Entropy

Specification Reference Topic 13B, points 12–22

Quick Notes

Entropy (S) measures disorder, higher entropy = more disorder.
Entropy increases during:
- Change from solid to liquid to gas.
- Dissolving solids.
- Increase in number of gas molecules.
ΔS_total = ΔS_system + ΔS_surroundings
ΔS_system = ΣS_products – ΣS_reactants
ΔS_surroundings = –ΔH / T (T in K)
Gibbs free energy (ΔG) determines feasibility
- ΔG = ΔH – TΔS_system
- If ΔG < 0, the reaction is thermodynamically feasible.
- Minimum temperature at which a reaction is feasibile (when ΔG = 0):
  T = ΔH / ΔS_system
ΔG = –RT ln K links Gibbs Free Energy and the equilibrium constant, K.
- A negative ΔG gives a large K and reaction favours forward direction (products).
Some feasible reactions (ΔG < 0) may not happen due to kinetic factors (e.g. high activation energy).

Full Notes

Entropy (symbol S) is a measure of the amount of disorder in a system, or the number of ways particles and their energy can be arranged.

A system with more possible arrangements (e.g. gases) has higher entropy than one with fewer (e.g. solids).

It has units of J mol⁻¹ K⁻¹.

The natural direction of change is towards greater disorder, so spontaneous processes involve an increase in total entropy.

In chemistry, we are interested in how entropy changes (ΔS) during a reaction or process.

Entropy change (ΔS)

We can predict entropy changes based on the type of process or reaction occuring:

Changes of state

Edexcel A-Level Chemistry diagram showing entropy increasing from solid to liquid to gas as particle disorder increases.

Going from a solid to a liquid or a gas increases entropy because particles move more freely and can be arranged in more possible ways.

Dissolving ionic solids

Edexcel A-Level Chemistry diagram showing entropy increase when an ionic solid dissolves into aqueous ions.

Dissolving a crystalline solid into solution increases disorder.

More gas particles formed

Edexcel A-Level Chemistry diagram highlighting larger entropy increases when the number of gaseous molecules rises.

Reactions where the number of gaseous molecules increases show a big jump in entropy.

The opposite is also true for all the above.

Total Entropy Change

The overall entropy change of a reaction or process is made of two parts:

ΔS_system , the entropy change from the reaction itself.
ΔS_surroundings , the entropy change due to heat energy exchanged with surroundings.

These combine as:
ΔS_total = ΔS_system + ΔS_surroundings

If ΔS_total > 0, the reaction is thermodynamically feasible (can happen).

Calculating ΔS_system

Every substance, in a given state, has a standard entropy value (S°).

We can use the standard entropy values (S°) for each substance in a reaction (usually given in tables) to determine the entropy change, ΔS, that occurs for the reacting system (ΔS_system).

Formula:

Edexcel A-Level Chemistry calculating entropy change ΔS° = Σ(entropies of products) − Σ(entropies of reactants).

Matt’s exam tip

Don’t forget the molar ratios of everything in the equation and keep workings very clear. Also, make sure you are using a substance’s S° value for the correct state (for example, H₂O(l) has a different S° to H₂O(g)).

Worked Example

Calculate ΔS_system for the following reaction:
2H₂(g) + O₂(g) → 2H₂O(l)

Given:
S°(H₂O(l)) = 70 J mol⁻¹ K⁻¹
S°(H₂(g)) = 131 J mol⁻¹ K⁻¹
S°(O₂(g)) = 205 J mol⁻¹ K⁻¹

Write the expression:
ΔS = ΣS°(products) – ΣS°(reactants)
Substitute values:
ΔS = [2 × 70] – [2 × 131 + 1 × 205]
Work out the totals:
ΔS = 140 – (262 + 205)
Final calculation:
ΔS = 140 – 467 = −327 J mol⁻¹ K⁻¹

Answer: Entropy decreases in this reaction.

How to Calculate ΔS_surroundings

When a chemical reaction releases or absorbs heat, it changes the entropy of the surroundings.

The greater the heat transfer and the lower the temperature, the bigger the entropy change in the surroundings. We can calculate the entropy change of surroundings using:

ΔS_surroundings = –ΔH / T

ΔH must be in J mol⁻¹ (convert from kJ if needed).
T is the temperature in Kelvin.

A large, exothermic ΔH (negative value) leads to a positive ΔS_surroundings, making the reaction more likely to be feasible.

Gibbs Free Energy (ΔG)

Both the enthalpy change (ΔH) and entropy change (ΔS) of a reaction have an impact on whether the reaction is feasibly (can happen) based on energy.

These can be linked, along with temperature (T), by something called Gibbs Free Energy Change (ΔG)

Edexcel A-Level Chemistry calculating Gibbs Free Energy Change ΔG = ΔH − TΔS with term meanings.

A reaction is feasible if ΔG is negative (ΔG < 0).
Even if ΔG < 0, a reaction may not occur if activation energy is too high.

units:

ΔG = Gibbs Free Energy change (kJ mol⁻¹)
ΔH = Enthalpy change (kJ mol⁻¹)
T = Temperature (K)
ΔS = Entropy change (J K⁻¹ mol⁻¹) (convert to kJ K⁻¹ mol⁻¹ by dividing by 1000)

Matt’s exam tip

Remember to check and convert units when using this equation! Entropy change (ΔS) is always given in J per K per mol, whereas Enthalpy change (ΔH) and Gibbs Free Energy Change (ΔG) and given in kJ per mol. Make sure to convert entropy to kJ per mol (divide by 1000).

Finding Temperature for Feasibility

You can rearrange the Gibbs equation to find the minimum temperature at which a reaction becomes feasible:

To find the minimum temperature (T) where a reaction is feasible, we can set ΔG = 0:

(If ΔG has to be zero for a reaction to be feasible, then the temperature when ΔG is zero is the minimum that it can be!)

T = ΔH / ΔS

If T is high, the reaction is only feasible at high temperatures.
If T is low, the reaction is feasible at lower temperatures.

Worked Example Minimum feasible temperature

Worked Example

A reaction has ΔH = +50 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹ (= +0.100 kJ K⁻¹ mol⁻¹). Find the minimum temperature T at which the reaction is feasible.

Use ΔG = ΔH − TΔS and set ΔG = 0 → 0 = ΔH − TΔS
Rearrange: T = ΔH / ΔS = 50 / 0.100 = 500 K

Answer: The reaction becomes feasible at T ≥ 500 K.

Link Between ΔG and K (Equilibrium Constant)

There’s a direct relationship between Gibbs free energy and equilibrium position:

ΔG = –RT ln K

Where

R = 8.314 J mol⁻¹ K⁻¹
T in K
If ΔG is very negative, K is large and the reaction lies to the right (mostly products).
If ΔG is positive, K is small and the reaction lies to the left (mostly reactants).

This shows how thermodynamic feasibility links to position of equilibrium.

Why Some Feasible Reactions Don’t Happen

Just because a reaction can happen (ΔG is negative) and is feasible doesn’t mean that it will happen - a high activation energy barrier may prevent a feasible reaction from occurring. The rate of reaction may be so slow that the reaction doesn’t happen fast enough to observe.

For example, the conversion of diamond into graphite has a negative ΔG, meaning it is thermodynamically feasible. In theory, this process should happen spontaneously.

Edexcel A-Level Chemistry diagram comparing diamond and graphite with note that conversion is feasible but kinetically hindered.

However, in reality, the transformation does not occur noticeably because it has a very high activation energy (Ea). This large energy barrier prevents the reaction from proceeding at a measurable rate under standard conditions.

This is the difference between thermodynamic feasibility (ΔG < 0) and kinetic feasibility (reaction rate fast enough to observe).

Summary

Entropy measures disorder and increases on melting, boiling and dissolving.
ΔS_total = ΔS_system + ΔS_surroundings and feasibility requires ΔS_total > 0.
ΔS_surroundings = −ΔH/T and is larger at lower temperatures for the same ΔH.
ΔG = ΔH − TΔS links enthalpy, entropy and temperature to feasibility.
At ΔG = 0, T = ΔH/ΔS gives the minimum feasible temperature.
ΔG = −RT ln K connects thermodynamics to equilibrium position.
Some thermodynamically feasible reactions do not occur because of high activation energy.