Entropy trends you must know

• Entropy usually increases when:

  • a solid melts

  • a liquid vaporizes

  • a solid sublimes

  • the number of gas moles increases

  • substances mix or dissolve

• Entropy usually decreases when:

  • a gas condenses

  • a liquid freezes

  • the number of gas moles decreases

  • a system becomes more ordered

• Biggest entropy clue in reactions: check change in gaseous particles first.

• Common exam shortcut:

  • more gas particles produced $\Rightarrow \Delta S > 0$

  • fewer gas particles produced $\Rightarrow \Delta S < 0$

This figure shows that entropy rises as temperature increases and rises sharply during phase changes because particles gain more possible arrangements. It is excellent for linking state changes to increasing disorder/energy dispersal. Source

Predicting the sign of $\Delta S^\circ$

• Use: $\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$

• $S^\circ$ values are given in the data booklet.

• Include stoichiometric coefficients in the calculation.

• Units of standard entropy, $S^\circ$: J K$^{-1}$ mol$^{-1}$.

• Units of standard entropy change, $\Delta S^\circ$: J K$^{-1}$ mol$^{-1}$.

• Exam habit: write the calculation clearly in two steps:

  • total products entropy

  • total reactants entropy

• Then subtract: products − reactants.

Gibbs energy equation

• Gibbs energy links enthalpy, entropy, and temperature:

  • $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

• Meanings:

  • $\Delta G^\circ$ = standard Gibbs energy change

  • $\Delta H^\circ$ = standard enthalpy change

  • $T$ = absolute temperature in kelvin

  • $\Delta S^\circ$ = standard entropy change

• Units:

  • $\Delta H^\circ$ in kJ mol$^{-1}$

  • $\Delta S^\circ$ usually in J K$^{-1}$ mol$^{-1}$

  • $\Delta G^\circ$ in kJ mol$^{-1}$

• So before substituting, usually convert $\Delta S^\circ$ from J to kJ:

  • divide by 1000

• Temperature must always be in K, never °C.

How to interpret $\Delta G$

• At constant pressure, a change is spontaneous if $\Delta G < 0$.

• If $\Delta G > 0$, the change is non-spontaneous in the forward direction.

• If $\Delta G = 0$, the system is at equilibrium.

• IB wording to use:

  • negative $\Delta G$ = thermodynamically feasible / spontaneous

  • positive $\Delta G$ = not spontaneous under those conditions

• A reaction can become more or less spontaneous as temperature changes because the $T\Delta S$ term changes.

Temperature dependence of spontaneity

• The sign combination of $\Delta H$ and $\Delta S$ tells you whether temperature matters.

• Four essential cases:

  • $\Delta H < 0$, $\Delta S > 0$ → always spontaneous

  • $\Delta H > 0$, $\Delta S < 0$ → never spontaneous

  • $\Delta H < 0$, $\Delta S < 0$ → spontaneous at low temperature

  • $\Delta H > 0$, $\Delta S > 0$ → spontaneous at high temperature

• Why?

  • negative $\Delta H$ helps make $\Delta G$ negative

  • positive $\Delta S$ makes $-T\Delta S$ more negative

• Strong exam technique: do a sign analysis first before calculating.

This page is useful for visualising why some reactions are spontaneous only at high or low temperature. It directly reinforces IB-style sign analysis for $\Delta G$ questions. Source

Finding the temperature where a reaction becomes spontaneous

• At the boundary between spontaneous and non-spontaneous:

  • $\Delta G = 0$

• Therefore:

  • $0 = \Delta H - T\Delta S$

  • $T = \dfrac{\Delta H}{\Delta S}$

• Use this only when asked for the temperature at which spontaneity changes.

• Important conditions:

  • units must match before dividing

  • temperature answer must be in K

• Interpretation:

  • if the question asks above or below this temperature, use the signs of $\Delta H$ and $\Delta S$ to decide.

Equilibrium and Gibbs energy

• As a reaction moves toward equilibrium, $\Delta G$ becomes less negative.

• At equilibrium: $\Delta G = 0$.

• Relationship with reaction quotient and equilibrium constant:

  • $\Delta G = \Delta G^\circ + RT\ln Q$

  • at equilibrium, $Q = K$, so $\Delta G^\circ = -RT\ln K$

• Meaning of $\Delta G^\circ = -RT\ln K$:

  • $\Delta G^\circ < 0$ → $K > 1$, products favoured

  • $\Delta G^\circ > 0$ → $K < 1$, reactants favoured

  • $\Delta G^\circ = 0$ → $K = 1$

• This links thermodynamics to equilibrium position.

Exam calculation traps

• Do not mix J and kJ.

• Do not use °C in the Gibbs equation.

• Do not forget stoichiometric coefficients in $\Delta S^\circ$ calculations.

• A negative enthalpy alone does not guarantee spontaneity.

• A reaction may be spontaneous but very slow.

• $\Delta G^\circ$ refers to standard conditions; $\Delta G$ can change with actual conditions via $Q$.

Fast exam strategy

• Step 1: identify what is being asked: sign, missing value, threshold temperature, or equilibrium link.

• Step 2: check units.

• Step 3: do a sign analysis before using numbers.

• Step 4: use the correct equation:

  • $\Delta S^\circ = \sum S^\circ(\text{products}) - \sum S^\circ(\text{reactants})$

  • $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

  • $\Delta G = \Delta G^\circ + RT\ln Q$

  • $\Delta G^\circ = -RT\ln K$

• Step 5: interpret the result in words, not just symbols.

This page is ideal for correcting a common IB error: confusing spontaneous with fast. It supports the exam language needed when explaining why feasibility and rate are not the same idea. Source

Memory box: absolute essentials

• Entropy: measure of dispersal of matter/energy.

• State order: solid < liquid < gas.

• Spontaneous at constant pressure: $\Delta G < 0$.

• Equilibrium: $\Delta G = 0$.

• Core equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.

• High $T$ matters most when $\Delta S$ is significant.

• Products favoured: $K > 1$, so $\Delta G^\circ < 0$.

This figure is helpful for the last part of the topic: connecting Gibbs energy to equilibrium position. It is especially useful for revision because it compresses several linked ideas into one diagram/table. Source