Entropy

What you'll learn

What entropy means, and why gases usually have high entropy.
How to predict whether entropy increases or decreases in a reaction.
How to calculate a standard entropy change using data.
How entropy links to feasibility using total entropy change and Gibbs energy.

The big idea: energy spreading out

In Energetics I, you met enthalpy change, $ΔH\Delta H$ , which tells you about heat energy transferred at constant pressure. Entropy adds another idea: chemical change is also affected by how spread out matter and energy become.

A reaction may be exothermic and release heat, but that is not the only reason reactions happen. Many processes are favoured because they lead to a greater spreading out of particles and energy.

Definition

Entropy

Entropy, symbol $S$ , is a measure of how dispersed energy and particles are in a system. At A-Level, you can think of higher entropy as more possible arrangements of particles and energy.

A system is the chemical reaction or physical change you are focusing on. The surroundings are everything outside the system. Together, system plus surroundings make up the universe.

Entropy and physical state

Particles in a solid are held in fixed positions, so there are relatively few ways to arrange them. In a liquid, particles can move around each other. In a gas, particles are far apart and move randomly, giving many more possible arrangements.

So, for the same substance:

solid has the lowest entropy, liquid is higher, and gas is highest.

Diagram showing entropy increasing from solid to liquid to gas

Key Idea

State changes and entropy

Processes that form gases, increase the number of gas particles, dissolve solids, or mix substances usually increase entropy.

Predicting the sign of an entropy change

An entropy change, $ΔS\Delta S$ , compares entropy after a change with entropy before it:

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

If $ΔS\Delta S$ is positive, entropy has increased. If $ΔS\Delta S$ is negative, entropy has decreased.

Example

Predicting entropy change from state symbols

For the thermal decomposition of calcium carbonate:

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)

Compare the states of the reactants and products. The reactant is one solid; the products include a solid and a gas.
Focus especially on gas particles, because gases usually dominate entropy changes. The reaction produces carbon dioxide gas from a solid.
Conclude that the entropy of the system increases, so $ΔSsystem\Delta S_{\text{system}}$ is positive.

Common Mistake

Only counting total moles

Do not just count total moles and stop. The number of gas particles usually matters much more than the number of solid or liquid particles.

Standard molar entropy

Unlike enthalpy, entropy values can be measured on an absolute scale. This is because a perfect crystal at 0 K is taken to have zero entropy.

Definition

Standard molar entropy

The standard molar entropy, $S∘S^\circ$ , is the entropy of one mole of a substance in its standard state, usually at 298 K and 100 kPa. Its units are J K⁻¹ mol⁻¹.

Standard molar entropy values depend on the state symbol. For example, steam has a much higher entropy than liquid water because gas particles have many more possible arrangements.

Calculating the entropy change of a system

For a reaction, the standard entropy change of the system is calculated using:

ΔSsystem∘=∑Sproducts∘−∑Sreactants∘\Delta S^\circ_{\text{system}} = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}

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

Example

Calculating standard entropy change

Calculate $ΔSsystem∘\Delta S^\circ_{\text{system}}$ for the combustion of methane:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)

Data:

$S∘S^\circ$ of CH₄(g) = 186.3 J K⁻¹ mol⁻¹
$S∘S^\circ$ of O₂(g) = 205.0 J K⁻¹ mol⁻¹
$S∘S^\circ$ of CO₂(g) = 213.7 J K⁻¹ mol⁻¹
$S∘S^\circ$ of H₂O(l) = 69.9 J K⁻¹ mol⁻¹

Add the entropy values for the products, using the equation coefficients:
$353.5\ \text{J K}^{-1}\text{ mol}^{-1}$
Add the entropy values for the reactants, again using the coefficients:
$596.3\ \text{J K}^{-1}\text{ mol}^{-1}$
Subtract reactants from products:
$mol−1\Delta S^\circ_{\text{system}} = 353.5 - 596.3 = -242.8\ \text{J K}^{-1}\text{ mol}^{-1}$

So the system entropy decreases. This makes sense because three moles of gaseous reactants form only one mole of gas plus liquid water.

Tip

A quick sense check

If your calculated $ΔSsystem∘\Delta S^\circ_{\text{system}}$ says entropy decreases, check whether gas moles decrease or a gas forms a liquid or solid. That often explains the sign.

Entropy of the surroundings

A reaction can decrease the entropy of the system but still be feasible because the surroundings gain entropy.

For the surroundings:

ΔSsurroundings=−ΔHT\Delta S_{\text{surroundings}} = \frac{-\Delta H}{T}

Here, $ΔH\Delta H$ must be in J mol⁻¹ if entropy is in J K⁻¹ mol⁻¹, and $T$ must be in K.

For an exothermic reaction, $ΔH\Delta H$ is negative. Therefore $−ΔH-\Delta H$ is positive, so the surroundings gain entropy.

For an endothermic reaction, $ΔH\Delta H$ is positive. Therefore $−ΔH-\Delta H$ is negative, so the surroundings lose entropy.

Example

Calculating entropy change of the surroundings

For methane combustion at 298 K:

mol−1\Delta H^\circ = -890\ \text{kJ mol}^{-1}

Use the value from the previous example:

mol−1\Delta S^\circ_{\text{system}} = -242.8\ \text{J K}^{-1}\text{ mol}^{-1}

Convert $ΔH∘\Delta H^\circ$ into J mol⁻¹:
$mol−1-890\ \text{kJ mol}^{-1} = -890000\ \text{J mol}^{-1}$
Substitute into the surroundings equation:
$ΔSsurroundings=−(−890000)298\Delta S_{\text{surroundings}} = \frac{-(-890000)}{298}$
Calculate:
$mol−1\Delta S_{\text{surroundings}} = +2987\ \text{J K}^{-1}\text{ mol}^{-1}$

The surroundings gain a large amount of entropy because heat is released into them.

Common Mistake

Mixing kJ and J

Entropy is usually in J K⁻¹ mol⁻¹, while enthalpy is often in kJ mol⁻¹. Convert carefully before using $ΔSsurroundings=−ΔH/T\Delta S_{\text{surroundings}} = -\Delta H/T$ .

Total entropy change and feasibility

The total entropy change combines the system and surroundings:

ΔStotal=ΔSsystem+ΔSsurroundings\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}

A reaction is thermodynamically feasible if the total entropy change is positive.

Definition

Feasible reaction

A feasible reaction is thermodynamically possible under the stated conditions. It does not necessarily happen quickly.

Example

Judging feasibility using total entropy

Using methane combustion at 298 K:

$ΔSsystem∘=−242.8\Delta S^\circ_{\text{system}} = -242.8$ J K⁻¹ mol⁻¹
$ΔSsurroundings=+2987\Delta S_{\text{surroundings}} = +2987$ J K⁻¹ mol⁻¹

Add the system and surroundings entropy changes:
$ΔStotal=−242.8+2987\Delta S_{\text{total}} = -242.8 + 2987$
Calculate the total entropy change:
$mol−1\Delta S_{\text{total}} = +2744\ \text{J K}^{-1}\text{ mol}^{-1}$
Compare with zero. Since $ΔStotal\Delta S_{\text{total}}$ is positive, methane combustion is thermodynamically feasible at 298 K.

Common Mistake

Feasible does not mean fast

A reaction can be thermodynamically feasible but have a very high activation energy, so it may be extremely slow without a catalyst or spark. Entropy tells you about thermodynamics, not rate.

Gibbs energy: a convenient shortcut

Instead of calculating the entropy change of the surroundings separately, chemists often use Gibbs energy change, $ΔG\Delta G$ :

ΔG=ΔH−TΔSsystem\Delta G = \Delta H - T\Delta S_{\text{system}}

Use consistent units: if $ΔH\Delta H$ is in kJ mol⁻¹, then $ΔSsystem\Delta S_{\text{system}}$ must be converted into kJ K⁻¹ mol⁻¹.

The rule is:

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

Example

Finding the temperature for feasibility

Calcium carbonate decomposes as follows:

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)

Suppose:

$mol−1\Delta H^\circ = +178\ \text{kJ mol}^{-1}$
$mol−1\Delta S^\circ_{\text{system}} = +161\ \text{J K}^{-1}\text{ mol}^{-1}$

Find the temperature above which the reaction becomes feasible.

At the threshold of feasibility, $ΔG=0\Delta G = 0$ :
$\Delta H - T\Delta S$
Rearrange for temperature:
$\frac{\Delta H}{\Delta S}$
Convert entropy into kJ K⁻¹ mol⁻¹:
$mol−1161\ \text{J K}^{-1}\text{ mol}^{-1} = 0.161\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute:
$\frac{178}{0.161} = 1106\ \text{K}$

So the decomposition becomes thermodynamically feasible above about 1110 K.

Key Idea

Temperature can change feasibility

If $ΔH\Delta H$ and $ΔS\Delta S$ have the same sign, temperature can determine whether a reaction is feasible because of the $TΔST\Delta S$ term in $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ .

Sign patterns for Gibbs energy

You can reason about feasibility without full calculation:

If $ΔH\Delta H$ is negative and $ΔS\Delta S$ is positive, $ΔG\Delta G$ is always negative, so the reaction is feasible at all temperatures.
If $ΔH\Delta H$ is positive and $ΔS\Delta S$ is negative, $ΔG\Delta G$ is always positive, so the reaction is not feasible at any temperature.
If both are positive, high temperature favours feasibility.
If both are negative, low temperature favours feasibility.

Be careful: this assumes $ΔH\Delta H$ and $ΔS\Delta S$ do not change significantly over the temperature range.

Exam technique

In the exam

Always include state symbols when judging entropy qualitatively; gas particles are usually the key evidence.
For calculations, write products minus reactants and multiply by equation coefficients.
Check units before using Gibbs energy: convert J to kJ, or kJ to J, so $ΔH\Delta H$ and $TΔST\Delta S$ match.

Self review

Check yourself

Why does forming a gas usually increase entropy?
How would you calculate $ΔSsystem∘\Delta S^\circ_{\text{system}}$ from standard molar entropy data?
What is the difference between a reaction being feasible and a reaction being fast?

What you'll learn

What entropy means, and why gases usually have high entropy.
How to predict whether entropy increases or decreases in a reaction.
How to calculate a standard entropy change using data.
How entropy links to feasibility using total entropy change and Gibbs energy.

The big idea: energy spreading out

A reaction may be exothermic and release heat, but that is not the only reason reactions happen. Many processes are favoured because they lead to a greater spreading out of particles and energy.

Definition

Entropy

Entropy, symbol $S$ , is a measure of how dispersed energy and particles are in a system. At A-Level, you can think of higher entropy as more possible arrangements of particles and energy.

A system is the chemical reaction or physical change you are focusing on. The surroundings are everything outside the system. Together, system plus surroundings make up the universe.

Entropy and physical state

So, for the same substance:

solid has the lowest entropy, liquid is higher, and gas is highest.

Diagram showing entropy increasing from solid to liquid to gas

Key Idea

State changes and entropy

Processes that form gases, increase the number of gas particles, dissolve solids, or mix substances usually increase entropy.

Predicting the sign of an entropy change

An entropy change, $ΔS\Delta S$ , compares entropy after a change with entropy before it:

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

If $ΔS\Delta S$ is positive, entropy has increased. If $ΔS\Delta S$ is negative, entropy has decreased.

Example

Predicting entropy change from state symbols

For the thermal decomposition of calcium carbonate:

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)

Compare the states of the reactants and products. The reactant is one solid; the products include a solid and a gas.
Focus especially on gas particles, because gases usually dominate entropy changes. The reaction produces carbon dioxide gas from a solid.
Conclude that the entropy of the system increases, so $ΔSsystem\Delta S_{\text{system}}$ is positive.

Common Mistake

Only counting total moles

Do not just count total moles and stop. The number of gas particles usually matters much more than the number of solid or liquid particles.

Standard molar entropy

Unlike enthalpy, entropy values can be measured on an absolute scale. This is because a perfect crystal at 0 K is taken to have zero entropy.

Definition

Standard molar entropy

The standard molar entropy, $S∘S^\circ$ , is the entropy of one mole of a substance in its standard state, usually at 298 K and 100 kPa. Its units are J K⁻¹ mol⁻¹.

Standard molar entropy values depend on the state symbol. For example, steam has a much higher entropy than liquid water because gas particles have many more possible arrangements.

Calculating the entropy change of a system

For a reaction, the standard entropy change of the system is calculated using:

ΔSsystem∘=∑Sproducts∘−∑Sreactants∘\Delta S^\circ_{\text{system}} = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}

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

Example

Calculating standard entropy change

Calculate $ΔSsystem∘\Delta S^\circ_{\text{system}}$ for the combustion of methane:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)\text{CH}_4(g) + 2\text{O}_2(g) \to \text{CO}_2(g) + 2\text{H}_2\text{O}(l)

Data:

$S∘S^\circ$ of CH₄(g) = 186.3 J K⁻¹ mol⁻¹
$S∘S^\circ$ of O₂(g) = 205.0 J K⁻¹ mol⁻¹
$S∘S^\circ$ of CO₂(g) = 213.7 J K⁻¹ mol⁻¹
$S∘S^\circ$ of H₂O(l) = 69.9 J K⁻¹ mol⁻¹

Add the entropy values for the products, using the equation coefficients:
$353.5\ \text{J K}^{-1}\text{ mol}^{-1}$
Add the entropy values for the reactants, again using the coefficients:
$596.3\ \text{J K}^{-1}\text{ mol}^{-1}$
Subtract reactants from products:
$mol−1\Delta S^\circ_{\text{system}} = 353.5 - 596.3 = -242.8\ \text{J K}^{-1}\text{ mol}^{-1}$

So the system entropy decreases. This makes sense because three moles of gaseous reactants form only one mole of gas plus liquid water.

Tip

A quick sense check

If your calculated $ΔSsystem∘\Delta S^\circ_{\text{system}}$ says entropy decreases, check whether gas moles decrease or a gas forms a liquid or solid. That often explains the sign.

Entropy of the surroundings

A reaction can decrease the entropy of the system but still be feasible because the surroundings gain entropy.

For the surroundings:

ΔSsurroundings=−ΔHT\Delta S_{\text{surroundings}} = \frac{-\Delta H}{T}

Here, $ΔH\Delta H$ must be in J mol⁻¹ if entropy is in J K⁻¹ mol⁻¹, and $T$ must be in K.

For an exothermic reaction, $ΔH\Delta H$ is negative. Therefore $−ΔH-\Delta H$ is positive, so the surroundings gain entropy.

For an endothermic reaction, $ΔH\Delta H$ is positive. Therefore $−ΔH-\Delta H$ is negative, so the surroundings lose entropy.

Example

Calculating entropy change of the surroundings

For methane combustion at 298 K:

mol−1\Delta H^\circ = -890\ \text{kJ mol}^{-1}

Use the value from the previous example:

mol−1\Delta S^\circ_{\text{system}} = -242.8\ \text{J K}^{-1}\text{ mol}^{-1}

Convert $ΔH∘\Delta H^\circ$ into J mol⁻¹:
$mol−1-890\ \text{kJ mol}^{-1} = -890000\ \text{J mol}^{-1}$
Substitute into the surroundings equation:
$ΔSsurroundings=−(−890000)298\Delta S_{\text{surroundings}} = \frac{-(-890000)}{298}$
Calculate:
$mol−1\Delta S_{\text{surroundings}} = +2987\ \text{J K}^{-1}\text{ mol}^{-1}$

The surroundings gain a large amount of entropy because heat is released into them.

Common Mistake

Mixing kJ and J

Entropy is usually in J K⁻¹ mol⁻¹, while enthalpy is often in kJ mol⁻¹. Convert carefully before using $ΔSsurroundings=−ΔH/T\Delta S_{\text{surroundings}} = -\Delta H/T$ .

Total entropy change and feasibility

The total entropy change combines the system and surroundings:

ΔStotal=ΔSsystem+ΔSsurroundings\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}

A reaction is thermodynamically feasible if the total entropy change is positive.

Definition

Feasible reaction

A feasible reaction is thermodynamically possible under the stated conditions. It does not necessarily happen quickly.

Example

Judging feasibility using total entropy

Using methane combustion at 298 K:

$ΔSsystem∘=−242.8\Delta S^\circ_{\text{system}} = -242.8$ J K⁻¹ mol⁻¹
$ΔSsurroundings=+2987\Delta S_{\text{surroundings}} = +2987$ J K⁻¹ mol⁻¹

Add the system and surroundings entropy changes:
$ΔStotal=−242.8+2987\Delta S_{\text{total}} = -242.8 + 2987$
Calculate the total entropy change:
$mol−1\Delta S_{\text{total}} = +2744\ \text{J K}^{-1}\text{ mol}^{-1}$
Compare with zero. Since $ΔStotal\Delta S_{\text{total}}$ is positive, methane combustion is thermodynamically feasible at 298 K.

Common Mistake

Feasible does not mean fast

A reaction can be thermodynamically feasible but have a very high activation energy, so it may be extremely slow without a catalyst or spark. Entropy tells you about thermodynamics, not rate.

Gibbs energy: a convenient shortcut

Instead of calculating the entropy change of the surroundings separately, chemists often use Gibbs energy change, $ΔG\Delta G$ :

ΔG=ΔH−TΔSsystem\Delta G = \Delta H - T\Delta S_{\text{system}}

Use consistent units: if $ΔH\Delta H$ is in kJ mol⁻¹, then $ΔSsystem\Delta S_{\text{system}}$ must be converted into kJ K⁻¹ mol⁻¹.

The rule is:

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

Example

Finding the temperature for feasibility

Calcium carbonate decomposes as follows:

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)

Suppose:

$mol−1\Delta H^\circ = +178\ \text{kJ mol}^{-1}$
$mol−1\Delta S^\circ_{\text{system}} = +161\ \text{J K}^{-1}\text{ mol}^{-1}$

Find the temperature above which the reaction becomes feasible.

At the threshold of feasibility, $ΔG=0\Delta G = 0$ :
$\Delta H - T\Delta S$
Rearrange for temperature:
$\frac{\Delta H}{\Delta S}$
Convert entropy into kJ K⁻¹ mol⁻¹:
$mol−1161\ \text{J K}^{-1}\text{ mol}^{-1} = 0.161\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute:
$\frac{178}{0.161} = 1106\ \text{K}$

So the decomposition becomes thermodynamically feasible above about 1110 K.

Key Idea

Temperature can change feasibility

Sign patterns for Gibbs energy

You can reason about feasibility without full calculation:

If $ΔH\Delta H$ is negative and $ΔS\Delta S$ is positive, $ΔG\Delta G$ is always negative, so the reaction is feasible at all temperatures.
If $ΔH\Delta H$ is positive and $ΔS\Delta S$ is negative, $ΔG\Delta G$ is always positive, so the reaction is not feasible at any temperature.
If both are positive, high temperature favours feasibility.
If both are negative, low temperature favours feasibility.

Be careful: this assumes $ΔH\Delta H$ and $ΔS\Delta S$ do not change significantly over the temperature range.

Exam technique

In the exam

Always include state symbols when judging entropy qualitatively; gas particles are usually the key evidence.
For calculations, write products minus reactants and multiply by equation coefficients.
Check units before using Gibbs energy: convert J to kJ, or kJ to J, so $ΔH\Delta H$ and $TΔST\Delta S$ match.

Self review

Check yourself

Why does forming a gas usually increase entropy?
How would you calculate $ΔSsystem∘\Delta S^\circ_{\text{system}}$ from standard molar entropy data?
What is the difference between a reaction being feasible and a reaction being fast?

What you'll learn

The big idea: energy spreading out

Entropy and physical state

Predicting the sign of an entropy change

Standard molar entropy

Calculating the entropy change of a system

Entropy of the surroundings

Total entropy change and feasibility

Gibbs energy: a convenient shortcut

Sign patterns for Gibbs energy

Review Entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy

What you'll learn

The big idea: energy spreading out

Entropy and physical state

Predicting the sign of an entropy change

Standard molar entropy

Calculating the entropy change of a system

Entropy of the surroundings

Total entropy change and feasibility

Gibbs energy: a convenient shortcut

Sign patterns for Gibbs energy

Review Entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy

What you'll learn

The big idea: energy spreading out

Entropy and physical state

Predicting the sign of an entropy change

Standard molar entropy

Calculating the entropy change of a system

Entropy of the surroundings

Total entropy change and feasibility

Gibbs energy: a convenient shortcut

Sign patterns for Gibbs energy

Review Entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy Revision Guide

Entropy

What you'll learn

The big idea: energy spreading out

Entropy and physical state

Predicting the sign of an entropy change

Standard molar entropy

Calculating the entropy change of a system

Entropy of the surroundings

Total entropy change and feasibility

Gibbs energy: a convenient shortcut

Sign patterns for Gibbs energy

Review Entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Entropy Revision Guide