Gibbs free energy

What you'll learn

What Gibbs free energy change means, and how it links enthalpy and entropy.
How to use $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ with correct units.
How temperature can make some reactions feasible or not feasible.
How to calculate $ΔG∘\Delta G^\circ$ from standard Gibbs free energies of formation.

The big idea: reactions need a thermodynamic “driving force”

In Energetics II, you move beyond simply asking whether a reaction is exothermic. Some endothermic reactions happen, and some exothermic reactions do not happen unless conditions are right.

To predict whether a reaction is thermodynamically feasible, you need to consider both:

enthalpy change, $ΔH\Delta H$
entropy change, $ΔS\Delta S$

Gibbs free energy combines these into one value: $ΔG\Delta G$ .

Definition

Thermodynamically feasible

A reaction is thermodynamically feasible under given conditions if its Gibbs free energy change is negative: $ΔG<0\Delta G < 0$ . This means the reaction is energetically possible, but it does not mean it will happen quickly.

Prerequisite 1: enthalpy change

Definition

Enthalpy change

The enthalpy change, $ΔH\Delta H$ , is the heat energy change of a reaction at constant pressure. Exothermic reactions have $ΔH<0\Delta H < 0$ ; endothermic reactions have $ΔH>0\Delta H > 0$ .

For example, combustion reactions are usually exothermic, so their $ΔH\Delta H$ values are negative. Thermal decompositions are often endothermic, so their $ΔH\Delta H$ values are positive.

Enthalpy alone is not enough to predict feasibility. A reaction can absorb heat and still be feasible if it causes a large enough increase in entropy.

Prerequisite 2: entropy change

Definition

Entropy

Entropy, $S$ , is a measure of how dispersed the energy and particles are in a system. A higher entropy means more possible arrangements of particles and energy.

For A-Level Chemistry, the most useful trends are:

gases usually have much higher entropy than liquids or solids
more moles of gas usually means higher entropy
dissolving or mixing often increases entropy
a more complex molecule often has higher entropy than a simpler molecule

The entropy change of a reaction, $ΔS\Delta S$ , is calculated from standard entropy values:

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 its balancing coefficient.

Common Mistake

Elements still have entropy

Standard entropies, $S∘S^\circ$ , of elements are not zero. This is different from standard enthalpies of formation and standard Gibbs free energies of formation, which are zero for elements in their standard states.

Example

Calculating an entropy change

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

Standard entropy values:

CaCO₃(s): 92.9 J K⁻¹ mol⁻¹
CaO(s): 39.8 J K⁻¹ mol⁻¹
CO₂(g): 213.7 J K⁻¹ mol⁻¹

Use products minus reactants, including the 1:1:1 molar ratio:
$ΔS∘=[39.8+213.7]−[92.9]\Delta S^\circ = [39.8 + 213.7] - [92.9]$
Calculate the entropy change:
$mol−1\Delta S^\circ = 253.5 - 92.9 = 160.6\ \text{J K}^{-1}\text{ mol}^{-1}$
Interpret the sign: $ΔS∘\Delta S^\circ$ is positive because one mole of gas is produced from a solid, so the particles become more dispersed.

Gibbs free energy change

Definition

Gibbs free energy change

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

The key equation is:

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

where:

$ΔG\Delta G$ is the Gibbs free energy change, usually in kJ mol⁻¹
$ΔH\Delta H$ is the enthalpy change, usually in kJ mol⁻¹
$T$ is the temperature in K
$ΔS\Delta S$ is the entropy change, usually first given in J K⁻¹ mol⁻¹

Key Idea

The sign of ΔG decides feasibility

If $ΔG<0\Delta G < 0$ , the reaction is thermodynamically feasible.
If $ΔG=0\Delta G = 0$ , the system is at equilibrium under those conditions.
If $ΔG>0\Delta G > 0$ , the reaction is not thermodynamically feasible as written.

Why the equation makes sense

Entropy change of the universe decides whether a process is feasible. The universe means:

the system: the reacting chemicals
the surroundings: everything outside the reacting chemicals

Gibbs free energy is useful because it lets you focus on the system only. It includes the entropy effect of the surroundings through the enthalpy term.

The essential unit conversion

This is the most common calculation trap in this topic.

$ΔH\Delta H$ is usually given in kJ mol⁻¹, but $ΔS\Delta S$ is usually given in J K⁻¹ mol⁻¹.

Before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , convert entropy into kJ K⁻¹ mol⁻¹:

mol−11000\Delta S\text{ in kJ K}^{-1}\text{ mol}^{-1} = \frac{\Delta S\text{ in J K}^{-1}\text{ mol}^{-1}}{1000}

Common Mistake

Forgetting to convert entropy units

Do not substitute $mol−1\Delta S = 160.6\ \text{J K}^{-1}\text{ mol}^{-1}$ directly next to $mol−1\Delta H = 178\ \text{kJ mol}^{-1}$ . Convert it to $mol−10.1606\ \text{kJ K}^{-1}\text{ mol}^{-1}$ first.

Example

Calculating Gibbs free energy

For the decomposition of calcium carbonate:

CaCO₃(s) → CaO(s) + CO₂(g)

Use $mol−1\Delta H^\circ = +178.3\ \text{kJ mol}^{-1}$ and $mol−1\Delta S^\circ = +160.6\ \text{J K}^{-1}\text{ mol}^{-1}$ at 298 K.

Convert the entropy change into kJ K⁻¹ mol⁻¹:
$mol−1\Delta S^\circ = \frac{160.6}{1000} = 0.1606\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute into the Gibbs equation:
$ΔG∘=178.3−(298)(0.1606)\Delta G^\circ = 178.3 - (298)(0.1606)$
Calculate the value:
$mol−1\Delta G^\circ = 178.3 - 47.9 = +130.4\ \text{kJ mol}^{-1}$
Decide feasibility: $ΔG∘\Delta G^\circ$ is positive, so the decomposition is not thermodynamically feasible under standard conditions at 298 K.

How temperature affects feasibility

The temperature term is $TΔST\Delta S$ , so temperature only affects $ΔG\Delta G$ if there is an entropy change.

The sign pattern is easiest to remember as a grid: combine whether the reaction is exothermic or endothermic with whether entropy increases or decreases.

Decision map for Gibbs free energy feasibility

The four cases

$ΔH<0\Delta H < 0$ and $ΔS>0\Delta S > 0$ : feasible at all temperatures.
$ΔH>0\Delta H > 0$ and $ΔS<0\Delta S < 0$ : not feasible at any temperature.
$ΔH<0\Delta H < 0$ and $ΔS<0\Delta S < 0$ : feasible at low temperature only.
$ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ : feasible at high temperature only.

Tip

Temperature logic

If $ΔS\Delta S$ is positive, increasing temperature makes $−TΔS-T\Delta S$ more negative, so high temperature helps feasibility. If $ΔS\Delta S$ is negative, increasing temperature makes $−TΔS-T\Delta S$ more positive, so high temperature works against feasibility.

Example

Finding the temperature for feasibility

For CaCO₃(s) → CaO(s) + CO₂(g), use:

$mol−1\Delta H^\circ = +178.3\ \text{kJ mol}^{-1}$
$mol−1\Delta S^\circ = +0.1606\ \text{kJ K}^{-1}\text{ mol}^{-1}$

Find the temperature where the reaction just becomes feasible.

At the boundary between feasible and not feasible, set $ΔG=0\Delta G = 0$ :
$\Delta H - T\Delta S$
Rearrange for temperature:
$\frac{\Delta H}{\Delta S}$
Substitute the values, keeping units consistent:
$\frac{178.3}{0.1606} = 1110\ \text{K}$
Interpret the result: because $ΔH\Delta H$ is positive and $ΔS\Delta S$ is positive, the reaction becomes feasible above about 1110 K.

Common Mistake

Threshold temperatures are approximate

Calculations like $\frac{\Delta H}{\Delta S}$ assume that $ΔH\Delta H$ and $ΔS\Delta S$ stay constant as temperature changes. In reality they can vary with temperature, but this approximation is normally expected at A-Level.

Standard conditions and standard Gibbs changes

A superscript ° means standard conditions. In A-Level Chemistry this usually means:

temperature of 298 K, unless stated otherwise
pressure of 100 kPa
solutions at concentration 1.00 mol dm⁻³
substances in their standard states

So $ΔG∘\Delta G^\circ$ means the standard Gibbs free energy change.

Using standard Gibbs free energies of formation

Definition

Standard Gibbs free energy of formation

The standard Gibbs free energy of formation, $ΔGf∘\Delta G_f^\circ$ , is the Gibbs free energy change when one mole of a compound is formed from its elements in their standard states under standard conditions.

You can calculate a reaction’s standard Gibbs free energy change using:

ΔG∘=∑ΔGf∘(products)−∑ΔGf∘(reactants)\Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})

Elements in their standard states have $ΔGf∘=0\Delta G_f^\circ = 0$ .

Example

Using Gibbs formation values

Calculate $ΔG∘\Delta G^\circ$ for the hydrogenation of ethene:

C₂H₄(g) + H₂(g) → C₂H₆(g)

Use:

$ΔGf∘\Delta G_f^\circ$ of C₂H₄(g) = +68.1 kJ mol⁻¹
$ΔGf∘\Delta G_f^\circ$ of H₂(g) = 0 kJ mol⁻¹
$ΔGf∘\Delta G_f^\circ$ of C₂H₆(g) = −32.9 kJ mol⁻¹

Apply products minus reactants:
$ΔG∘=[−32.9]−[68.1+0]\Delta G^\circ = [-32.9] - [68.1 + 0]$
Calculate the Gibbs free energy change:
$mol−1\Delta G^\circ = -101.0\ \text{kJ mol}^{-1}$
Interpret the sign: the value is negative, so the hydrogenation of ethene is thermodynamically feasible under standard conditions.

Feasible does not always mean observable

A negative $ΔG\Delta G$ tells you a reaction is thermodynamically possible. It does not tell you about the rate.

A reaction may be feasible but extremely slow if it has a high activation energy. For example, many fuels are thermodynamically feasible to burn in oxygen, but they do not ignite without a spark or flame.

Common Mistake

Confusing thermodynamics with kinetics

Gibbs free energy predicts whether a reaction is possible. It does not predict how fast it happens. Rate depends on activation energy, catalysts, concentration, temperature and other kinetic factors.

Exam technique

In the exam

Check the units before substituting: $ΔH\Delta H$ and $ΔG\Delta G$ are usually in kJ mol⁻¹, while $ΔS\Delta S$ is usually in J K⁻¹ mol⁻¹.
Always use temperature in K, not °C.
After calculating $ΔG\Delta G$ , state the sign and link it clearly to feasibility under the stated conditions.

Self review

Check yourself

Why can an endothermic reaction become feasible at high temperature?
What unit conversion is usually needed before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ ?
What is the difference between a reaction being thermodynamically feasible and being fast?

What you'll learn

What Gibbs free energy change means, and how it links enthalpy and entropy.
How to use $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ with correct units.
How temperature can make some reactions feasible or not feasible.
How to calculate $ΔG∘\Delta G^\circ$ from standard Gibbs free energies of formation.

The big idea: reactions need a thermodynamic “driving force”

In Energetics II, you move beyond simply asking whether a reaction is exothermic. Some endothermic reactions happen, and some exothermic reactions do not happen unless conditions are right.

To predict whether a reaction is thermodynamically feasible, you need to consider both:

enthalpy change, $ΔH\Delta H$
entropy change, $ΔS\Delta S$

Gibbs free energy combines these into one value: $ΔG\Delta G$ .

Definition

Thermodynamically feasible

Prerequisite 1: enthalpy change

Definition

Enthalpy change

The enthalpy change, $ΔH\Delta H$ , is the heat energy change of a reaction at constant pressure. Exothermic reactions have $ΔH<0\Delta H < 0$ ; endothermic reactions have $ΔH>0\Delta H > 0$ .

For example, combustion reactions are usually exothermic, so their $ΔH\Delta H$ values are negative. Thermal decompositions are often endothermic, so their $ΔH\Delta H$ values are positive.

Enthalpy alone is not enough to predict feasibility. A reaction can absorb heat and still be feasible if it causes a large enough increase in entropy.

Prerequisite 2: entropy change

Definition

Entropy

Entropy, $S$ , is a measure of how dispersed the energy and particles are in a system. A higher entropy means more possible arrangements of particles and energy.

For A-Level Chemistry, the most useful trends are:

gases usually have much higher entropy than liquids or solids
more moles of gas usually means higher entropy
dissolving or mixing often increases entropy
a more complex molecule often has higher entropy than a simpler molecule

The entropy change of a reaction, $ΔS\Delta S$ , is calculated from standard entropy values:

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 its balancing coefficient.

Common Mistake

Elements still have entropy

Example

Calculating an entropy change

Calcium carbonate decomposes when heated:

CaCO₃(s) → CaO(s) + CO₂(g)

Standard entropy values:

CaCO₃(s): 92.9 J K⁻¹ mol⁻¹
CaO(s): 39.8 J K⁻¹ mol⁻¹
CO₂(g): 213.7 J K⁻¹ mol⁻¹

Use products minus reactants, including the 1:1:1 molar ratio:
$ΔS∘=[39.8+213.7]−[92.9]\Delta S^\circ = [39.8 + 213.7] - [92.9]$
Calculate the entropy change:
$mol−1\Delta S^\circ = 253.5 - 92.9 = 160.6\ \text{J K}^{-1}\text{ mol}^{-1}$
Interpret the sign: $ΔS∘\Delta S^\circ$ is positive because one mole of gas is produced from a solid, so the particles become more dispersed.

Gibbs free energy change

Definition

Gibbs free energy change

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

The key equation is:

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

where:

$ΔG\Delta G$ is the Gibbs free energy change, usually in kJ mol⁻¹
$ΔH\Delta H$ is the enthalpy change, usually in kJ mol⁻¹
$T$ is the temperature in K
$ΔS\Delta S$ is the entropy change, usually first given in J K⁻¹ mol⁻¹

Key Idea

The sign of ΔG decides feasibility

If $ΔG<0\Delta G < 0$ , the reaction is thermodynamically feasible.
If $ΔG=0\Delta G = 0$ , the system is at equilibrium under those conditions.
If $ΔG>0\Delta G > 0$ , the reaction is not thermodynamically feasible as written.

Why the equation makes sense

Entropy change of the universe decides whether a process is feasible. The universe means:

the system: the reacting chemicals
the surroundings: everything outside the reacting chemicals

Gibbs free energy is useful because it lets you focus on the system only. It includes the entropy effect of the surroundings through the enthalpy term.

The essential unit conversion

This is the most common calculation trap in this topic.

$ΔH\Delta H$ is usually given in kJ mol⁻¹, but $ΔS\Delta S$ is usually given in J K⁻¹ mol⁻¹.

Before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , convert entropy into kJ K⁻¹ mol⁻¹:

mol−11000\Delta S\text{ in kJ K}^{-1}\text{ mol}^{-1} = \frac{\Delta S\text{ in J K}^{-1}\text{ mol}^{-1}}{1000}

Common Mistake

Forgetting to convert entropy units

Example

Calculating Gibbs free energy

For the decomposition of calcium carbonate:

CaCO₃(s) → CaO(s) + CO₂(g)

Use $mol−1\Delta H^\circ = +178.3\ \text{kJ mol}^{-1}$ and $mol−1\Delta S^\circ = +160.6\ \text{J K}^{-1}\text{ mol}^{-1}$ at 298 K.

Convert the entropy change into kJ K⁻¹ mol⁻¹:
$mol−1\Delta S^\circ = \frac{160.6}{1000} = 0.1606\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute into the Gibbs equation:
$ΔG∘=178.3−(298)(0.1606)\Delta G^\circ = 178.3 - (298)(0.1606)$
Calculate the value:
$mol−1\Delta G^\circ = 178.3 - 47.9 = +130.4\ \text{kJ mol}^{-1}$
Decide feasibility: $ΔG∘\Delta G^\circ$ is positive, so the decomposition is not thermodynamically feasible under standard conditions at 298 K.

How temperature affects feasibility

The temperature term is $TΔST\Delta S$ , so temperature only affects $ΔG\Delta G$ if there is an entropy change.

The sign pattern is easiest to remember as a grid: combine whether the reaction is exothermic or endothermic with whether entropy increases or decreases.

Decision map for Gibbs free energy feasibility

The four cases

$ΔH<0\Delta H < 0$ and $ΔS>0\Delta S > 0$ : feasible at all temperatures.
$ΔH>0\Delta H > 0$ and $ΔS<0\Delta S < 0$ : not feasible at any temperature.
$ΔH<0\Delta H < 0$ and $ΔS<0\Delta S < 0$ : feasible at low temperature only.
$ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ : feasible at high temperature only.

Tip

Temperature logic

Example

Finding the temperature for feasibility

For CaCO₃(s) → CaO(s) + CO₂(g), use:

$mol−1\Delta H^\circ = +178.3\ \text{kJ mol}^{-1}$
$mol−1\Delta S^\circ = +0.1606\ \text{kJ K}^{-1}\text{ mol}^{-1}$

Find the temperature where the reaction just becomes feasible.

At the boundary between feasible and not feasible, set $ΔG=0\Delta G = 0$ :
$\Delta H - T\Delta S$
Rearrange for temperature:
$\frac{\Delta H}{\Delta S}$
Substitute the values, keeping units consistent:
$\frac{178.3}{0.1606} = 1110\ \text{K}$
Interpret the result: because $ΔH\Delta H$ is positive and $ΔS\Delta S$ is positive, the reaction becomes feasible above about 1110 K.

Common Mistake

Threshold temperatures are approximate

Standard conditions and standard Gibbs changes

A superscript ° means standard conditions. In A-Level Chemistry this usually means:

temperature of 298 K, unless stated otherwise
pressure of 100 kPa
solutions at concentration 1.00 mol dm⁻³
substances in their standard states

So $ΔG∘\Delta G^\circ$ means the standard Gibbs free energy change.

Using standard Gibbs free energies of formation

Definition

Standard Gibbs free energy of formation

You can calculate a reaction’s standard Gibbs free energy change using:

ΔG∘=∑ΔGf∘(products)−∑ΔGf∘(reactants)\Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})

Elements in their standard states have $ΔGf∘=0\Delta G_f^\circ = 0$ .

Example

Using Gibbs formation values

Calculate $ΔG∘\Delta G^\circ$ for the hydrogenation of ethene:

C₂H₄(g) + H₂(g) → C₂H₆(g)

Use:

$ΔGf∘\Delta G_f^\circ$ of C₂H₄(g) = +68.1 kJ mol⁻¹
$ΔGf∘\Delta G_f^\circ$ of H₂(g) = 0 kJ mol⁻¹
$ΔGf∘\Delta G_f^\circ$ of C₂H₆(g) = −32.9 kJ mol⁻¹

Apply products minus reactants:
$ΔG∘=[−32.9]−[68.1+0]\Delta G^\circ = [-32.9] - [68.1 + 0]$
Calculate the Gibbs free energy change:
$mol−1\Delta G^\circ = -101.0\ \text{kJ mol}^{-1}$
Interpret the sign: the value is negative, so the hydrogenation of ethene is thermodynamically feasible under standard conditions.

Feasible does not always mean observable

A negative $ΔG\Delta G$ tells you a reaction is thermodynamically possible. It does not tell you about the rate.

Common Mistake

Confusing thermodynamics with kinetics

Gibbs free energy predicts whether a reaction is possible. It does not predict how fast it happens. Rate depends on activation energy, catalysts, concentration, temperature and other kinetic factors.

Exam technique

In the exam

Check the units before substituting: $ΔH\Delta H$ and $ΔG\Delta G$ are usually in kJ mol⁻¹, while $ΔS\Delta S$ is usually in J K⁻¹ mol⁻¹.
Always use temperature in K, not °C.
After calculating $ΔG\Delta G$ , state the sign and link it clearly to feasibility under the stated conditions.

Self review

Check yourself

Why can an endothermic reaction become feasible at high temperature?
What unit conversion is usually needed before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ ?
What is the difference between a reaction being thermodynamically feasible and being fast?

What you'll learn

The big idea: reactions need a thermodynamic “driving force”

Prerequisite 1: enthalpy change

Prerequisite 2: entropy change

Gibbs free energy change

Why the equation makes sense

The essential unit conversion

How temperature affects feasibility

The four cases

Standard conditions and standard Gibbs changes

Using standard Gibbs free energies of formation

Feasible does not always mean observable

Review Gibbs free energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gibbs free energy

What you'll learn

The big idea: reactions need a thermodynamic “driving force”

Prerequisite 1: enthalpy change

Prerequisite 2: entropy change

Gibbs free energy change

Why the equation makes sense

The essential unit conversion

How temperature affects feasibility

The four cases

Standard conditions and standard Gibbs changes

Using standard Gibbs free energies of formation

Feasible does not always mean observable

Review Gibbs free energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gibbs free energy

What you'll learn

The big idea: reactions need a thermodynamic “driving force”

Prerequisite 1: enthalpy change

Prerequisite 2: entropy change

Gibbs free energy change

Why the equation makes sense

The essential unit conversion

How temperature affects feasibility

The four cases

Standard conditions and standard Gibbs changes

Using standard Gibbs free energies of formation

Feasible does not always mean observable

Review Gibbs free energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gibbs free energy Revision Guide

Gibbs free energy

What you'll learn

The big idea: reactions need a thermodynamic “driving force”

Prerequisite 1: enthalpy change

Prerequisite 2: entropy change

Gibbs free energy change

Why the equation makes sense

The essential unit conversion

How temperature affects feasibility

The four cases

Standard conditions and standard Gibbs changes

Using standard Gibbs free energies of formation

Feasible does not always mean observable

Review Gibbs free energy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Gibbs free energy Revision Guide