Entropy and Free Energy | A-Level Chemistry

Last updated: 17 August 2026

Diagnose → Learn → Master: Every study-note section includes 15 free MCQs — start with 5 Diagnostic MCQs to identify gaps in your understanding, then complete 10 Mastery MCQs near the end to check what you have learned.

Key Takeaway

Table of Contents

Entropy and Gibbs free energy help us predict whether a change is thermodynamically feasible

Enthalpy tells us whether heat is absorbed or released, but enthalpy alone cannot determine whether a process is thermodynamically feasible. We must also consider entropy, which describes how widely matter and energy can be distributed within a system.

Entropy, S, increases as matter and energy become more widely dispersed.
For the same substance, entropy generally increases in the order solid < liquid < gas.
An increase in the number of moles of gaseous particles usually gives ΔS > 0.
Gibbs free energy combines enthalpy and entropy: ΔG = ΔH − TΔS.
If ΔG < 0, the forward process is thermodynamically feasible under the stated conditions.
If ΔG > 0, the forward process is not thermodynamically feasible under those conditions.
If ΔG = 0, the system is at equilibrium.
Temperature can change the sign of ΔG because it changes the magnitude of the TΔS term.
A negative ΔG does not mean that a reaction is fast; reaction rate is a kinetic question.

Some changes seem to have a natural direction. A gas released into a container spreads throughout the available space. Ice melts when it is sufficiently warm. Two gases mix together and do not normally separate themselves again. Yet not every energetically possible reaction proceeds readily, and simply knowing whether a reaction is exothermic is not enough to predict what will happen.

To understand this, we need a second thermodynamic quantity: entropy. Entropy describes how matter and energy are distributed in a system. When entropy is combined with enthalpy through Gibbs free energy, we can determine whether a process is thermodynamically feasible under particular conditions.

This also explains an important feature of Chemistry: the same reaction can be thermodynamically feasible at one temperature but not at another. The key is to understand the competition between ΔH and TΔS, rather than memorising isolated rules.

Diagnostic Quiz

What do you already know about entropy and free energy?

Try these five questions before reading the chapter. They check the particle ideas, signs and equations that will form the foundation of this topic.

5 short questions
Instant explanation for every option
No marks are lost — this is a starting-point check

Question 1 of 5

For the same substance at the same temperature and pressure where comparison is possible, which physical state generally has the highest entropy?

Solid

In a solid, particles are strongly constrained around fixed positions, so there are relatively few possible spatial arrangements.

Liquid

A liquid has greater entropy than the corresponding solid because its particles can move past one another, but its particles are still much less dispersed than in a gas.

Gas

Correct. Gas particles can move throughout the available volume and can be arranged in far more ways than particles in a liquid or solid.

All three states must have the same entropy

The freedom of movement and number of possible arrangements are very different in the three physical states.

For the same substance, the usual order is S(g) > S(l) > S(s). Gas particles have the greatest freedom of movement and the greatest number of possible arrangements.

What is the most likely sign of ΔS for N₂O₄(g) → 2NO₂(g)?

N₂O₄(g) → 2NO₂(g)

Positive

Correct. One mole of gaseous particles becomes two moles of gaseous particles, so the number of possible arrangements increases.

Negative

A negative entropy change would be expected if the gaseous particles became less numerous or more constrained. The opposite occurs here.

Zero

The number of gaseous particles changes substantially, so there is no reason to expect the entropy change to be zero.

The sign cannot be predicted because all species are gases

The fact that all species are gases does not prevent a prediction. The number of moles of gas increases from one to two.

An increase in the number of moles of gaseous particles is a strong indication that entropy increases. Hence ΔS is positive.

Which equation relates Gibbs free energy change, enthalpy change and entropy change?

ΔG = ΔH + TΔS

The entropy term is subtracted, not added.

ΔG = ΔH − TΔS

Correct. Gibbs free energy combines the enthalpy contribution with the temperature-dependent entropy contribution.

ΔG = TΔH − ΔS

Temperature multiplies ΔS, not ΔH.

ΔG = ΔS − TΔH

This does not have the correct thermodynamic relationship or consistent units.

The central equation for this topic is ΔG = ΔH − TΔS. The minus sign before TΔS is essential when deciding how entropy affects feasibility.

A reaction has ΔG < 0 under a particular set of conditions. What can you conclude?

The reaction must be extremely fast

Reaction rate depends on kinetics and activation energy. ΔG does not tell us how rapidly a reaction occurs.

The forward reaction is thermodynamically feasible under those conditions

Correct. A negative Gibbs free energy change indicates thermodynamic feasibility in the forward direction under the stated conditions.

The reaction must be exothermic

Not necessarily. An endothermic reaction can have ΔG < 0 if a favourable entropy term is large enough.

The activation energy must be zero

Thermodynamic feasibility does not remove the activation-energy barrier.

ΔG answers a thermodynamic question: whether the forward process is feasible. It does not answer the kinetic question of how quickly the process occurs.

Which temperature should be substituted into ΔG = ΔH − TΔS for a reaction carried out at 25 °C?

25 K

A Celsius value cannot simply be relabelled as a kelvin value.

273 K

273 K is approximately 0 °C, not 25 °C.

298 K

Correct. 25 °C + 273 = 298 K.

325 K

This is not the correct Celsius-to-kelvin conversion.

Gibbs free energy calculations require absolute temperature in kelvin: T/K = temperature/°C + 273.

What is entropy?

Entropy, S, describes the extent to which matter and energy are dispersed among the possible arrangements available to a system. At this level, entropy is often introduced as a measure of disorder or randomness.

The word “disorder” can be useful, but it should not be interpreted as ordinary untidiness. A better way to think about entropy is to ask: how many different ways can the particles and their energy be arranged?

If matter and energy can be distributed in many different ways, the system has relatively high entropy. If the particles are strongly restricted and only a smaller number of arrangements are available, the entropy is relatively low.

Think about dispersal, not simply “messiness”.

A higher-entropy state allows matter and energy to be distributed among more possible arrangements. This particle-level interpretation is more useful in Chemistry than imagining whether something looks visually tidy or untidy.

Entropy and the three physical states

The relationship between entropy and particle freedom becomes particularly clear when we compare the three physical states of the same substance.

Physical state	Particle behaviour	Relative entropy
Solid	Particles are held close together and vibrate about relatively fixed positions.	Lowest of the three states
Liquid	Particles remain close together but can move past one another.	Higher than the solid
Gas	Particles move freely throughout the available volume.	Highest of the three states

For the same substance, we therefore use the general order:

S(g) > S(l) > S(s)

Melting increases entropy because the particles become less restricted. Vaporisation causes an even larger increase because the particles become widely separated and can move throughout the entire available volume.

Factors that affect entropy

To predict whether entropy increases or decreases, it is useful to look for changes in temperature, physical state and, especially in chemical equations, the number of moles of gas.

Increasing temperature

When a substance is heated, energy is transferred to its particles. Depending on the state of the substance, its particles can vibrate, rotate or move more energetically.

There are then more ways in which the available energy can be distributed among the particles. Therefore, for a particular substance in a particular physical state:

increasing temperature → increasing entropy

Changing physical state

A change of physical state changes how restricted the particles are. This usually gives a clear prediction for the sign of the entropy change.

Change	Particle-level effect	Sign of ΔS
Solid → liquid	Particles become more mobile.	Positive
Liquid → gas	Particles become much more widely dispersed.	Positive
Gas → liquid	Particles become less widely dispersed.	Negative
Liquid → solid	Particles become more restricted in position.	Negative

For example:

H₂O(s) → H₂O(l)

The particles in liquid water have greater freedom of movement than those in solid ice, so:

ΔS > 0

The number of moles of gas

In examination questions, the number of gaseous particles is often the most useful clue. Gas particles have much greater freedom of movement than particles in liquids or solids, so changes in the amount of gas often have a large effect on entropy.

Consider:

N₂O₄(g) → 2NO₂(g)

One mole of gaseous particles becomes two moles of gaseous particles. There are now more independently moving particles and many more possible arrangements. Hence:

ΔS > 0

Now consider:

2SO₂(g) + O₂(g) → 2SO₃(g)

The reactants contain a total of three moles of gas, while the products contain two moles of gas. The number of gaseous particles decreases, so the system becomes less dispersed.

ΔS < 0

For reactions involving gases, count the gaseous particles first.

An increase in the number of moles of gas usually indicates an increase in entropy, while a decrease in the number of moles of gas usually indicates a decrease in entropy.

What if the number of moles of gas is unchanged?

The gas-mole rule is useful, but it should not be turned into an absolute rule. Consider:

H₂(g) + Cl₂(g) → 2HCl(g)

There are two moles of gaseous particles on each side. Counting gas moles therefore gives no clear prediction about the sign of ΔS.

This does not mean that ΔS must be zero. It means only that the simple gas-counting argument cannot decide the sign. Other differences between reactants and products may then need to be considered.

What happens when a solid dissolves?

When many solids dissolve, their particles become dispersed throughout the solution. For example, ions from an ionic solid can separate and spread through the solvent.

This often leads to an increase in entropy because the particles become more dispersed. However, it is safer to regard this as a general tendency rather than an absolute rule, because the interactions between solute particles and solvent molecules can also affect the overall entropy change.

Oversimplified idea: More particles always means higher entropy.

Better reasoning: Consider how widely the particles and energy are dispersed and how much freedom of movement the particles have. Changes in gaseous particles are particularly important.

Predicting the sign of ΔS

The change in entropy of a system is represented by ΔS.

ΔS = S_final − S_initial

For a chemical reaction, we compare the entropy of the products with that of the reactants.

Sign of ΔS	Meaning	Typical situation
ΔS > 0	The products have higher entropy than the reactants.	More gas is formed, melting or vaporisation occurs.
ΔS < 0	The products have lower entropy than the reactants.	Gas moles decrease, condensation or freezing occurs.

A reliable method for predicting ΔS

Identify all physical states. Look for changes between solid, liquid and gas.
Count the moles of gas. Compare the total amount of gaseous reactants with the total amount of gaseous products.
Explain the particle change. State whether matter becomes more or less dispersed and therefore whether entropy increases or decreases.

Example 1: a gas is produced from solids

Consider:

CaCO₃(s) → CaO(s) + CO₂(g)

The reactant is entirely solid, whereas a gaseous product is formed. The CO₂ molecules have much greater freedom of movement than particles held within a solid.

Prediction: ΔS is positive because gas is produced and matter becomes more widely dispersed.

Example 2: the amount of gas decreases

Consider:

N₂(g) + 3H₂(g) → 2NH₃(g)

There are four moles of gaseous particles on the reactant side but only two moles on the product side.

Prediction: ΔS is negative because the number of moles of gaseous particles decreases from four to two.

Weak explanation: “The reaction becomes more ordered.”

Stronger explanation: “The number of moles of gaseous particles decreases, so matter is less widely dispersed and fewer arrangements are available. Hence ΔS is negative.”

Gibbs free energy

Entropy alone does not determine whether a reaction is thermodynamically feasible. A reaction may increase entropy but require a large input of heat. Another reaction may release heat but produce a lower-entropy state.

The contributions from enthalpy and entropy are combined using Gibbs free energy:

ΔG = ΔH − TΔS

where:

ΔG is the Gibbs free energy change;
ΔH is the enthalpy change;
T is the absolute temperature in kelvin; and
ΔS is the entropy change.

What does the sign of ΔG mean?

Value of ΔG	Meaning	Forward process
ΔG < 0	The Gibbs free energy decreases.	Thermodynamically feasible
ΔG > 0	The Gibbs free energy would increase.	Not thermodynamically feasible under those conditions
ΔG = 0	There is no net thermodynamic driving force.	The system is at equilibrium

A process with ΔG < 0 is sometimes described as spontaneous. This means that it is thermodynamically favourable in the forward direction under the specified conditions. It does not mean that the process must be rapid.

Thermodynamically feasible does not mean fast.

ΔG tells us about the thermodynamic direction of a process. Reaction rate depends on kinetic factors such as activation energy and reaction mechanism.

Using ΔG = ΔH − TΔS

Gibbs free energy calculations are straightforward once the signs and units are handled carefully.

Temperature must be in kelvin

The temperature in the Gibbs equation is the absolute temperature, so a Celsius temperature must first be converted to kelvin.

T / K = temperature / °C + 273

For example:

25 °C = 25 + 273 = 298 K

Make the energy units consistent

ΔH is commonly expressed in kJ mol⁻¹, whereas ΔS may be given in J K⁻¹ mol⁻¹. These units cannot be inserted directly into the same equation without conversion.

If ΔH is kept in kJ mol⁻¹, convert ΔS into kJ K⁻¹ mol⁻¹:

ΔS in kJ K⁻¹ mol⁻¹ = ΔS in J K⁻¹ mol⁻¹ ÷ 1000

Before substituting, check T and the units of ΔH and ΔS.

Use kelvin for temperature and make sure the energy unit in TΔS matches the energy unit used for ΔH.

Worked example: calculating ΔG

A reaction has:

ΔH = −95.0 kJ mol⁻¹;
ΔS = −120 J K⁻¹ mol⁻¹;
T = 298 K.

First convert the entropy change:

ΔS = −120 ÷ 1000 = −0.120 kJ K⁻¹ mol⁻¹

Then:

TΔS = 298(−0.120) = −35.76 kJ mol⁻¹

Substituting into the Gibbs equation:

ΔG = ΔH − TΔS

ΔG = −95.0 − (−35.76)

ΔG = −59.24 kJ mol⁻¹

Answer: ΔG = −59.2 kJ mol⁻¹ to 3 significant figures. Since ΔG is negative, the forward reaction is thermodynamically feasible under the stated conditions.

Temperature and thermodynamic feasibility

The equation

ΔG = ΔH − TΔS

shows why temperature can change whether a reaction is thermodynamically feasible. As T increases, the magnitude of the entropy term TΔS increases.

There are four combinations of signs for ΔH and ΔS.

ΔH	ΔS	Effect on ΔG	Thermodynamic feasibility
Negative	Positive	ΔH and −TΔS are both negative.	Feasible at all temperatures
Positive	Negative	ΔH and −TΔS are both positive.	Not feasible at any positive temperature
Negative	Negative	Negative ΔH competes with positive −TΔS.	Feasible at sufficiently low temperatures
Positive	Positive	Positive ΔH competes with negative −TΔS.	Feasible at sufficiently high temperatures

Case 1: ΔH < 0 and ΔS > 0

ΔH is negative, which helps to make ΔG negative. Because ΔS is positive, TΔS is positive and therefore −TΔS is also negative.

ΔG = negative − positive

Both contributions favour a negative ΔG. The reaction is therefore thermodynamically feasible at all temperatures, provided the signs of ΔH and ΔS remain the same.

Case 2: ΔH > 0 and ΔS < 0

ΔH is positive. Since ΔS is negative, TΔS is negative, so subtracting it gives another positive contribution:

ΔG = positive − negative = positive + positive

Both contributions make ΔG positive. The forward reaction is therefore not thermodynamically feasible at any positive temperature under this simplified treatment.

Case 3: ΔH < 0 and ΔS < 0

The negative ΔH favours feasibility, but the negative ΔS makes −TΔS positive. The two terms therefore oppose one another.

At low temperatures, the magnitude of TΔS is relatively small. The negative ΔH term can dominate and ΔG can remain negative.

As temperature increases, the positive −TΔS term becomes larger. At sufficiently high temperature, ΔG becomes positive.

ΔH < 0 and ΔS < 0: thermodynamically feasible at sufficiently low temperatures.

Case 4: ΔH > 0 and ΔS > 0

The positive ΔH opposes feasibility, but the positive ΔS makes −TΔS negative.

At low temperature, the favourable entropy term may be too small to overcome the positive ΔH. As temperature increases, however, −TΔS becomes increasingly negative.

At sufficiently high temperature, the entropy contribution can outweigh the positive enthalpy contribution and make ΔG negative.

ΔH > 0 and ΔS > 0: thermodynamically feasible at sufficiently high temperatures.

Temperature becomes decisive when the enthalpy and entropy contributions compete.

For ΔH < 0 and ΔS < 0, low temperature favours feasibility. For ΔH > 0 and ΔS > 0, high temperature favours feasibility.

ΔH < 0, ΔS < 0: increasing temperature works against feasibility because −TΔS becomes increasingly positive.

ΔH > 0, ΔS > 0: increasing temperature favours feasibility because −TΔS becomes increasingly negative.

Finding the temperature at which ΔG = 0

For a reaction whose feasibility changes with temperature, we can calculate the temperature at which the change occurs.

At the boundary between ΔG < 0 and ΔG > 0:

ΔG = 0

Therefore:

0 = ΔH − TΔS

TΔS = ΔH

T = ΔH ÷ ΔS

Worked example: finding the boundary temperature

Suppose:

ΔH = +42.0 kJ mol⁻¹;
ΔS = +120 J K⁻¹ mol⁻¹.

Convert ΔS:

ΔS = 0.120 kJ K⁻¹ mol⁻¹

Then:

T = 42.0 ÷ 0.120

T = 350 K

At 350 K, ΔG = 0.

Since both ΔH and ΔS are positive, increasing temperature makes −TΔS more negative. The reaction is therefore thermodynamically feasible when:

T > 350 K

The calculated temperature is the boundary. Always decide whether the reaction is feasible above or below this temperature rather than stopping immediately after calculating T.

Thermodynamics and kinetics are different

Gibbs free energy tells us whether a process is thermodynamically feasible. It does not tell us whether the process will occur at an observable rate.

For a reaction to occur, reacting particles usually need to overcome an activation-energy barrier. If this barrier is large, the reaction may be extremely slow even when ΔG is strongly negative.

Thermodynamics asks: Is the process thermodynamically feasible under these conditions?

Kinetics asks: How quickly does the process occur?

What does a catalyst change?

A catalyst provides an alternative reaction pathway with a lower activation energy. It can therefore make a slow reaction proceed more rapidly.

A catalyst does not change ΔG, ΔH or ΔS. It therefore does not change whether a reaction is thermodynamically feasible. It changes only the rate at which the system moves towards equilibrium.

A negative ΔG does not remove the activation-energy barrier.

Thermodynamic feasibility and reaction rate must always be treated as separate ideas. A reaction can be feasible but kinetically very slow.

Common mistakes with entropy and Gibbs free energy

Common mistake	Why it is wrong	Correct approach
Defining entropy only as “messiness”	This gives no particle-level explanation.	Relate entropy to dispersal of matter and energy and the number of possible arrangements.
Assuming more total particles must mean ΔS > 0	The physical states and freedom of the particles matter.	Look especially for changes in the number of moles of gas.
Assuming equal gas moles mean ΔS = 0	Gas counting then gives no clear prediction, but other entropy differences may still exist.	State that the sign cannot be determined from gas moles alone.
Assuming dissolution always has ΔS > 0	Solute–solvent interactions can also affect entropy.	Treat increased dispersal on dissolving as a general tendency rather than an absolute rule.
Using °C directly in ΔG = ΔH − TΔS	T must be an absolute temperature.	Convert the temperature to kelvin.
Combining ΔH in kJ mol⁻¹ with ΔS in J K⁻¹ mol⁻¹	The energy units are inconsistent.	Convert one quantity so both use the same energy unit.
Forgetting the minus sign in −TΔS	This reverses the effect of entropy on ΔG.	Write the complete Gibbs equation before substituting values.
Saying ΔG < 0 means the reaction must be fast	ΔG describes thermodynamic feasibility, not rate.	Consider activation energy and kinetics separately.
Saying a catalyst makes ΔG more negative	A catalyst changes activation energy, not the Gibbs free energy change.	State that the catalyst changes reaction rate without changing ΔG.
Calculating T = ΔH/ΔS and stopping	The calculated value is only the temperature at which ΔG = 0.	Use the signs of ΔH and ΔS to determine whether feasibility occurs above or below that temperature.

Entropy and free-energy checklist

Predict entropy

Identify the physical states.
Look for melting, vaporisation, condensation or freezing.
Count the moles of gaseous particles.
Explain whether matter and energy become more or less dispersed.

Use the Gibbs equation

Write ΔG = ΔH − TΔS.
Convert temperature to kelvin.
Make the energy units of ΔH and ΔS consistent.
Keep the minus sign before TΔS.

Interpret ΔG

ΔG < 0 means thermodynamically feasible.
ΔG > 0 means not feasible in the forward direction.
ΔG = 0 corresponds to equilibrium.
Use ΔH and ΔS together when considering temperature.

Keep kinetics separate

Do not use ΔG to predict reaction speed.
A large activation energy can make a feasible reaction very slow.
A catalyst lowers activation energy.
A catalyst does not change ΔG, ΔH or ΔS.

Mastery Quiz

Entropy and Free Energy Mastery Quiz

Now apply the chapter to reactions, calculations and unfamiliar situations. These questions focus on reasoning rather than simple definition recall.

10 application-focused questions
Feedback explains every answer choice
Use your score to identify what to revise

Question 1 of 10

What is the most likely sign of ΔS for the following reaction?

2SO₂(g) + O₂(g) → 2SO₃(g)

Positive, because three molecules form two molecules

A decrease in the number of gaseous particles normally decreases entropy, not increases it.

Negative, because the number of moles of gas decreases

Correct. The total gaseous amount decreases from three moles to two moles, so there are fewer independently moving gaseous particles.

Zero, because every species is a gas

Being in the same physical state does not mean the entropy change is zero. The number of gaseous particles changes.

Positive, because oxygen is consumed

Consuming oxygen does not itself determine the sign of ΔS. The important clue here is the decrease in total moles of gas.

Three moles of gaseous particles become two moles of gaseous particles. The system becomes less dispersed, so ΔS is expected to be negative.

Why is ΔS expected to be positive for CaCO₃(s) → CaO(s) + CO₂(g)?

CaCO₃(s) → CaO(s) + CO₂(g)

Because every decomposition reaction has positive ΔS

Decomposition alone does not guarantee a positive entropy change. The physical states and particle distribution must be considered.

Because CaO has a larger formula mass than CO₂

Formula mass is not the main reason for the predicted sign in this reaction.

Because the number of solid substances increases

The formation of an extra solid is much less important than the production of a gaseous product.

Because a gaseous product is formed from an initially solid reactant

Correct. The CO₂ molecules are much more widely dispersed and have much greater freedom of movement than particles in a solid.

Formation of gas from an initially solid system greatly increases the possible particle arrangements, so the entropy change is positive.

A reaction has ΔH = +25.0 kJ mol⁻¹ and ΔS = +80.0 J K⁻¹ mol⁻¹ at 298 K. What is the approximate value of ΔG?

ΔG = ΔH − TΔS

+1.16 kJ mol⁻¹

Correct. ΔS = 0.0800 kJ K⁻¹ mol⁻¹, so TΔS = 23.84 kJ mol⁻¹ and ΔG = 25.0 − 23.84 = +1.16 kJ mol⁻¹.

−1.16 kJ mol⁻¹

The magnitude is obtained from the correct calculation, but the sign is wrong. 25.0 − 23.84 is positive.

+48.8 kJ mol⁻¹

This effectively adds the entropy contribution rather than subtracting TΔS.

−23 815 kJ mol⁻¹

This results from mixing joules and kilojoules without first making the units consistent.

Convert ΔS to 0.0800 kJ K⁻¹ mol⁻¹ first. Then ΔG = 25.0 − 298(0.0800) = +1.16 kJ mol⁻¹. The positive value means that the forward reaction is not thermodynamically feasible at 298 K.

A reaction has ΔH = +42.0 kJ mol⁻¹ and ΔS = +120 J K⁻¹ mol⁻¹. At what temperature is ΔG = 0?

ΔG = ΔH − TΔS

0.350 K

This results from using inconsistent kJ and J units.

286 K

This is not obtained from ΔH/ΔS using consistent units.

350 K

Correct. ΔS = 0.120 kJ K⁻¹ mol⁻¹, so T = 42.0/0.120 = 350 K.

5040 K

ΔH and ΔS should be divided, not multiplied, when ΔG = 0.

At ΔG = 0, ΔH = TΔS, so T = ΔH/ΔS. After converting the units, T = 350 K.

A reaction has ΔH > 0 and ΔS > 0. Which statement best describes its temperature dependence?

It is feasible only at sufficiently low temperatures

Low temperature reduces the favourable magnitude of the −TΔS term, so this is the opposite of the correct trend.

It is feasible at all temperatures

The positive ΔH opposes feasibility, so the favourable entropy term must become sufficiently large before ΔG becomes negative.

It is feasible only at sufficiently high temperatures

Correct. Since ΔS is positive, increasing T makes −TΔS increasingly negative until it can overcome the positive ΔH.

It can never be feasible because ΔH is positive

A positive enthalpy change does not automatically prevent feasibility. A sufficiently favourable entropy contribution can make ΔG negative.

For ΔH > 0 and ΔS > 0, high temperature favours feasibility because the negative −TΔS term becomes increasingly important.

A reaction has ΔH < 0 and ΔS < 0. Which change is most likely to favour thermodynamic feasibility?

Increasing the temperature

Since ΔS is negative, −TΔS is positive. Increasing temperature makes this unfavourable positive term larger.

Decreasing the temperature

Correct. Lower temperature reduces the positive contribution from −TΔS, allowing the favourable negative ΔH term to dominate more easily.

Adding a catalyst to make ΔG more negative

A catalyst does not change ΔG. It changes activation energy and therefore reaction rate.

Changing the temperature cannot affect ΔG

Temperature appears directly in the TΔS term, so changing temperature can alter ΔG.

For ΔH < 0 and ΔS < 0, low temperatures favour feasibility because they minimise the unfavourable positive −TΔS contribution.

A thermodynamically feasible reaction is extremely slow at room temperature. A catalyst is added. Which statement is correct?

The catalyst makes ΔG more negative

A catalyst does not alter the Gibbs free energy difference between reactants and products.

The catalyst changes ΔS from negative to positive

A catalyst does not change the entropy difference between the initial and final thermodynamic states.

The catalyst changes an endothermic reaction into an exothermic one

A catalyst does not change ΔH.

The catalyst lowers the activation energy without changing ΔG

Correct. A catalyst changes the kinetic pathway, allowing the reaction to proceed faster without altering its thermodynamic feasibility.

Catalysts affect kinetics, not thermodynamics. They lower activation energy but do not change ΔG, ΔH or ΔS.

A reaction has ΔG = −150 kJ mol⁻¹ but is barely observable at room temperature. Which explanation is most appropriate?

The reaction may have a large activation-energy barrier

Correct. A negative ΔG indicates thermodynamic feasibility, but a high activation energy can still make the reaction kinetically very slow.

A negative ΔG proves that no reaction can occur

This reverses the meaning of the sign. A negative ΔG favours the forward process thermodynamically.

The reaction must have ΔH > 0

The sign of ΔG alone does not uniquely determine the sign of ΔH.

The entropy change must be zero

Slow reaction rate gives no such information about the entropy change.

Thermodynamic feasibility and kinetic rate are separate. ΔG may be strongly negative while a large activation energy prevents the reaction from occurring rapidly.

Consider H₂(g) + Cl₂(g) → 2HCl(g). What can be concluded about ΔS by counting moles of gas alone?

H₂(g) + Cl₂(g) → 2HCl(g)

ΔS must be positive

The number of moles of gas does not increase.

ΔS must be negative

The number of moles of gas does not decrease.

ΔS must equal zero

Equal gas moles do not prove that reactants and products have identical entropy.

The sign cannot be determined from the gas-mole count alone

Correct. There are two moles of gas on each side, so the gas-counting shortcut provides no clear direction.

The gas-mole rule is most useful when the number of moles of gas changes. If the gaseous amount is unchanged, other entropy information is required.

Which statement correctly describes a system at equilibrium?

ΔG must be negative because both forward and reverse reactions continue

At equilibrium there is no net thermodynamic driving force in either direction, so ΔG is not negative.

ΔG must be positive because products are still reacting

Continuing microscopic reaction does not imply a positive Gibbs free energy change.

ΔG = 0 because there is no net thermodynamic driving force

Correct. At equilibrium the system has no net tendency to change composition in either direction.

ΔH and ΔS must both equal zero

Equilibrium requires ΔG = 0 under the actual conditions. ΔH and ΔS do not individually have to be zero.

At equilibrium, the actual Gibbs free energy change for further net reaction is zero. This does not require either the enthalpy change or entropy change individually to be zero.

Frequently Asked Questions

Questions students often ask about entropy and free energy

These questions extend the main ideas and clear up some of the distinctions that commonly cause confusion when students first use entropy and Gibbs free energy.

Does higher entropy simply mean that a system is more messy?

“More disordered” can be a useful starting description, but entropy is not really about visual untidiness. A better interpretation is that matter and energy can be distributed among a greater number of possible arrangements in a higher-entropy state.

Why do gases usually have much higher entropy than solids?

Particles in a solid are restricted to positions within a lattice and mainly vibrate about those positions. Gas particles can move throughout the available volume, so there are enormously more possible ways to arrange them.

If the number of moles of gas is the same on both sides, is ΔS equal to zero?

Not necessarily. Equal gaseous amounts simply mean that the gas-mole shortcut cannot decide the sign. The reactants and products may still have different entropies for other reasons.

Does dissolving a solid always increase entropy?

Dissolving often increases dispersal because the solute particles become distributed throughout the solvent, so a positive entropy change is often expected.

However, interactions between the solute and solvent can also change the arrangement of solvent molecules. It is therefore better to regard this as a general tendency rather than an absolute rule.

Why can an endothermic reaction be thermodynamically feasible?

An endothermic reaction has ΔH > 0, which works against a negative ΔG. However, if ΔS > 0, the term −TΔS is negative. At sufficiently high temperature, this favourable entropy contribution can outweigh the positive enthalpy term and make ΔG negative.

Can an exothermic reaction be thermodynamically non-feasible?

Yes. If ΔH < 0 but ΔS < 0, the enthalpy and entropy contributions oppose one another. At sufficiently high temperature, the positive −TΔS term can become larger than the magnitude of the negative ΔH, making ΔG positive.

Why do we use kelvin rather than degrees Celsius in the Gibbs equation?

The entropy term depends on absolute temperature. The kelvin scale begins at absolute zero, so T must be expressed in kelvin when calculating TΔS.

Why do I sometimes have to divide ΔS by 1000?

ΔS is commonly supplied in J K⁻¹ mol⁻¹ while ΔH is commonly given in kJ mol⁻¹. Dividing ΔS by 1000 converts it to kJ K⁻¹ mol⁻¹, allowing the two terms in ΔG = ΔH − TΔS to use consistent energy units.

Does ΔG < 0 mean that a reaction starts immediately when the reactants are mixed?

No. A negative ΔG tells us that the reaction is thermodynamically feasible, but the reaction may still have a large activation-energy barrier. If very few particles can overcome that barrier, the reaction may occur extremely slowly.

Can a catalyst make a thermodynamically non-feasible reaction feasible?

No. A catalyst changes the reaction pathway and lowers activation energy, but it does not change ΔG, ΔH or ΔS. It can speed up a feasible reaction, but it does not change its thermodynamic driving force.

What exactly does the temperature from T = ΔH/ΔS represent?

It is the temperature at which ΔG = 0, assuming ΔH and ΔS remain approximately constant over the temperature range considered.

It marks the boundary between conditions where ΔG is positive and conditions where ΔG is negative. You must still use the signs of ΔH and ΔS to decide whether the reaction is feasible above or below that temperature.

Entropy describes the tendency for matter and energy to become more widely dispersed, while Gibbs free energy combines this tendency with the enthalpy change of a process. The relationship ΔG = ΔH − TΔS explains why some reactions are thermodynamically feasible at all temperatures, some at none, and others only above or below a particular temperature. Once this thermodynamic picture is clear, the final distinction is equally important: ΔG tells us whether a process is feasible, while kinetics determines how quickly it actually occurs.