Entropy and reaction spontaneity

IB Chemistry HL· IB Chemistry HL Topic 16: Thermodynamics· 25 min read

1. Entropy: Definition and Predicting Entropy Changes★★☆☆☆⏱ 6 min

📘 Definition

Entropy

S

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.

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 $Δ S$ (entropy change) by analyzing the change in disorder between reactants and products.

For the same substance at the same temperature: $S_{solid} < S_{liquid} < S_{gas}$
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 $Δ S$ for each process: (a) $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$ (b) $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$ (c) $H_{2} O (l) \to H_{2} O (s)$ 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: $Δ S$ 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: $Δ S$ is positive
5
Process (c) is freezing: liquid water forms solid, molecular movement is restricted and disorder decreases.
6
Result: $Δ S$ is negative

Exam tip:

Always check the number of moles of gas first when predicting $Δ S$ , as gas molecules contribute far more to total entropy than solids or liquids.

2. Calculating Standard Entropy Change of Reaction★★★☆☆⏱ 7 min

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.

📘 Definition

Standard entropy change of reaction

$Δ S_{rxn}^{\circ}$

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

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})

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📐 Worked Example

Calculate $Δ S_{rxn}^{\circ}$ for $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$ , given: $S^{\circ} (N_{2}) = 191.6$ J K⁻¹ mol⁻¹, $S^{\circ} (H_{2}) = 130.7$ J K⁻¹ mol⁻¹, $S^{\circ} (NH_{3}) = 192.8$ 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.

✓ Quick check

Check your understanding:

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
Reveal answer
0 J K⁻¹ —
This is the third law of thermodynamics: a perfect crystal at absolute zero has zero entropy, as there is no molecular disorder.

3. Gibbs Free Energy and Reaction Spontaneity★★★☆☆⏱ 6 min

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 $Δ H$ and $Δ S$ of the system into a single value that predicts spontaneity at constant pressure and temperature.

📘 Definition

Gibbs Free Energy Change

$Δ G$

A thermodynamic function that predicts spontaneity at constant T and P: negative $Δ G$ = spontaneous, positive $Δ G$ = non-spontaneous, $Δ G = 0$ = equilibrium.

Example:

Given by the relationship $Δ G = Δ H - T Δ S$

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

📐 Worked Example

A reaction has $Δ H = - 100$ kJ mol⁻¹ and $Δ S = - 150$ J K⁻¹ mol⁻¹ at 298 K. Is the reaction spontaneous at this temperature?

1
Convert $Δ S$ to kJ to match the units of $Δ H$ :
2
$Δ S = - 150$ 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
$Δ G$ is negative, so the reaction is spontaneous at 298 K.

4. Temperature Dependence of Spontaneity★★★★☆⏱ 6 min

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

$Δ H$ sign	$Δ S$ sign	Spontaneous when	Sign of $Δ G$
		All temperatures	Always negative
		Low temperatures	Negative at low T
		High temperatures	Negative at high T
		Never	Always positive

📐 Worked Example

A reaction has $Δ H = + 150$ kJ mol⁻¹ and $Δ S = + 450$ J K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous?

1
A reaction becomes spontaneous when $Δ G < 0$ . Find the threshold temperature where $Δ G = 0$ :
2
$0 = \Delta H - T\Delta S \implies T = \frac{\Delta H}{\Delta S}$
3
Convert $Δ S$ to kJ:
4
$Δ S = 0.450$ 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 $Δ H > 0$ and $Δ S > 0$ , the reaction is spontaneous above this temperature. So the reaction is spontaneous when $T > 333$ K.

5. Common Pitfalls

Wrong move:

Predicting a positive $Δ S$ because the total number of product moles is higher than reactant moles, even when the number of gas moles decreases.

Why:

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 move:

Count only the change in the number of moles of gas when predicting the sign of $Δ S$ .

Wrong move:

Assuming $S^{\circ}$ for elements in their standard state is zero, like $Δ H_{f}^{\circ}$ .

Why:

Unlike enthalpy of formation, entropy measures disorder. All substances have positive entropy at temperatures above 0 K.

Correct move:

Expect all $S^{\circ}$ values given in exams to be positive, and do not adjust your calculation for zero values.

Wrong move:

Forgetting to convert $Δ S$ from J to kJ when calculating $Δ G$ .

Why:

$Δ H$ is almost always given in kJ, so mismatched units give a final answer off by a factor of 1000, which is wrong.

Correct move:

Always convert $Δ S$ to kJ K⁻¹ mol⁻¹ before substituting into $Δ G = Δ H - T Δ S$ .

Wrong move:

Claiming that a non-spontaneous reaction can never occur under any conditions.

Why:

$Δ G$ only predicts spontaneity without external energy input. Non-spontaneous reactions can proceed with energy input.

Correct move:

State that non-spontaneous reactions do not occur on their own, but can occur with an external input of energy.

Wrong move:

Predicting that $Δ H > 0$ , $Δ S < 0$ reactions are spontaneous at high temperatures.

Why:

The $T Δ S$ term becomes more negative as temperature increases, so $Δ G = Δ H - T Δ S = positive - (T \times negative) = positive + positive$ which is always positive.

Correct move:

Remember that positive $Δ H$ and negative $Δ S$ means the reaction is never spontaneous.

6. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Entropy trend	solid < liquid < gas	Gas moles dominate $Δ S$ sign
Standard $Δ S_{rxn}^{\circ}$	$\sum n S^{\circ} (p r o d u c t s) - \sum m S^{\circ} (r e a c t an t s)$	All $S^{\circ}$ are positive at 298 K
Gibbs free energy	$Δ G = Δ H - T Δ S$	$Δ G < 0$ = spontaneous
$- Δ H$ , $+ Δ S$	Spontaneous at all T	$Δ G$ always negative
$- Δ H$ , $- Δ S$	Spontaneous at low T	Low T favors exothermic spontaneity
$+ Δ H$ , $+ Δ S$	Spontaneous at high T	High T favors entropy-driven spontaneity
$+ Δ H$ , $- Δ S$	Never spontaneous	$Δ G$ always positive

7. Frequently Asked

Can a non-spontaneous reaction ever occur?

Yes. Non-spontaneous reactions do not occur on their own, but can proceed with external energy input (e.g. electrolysis of water) or by coupling to a highly spontaneous reaction.

Is a negative entropy change always non-spontaneous?

No. Spontaneity depends on the combination of $Δ H$ , $Δ S$ and temperature. Exothermic reactions with negative $Δ S$ can be spontaneous at low temperatures.

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.

2025 · Paper 1
Entropy change prediction
2024 · Paper 2
Gibbs free energy calculation
2023 · Paper 1
Temperature dependence of spontaneity

Going deeper

syllabusOfficial IB Chemistry HL SyllabusFull topic 16 learning outcomes

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 $Δ G$ 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.