# Entropy

> CIE A-Level Chemistry · Further chemical energetics
> Source: https://www.owlsprep.com/study/cie-9701-u17-entropy/

This sub-topic introduces entropy, the thermodynamic measure of disorder in chemical systems. You will learn to predict and calculate entropy changes for reactions, and connect entropy to the spontaneity of chemical processes, a core foundation for Gibbs free energy.

**Prerequisites:** [Enthalpy changes and Hess' Law](https://www.owlsprep.com/study/cie-9701-u16-enthalpy-changes/)

## Learning objectives

- Define entropy and explain its physical meaning
- Predict the sign of entropy changes for physical and chemical processes
- Calculate standard entropy changes from standard molar entropies
- Calculate total entropy change and use it to determine reaction spontaneity

## Definition and Physical Meaning of Entropy

**Entropy** — 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.

*Notation:* S

*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.

> **info**
>
> Calling entropy 'disorder' is a useful simplification for exams, but it more accurately describes the spread of energy and matter across all available possible arrangements.

**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): $H_2O(g)$ 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 $C_6H_{12}O_6(s)$ has higher entropy.

## Predicting Entropy Changes

For any process, the entropy change is calculated as $\Delta S = S_{\text{final}} - S_{\text{initial}}$. A positive $\Delta S$ means entropy increases (the system becomes more disordered), while a negative $\Delta S$ 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 $\Delta S_{\text{system}}$ for each reaction: (a) $CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$ (b) $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$

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 $\Delta S_{\text{system}}$ 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 $\Delta S_{\text{system}}$ is negative.

## Calculating Standard Entropy Changes

**Standard Molar Entropy** — The entropy of 1 mole of a substance under standard conditions (298 K, 1 atm), with units $J\ K^{-1}\ mol^{-1}$. Unlike standard enthalpy of formation, $S^0$ is always positive for any pure substance above 0 K.

*Notation:* $S^0$

*Example:* $S^0(H_2O(g)) = 189\ J\ K^{-1}\ mol^{-1}$, $S^0(H_2O(l)) = 70\ J\ K^{-1}\ mol^{-1}$

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 $\Delta S^0_{\text{system}}$ for $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$, given $S^0(SO_2(g)) = 248\ J\ K^{-1}\ mol^{-1}$, $S^0(O_2(g)) = 205\ J\ K^{-1}\ mol^{-1}$, $S^0(SO_3(g)) = 257\ J\ K^{-1}\ mol^{-1}$.

1. Substitute values into the formula for $\Delta S^0_{\text{system}}$:
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 $\Delta S^0$ should be negative, which matches our calculation.

## Total Entropy and Spontaneity

To determine if a reaction is spontaneous, we need the total entropy change, which adds the entropy change of the system ($\Delta S_{\text{system}}$) and the entropy change of the surroundings ($\Delta S_{\text{surroundings}}$). 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 $\Delta H$ 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 $C(s) + O_2(g) \rightarrow CO_2(g)$ has $\Delta H = -393\ kJ\ mol^{-1}$ and $\Delta S_{\text{system}} = +3\ J\ K^{-1}\ mol^{-1}$. Show the reaction is spontaneous at 298 K.

1. Convert $\Delta H$ to $J\ mol^{-1}$ to match the units of entropy:
2. $$\Delta H = -393000\ J\ mol^{-1}$$
3. Calculate $\Delta S_{\text{surroundings}}$ 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 $\Delta S_{\text{total}}$ is positive, the reaction is spontaneous at 298 K.

## Common pitfalls

- **Wrong:** Forgetting the negative sign in $\Delta S_{\text{surroundings}} = -\Delta H/T$, writing $\Delta H/T$ instead
  - Why it fails: The formula accounts for heat transferred from the system to the surroundings: an exothermic reaction (negative $\Delta H$) releases heat to the surroundings, increasing its entropy
  - Correct: Always write the formula with the negative sign: exothermic reactions give positive $\Delta S_{\text{surroundings}}$
- **Wrong:** Using $\Delta H$ in $kJ\ mol^{-1}$ directly with entropy in $J\ K^{-1}\ mol^{-1}$ without unit conversion
  - Why it fails: This leads to calculation errors that are orders of magnitude wrong, which are commonly penalized in exams
  - Correct: Always convert $\Delta H$ from kJ to J by multiplying by 1000 before calculating $\Delta S_{\text{surroundings}}$
- **Wrong:** Claiming any reaction with a negative $\Delta S_{\text{system}}$ cannot be spontaneous
  - Why it fails: Spontaneity depends on total entropy change, not just the entropy change of the system
  - Correct: Always calculate $\Delta S_{\text{total}}$: if $\Delta S_{\text{surroundings}}$ is large enough positive, the total can still be positive even if $\Delta S_{\text{system}}$ is negative
- **Wrong:** Assuming all solids have lower entropy than all liquids regardless of molecular size
  - Why it fails: Entropy depends on both physical state and molecular complexity: a large complex solid can have higher entropy than a small simple liquid
  - Correct: Prioritize state when predicting entropy, but for same-state comparisons, larger molecules have higher entropy than smaller molecules

## 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 | $\Delta S^0 = \sum nS^0(\text{products}) - \sum mS^0(\text{reactants})$ | Units: $J\ K^{-1}\ mol^{-1}$ |
| ΔS surroundings | $\Delta S_{\text{surr}} = -\Delta H/T$ | Convert ΔH to J to match units |
| Spontaneity condition | $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$ | Positive total ΔS = spontaneous |

## 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.

- [Gibbs Free Energy](https://www.owlsprep.com/study/cie-9701-u17-gibbs-free-energy/)
- [Further chemical equilibria](https://www.owlsprep.com/study/cie-9701-u18-overview/)
- [Acid-base equilibria](https://www.owlsprep.com/study/cie-9701-u18-acid-base-equilibria/)

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