaneous dissolution of the starting salt.

Question 5

Two idealised reactions produce hydrogen gas from different metals in excess aqueous acid.

Reaction M uses manganese:

\[ \ce{Mn(s) + 2H+(aq) -> Mn^2+(aq) + H2(g)} \]

Reaction Z uses zinc:

\[ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} \]

Suppose separate gas syringes each collect 24.8 mL of hydrogen at 298 K and 100 kPa.

A student claims:

“Both reactions produce exactly the same amount of gas from one mole of metal, so their entropy changes must have the same positive sign. If either has a negative entropy change, it cannot be thermodynamically favourable.”

Use the following rounded standard thermodynamic data.

The zero for aqueous hydrogen ions is a thermodynamic reference convention.

The standard enthalpy changes are:

• Reaction M: \(\Delta_{\mathrm r}H^\circ=-220.8\ \mathrm{kJ\,mol^{-1}}\)
• Reaction Z: \(\Delta_{\mathrm r}H^\circ=-153.9\ \mathrm{kJ\,mol^{-1}}\)

Assume 298 K, ideal gas behaviour, and standard thermodynamic states for the reaction calculations.

Determine:

(a) The standard reaction entropy change for each reaction.

(b) The amount of hydrogen in each gas syringe, and the corresponding entropy-change contribution for that reaction extent using the standard-state approximation.

(c) Whether each reaction is thermodynamically favourable under standard conditions at 298 K.

(d) Why two reactions producing the same gas volume can have opposite entropy signs, and whether the thermodynamic calculations establish which reaction is faster.

Solution 5

Reaction M has a positive standard entropy change of \(+25.1\ \mathrm{J\,mol^{-1}\,K^{-1}}\), whereas Reaction Z has a negative change of \(-23.0\ \mathrm{J\,mol^{-1}\,K^{-1}}\).

Both reactions are thermodynamically favourable under the stated standard conditions at 298 K.

The student’s claim fails because the entropy of the hydrogen gas is only one part of each reaction’s total entropy change. The dissolved metal ions and their interactions with water also matter.

Part (a): Calculate the standard entropy changes.

For manganese:

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=[-73.6+130.7]\\ &\quad -[32.0+2(0.0)]\\ &=+25.1\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

For zinc:

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=[-112.1+130.7]\\ &\quad -[41.6+2(0.0)]\\ &=-23.0\ \mathrm{J\,mol^{-1}\,K^{-1}} \end{aligned} \]

Both reactions have:

\[ \Delta n_{\mathrm{gas}}=1-0=+1 \]

But their standard entropy changes have opposite signs.

This is a direct counterexample to treating increased gaseous moles as a guaranteed positive entropy change.

Part (b): Relate the reaction entropy to the gas collected.

Use:

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

Converting 24.8 mL to 0.0248 L:

\[ \begin{aligned} n&=\frac{100(0.0248)}{8.314(298)}\\ &=1.00\times10^{-3}\ \mathrm{mol} \end{aligned} \]

Each syringe therefore contains approximately 0.00100 mol of hydrogen.

Both balanced equations produce one mole of hydrogen per mole of reaction, so each experiment corresponds to approximately 0.00100 mol of reaction.

Scaling the standard entropy changes gives:

For manganese:

\[ \begin{aligned} \Delta S_{\mathrm M} &=(0.00100)(25.1)\\ &=+0.0251\ \mathrm{J\,K^{-1}} \end{aligned} \]

For zinc:

\[ \begin{aligned} \Delta S_{\mathrm Z} &=(0.00100)(-23.0)\\ &=-0.0230\ \mathrm{J\,K^{-1}} \end{aligned} \]

These are standard-state scaling estimates for the reaction contributions. They aren’t necessarily the exact total entropy changes in real gas-syringe experiments, where concentrations, gas pressures, and temperature changes may differ from standard conditions.

Part (c): Determine thermodynamic favourability.

Use the Gibbs free energy equation:

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

Entropy must first be converted from joules to kilojoules so its units match those of enthalpy.

For manganese:

\[ \Delta_{\mathrm r}S^\circ =+0.0251\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Therefore:

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=-220.8-298(0.0251)\\ &=-228.3\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

For zinc:

\[ \Delta_{\mathrm r}S^\circ =-0.0230\ \mathrm{kJ\,mol^{-1}\,K^{-1}} \]

Therefore:

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=-153.9-298(-0.0230)\\ &=-147.0\ \mathrm{kJ\,mol^{-1}} \end{aligned} \]

Both standard Gibbs free energy changes are negative.

Therefore, both forward reactions are thermodynamically favourable under standard conditions at 298 K.

Zinc’s negative entropy change does not prevent its reaction from being favourable because the negative enthalpy change is sufficiently large.

Part (d): Explain the apparent contradiction.

Both reactions form one mole of hydrogen gas per mole of metal.

If you looked only at hydrogen gas formation, you would predict increasing entropy in both cases.

But the reactions also convert solid metals into hydrated aqueous ions.

The standard entropy contribution associated with producing \(\ce{Zn^2+(aq)}\) is different from that associated with producing \(\ce{Mn^2+(aq)}\).

Hydration and the associated constraints on surrounding water molecules are part of the full thermodynamic description.

For manganese, the total standard entropy change remains positive.

For zinc, the competing changes are sufficiently large to make the total standard entropy change negative.

The negative tabulated entropy of a single aqueous ion must not be interpreted as a negative absolute physical entropy. Single-ion values are expressed using the conventional zero assigned to \(\ce{H+(aq)}\). The entropy change for the complete charge-balanced reaction is physically meaningful.

Finally, neither Gibbs free energy result tells us which reaction is faster.

Reaction rates depend on kinetic factors such as activation energy, surface condition, and reactant accessibility. A metal surface covered by an oxide layer may react slowly even when the overall reaction is thermodynamically favourable.

The deeper lesson is that a chemical equation gives you a powerful first prediction, but not an infallible entropy measurement.

Gas production often dominates the result, yet the complete entropy change accounts for every reactant and product, including dissolved species.

06What the entropy sign lets you decide next

For an exam question, first identify whether a clear gas or phase change makes the sign predictable. If the evidence is mixed, stop guessing and use standard molar entropy values when they are provided.

Once you’ve established the sign of \(\Delta S\), the next decision is whether that entropy change helps or opposes thermodynamic feasibility. That requires combining it with \(\Delta H\) and temperature through Gibbs free energy, while keeping reaction rate as a separate question.