Enthalpy and entropy

What you'll learn

What entropy means, and why gases usually have much higher entropy than solids.
How to calculate an entropy change, $ΔS\Delta S$ , from entropy data.
How enthalpy, entropy and temperature combine in the Gibbs equation.
Why a negative $ΔG\Delta G$ predicts thermodynamic feasibility, but not necessarily a fast reaction.

The starting point: enthalpy changes

You already know that reactions involve energy transfers. The enthalpy change, $ΔH\Delta H$ , is the heat energy change of a reaction at constant pressure.

$ΔH<0\Delta H < 0$ : exothermic, energy is released to the surroundings.
$ΔH>0\Delta H > 0$ : endothermic, energy is taken in from the surroundings.

For this topic, enthalpy is only part of the story. Some endothermic processes still happen, so we need another idea: entropy.

Definition

Enthalpy change

The enthalpy change, $ΔH\Delta H$ , is the heat energy change of a system at constant pressure, usually measured in kJ mol⁻¹.

Entropy: energy dispersal and disorder

Entropy is about how spread out energy is, and how many ways particles and energy can be arranged.

Definition

Entropy

Entropy, $S$ , is a measure of the dispersal of energy in a system. A system has greater entropy when it is more disordered.

A “system” means the chemicals or process you are focusing on. For example, in a reaction mixture, the reactants and products are the system.

The key model is: more possible arrangements = greater entropy. Gas particles can spread out through a large volume and move freely, so they have many more possible arrangements than particles locked into a solid lattice.

Entropy comparison for solids, liquids, gases and gas mole changes

Entropy of solids, liquids and gases

For the same substance, entropy usually increases in this order:

Ssolid<Sliquid<SgasS_{\text{solid}} < S_{\text{liquid}} < S_{\text{gas}}

This is why melting and boiling have positive entropy changes: particles become more free to move, and energy becomes more dispersed.

Key Idea

State and entropy

Gases have much larger entropy values than liquids or solids because gas particles are much more spread out and have far more possible arrangements.

Example

Predicting the sign of an entropy change

Predict the sign of $ΔS\Delta S$ for:

$H2O(l)→H2O(g)\mathrm{H_2O(l)} \to \mathrm{H_2O(g)}$

Compare the physical states: the reactant is a liquid and the product is a gas.
A gas has much greater entropy than a liquid because its particles are more spread out and move more freely.
Therefore the entropy change is positive: $ΔS>0\Delta S > 0$ .

Entropy changes in reactions involving gases

A very important A-Level shortcut is to look at the number of gaseous molecules.

If a reaction produces more moles of gas than it uses up, entropy usually increases. If it uses up more moles of gas than it produces, entropy usually decreases.

For example:

CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}

This produces a gas from solids, so $ΔS\Delta S$ is usually positive.

Tip

Fast entropy sign check

When gases are involved, count moles of gas on each side first. A change from fewer gas molecules to more gas molecules usually means $ΔS>0\Delta S > 0$ .

Common Mistake

Ignoring state symbols

Do not just count total moles in the equation. Entropy changes are often dominated by gases, so state symbols such as (s), (l), (g) and (aq) matter.

Example

Using gaseous moles to predict entropy

Predict the likely sign of $ΔS\Delta S$ for the Haber process:

$N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$

Count gaseous moles on the reactant side: 1 mol of $N2\mathrm{N_2}$ plus 3 mol of $H2\mathrm{H_2}$ gives 4 mol of gas.
Count gaseous moles on the product side: 2 mol of $NH3\mathrm{NH_3}$ gives 2 mol of gas.
The reaction goes from 4 mol of gas to 2 mol of gas, so disorder decreases and $ΔS\Delta S$ is likely to be negative.

Calculating entropy change, $ΔS\Delta S$

Entropy values are usually given as standard molar entropies, $S∘S^\circ$ , in J K⁻¹ mol⁻¹.

Definition

Standard molar entropy

The standard molar entropy, $S∘S^\circ$ , is the entropy of one mole of a substance under standard conditions, measured in J K⁻¹ mol⁻¹.

To calculate the entropy change of a reaction:

ΔS=∑Sproducts−∑Sreactants\Delta S = \sum S_{\text{products}} - \sum S_{\text{reactants}}

You must multiply each entropy value by the balancing number in the equation.

Example

Calculating entropy change from entropy data

Calculate $ΔS\Delta S$ for:

$N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$

Use these standard entropy values:

$N2(g)\mathrm{N_2(g)}$ : 192 J K⁻¹ mol⁻¹
$H2(g)\mathrm{H_2(g)}$ : 131 J K⁻¹ mol⁻¹
$NH3(g)\mathrm{NH_3(g)}$ : 193 J K⁻¹ mol⁻¹

Calculate the total entropy of the products:
$mol−1\sum S_{\text{products}} = 2 \times 193 = 386\ \text{J K}^{-1}\text{ mol}^{-1}$
Calculate the total entropy of the reactants:
$mol−1\sum S_{\text{reactants}} = 192 + \left(3 \times 131\right) = 585\ \text{J K}^{-1}\text{ mol}^{-1}$
Subtract reactants from products:
$mol−1\Delta S = 386 - 585 = -199\ \text{J K}^{-1}\text{ mol}^{-1}$
Interpret the sign: $ΔS\Delta S$ is negative, which matches the reaction going from 4 mol of gas to 2 mol of gas.

Free energy and feasibility

A process is feasible if it is thermodynamically possible under the stated conditions. OCR links feasibility to the Gibbs free energy change, $ΔG\Delta G$ .

Definition

Gibbs free energy change

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

The equation is:

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

where:

$ΔG\Delta G$ is in kJ mol⁻¹
$ΔH\Delta H$ is in kJ mol⁻¹
$T$ is temperature in K
$ΔS\Delta S$ must be in kJ K⁻¹ mol⁻¹ when used in the equation

A process is feasible when:

ΔG<0\Delta G < 0

At $ΔG=0\Delta G = 0$ , the system is at equilibrium under those conditions.

Gibbs free energy sign cases and temperature dependence

Common Mistake

Mixing entropy units

Entropy data are usually given in J K⁻¹ mol⁻¹, but $ΔH\Delta H$ and $ΔG\Delta G$ are usually in kJ mol⁻¹. Divide $ΔS\Delta S$ by 1000 before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ .

Example

Calculating Gibbs free energy

For the Haber process at 298 K:

$N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$

Use $mol−1\Delta H = -92.0\ \text{kJ mol}^{-1}$ and $mol−1\Delta S = -199\ \text{J K}^{-1}\text{ mol}^{-1}$ . Calculate $ΔG\Delta G$ and decide whether the reaction is feasible.

Convert entropy into kJ K⁻¹ mol⁻¹:
$mol−1\Delta S = \frac{-199}{1000} = -0.199\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute into the Gibbs equation:
$ΔG=−92.0−(298×−0.199)\Delta G = -92.0 - \left(298 \times -0.199\right)$
Calculate the temperature–entropy term and final value:
$mol−1\Delta G = -92.0 + 59.3 = -32.7\ \text{kJ mol}^{-1}$
Interpret the sign: $ΔG\Delta G$ is negative, so the reaction is thermodynamically feasible at 298 K.

How temperature affects feasibility

The term $TΔST\Delta S$ means temperature can change whether a process is feasible.

There are four common sign combinations:

$ΔH<0\Delta H < 0$ and $ΔS>0\Delta S > 0$ : always feasible.
$ΔH>0\Delta H > 0$ and $ΔS<0\Delta S < 0$ : never feasible.
$ΔH<0\Delta H < 0$ and $ΔS<0\Delta S < 0$ : feasible at low temperature.
$ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ : feasible at high temperature.

This explains why some endothermic reactions can happen: a large positive entropy change can outweigh the positive enthalpy change at high temperature.

Example

Finding the temperature for feasibility

Calcium carbonate decomposes as follows:

$CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}$

Use $mol−1\Delta H = +178\ \text{kJ mol}^{-1}$ and $mol−1\Delta S = +161\ \text{J K}^{-1}\text{ mol}^{-1}$ . Estimate the temperature above which the reaction becomes feasible.

At the boundary between feasible and not feasible, set $ΔG=0\Delta G = 0$ :
$\Delta H - T\Delta S$
Rearrange for temperature:
$\frac{\Delta H}{\Delta S}$
Convert $ΔS\Delta S$ into kJ K⁻¹ mol⁻¹:
$mol−1\Delta S = \frac{161}{1000} = 0.161\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute values:
$\frac{178}{0.161} = 1106\ \text{K}$
Interpret the result: because $ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ , the reaction is feasible at temperatures above about 1110 K.

Limitations: feasibility is not the same as rate

A negative $ΔG\Delta G$ tells you that a process is thermodynamically feasible. It does not tell you whether it happens quickly.

Many feasible reactions are extremely slow because they have a high activation energy, $E_a$ . Particles may not have enough energy to start the reaction, even if the products are more stable overall.

Definition

Activation energy

Activation energy, $E_a$ , is the minimum energy particles need for a successful reaction to occur.

Example

Feasibility versus rate

Hydrogen and oxygen can react to form water:

$2H2(g)+O2(g)→2H2O(l)2\mathrm{H_2(g)} + \mathrm{O_2(g)} \to 2\mathrm{H_2O(l)}$

The reaction has a large negative $ΔH\Delta H$ , so forming water is energetically favourable.
The overall $ΔG\Delta G$ is negative under normal conditions, so the reaction is thermodynamically feasible.
However, a mixture of hydrogen and oxygen can remain unchanged until a spark is provided because the activation energy is high.
The spark helps particles overcome $E_a$ , so the reaction then occurs rapidly.

Common Mistake

What ΔG cannot tell you

$ΔG\Delta G$ predicts thermodynamic feasibility, not reaction speed. For rate, you need kinetic ideas such as activation energy, collision frequency and catalysts.

Exam technique

In the exam

Always check units before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ : 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.
For entropy predictions, look first at state symbols and the change in moles of gas.
After calculating $ΔG\Delta G$ , state the conclusion clearly: negative means feasible; positive means not feasible under those conditions; zero means equilibrium.

Self review

Check yourself

Why does a gas usually have a much higher entropy than a solid?
In $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , why must temperature be in K rather than °C?
Give one reason why a reaction with negative $ΔG\Delta G$ might still appear not to happen.

What you'll learn

What entropy means, and why gases usually have much higher entropy than solids.
How to calculate an entropy change, $ΔS\Delta S$ , from entropy data.
How enthalpy, entropy and temperature combine in the Gibbs equation.
Why a negative $ΔG\Delta G$ predicts thermodynamic feasibility, but not necessarily a fast reaction.

The starting point: enthalpy changes

You already know that reactions involve energy transfers. The enthalpy change, $ΔH\Delta H$ , is the heat energy change of a reaction at constant pressure.

$ΔH<0\Delta H < 0$ : exothermic, energy is released to the surroundings.
$ΔH>0\Delta H > 0$ : endothermic, energy is taken in from the surroundings.

For this topic, enthalpy is only part of the story. Some endothermic processes still happen, so we need another idea: entropy.

Definition

Enthalpy change

The enthalpy change, $ΔH\Delta H$ , is the heat energy change of a system at constant pressure, usually measured in kJ mol⁻¹.

Entropy: energy dispersal and disorder

Entropy is about how spread out energy is, and how many ways particles and energy can be arranged.

Definition

Entropy

Entropy, $S$ , is a measure of the dispersal of energy in a system. A system has greater entropy when it is more disordered.

A “system” means the chemicals or process you are focusing on. For example, in a reaction mixture, the reactants and products are the system.

Entropy comparison for solids, liquids, gases and gas mole changes

Entropy of solids, liquids and gases

For the same substance, entropy usually increases in this order:

Ssolid<Sliquid<SgasS_{\text{solid}} < S_{\text{liquid}} < S_{\text{gas}}

This is why melting and boiling have positive entropy changes: particles become more free to move, and energy becomes more dispersed.

Key Idea

State and entropy

Gases have much larger entropy values than liquids or solids because gas particles are much more spread out and have far more possible arrangements.

Example

Predicting the sign of an entropy change

Predict the sign of $ΔS\Delta S$ for:

$H2O(l)→H2O(g)\mathrm{H_2O(l)} \to \mathrm{H_2O(g)}$

Compare the physical states: the reactant is a liquid and the product is a gas.
A gas has much greater entropy than a liquid because its particles are more spread out and move more freely.
Therefore the entropy change is positive: $ΔS>0\Delta S > 0$ .

Entropy changes in reactions involving gases

A very important A-Level shortcut is to look at the number of gaseous molecules.

If a reaction produces more moles of gas than it uses up, entropy usually increases. If it uses up more moles of gas than it produces, entropy usually decreases.

For example:

CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}

This produces a gas from solids, so $ΔS\Delta S$ is usually positive.

Tip

Fast entropy sign check

When gases are involved, count moles of gas on each side first. A change from fewer gas molecules to more gas molecules usually means $ΔS>0\Delta S > 0$ .

Common Mistake

Ignoring state symbols

Do not just count total moles in the equation. Entropy changes are often dominated by gases, so state symbols such as (s), (l), (g) and (aq) matter.

Example

Using gaseous moles to predict entropy

Predict the likely sign of $ΔS\Delta S$ for the Haber process:

$N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$

Count gaseous moles on the reactant side: 1 mol of $N2\mathrm{N_2}$ plus 3 mol of $H2\mathrm{H_2}$ gives 4 mol of gas.
Count gaseous moles on the product side: 2 mol of $NH3\mathrm{NH_3}$ gives 2 mol of gas.
The reaction goes from 4 mol of gas to 2 mol of gas, so disorder decreases and $ΔS\Delta S$ is likely to be negative.

Calculating entropy change, $ΔS\Delta S$

Entropy values are usually given as standard molar entropies, $S∘S^\circ$ , in J K⁻¹ mol⁻¹.

Definition

Standard molar entropy

The standard molar entropy, $S∘S^\circ$ , is the entropy of one mole of a substance under standard conditions, measured in J K⁻¹ mol⁻¹.

To calculate the entropy change of a reaction:

ΔS=∑Sproducts−∑Sreactants\Delta S = \sum S_{\text{products}} - \sum S_{\text{reactants}}

You must multiply each entropy value by the balancing number in the equation.

Example

Calculating entropy change from entropy data

Calculate $ΔS\Delta S$ for:

$N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$

Use these standard entropy values:

$N2(g)\mathrm{N_2(g)}$ : 192 J K⁻¹ mol⁻¹
$H2(g)\mathrm{H_2(g)}$ : 131 J K⁻¹ mol⁻¹
$NH3(g)\mathrm{NH_3(g)}$ : 193 J K⁻¹ mol⁻¹

Calculate the total entropy of the products:
$mol−1\sum S_{\text{products}} = 2 \times 193 = 386\ \text{J K}^{-1}\text{ mol}^{-1}$
Calculate the total entropy of the reactants:
$mol−1\sum S_{\text{reactants}} = 192 + \left(3 \times 131\right) = 585\ \text{J K}^{-1}\text{ mol}^{-1}$
Subtract reactants from products:
$mol−1\Delta S = 386 - 585 = -199\ \text{J K}^{-1}\text{ mol}^{-1}$
Interpret the sign: $ΔS\Delta S$ is negative, which matches the reaction going from 4 mol of gas to 2 mol of gas.

Free energy and feasibility

A process is feasible if it is thermodynamically possible under the stated conditions. OCR links feasibility to the Gibbs free energy change, $ΔG\Delta G$ .

Definition

Gibbs free energy change

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

The equation is:

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

where:

$ΔG\Delta G$ is in kJ mol⁻¹
$ΔH\Delta H$ is in kJ mol⁻¹
$T$ is temperature in K
$ΔS\Delta S$ must be in kJ K⁻¹ mol⁻¹ when used in the equation

A process is feasible when:

ΔG<0\Delta G < 0

At $ΔG=0\Delta G = 0$ , the system is at equilibrium under those conditions.

Gibbs free energy sign cases and temperature dependence

Common Mistake

Mixing entropy units

Example

Calculating Gibbs free energy

For the Haber process at 298 K:

$N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g)} + 3\mathrm{H_2(g)} \to 2\mathrm{NH_3(g)}$

Use $mol−1\Delta H = -92.0\ \text{kJ mol}^{-1}$ and $mol−1\Delta S = -199\ \text{J K}^{-1}\text{ mol}^{-1}$ . Calculate $ΔG\Delta G$ and decide whether the reaction is feasible.

Convert entropy into kJ K⁻¹ mol⁻¹:
$mol−1\Delta S = \frac{-199}{1000} = -0.199\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute into the Gibbs equation:
$ΔG=−92.0−(298×−0.199)\Delta G = -92.0 - \left(298 \times -0.199\right)$
Calculate the temperature–entropy term and final value:
$mol−1\Delta G = -92.0 + 59.3 = -32.7\ \text{kJ mol}^{-1}$
Interpret the sign: $ΔG\Delta G$ is negative, so the reaction is thermodynamically feasible at 298 K.

How temperature affects feasibility

The term $TΔST\Delta S$ means temperature can change whether a process is feasible.

There are four common sign combinations:

$ΔH<0\Delta H < 0$ and $ΔS>0\Delta S > 0$ : always feasible.
$ΔH>0\Delta H > 0$ and $ΔS<0\Delta S < 0$ : never feasible.
$ΔH<0\Delta H < 0$ and $ΔS<0\Delta S < 0$ : feasible at low temperature.
$ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ : feasible at high temperature.

This explains why some endothermic reactions can happen: a large positive entropy change can outweigh the positive enthalpy change at high temperature.

Example

Finding the temperature for feasibility

Calcium carbonate decomposes as follows:

$CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \to \mathrm{CaO(s)} + \mathrm{CO_2(g)}$

Use $mol−1\Delta H = +178\ \text{kJ mol}^{-1}$ and $mol−1\Delta S = +161\ \text{J K}^{-1}\text{ mol}^{-1}$ . Estimate the temperature above which the reaction becomes feasible.

At the boundary between feasible and not feasible, set $ΔG=0\Delta G = 0$ :
$\Delta H - T\Delta S$
Rearrange for temperature:
$\frac{\Delta H}{\Delta S}$
Convert $ΔS\Delta S$ into kJ K⁻¹ mol⁻¹:
$mol−1\Delta S = \frac{161}{1000} = 0.161\ \text{kJ K}^{-1}\text{ mol}^{-1}$
Substitute values:
$\frac{178}{0.161} = 1106\ \text{K}$
Interpret the result: because $ΔH>0\Delta H > 0$ and $ΔS>0\Delta S > 0$ , the reaction is feasible at temperatures above about 1110 K.

Limitations: feasibility is not the same as rate

A negative $ΔG\Delta G$ tells you that a process is thermodynamically feasible. It does not tell you whether it happens quickly.

Definition

Activation energy

Activation energy, $E_a$ , is the minimum energy particles need for a successful reaction to occur.

Example

Feasibility versus rate

Hydrogen and oxygen can react to form water:

$2H2(g)+O2(g)→2H2O(l)2\mathrm{H_2(g)} + \mathrm{O_2(g)} \to 2\mathrm{H_2O(l)}$

The reaction has a large negative $ΔH\Delta H$ , so forming water is energetically favourable.
The overall $ΔG\Delta G$ is negative under normal conditions, so the reaction is thermodynamically feasible.
However, a mixture of hydrogen and oxygen can remain unchanged until a spark is provided because the activation energy is high.
The spark helps particles overcome $E_a$ , so the reaction then occurs rapidly.

Common Mistake

What ΔG cannot tell you

$ΔG\Delta G$ predicts thermodynamic feasibility, not reaction speed. For rate, you need kinetic ideas such as activation energy, collision frequency and catalysts.

Exam technique

In the exam

Always check units before using $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ : 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.
For entropy predictions, look first at state symbols and the change in moles of gas.
After calculating $ΔG\Delta G$ , state the conclusion clearly: negative means feasible; positive means not feasible under those conditions; zero means equilibrium.

Self review

Check yourself

Why does a gas usually have a much higher entropy than a solid?
In $ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S$ , why must temperature be in K rather than °C?
Give one reason why a reaction with negative $ΔG\Delta G$ might still appear not to happen.

What you'll learn

The starting point: enthalpy changes

Entropy: energy dispersal and disorder

Entropy of solids, liquids and gases

Entropy changes in reactions involving gases

Calculating entropy change, $ΔS\Delta S$

Free energy and feasibility

How temperature affects feasibility

Limitations: feasibility is not the same as rate

Recap questions

Review Enthalpy and entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Enthalpy and entropy

What you'll learn

The starting point: enthalpy changes

Entropy: energy dispersal and disorder

Entropy of solids, liquids and gases

Entropy changes in reactions involving gases

Calculating entropy change, $ΔS\Delta S$

Free energy and feasibility

How temperature affects feasibility

Limitations: feasibility is not the same as rate

Recap questions

Review Enthalpy and entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Enthalpy and entropy

What you'll learn

The starting point: enthalpy changes

Entropy: energy dispersal and disorder

Entropy of solids, liquids and gases

Entropy changes in reactions involving gases

Calculating entropy change, ΔS\Delta SΔS

Free energy and feasibility

How temperature affects feasibility

Limitations: feasibility is not the same as rate

Recap questions

Review Enthalpy and entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Enthalpy and entropy Revision Guide

Enthalpy and entropy

What you'll learn

The starting point: enthalpy changes

Entropy: energy dispersal and disorder

Entropy of solids, liquids and gases

Entropy changes in reactions involving gases

Calculating entropy change, ΔS\Delta SΔS

Free energy and feasibility

How temperature affects feasibility

Limitations: feasibility is not the same as rate

Recap questions

Review Enthalpy and entropy by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Enthalpy and entropy Revision Guide

Calculating entropy change, $ΔS\Delta S$

Calculating entropy change, $ΔS\Delta S$