Entropy

CIE A-Level Chemistry· Unit 17: Further chemical energetics· 15 min read

1. Definition and Physical Meaning of Entropy★★☆☆☆⏱ 4 min

📘 Definition

Entropy

S

A state function that measures the number of possible microstates (ways to arrange particles and energy) in a system, commonly simplified as a measure of disorder. Higher entropy = more possible arrangements = greater disorder.

Example:

1 mole of gaseous water has higher entropy than 1 mole of liquid water, because gas particles have much more freedom of movement.

Entropy is a state function, so its value depends only on the current state of the system, not the path taken to reach that state. The second law of thermodynamics states that the total entropy of an isolated system always increases for any spontaneous process.

📐 Worked Example

Predict which substance in each pair has higher entropy: (a) 1 mol H₂O(l) at 25°C vs 1 mol H₂O(g) at 25°C; (b) 1 mol C(s, graphite) vs 1 mol C₆H₁₂O₆(s) glucose.

1
For part (a), compare physical states: entropy always increases from solid → liquid → gas. Gaseous particles have far more freedom of movement than liquid particles, so they have more possible microstates.
2
Conclusion for (a): $H_{2} O (g)$ has higher entropy.
3
For part (b), both are solid, so compare molecular size. Larger, more complex molecules have more bonds and more ways to distribute vibrational energy, so they have higher entropy than smaller simpler molecules.
4
Conclusion for (b): Glucose $C_{6} H_{12} O_{6} (s)$ has higher entropy.

2. Predicting Entropy Changes★★☆☆☆⏱ 4 min

For any process, the entropy change is calculated as $Δ S = S_{final} - S_{initial}$ . A positive $Δ S$ means entropy increases (the system becomes more disordered), while a negative $Δ S$ means entropy decreases.

Entropy increases (ΔS positive): solid → liquid → gas, increase in moles of gas, temperature increase, solid dissolves into solution
Entropy decreases (ΔS negative): gas → liquid → solid, decrease in moles of gas, solid precipitates from solution

📐 Worked Example

Predict the sign of $Δ S_{system}$ for each reaction: (a) $C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$ (b) $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$

1
For reaction (a), count moles of gas on each side: 0 mol gas on the reactant side, 1 mol gas on the product side.
2
An increase in the number of moles of gas leads to a large increase in entropy, so $Δ S_{system}$ is positive.
3
For reaction (b), count moles of gas: 1 + 3 = 4 mol gas on reactant side, 2 mol gas on product side.
4
The number of moles of gas decreases, so entropy decreases, so $Δ S_{system}$ is negative.

3. Calculating Standard Entropy Changes★★★☆☆⏱ 5 min

📘 Definition

Standard Molar Entropy

$S^{0}$

The entropy of 1 mole of a substance under standard conditions (298 K, 1 atm), with units $J K^{- 1} m o l^{- 1}$ . Unlike standard enthalpy of formation, $S^{0}$ is always positive for any pure substance above 0 K.

Example:

$S^{0} (H_{2} O (g)) = 189 J K^{- 1} m o l^{- 1}$ , $S^{0} (H_{2} O (l)) = 70 J K^{- 1} m o l^{- 1}$

The standard entropy change of a system for a reaction is calculated by subtracting the total standard entropy of reactants from the total standard entropy of products, weighted by their stoichiometric coefficients:

\Delta S^0_{\text{system}} = \sum n S^0 (\text{products}) - \sum m S^0 (\text{reactants})

📐 Worked Example

Calculate $Δ S_{system}^{0}$ for $2 S O_{2} (g) + O_{2} (g) \to 2 S O_{3} (g)$ , given $S^{0} (S O_{2} (g)) = 248 J K^{- 1} m o l^{- 1}$ , $S^{0} (O_{2} (g)) = 205 J K^{- 1} m o l^{- 1}$ , $S^{0} (S O_{3} (g)) = 257 J K^{- 1} m o l^{- 1}$ .

1
Substitute values into the formula for $Δ S_{system}^{0}$ :
2
$\Delta S^0_{\text{system}} = [2 \times S^0(SO_3)] - [2 \times S^0(SO_2) + 1 \times S^0(O_2)]$
3
Plug in the given standard entropy values:
4
$= [2(257)] - [2(248) + 205]$
5
Calculate the final result:
6
$= 514 - 701 = -187\ J\ K^{-1}\ mol^{-1}$
7
Check the result against our prediction rule: moles of gas decrease from 3 to 2, so $Δ S^{0}$ should be negative, which matches our calculation.

4. Total Entropy and Spontaneity★★★☆☆⏱ 5 min

To determine if a reaction is spontaneous, we need the total entropy change, which adds the entropy change of the system ( $Δ S_{system}$ ) and the entropy change of the surroundings ( $Δ S_{surroundings}$ ). The entropy change of the surroundings is related to the enthalpy change of the reaction:

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

Where $T$ is absolute temperature in Kelvin, and $Δ H$ is the enthalpy change of the reaction. A reaction is spontaneous if the total entropy change is positive:

\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0

📐 Worked Example

At 298 K, the reaction $C (s) + O_{2} (g) \to C O_{2} (g)$ has $Δ H = - 393 k J m o l^{- 1}$ and $Δ S_{system} = + 3 J K^{- 1} m o l^{- 1}$ . Show the reaction is spontaneous at 298 K.

1
Convert $Δ H$ to $J m o l^{- 1}$ to match the units of entropy:
2
$\Delta H = -393000\ J\ mol^{-1}$
3
Calculate $Δ S_{surroundings}$ using the formula:
4
$\Delta S_{\text{surroundings}} = -\frac{(-393000)}{298} = +1319\ J\ K^{-1}\ mol^{-1}$
5
Calculate total entropy change:
6
$\Delta S_{\text{total}} = 3 + 1319 = +1322\ J\ K^{-1}\ mol^{-1}$
7
Since $Δ S_{total}$ is positive, the reaction is spontaneous at 298 K.

5. Common Pitfalls

Wrong move:

Forgetting the negative sign in $Δ S_{surroundings} = - Δ H / T$ , writing $Δ H / T$ instead

Why:

The formula accounts for heat transferred from the system to the surroundings: an exothermic reaction (negative $Δ H$ ) releases heat to the surroundings, increasing its entropy

Correct move:

Always write the formula with the negative sign: exothermic reactions give positive $Δ S_{surroundings}$

Wrong move:

Using $Δ H$ in $k J m o l^{- 1}$ directly with entropy in $J K^{- 1} m o l^{- 1}$ without unit conversion

Why:

This leads to calculation errors that are orders of magnitude wrong, which are commonly penalized in exams

Correct move:

Always convert $Δ H$ from kJ to J by multiplying by 1000 before calculating $Δ S_{surroundings}$

Wrong move:

Claiming any reaction with a negative $Δ S_{system}$ cannot be spontaneous

Why:

Spontaneity depends on total entropy change, not just the entropy change of the system

Correct move:

Always calculate $Δ S_{total}$ : if $Δ S_{surroundings}$ is large enough positive, the total can still be positive even if $Δ S_{system}$ is negative

Wrong move:

Assuming all solids have lower entropy than all liquids regardless of molecular size

Why:

Entropy depends on both physical state and molecular complexity: a large complex solid can have higher entropy than a small simple liquid

Correct move:

Prioritize state when predicting entropy, but for same-state comparisons, larger molecules have higher entropy than smaller molecules

6. Quick Reference Cheatsheet

Concept	Key Formula/Rule	Exam Notes
Entropy (S)		Measure of disorder/microstates, always positive above 0 K
Entropy change prediction		ΔS positive if moles of gas increase
Standard ΔS calculation	$Δ S^{0} = \sum n S^{0} (products) - \sum m S^{0} (reactants)$	Units: $J K^{- 1} m o l^{- 1}$
ΔS surroundings	$Δ S_{surr} = - Δ H / T$	Convert ΔH to J to match units
Spontaneity condition	$Δ S_{total} = Δ S_{sys} + Δ S_{surr} > 0$	Positive total ΔS = spontaneous

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · P1
Predict sign of entropy change
2023 · P2
Calculate standard entropy change
2021 · P2
Check reaction spontaneity

Going deeper

syllabusCIE 9701 2025-2027 SyllabusOfficial syllabus document

What's Next

Entropy is the foundational concept for understanding why reactions occur spontaneously, and it is directly used to derive Gibbs free energy, the most commonly used tool for predicting spontaneity in A-level chemistry. Mastery of entropy predictions and calculations is required for almost all physical chemistry topics that follow, from chemical equilibrium to electrode potentials. Entropy also explains the observation that some endothermic reactions occur spontaneously, a question that cannot be answered by enthalpy alone. Next, you will build on this knowledge to learn about Gibbs free energy, a core heavily tested topic in CIE A-level Chemistry.