Entropy — AQA A-level Chemistry

SPEC 3.1.8.2 - THERMODYNAMICS

Entropy & Gibbs Free Energy

Understanding disorder, spontaneity, and predicting whether reactions are thermodynamically feasible

Introduction

You may have wondered why endothermic reactions occur at all. After all, what can be the driving force behind endothermic reactions if the products end up in a less stable, higher energy state?

The driving force behind chemical reactions cannot be explained by enthalpy changes alone - it makes no sense for a chemical to end up in a less stable, higher energy state. The answer is entropy.

Learning Objectives

Define entropy and understand what it measures

Predict whether entropy increases or decreases in chemical and physical changes

Calculate entropy change using standard molar entropy values

Apply the Gibbs free energy equation: ΔG = ΔH − TΔS

Determine whether a reaction is thermodynamically feasible

Calculate the temperature at which a reaction becomes feasible

What is Entropy?

Definition

Entropy (S) is the number of possible arrangements of the particles and their energy in a given system. In other words, it is a measure of how disordered or chaotic a system is.

When a system becomes more disordered, its entropy will increase. Crucially, an increase in entropy means that the system becomes energetically more stable.

The Second Law of Thermodynamics

The entropy of the universe is always increasing. This is why:

Cups fall off tables and break into many pieces - never the other way around

Hot objects cool and spread their heat into the surroundings - never the reverse

Living things die and decompose into disordered simple molecules

The system with higher entropy will be energetically favourable because energy is more spread out.

Why Entropy Matters

This drive for disorder can make certain reactions feasible even when the enthalpy change is endothermic. For example:

NaHCO₃(s) + HCl(aq) → NaCl(aq) + CO₂(g) + H₂O(l)

This reaction is endothermic but still feasible because producing gaseous CO₂ greatly increases entropy, overcoming the unfavourable enthalpy change.

Units

Entropy is measured in J K⁻¹ mol⁻¹ (joules per kelvin per mole).

Note: This is joules, NOT kilojoules! This becomes critically important when we do Gibbs free energy calculations later.

Entropy and States of Matter

The physical state of a substance has the biggest effect on its entropy. This is because particles have different freedom of movement in each state.

Key Pattern

S_{solid} < S_{liquid} ≪ S_{gas}

Gases have much higher entropy than solids or liquids because particles can move freely throughout the entire container with countless possible arrangements.

Phase Changes

When a substance melts or evaporates, there is a sudden increase in entropy:

ΔS_vaporisation >> ΔS_fusion

The entropy change of vaporisation is much greater than fusion because a gas is much more disordered than a liquid or a solid.

Entropy vs Temperature

The graph below shows how entropy changes with temperature for a substance. Notice the discontinuous jumps at phase transitions.

Key Observations

Within each phase: Entropy increases gradually as temperature rises (particles gain kinetic energy)

At melting point: Small jump in entropy (ΔS_fus) - particles gain some freedom

At boiling point: Large jump in entropy (ΔS_vap) - particles become completely free

ΔS_vap >> ΔS_fus because the change from liquid to gas involves much greater increase in disorder

Factors Affecting Entropy

Several factors determine the entropy of a system. Understanding these helps you predict whether entropy increases or decreases in a reaction.

FactorEffect on EntropyExample

State change (s → l → g)Entropy increasesH₂O(l) → H₂O(g)

Increase in temperatureEntropy increasesMore particle movement

More moles of gas producedEntropy increases2H₂O₂ → 2H₂O + O₂

More particles producedEntropy increasesCaCO₃ → CaO + CO₂

Dissolving ionic solidUsually increasesNaCl(s) → Na⁺(aq) + Cl⁻(aq)

Examiner Tip

Don't confuse the system with the surroundings! The system consists of the molecules that are reacting in a chemical reaction. The surroundings are everything else such as the solvent, the air around the reaction, test-tube, etc.

Dissolution and Entropy

When an ionic solid dissolves in water, the ions become dispersed throughout the solution:

NaCl(s) → Na⁺(aq) + Cl⁻(aq)

The ions in solution have more freedom of movement and more possible arrangements than in the rigid crystal lattice, so entropy usually increases. This positive ΔS can make dissolution feasible even when ΔH is endothermic.

Important

When counting moles of gas, only count gaseous species. Solids and liquids have much lower entropy than gases, so changes in moles of gas usually dominate the overall entropy change.

01Worked Example

Predict whether entropy increases or decreases for each reaction:

(a) CaCO₃(s) → CaO(s) + CO₂(g)

(b) 2Mg(s) + O₂(g) → 2MgO(s)

(c) N₂(g) + 3H₂(g) → 2NH₃(g)

(a)

ΔS is positive (entropy increases)

1 mol solid → 1 mol solid + 1 mol gas. Producing a gas greatly increases entropy.

(b)

ΔS is negative (entropy decreases)

Moles of gas: 1 → 0. Gas is consumed and no gases produced.

(c)

ΔS is negative (entropy decreases)

Moles of gas: 4 → 2. Fewer gas molecules = fewer arrangements = lower entropy.

Calculating Entropy Change

Standard Entropy

Standard entropy (S°) is the entropy of 1 mole of a substance under standard conditions (100 kPa pressure and 298 K temperature). Its units are J K⁻¹ mol⁻¹.

Entropy changes are an order of magnitude smaller than enthalpy changes, so entropy is measured in joules rather than kilojoules.

The standard entropy change (ΔS°) for a reaction can be calculated using:

Formula

Δ S ° = \sum S °_{products} - \sum S °_{reactants}

Steps for Calculating ΔS

Find the sum of standard entropies for the products (ΣS°_products)

Find the sum of standard entropies for the reactants (ΣS°_reactants)

Subtract the sum of reactant entropies from the sum of product entropies

If the balanced equation has coefficients other than 1, multiply the standard entropy of each substance by its coefficient.

Important Difference from Enthalpy

Unlike enthalpy of formation, entropy for elements is NOT zero. You can find entropy values for elements AND compounds in data books. This is because all substances have some disorder at temperatures above absolute zero.

Typical S° Values

Gases: 150–250 J K⁻¹ mol⁻¹ | Liquids: 60–150 J K⁻¹ mol⁻¹ | Solids: 20–80 J K⁻¹ mol⁻¹

Use these ranges to check your answers make sense!

02Worked Example

Calculate the entropy change for the reaction between hydrogen chloride and ammonia:

HCl(g) + NH₃(g) → NH₄Cl(s)

SubstanceS° / J K⁻¹ mol⁻¹

HCl(g)186.9

NH₃(g)192.8

NH₄Cl(s)94.6

1

Calculate entropy of products:

S°_products = S°[NH₄Cl] = 94.6 J K⁻¹ mol⁻¹

2

Calculate entropy of reactants:

S°_reactants = S°[HCl] + S°[NH₃]

S°_reactants = 186.9 + 192.8 = 379.7 J K⁻¹ mol⁻¹

3

Apply the formula:

ΔS = S°_products − S°_reactants

ΔS = 94.6 − 379.7

Answer: ΔS = −285.1 J K⁻¹ mol⁻¹

The negative ΔS indicates a decrease in entropy, as expected when two gases combine to form a solid.

03Worked Example

Calculate ΔS for the Haber process:

N₂(g) + 3H₂(g) → 2NH₃(g)

SubstanceS° / J K⁻¹ mol⁻¹

N₂(g)192

H₂(g)131

NH₃(g)193

1

Account for stoichiometric coefficients:

Products: 2 mol NH₃

Reactants: 1 mol N₂ + 3 mol H₂

2

Apply the formula:

ΔS = [2 × S°(NH₃)] − [S°(N₂) + 3 × S°(H₂)]

ΔS = [2 × 193] − [192 + 3 × 131]

ΔS = 386 − [192 + 393]

ΔS = 386 − 585

Answer: ΔS = −199 J K⁻¹ mol⁻¹

Negative ΔS because moles of gas decrease from 4 to 2.

Gibbs Free Energy

The feasibility of a reaction is determined by two factors: the enthalpy change and the entropy change. These come together in a fundamental thermodynamic concept called the Gibbs free energy (G).

The Gibbs Equation

Δ G ° = Δ H °_{reaction} - T Δ S °_{system}

SymbolQuantityUnit

ΔG°Standard Gibbs free energy changekJ mol⁻¹

ΔH°Standard enthalpy changekJ mol⁻¹

TTemperatureK (Kelvin)

ΔS°Standard entropy changekJ K⁻¹ mol⁻¹ (convert!)

Two Ways to Calculate ΔG°

Method 1: From ΔH° and ΔS° values using ΔG° = ΔH° − TΔS°

Method 2: From ΔG°_f values: ΔG° = ΣΔG°_f(products) − ΣΔG°_f(reactants)

Critical Unit Conversion

Most Common Exam Error

ΔS is usually given in J K⁻¹ mol⁻¹ but must be converted to kJ K⁻¹ mol⁻¹ before using in the Gibbs equation!

Divide ΔS by 1000

QuantityUsually Given AsNeeded in EquationConversion

ΔHkJ mol⁻¹kJ mol⁻¹None needed

ΔSJ K⁻¹ mol⁻¹kJ K⁻¹ mol⁻¹÷ 1000

T°C or KK+ 273 (if in °C)

Checklist

Convert ΔS from J to kJ (÷ 1000)

Convert temperature to Kelvin (+ 273 if in °C)

Substitute into ΔG = ΔH − TΔS

Check the sign of your answer

Determining Feasibility

The sign of ΔG tells us whether a reaction can happen spontaneously under the given conditions.

Value of ΔGFeasibilityMeaning

ΔG < 0FEASIBLEReaction can occur spontaneously

ΔG > 0NOT FEASIBLEReaction cannot occur spontaneously

ΔG = 0EQUILIBRIUMSystem is balanced, no net change

Feasible vs Spontaneous

Feasible tends to describe reactions which are energetically favourable - reactions that should go.

Spontaneous tends to describe reactions that go of their own accord.

Critical Understanding

Feasibility takes no account of the rate of reaction and states only what is possible, not what actually happens. A feasible reaction might be incredibly slow, such as the rusting of iron.

04Worked Example

Calculate ΔG at 298 K for the decomposition of calcium carbonate and determine if it is feasible.

Given: ΔH = +178 kJ mol⁻¹, ΔS = +161 J K⁻¹ mol⁻¹

1

Convert ΔS to kJ K⁻¹ mol⁻¹:

ΔS = +161 J K⁻¹ mol⁻¹ = +0.161 kJ K⁻¹ mol⁻¹

2

Substitute into the equation:

ΔG = ΔH − TΔS

ΔG = (+178) − (298 × 0.161)

ΔG = 178 − 48.0

3

ΔG = +130 kJ mol⁻¹

This explains why limestone doesn't spontaneously decompose at room temperature!

AnswerΔG is positive, so the reaction is NOT feasible at 298 K.

05Worked Example

For the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

ΔH = −890 kJ mol⁻¹, ΔS = −243 J K⁻¹ mol⁻¹. Is this feasible at 298 K?

1

Convert ΔS: −243 J K⁻¹ mol⁻¹ = −0.243 kJ K⁻¹ mol⁻¹

2

Calculate ΔG:

ΔG = ΔH − TΔS

ΔG = (−890) − (298 × −0.243)

ΔG = −890 − (−72.4)

ΔG = −890 + 72.4 = −818 kJ mol⁻¹

Despite the decrease in entropy, the highly exothermic enthalpy (−890 kJ) dominates.

AnswerΔG is negative, so the reaction IS feasible at 298 K.

Temperature and Feasibility

The temperature term (TΔS) in the Gibbs equation means that feasibility can change with temperature. We can find the minimum temperature at which a reaction becomes feasible.

At the Boundary

At the exact temperature where a reaction becomes feasible, ΔG = 0

Deriving the Temperature Formula

Starting from: ΔG = ΔH − TΔS

Set ΔG = 0: 0 = ΔH − TΔS

Rearrange: TΔS = ΔH

Formula

T = \frac{Δ H}{Δ S}

Where: T = temperature at which reaction becomes feasible (K) ΔH = enthalpy change (J mol⁻¹) ΔS = entropy change (J K⁻¹ mol⁻¹)

Critical: Unit Consistency

Both ΔH and ΔS must be in the same energy unit before dividing:

Option 1: Convert ΔH from kJ to J (×1000), keep ΔS in J K⁻¹ mol⁻¹

Option 2: Keep ΔH in kJ, convert ΔS from J to kJ (÷1000)

Either approach gives T in Kelvin. This calculation only makes sense when ΔH and ΔS have the same sign.

06Worked Example

Calculate the minimum temperature at which the decomposition of calcium carbonate becomes feasible.

CaCO₃(s) → CaO(s) + CO₂(g)

ΔH = +178 kJ mol⁻¹, ΔS = +161 J K⁻¹ mol⁻¹

1

Convert units to be consistent:

ΔS = +161 J K⁻¹ mol⁻¹ = +0.161 kJ K⁻¹ mol⁻¹

2

Apply the formula:

T = ΔH / ΔS

T = 178 / 0.161

T = 1106 K

3

Convert to °C if needed:

T = 1106 − 273 = 833°C

This is why limestone is heated in kilns at high temperatures to make quicklime!

AnswerThe reaction becomes feasible above 1106 K (833°C).

Graphing the Gibbs Equation

Rearranging the Gibbs equation allows you to determine the temperature at which a non-spontaneous reaction becomes feasible.

Straight Line Form

The Gibbs equation can be expressed as the equation for a straight line:

Δ G ° = Δ H ° - T Δ S °

Δ G ° = - Δ S ° \cdot T + Δ H °

y = m x + c

A graph of ΔG° (y-axis) versus temperature in K (x-axis) will give a straight line with:

FeatureValueMeaning

Gradient (m)−ΔS°Negative of entropy change

Y-intercept (c)ΔH°Enthalpy change

X-interceptT where ΔG = 0Temperature of feasibility change

Reading the Graph

Where the line crosses the x-axis (ΔG = 0), the reaction changes from feasible to non-feasible (or vice versa).

Example: Ammonia Synthesis

The graph below shows ΔG° vs temperature for the Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Given: ΔH° = −92 kJ mol⁻¹, ΔS° = −199 J K⁻¹ mol⁻¹

Key Observations

Y-intercept (−92 kJ mol⁻¹): This equals ΔH° — the reaction is exothermic.

Positive gradient (+0.199): Since gradient = −ΔS°, this means ΔS° is negative (4 mol gas → 2 mol gas).

X-intercept (462 K): Above this temperature, ΔG° becomes positive and the forward reaction is not feasible.

The Four Scenarios

The signs of ΔH and ΔS determine whether a reaction is always feasible, never feasible, or depends on temperature.

ΔHΔSSpontaneous?Example

Negative (−) exothermicPositive (+) more disorderAlwaysCombustion reactions

Positive (+) endothermicNegative (−) more orderNeverReverse spontaneous reactions

Negative (−) exothermicNegative (−) more orderAt LOW TH₂O(l) → H₂O(s) Freezing of water

Positive (+) endothermicPositive (+) more disorderAt HIGH TCaCO₃ → CaO + CO₂ Thermal decomposition

Understanding with Numbers

CaCO₃ → CaO + CO₂: ΔH = +178 kJ mol⁻¹, ΔS = +161 J K⁻¹ mol⁻¹

At 300 K: ΔG = +178 − (300 × 0.161) = +130 kJ mol⁻¹ → Not feasible

At 1200 K: ΔG = +178 − (1200 × 0.161) = −15 kJ mol⁻¹ → Feasible

Limitations of Gibbs Free Energy

While ΔG tells us if a reaction can happen, it doesn't tell us if it will happen or how fast.

Key Distinction

Thermodynamic feasibility (ΔG < 0) means a reaction is energetically possible.

Kinetic feasibility depends on activation energy and reaction rate.

Classic Example: Diamond → Graphite

For: C(diamond) → C(graphite), ΔG = −3 kJ mol⁻¹

Thermodynamics says:Reality shows:

ΔG is negative Reaction IS feasible Graphite is more stableDiamonds don't turn into graphite The reaction is incredibly slow Activation energy is too high

Exam Point

If asked why a feasible reaction doesn't occur, answer: "The activation energy is too high" or "The rate of reaction is too slow to observe."

Practice

Question 1:

2H₂(g) + O₂(g) → 2H₂O(l)

SubstanceS° / J K⁻¹ mol⁻¹

H₂(g)131

O₂(g)205

H₂O(l)70

1

ΔS = ΣS°(products) − ΣS°(reactants)

2

ΔS = [2 × 70] − [2 × 131 + 205]

ΔS = 140 − [262 + 205]

ΔS = 140 − 467

Answer: ΔS = −327 J K⁻¹ mol⁻¹

Entropy decreases because 3 moles of gas become 2 moles of liquid.

Practice

Question 2:

ΔH = +57.2 kJ mol⁻¹ and ΔS = +176 J K⁻¹ mol⁻¹

(a) Calculate ΔG at 298 K and state whether the reaction is feasible.

(b) Calculate the temperature above which the reaction becomes feasible.

(a)

Convert ΔS: +176 J K⁻¹ mol⁻¹ = +0.176 kJ K⁻¹ mol⁻¹

ΔG = ΔH − TΔS = 57.2 − (298 × 0.176) = 57.2 − 52.4

ΔG = +4.8 kJ mol⁻¹ → Not feasible at 298 K

(b)

At the boundary, ΔG = 0:

T = ΔH / ΔS = 57.2 / 0.176

T = 325 K (or 52°C)

The reaction becomes feasible above 325 K.

Practice

Question 3:

2Fe₂O₃(s) + 3C(s) → 4Fe(s) + 3CO₂(g)

ΔH = +467 kJ mol⁻¹, ΔS = +560 J K⁻¹ mol⁻¹

(a) Calculate the minimum temperature for this reaction to be feasible.

(b) Explain why this reaction is carried out in a blast furnace at high temperatures.

(a)

Convert ΔS: +560 J K⁻¹ mol⁻¹ = +0.560 kJ K⁻¹ mol⁻¹

T = ΔH / ΔS = 467 / 0.560

T = 834 K (or 561°C)

(b)

The reaction is endothermic (+ΔH) but has positive entropy change (+ΔS) due to gas production. At high temperatures, the TΔS term becomes large enough to make ΔG negative. The blast furnace operates above 834 K to ensure thermodynamic feasibility.

Summary

Key Formulas

FormulaUse

Δ S = \sum S °_{products} - \sum S °_{reactants}

Calculate entropy change

Δ G = Δ H - T Δ S

Calculate Gibbs free energy

T = \frac{Δ H}{Δ S}

Find temperature for feasibility

Key Exam Skills

Always convert ΔS from J to kJ (÷1000) before using ΔG = ΔH − TΔS

Temperature must be in Kelvin (+273 if given in °C)

ΔG < 0 means feasible; ΔG > 0 means not feasible

Feasible does not mean fast - activation energy determines rate

S(solid) < S(liquid) < S(gas)