Entropy and feasibility of reactions

What you'll learn

What entropy means in terms of particle freedom.
How to predict and calculate entropy changes using absolute entropy data.
How to use $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ to decide whether a reaction is feasible.
Why some endothermic processes can still happen spontaneously.

Why enthalpy is not enough

From earlier work on energetics, you know that an enthalpy change, $ΔH\Delta H$ , is the heat energy change at constant pressure.

$ΔH<0\Delta H < 0$ : exothermic, heat released.
$ΔH>0\Delta H > 0$ : endothermic, heat absorbed.

It is tempting to think “exothermic means it happens” and “endothermic means it does not happen”. That is not reliable. To predict feasibility, we also need entropy.

Entropy: a measure of particle freedom

Definition

Entropy

Entropy, $S$ , is a measure of the freedom possessed by particles in a system. More freedom means more possible ways for particles and energy to be arranged.

A system is the chemical reaction or physical process being studied. The surroundings are everything outside the system.

Entropy is often described loosely as “disorder”, but for A-Level Chemistry it is better to think in terms of particle freedom and the number of possible arrangements.

Entropy and states of matter

Particles in a solid are held in fixed positions and mainly vibrate. Particles in a liquid can move past each other. Particles in a gas are far apart and move freely. So, other factors being equal:

S(gas)>S(liquid)>S(solid)S(\text{gas}) > S(\text{liquid}) > S(\text{solid})

Particle freedom and entropy increase from solid to liquid to gas

Key Idea

State and entropy

Changing from solid → liquid → gas increases entropy because particle freedom increases.

Natural changes and maximum entropy

A natural change is one that occurs without continuous external forcing once suitable conditions are provided. For an isolated system, natural changes occur in the direction that increases total entropy towards a maximum.

At equilibrium, entropy has reached a maximum for the conditions, so there is no further net change.

Key Idea

The direction of natural change

Natural changes are associated with an increase in total entropy, but the entropy of the chemical system alone can decrease if the surroundings compensate.

Predicting the sign of an entropy change

For reactions, a useful first check is to look at the number and state of particles.

Entropy usually increases when:

a gas is formed from a solid or liquid
the number of moles of gas increases
a solid dissolves into many mobile ions or molecules

Entropy usually decreases when:

gas particles are removed
fewer gas molecules are produced
a precipitate forms from aqueous ions

Example

Predicting the sign of an entropy change

Consider the thermal decomposition:

CaCO₃(s) → CaO(s) + CO₂(g)

Compare the initial and final states. Initially there is one solid reactant; finally there is one solid product and one gaseous product.
Focus on the gas because gas particles have much more freedom than solid particles. CO₂(g) has far more freedom than particles locked in a solid lattice.
The products have greater particle freedom overall, so the entropy change is positive: $ΔS>0\Delta S > 0$ .

Tip

Quick gas check

When gases are involved, the change in the number of moles of gas often dominates the sign of $ΔS\Delta S$ .

Calculating entropy change from absolute entropy values

Definition

Absolute entropy

An absolute entropy, usually written as $S∘S^\circ$ , is the entropy of one mole of a substance under standard conditions, commonly quoted in J K⁻¹ mol⁻¹.

Unlike standard enthalpies of formation, standard entropy values for elements are not zero. Even pure elements have particles with energy and possible arrangements.

For a reaction:

ΔS∘=Sfinal−Sinitial\Delta S^\circ = S_{\text{final}} - S_{\text{initial}}

For chemical equations, this means:

ΔS∘=∑S∘(products)−∑S∘(reactants)\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})

Remember to multiply each entropy value by the balancing coefficient in the equation.

Common Mistake

Forgetting the coefficients

If the equation has 3H₂(g), you must use three times the entropy of H₂(g), not just one value.

Example

Calculating an entropy change from entropy data

For the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Use these standard entropy values:

N₂(g): 191.5 J K⁻¹ mol⁻¹
H₂(g): 130.7 J K⁻¹ mol⁻¹
NH₃(g): 192.5 J K⁻¹ mol⁻¹

Calculate the total entropy of the products: $\times 192.5 = 385.0 \ \text{J K}^{-1}\text{ mol}^{-1}$ .
Calculate the total entropy of the reactants: $\times 130.7) = 583.6 \ \text{J K}^{-1}\text{ mol}^{-1}$ .
Apply products minus reactants: $mol−1\Delta S^\circ = 385.0 - 583.6 = -198.6 \ \text{J K}^{-1}\text{ mol}^{-1}$ .
Interpret the sign. Four moles of gas become two moles of gas, so particle freedom decreases; the negative value makes chemical sense.

Gibbs free energy change

Definition

Gibbs free energy change

The Gibbs free energy change, $ΔG\Delta G$ , combines enthalpy and entropy to predict whether a reaction is thermodynamically feasible at a particular temperature.

The Eduqas relationship is:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

where:

$ΔG\Delta G$ is usually in kJ mol⁻¹
$ΔH\Delta H$ is usually in kJ mol⁻¹
$T$ is temperature in K
$ΔS\Delta S$ must be in kJ K⁻¹ mol⁻¹ if $ΔH\Delta H$ is in kJ mol⁻¹

Tip

Unit check for Gibbs calculations

Entropy data are usually given in J K⁻¹ mol⁻¹, but enthalpy is usually in kJ mol⁻¹. Convert entropy by dividing by 1000 before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ .

What the sign of Gibbs energy means

At a stated temperature:

$ΔG<0\Delta G < 0$ : reaction is thermodynamically feasible, or spontaneous.
$ΔG=0\Delta G = 0$ : system is at equilibrium.
$ΔG>0\Delta G > 0$ : forward reaction is not thermodynamically feasible under those conditions.

Common Mistake

Feasible does not mean fast

A reaction with $ΔG<0\Delta G < 0$ may still be very slow if it has a high activation energy. Gibbs energy tells you about thermodynamic feasibility, not rate.

Example

Calculating Gibbs free energy

Calcium carbonate decomposes as follows:

CaCO₃(s) → CaO(s) + CO₂(g)

For this reaction, $mol−1\Delta H = +178 \ \text{kJ mol}^{-1}$ and $mol−1\Delta S = +161 \ \text{J K}^{-1}\text{ mol}^{-1}$ . Calculate $ΔG\Delta G$ at 1200 K.

Convert entropy into kJ K⁻¹ mol⁻¹: $161 \div 1000 = 0.161 \ \text{kJ K}^{-1}\text{ mol}^{-1}$ .
Substitute into the Gibbs equation: $ΔG=178−(1200×0.161)\Delta G = 178 - (1200 \times 0.161)$ .
Calculate the entropy term: $1200 \times 0.161 = 193.2 \ \text{kJ mol}^{-1}$ .
Find $ΔG\Delta G$ : $mol−1\Delta G = 178 - 193.2 = -15.2 \ \text{kJ mol}^{-1}$ .
Interpret the result. Since $ΔG<0\Delta G < 0$ , the decomposition is thermodynamically feasible at 1200 K.

How temperature affects feasibility

The equation $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ shows that temperature controls the size of the entropy term, $TΔST\Delta S$ .

If $ΔH<0\Delta H < 0$ and $ΔS>0\Delta S > 0$ , then $ΔG\Delta G$ is always negative.
If $ΔH>0\Delta H > 0$ and $ΔS<0\Delta S < 0$ , then $ΔG\Delta G$ is always positive.
If $ΔH<0\Delta H < 0$ and $ΔS<0\Delta S < 0$ , the reaction is more feasible at low temperature.
If $ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ , the reaction is more feasible at high temperature.

The graph below summarises how different signs of $ΔH\Delta H$ and $ΔS\Delta S$ affect $ΔG\Delta G$ as temperature changes.

Gibbs free energy against temperature for different signs of enthalpy and entropy change

Finding the temperature where feasibility changes

The boundary between feasible and not feasible occurs when $ΔG=0\Delta G = 0$ .

Starting from:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

At the boundary:

\Delta H - T\Delta S

So:

\frac{\Delta H}{\Delta S}

This only works if $ΔH\Delta H$ and $ΔS\Delta S$ are in compatible units.

Example

Finding the feasibility temperature

For CaCO₃(s) → CaO(s) + CO₂(g), use $mol−1\Delta H = +178 \ \text{kJ mol}^{-1}$ and $mol−1\Delta S = +161 \ \text{J K}^{-1}\text{ mol}^{-1}$ .

Set the boundary condition: at the changeover point, $ΔG=0\Delta G = 0$ , so $\frac{\Delta H}{\Delta S}$ .
Convert entropy into kJ K⁻¹ mol⁻¹: $\ \text{J K}^{-1}\text{ mol}^{-1} = 0.161 \ \text{kJ K}^{-1}\text{ mol}^{-1}$ .
Substitute values: $\frac{178}{0.161} = 1106 \ \text{K}$ .
Interpret the result. Since this reaction has $ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ , it becomes feasible above about 1100 K.

Why endothermic processes can be spontaneous

An endothermic process has $ΔH>0\Delta H > 0$ , which seems unfavourable. But if entropy increases enough, then $TΔST\Delta S$ can be larger than $ΔH\Delta H$ .

That makes:

ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0

So an endothermic process can be spontaneous when it has a sufficiently positive entropy change, especially at high temperature. This explains processes such as some thermal decompositions and some dissolving processes.

Common Mistake

Endothermic does not mean impossible

Endothermic means heat is absorbed. It does not automatically mean the process is not feasible; entropy and temperature must also be considered.

Exam technique

In the exam

For entropy calculations, do products minus reactants and multiply each $S∘S^\circ$ value by its balancing coefficient.
In Gibbs calculations, convert $ΔS\Delta S$ from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ if $ΔH\Delta H$ is in kJ mol⁻¹, and always use temperature in K.
Interpret the sign clearly: $ΔG<0\Delta G < 0$ means feasible, $ΔG=0\Delta G = 0$ means equilibrium, and $ΔG>0\Delta G > 0$ means not feasible in the forward direction.

Self review

Check yourself

Why does forming a gas usually increase entropy?
In $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , why must temperature be measured in K?
How can an endothermic reaction become feasible at high temperature?

What you'll learn

What entropy means in terms of particle freedom.
How to predict and calculate entropy changes using absolute entropy data.
How to use $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ to decide whether a reaction is feasible.
Why some endothermic processes can still happen spontaneously.

Why enthalpy is not enough

From earlier work on energetics, you know that an enthalpy change, $ΔH\Delta H$ , is the heat energy change at constant pressure.

$ΔH<0\Delta H < 0$ : exothermic, heat released.
$ΔH>0\Delta H > 0$ : endothermic, heat absorbed.

It is tempting to think “exothermic means it happens” and “endothermic means it does not happen”. That is not reliable. To predict feasibility, we also need entropy.

Entropy: a measure of particle freedom

Definition

Entropy

Entropy, $S$ , is a measure of the freedom possessed by particles in a system. More freedom means more possible ways for particles and energy to be arranged.

A system is the chemical reaction or physical process being studied. The surroundings are everything outside the system.

Entropy is often described loosely as “disorder”, but for A-Level Chemistry it is better to think in terms of particle freedom and the number of possible arrangements.

Entropy and states of matter

Particles in a solid are held in fixed positions and mainly vibrate. Particles in a liquid can move past each other. Particles in a gas are far apart and move freely. So, other factors being equal:

S(gas)>S(liquid)>S(solid)S(\text{gas}) > S(\text{liquid}) > S(\text{solid})

Particle freedom and entropy increase from solid to liquid to gas

Key Idea

State and entropy

Changing from solid → liquid → gas increases entropy because particle freedom increases.

Natural changes and maximum entropy

At equilibrium, entropy has reached a maximum for the conditions, so there is no further net change.

Key Idea

The direction of natural change

Natural changes are associated with an increase in total entropy, but the entropy of the chemical system alone can decrease if the surroundings compensate.

Predicting the sign of an entropy change

For reactions, a useful first check is to look at the number and state of particles.

Entropy usually increases when:

a gas is formed from a solid or liquid
the number of moles of gas increases
a solid dissolves into many mobile ions or molecules

Entropy usually decreases when:

gas particles are removed
fewer gas molecules are produced
a precipitate forms from aqueous ions

Example

Predicting the sign of an entropy change

Consider the thermal decomposition:

CaCO₃(s) → CaO(s) + CO₂(g)

Compare the initial and final states. Initially there is one solid reactant; finally there is one solid product and one gaseous product.
Focus on the gas because gas particles have much more freedom than solid particles. CO₂(g) has far more freedom than particles locked in a solid lattice.
The products have greater particle freedom overall, so the entropy change is positive: $ΔS>0\Delta S > 0$ .

Tip

Quick gas check

When gases are involved, the change in the number of moles of gas often dominates the sign of $ΔS\Delta S$ .

Calculating entropy change from absolute entropy values

Definition

Absolute entropy

An absolute entropy, usually written as $S∘S^\circ$ , is the entropy of one mole of a substance under standard conditions, commonly quoted in J K⁻¹ mol⁻¹.

Unlike standard enthalpies of formation, standard entropy values for elements are not zero. Even pure elements have particles with energy and possible arrangements.

For a reaction:

ΔS∘=Sfinal−Sinitial\Delta S^\circ = S_{\text{final}} - S_{\text{initial}}

For chemical equations, this means:

ΔS∘=∑S∘(products)−∑S∘(reactants)\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})

Remember to multiply each entropy value by the balancing coefficient in the equation.

Common Mistake

Forgetting the coefficients

If the equation has 3H₂(g), you must use three times the entropy of H₂(g), not just one value.

Example

Calculating an entropy change from entropy data

For the reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

Use these standard entropy values:

N₂(g): 191.5 J K⁻¹ mol⁻¹
H₂(g): 130.7 J K⁻¹ mol⁻¹
NH₃(g): 192.5 J K⁻¹ mol⁻¹

Calculate the total entropy of the products: $\times 192.5 = 385.0 \ \text{J K}^{-1}\text{ mol}^{-1}$ .
Calculate the total entropy of the reactants: $\times 130.7) = 583.6 \ \text{J K}^{-1}\text{ mol}^{-1}$ .
Apply products minus reactants: $mol−1\Delta S^\circ = 385.0 - 583.6 = -198.6 \ \text{J K}^{-1}\text{ mol}^{-1}$ .
Interpret the sign. Four moles of gas become two moles of gas, so particle freedom decreases; the negative value makes chemical sense.

Gibbs free energy change

Definition

Gibbs free energy change

The Gibbs free energy change, $ΔG\Delta G$ , combines enthalpy and entropy to predict whether a reaction is thermodynamically feasible at a particular temperature.

The Eduqas relationship is:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

where:

$ΔG\Delta G$ is usually in kJ mol⁻¹
$ΔH\Delta H$ is usually in kJ mol⁻¹
$T$ is temperature in K
$ΔS\Delta S$ must be in kJ K⁻¹ mol⁻¹ if $ΔH\Delta H$ is in kJ mol⁻¹

Tip

Unit check for Gibbs calculations

Entropy data are usually given in J K⁻¹ mol⁻¹, but enthalpy is usually in kJ mol⁻¹. Convert entropy by dividing by 1000 before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ .

What the sign of Gibbs energy means

At a stated temperature:

$ΔG<0\Delta G < 0$ : reaction is thermodynamically feasible, or spontaneous.
$ΔG=0\Delta G = 0$ : system is at equilibrium.
$ΔG>0\Delta G > 0$ : forward reaction is not thermodynamically feasible under those conditions.

Common Mistake

Feasible does not mean fast

A reaction with $ΔG<0\Delta G < 0$ may still be very slow if it has a high activation energy. Gibbs energy tells you about thermodynamic feasibility, not rate.

Example

Calculating Gibbs free energy

Calcium carbonate decomposes as follows:

CaCO₃(s) → CaO(s) + CO₂(g)

For this reaction, $mol−1\Delta H = +178 \ \text{kJ mol}^{-1}$ and $mol−1\Delta S = +161 \ \text{J K}^{-1}\text{ mol}^{-1}$ . Calculate $ΔG\Delta G$ at 1200 K.

Convert entropy into kJ K⁻¹ mol⁻¹: $161 \div 1000 = 0.161 \ \text{kJ K}^{-1}\text{ mol}^{-1}$ .
Substitute into the Gibbs equation: $ΔG=178−(1200×0.161)\Delta G = 178 - (1200 \times 0.161)$ .
Calculate the entropy term: $1200 \times 0.161 = 193.2 \ \text{kJ mol}^{-1}$ .
Find $ΔG\Delta G$ : $mol−1\Delta G = 178 - 193.2 = -15.2 \ \text{kJ mol}^{-1}$ .
Interpret the result. Since $ΔG<0\Delta G < 0$ , the decomposition is thermodynamically feasible at 1200 K.

How temperature affects feasibility

The equation $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ shows that temperature controls the size of the entropy term, $TΔST\Delta S$ .

If $ΔH<0\Delta H < 0$ and $ΔS>0\Delta S > 0$ , then $ΔG\Delta G$ is always negative.
If $ΔH>0\Delta H > 0$ and $ΔS<0\Delta S < 0$ , then $ΔG\Delta G$ is always positive.
If $ΔH<0\Delta H < 0$ and $ΔS<0\Delta S < 0$ , the reaction is more feasible at low temperature.
If $ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ , the reaction is more feasible at high temperature.

The graph below summarises how different signs of $ΔH\Delta H$ and $ΔS\Delta S$ affect $ΔG\Delta G$ as temperature changes.

Gibbs free energy against temperature for different signs of enthalpy and entropy change

Finding the temperature where feasibility changes

The boundary between feasible and not feasible occurs when $ΔG=0\Delta G = 0$ .

Starting from:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

At the boundary:

\Delta H - T\Delta S

So:

\frac{\Delta H}{\Delta S}

This only works if $ΔH\Delta H$ and $ΔS\Delta S$ are in compatible units.

Example

Finding the feasibility temperature

For CaCO₃(s) → CaO(s) + CO₂(g), use $mol−1\Delta H = +178 \ \text{kJ mol}^{-1}$ and $mol−1\Delta S = +161 \ \text{J K}^{-1}\text{ mol}^{-1}$ .

Set the boundary condition: at the changeover point, $ΔG=0\Delta G = 0$ , so $\frac{\Delta H}{\Delta S}$ .
Convert entropy into kJ K⁻¹ mol⁻¹: $\ \text{J K}^{-1}\text{ mol}^{-1} = 0.161 \ \text{kJ K}^{-1}\text{ mol}^{-1}$ .
Substitute values: $\frac{178}{0.161} = 1106 \ \text{K}$ .
Interpret the result. Since this reaction has $ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ , it becomes feasible above about 1100 K.

Why endothermic processes can be spontaneous

An endothermic process has $ΔH>0\Delta H > 0$ , which seems unfavourable. But if entropy increases enough, then $TΔST\Delta S$ can be larger than $ΔH\Delta H$ .

That makes:

ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0

Common Mistake

Endothermic does not mean impossible

Endothermic means heat is absorbed. It does not automatically mean the process is not feasible; entropy and temperature must also be considered.

Exam technique

In the exam

For entropy calculations, do products minus reactants and multiply each $S∘S^\circ$ value by its balancing coefficient.
In Gibbs calculations, convert $ΔS\Delta S$ from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ if $ΔH\Delta H$ is in kJ mol⁻¹, and always use temperature in K.
Interpret the sign clearly: $ΔG<0\Delta G < 0$ means feasible, $ΔG=0\Delta G = 0$ means equilibrium, and $ΔG>0\Delta G > 0$ means not feasible in the forward direction.

Self review

Check yourself

Why does forming a gas usually increase entropy?
In $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , why must temperature be measured in K?
How can an endothermic reaction become feasible at high temperature?

What you'll learn

Why enthalpy is not enough

Entropy: a measure of particle freedom

Entropy and states of matter

Natural changes and maximum entropy

Predicting the sign of an entropy change

Calculating entropy change from absolute entropy values

Gibbs free energy change

What the sign of Gibbs energy means

How temperature affects feasibility

Finding the temperature where feasibility changes

Why endothermic processes can be spontaneous

Review Entropy and feasibility of reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy and feasibility of reactions

What you'll learn

Why enthalpy is not enough

Entropy: a measure of particle freedom

Entropy and states of matter

Natural changes and maximum entropy

Predicting the sign of an entropy change

Calculating entropy change from absolute entropy values

Gibbs free energy change

What the sign of Gibbs energy means

How temperature affects feasibility

Finding the temperature where feasibility changes

Why endothermic processes can be spontaneous

Review Entropy and feasibility of reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy and feasibility of reactions

What you'll learn

Why enthalpy is not enough

Entropy: a measure of particle freedom

Entropy and states of matter

Natural changes and maximum entropy

Predicting the sign of an entropy change

Calculating entropy change from absolute entropy values

Gibbs free energy change

What the sign of Gibbs energy means

How temperature affects feasibility

Finding the temperature where feasibility changes

Why endothermic processes can be spontaneous

Review Entropy and feasibility of reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy and feasibility of reactions Revision Guide

Entropy and feasibility of reactions

What you'll learn

Why enthalpy is not enough

Entropy: a measure of particle freedom

Entropy and states of matter

Natural changes and maximum entropy

Predicting the sign of an entropy change

Calculating entropy change from absolute entropy values

Gibbs free energy change

What the sign of Gibbs energy means

How temperature affects feasibility

Finding the temperature where feasibility changes

Why endothermic processes can be spontaneous

Review Entropy and feasibility of reactions by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy and feasibility of reactions Revision Guide