# Entropy and reaction spontaneity

> IB Chemistry HL · R1: What drives chemical reactions?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u4-entropy-and-reaction-spontaneity/

This module covers entropy as a measure of molecular disorder, how to predict and calculate entropy changes, and how to use Gibbs free energy to predict reaction spontaneity and its dependence on temperature.

**Prerequisites:** [Enthalpy and thermochemical calculations](https://www.owlsprep.com/study/ib-chemistry-hl-u3-enthalpy-and-thermochemistry/)

## Learning objectives

- Define entropy and relate it to molecular disorder
- Predict entropy changes for physical and chemical processes
- Calculate standard entropy changes for chemical reactions
- Use Gibbs free energy to determine reaction spontaneity
- Predict how spontaneity depends on reaction temperature

## Entropy: Definition and Predicting Entropy Changes

**Entropy** — A thermodynamic quantity that measures the degree of randomness or disorder in a system. Higher disorder corresponds to higher entropy, with units of J K⁻¹ mol⁻¹.

*Notation:* S

*Example:* Gaseous water has higher entropy than liquid water because gas molecules have much greater freedom of movement.

Entropy follows predictable trends for physical and chemical processes. You can usually predict the sign of $\Delta S$ (entropy change) by analyzing the change in disorder between reactants and products.

- For the same substance at the same temperature: $S_{\text{solid}} < S_{\text{liquid}} < S_{\text{gas}}$
- Entropy increases when the number of moles of gas increases in a reaction
- Entropy increases with increasing temperature and when a solute dissolves in a solvent
- More complex molecules have higher entropy than simpler molecules at the same state and temperature

**Worked example:** Predict the sign of $\Delta S$ for each process: (a) $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ (b) $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ (c) $\text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{O}(s)$ at 0°C

1. Count moles of gas for reaction (a): 4 moles of gaseous reactants form 2 moles of gaseous products. Disorder decreases.
2. Result: $\Delta S$ is **negative**
3. Count moles of gas for reaction (b): 0 moles of gaseous reactants form 1 mole of gaseous product. Disorder increases.
4. Result: $\Delta S$ is **positive**
5. Process (c) is freezing: liquid water forms solid, molecular movement is restricted and disorder decreases.
6. Result: $\Delta S$ is **negative**

> **Exam tip:** Always check the number of moles of gas first when predicting $\Delta S$, as gas molecules contribute far more to total entropy than solids or liquids.

## Calculating Standard Entropy Change of Reaction

Standard molar entropy ($S^\circ$) is the entropy of 1 mole of a substance under standard conditions (1 atm, 298 K). Unlike standard enthalpy of formation, $S^\circ$ is always positive for all substances at temperatures above 0 K.

**Standard entropy change of reaction** — The total change in entropy for a reaction when all reactants and products are in their standard states.

*Notation:* $\Delta S^\circ_{\text{rxn}}$

*Example:* Calculated from standard molar entropy values of reactants and products.

$$\Delta S^\circ_{\text{rxn}} = \sum n S^\circ(\text{products}) - \sum m S^\circ(\text{reactants})$$

where $n$ and $m$ are the stoichiometric coefficients of products and reactants, respectively.

**Worked example:** Calculate $\Delta S^\circ_{\text{rxn}}$ for $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$, given: $S^\circ(\text{N}_2) = 191.6$ J K⁻¹ mol⁻¹, $S^\circ(\text{H}_2) = 130.7$ J K⁻¹ mol⁻¹, $S^\circ(\text{NH}_3) = 192.8$ J K⁻¹ mol⁻¹.

1. Calculate total entropy of products:
2. $$2 \times 192.8 = 385.6 \text{ J K}^{-1}$$
3. Calculate total entropy of reactants:
4. $$(1 \times 191.6) + (3 \times 130.7) = 191.6 + 392.1 = 583.7 \text{ J K}^{-1}$$
5. Subtract reactant entropy from product entropy:
6. $$\Delta S^\circ_{\text{rxn}} = 385.6 - 583.7 = -198.1 \text{ J K}^{-1} \text{ mol}^{-1}$$
7. This negative sign matches our earlier prediction, so the result makes sense.

**Check your understanding**

Check your understanding:

1. What is the entropy of a perfect crystal at 0 K?

   - 0 J K⁻¹
   - 1 J K⁻¹
   - Depends on the crystal
   - Equal to its enthalpy

   *Why:* This is the third law of thermodynamics: a perfect crystal at absolute zero has zero entropy, as there is no molecular disorder.

## Gibbs Free Energy and Reaction Spontaneity

The entropy of the system alone does not predict spontaneity. The second law of thermodynamics requires that total entropy (system + surroundings) increases for a spontaneous process. Gibbs free energy combines $\Delta H$ and $\Delta S$ of the system into a single value that predicts spontaneity at constant pressure and temperature.

**Gibbs Free Energy Change** — A thermodynamic function that predicts spontaneity at constant T and P: negative $\Delta G$ = spontaneous, positive $\Delta G$ = non-spontaneous, $\Delta G = 0$ = equilibrium.

*Notation:* $\Delta G$

*Example:* Given by the relationship $\Delta G = \Delta H - T\Delta S$

$$\Delta G = \Delta H - T\Delta S$$

**Worked example:** A reaction has $\Delta H = -100$ kJ mol⁻¹ and $\Delta S = -150$ J K⁻¹ mol⁻¹ at 298 K. Is the reaction spontaneous at this temperature?

1. Convert $\Delta S$ to kJ to match the units of $\Delta H$:
2. $\Delta S = -150$ J K⁻¹ mol⁻¹ = $-0.150$ kJ K⁻¹ mol⁻¹
3. Substitute into the Gibbs free energy equation:
4. $$\Delta G = (-100) - (298 \times -0.150) = -100 + 44.7 = -55.3 \text{ kJ mol}^{-1}$$
5. $\Delta G$ is negative, so the reaction is **spontaneous** at 298 K.

> **tip**
>
> Always check units before calculating $\Delta G$. $\Delta H$ is almost always given in kJ, while $\Delta S$ is given in J. Unit inconsistency is one of the most common sources of lost marks in exam questions.

## Temperature Dependence of Spontaneity

The sign of $\Delta G$ depends on the combination of signs of $\Delta H$ and $\Delta S$, and how the $T\Delta S$ term changes with temperature. The table below summarizes the four possible combinations:

| $\Delta H$ sign | $\Delta S$ sign | Spontaneous when | Sign of $\Delta G$ |
| --- | --- | --- | --- |
| - | + | All temperatures | Always negative |
| - | - | Low temperatures | Negative at low T |
| + | + | High temperatures | Negative at high T |
| + | - | Never | Always positive |

**Worked example:** A reaction has $\Delta H = +150$ kJ mol⁻¹ and $\Delta S = +450$ J K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous?

1. A reaction becomes spontaneous when $\Delta G < 0$. Find the threshold temperature where $\Delta G = 0$:
2. $$0 = \Delta H - T\Delta S \implies T = \frac{\Delta H}{\Delta S}$$
3. Convert $\Delta S$ to kJ:
4. $\Delta S = 0.450$ kJ K⁻¹ mol⁻¹
5. Substitute values:
6. $$T = \frac{150 \text{ kJ mol}^{-1}}{0.450 \text{ kJ K}^{-1} \text{ mol}^{-1}} = 333 \text{ K}$$
7. For $\Delta H > 0$ and $\Delta S > 0$, the reaction is spontaneous above this temperature. So the reaction is spontaneous when $T > 333$ K.

## Common pitfalls

- **Wrong:** Predicting a positive $\Delta S$ because the total number of product moles is higher than reactant moles, even when the number of gas moles decreases.
  - Why it fails: Gas molecules contribute ~1000 times more entropy than solids or liquids, so total moles of all species is irrelevant. Only the change in moles of gas matters.
  - Correct: Count only the change in the number of moles of gas when predicting the sign of $\Delta S$.
- **Wrong:** Assuming $S^\circ$ for elements in their standard state is zero, like $\Delta H^\circ_f$.
  - Why it fails: Unlike enthalpy of formation, entropy measures disorder. All substances have positive entropy at temperatures above 0 K.
  - Correct: Expect all $S^\circ$ values given in exams to be positive, and do not adjust your calculation for zero values.
- **Wrong:** Forgetting to convert $\Delta S$ from J to kJ when calculating $\Delta G$.
  - Why it fails: $\Delta H$ is almost always given in kJ, so mismatched units give a final answer off by a factor of 1000, which is wrong.
  - Correct: Always convert $\Delta S$ to kJ K⁻¹ mol⁻¹ before substituting into $\Delta G = \Delta H - T\Delta S$.
- **Wrong:** Claiming that a non-spontaneous reaction can never occur under any conditions.
  - Why it fails: $\Delta G$ only predicts spontaneity without external energy input. Non-spontaneous reactions can proceed with energy input.
  - Correct: State that non-spontaneous reactions do not occur on their own, but can occur with an external input of energy.
- **Wrong:** Predicting that $\Delta H > 0$, $\Delta S < 0$ reactions are spontaneous at high temperatures.
  - Why it fails: The $T\Delta S$ term becomes more negative as temperature increases, so $\Delta G = \Delta H - T\Delta S = \text{positive} - (T \times \text{negative}) = \text{positive} + \text{positive}$ which is always positive.
  - Correct: Remember that positive $\Delta H$ and negative $\Delta S$ means the reaction is never spontaneous.

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Entropy trend | solid < liquid < gas | Gas moles dominate $\Delta S$ sign |
| Standard $\Delta S^\circ_{\text{rxn}}$ | $\sum nS^\circ(products) - \sum mS^\circ(reactants)$ | All $S^\circ$ are positive at 298 K |
| Gibbs free energy | $\Delta G = \Delta H - T\Delta S$ | $\Delta G < 0$ = spontaneous |
| $-\Delta H$, $+\Delta S$ | Spontaneous at all T | $\Delta G$ always negative |
| $-\Delta H$, $-\Delta S$ | Spontaneous at low T | Low T favors exothermic spontaneity |
| $+\Delta H$, $+\Delta S$ | Spontaneous at high T | High T favors entropy-driven spontaneity |
| $+\Delta H$, $-\Delta S$ | Never spontaneous | $\Delta G$ always positive |

## What's next

Entropy and spontaneity form the foundation of all thermodynamic predictions in IB Chemistry HL, connecting the energy changes of reactions to their direction. This concept directly builds on enthalpy and lays the groundwork for understanding equilibrium, electrochemistry, and phase changes. The next logical step is to connect Gibbs free energy to equilibrium constants, which explains how $\Delta G$ relates to the position of equilibrium and how to calculate equilibrium constants from thermodynamic data. Mastery of this sub-topic is essential for all subsequent thermodynamics and electrochemistry topics in the IB syllabus.

- [Bond enthalpies](https://www.owlsprep.com/study/ib-chemistry-hl-u4-bond-enthalpies/)
- [Energy cycles](https://www.owlsprep.com/study/ib-chemistry-hl-u4-energy-cycles/)
- [AHL: Extended enthalpy and entropy calculations](https://www.owlsprep.com/study/ib-chemistry-hl-u4-ahl-extended-enthalpy-and-entropy/)

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