Gibbs free energy links enthalpy, entropy, and temperature into one criterion for thermodynamic driving force under standard conditions. This page focuses on using $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ at a specified temperature.

When and why to use $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

You use this relationship when you are given (or can determine) standard enthalpy change and standard entropy change for a process and need the standard Gibbs free energy change at a particular temperature.

This approach is especially useful because:

• Enthalpy captures heat/bond-energy effects.

• Entropy captures energy dispersal and matter/energy distribution effects.

• Temperature controls how strongly the entropy term contributes.

A key idea is that the entropy contribution is scaled by temperature, so the same $\Delta S^\circ$ matters more at higher $T$.

These lines show how $\Delta G$ varies with temperature for different sign combinations of $\Delta H$ and $\Delta S$. The vertical position relative to $\Delta G=0$ indicates whether the process is thermodynamically favored, and the temperature where a line crosses zero corresponds to the point where $\Delta H = T\Delta S$ (equilibrium for that process). Source

Key quantities (standard-state context)

Using $\Delta G^\circ$ is only meaningful when your $\Delta H^\circ$ and $\Delta S^\circ$ values refer to the same reaction as written (including physical states) and the same standard-state basis. In typical AP Chemistry usage, standard-state thermodynamic data are tabulated for a stated reference pressure and specified concentrations for solutes; the crucial point here is internal consistency among the values you use.

The equation and what each term means

This equation is a bookkeeping tool: it combines an enthalpy “push” ($\Delta H^\circ$) with an entropy “push” ($T\Delta S^\circ$) to determine the net free-energy change under standard conditions at that temperature.

Unit discipline (the most common source of mistakes)

Because $\Delta H^\circ$ and $T\Delta S^\circ$ must be subtracted, they must be in the same energy units.

Practical rules:

• Always use Kelvin for temperature: $T(\text{K}) = T(^{\circ}\text{C}) + 273.15$.

• If $\Delta S^\circ$ is given in $\text{J mol}^{-1}\text{K}^{-1}$ and $\Delta H^\circ$ is in $\text{kJ mol}^{-1}$, convert one so both match.

  • Commonly: convert $\Delta S^\circ$ to $\text{kJ mol}^{-1}\text{K}^{-1}$ by dividing by 1000.

• Keep track of the “per mole of reaction” basis:

  • If you scale the balanced equation, $\Delta H^\circ$, $\Delta S^\circ$, and therefore $\Delta G^\circ$ all scale by the same factor.

Interpreting the result at the given temperature (without overgeneralising)

After calculating $\Delta G^\circ$ at the stated temperature:

• A negative $\Delta G^\circ$ indicates the reaction/process is thermodynamically favoured under standard conditions at that temperature.

• A positive $\Delta G^\circ$ indicates it is not thermodynamically favoured under standard conditions at that temperature.

• A value near zero suggests little driving force under standard conditions at that temperature.

This interpretation is tied to the specific temperature used in the calculation; changing $T$ changes the magnitude of the entropy contribution ($T\Delta S^\circ$), which can shift $\Delta G^\circ$ even when $\Delta H^\circ$ and $\Delta S^\circ$ are treated as constant over the temperature range.

This table summarizes the four possible sign-combinations of $\Delta H$ and $\Delta S$ and the resulting temperature dependence of $\Delta G$. It makes the role of the $-T\Delta S$ term explicit: increasing $T$ strengthens the entropy contribution and can flip the sign of $\Delta G$ when $\Delta H$ and $\Delta S$ have the same sign. Source

Assumptions and limitations students should state or recognise

Using $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ cleanly at a “given temperature” rests on assumptions that are usually acceptable in AP-level problems but still worth knowing:

• $\Delta H^\circ$ and $\Delta S^\circ$ are treated as temperature-independent over the range considered (often around 298 K unless stated otherwise).

• The computed value is standard free energy, not the free energy change for arbitrary (nonstandard) concentrations/pressures.

• Physical states matter: $\Delta H^\circ$, $\Delta S^\circ$, and $\Delta G^\circ$ depend on whether substances are (s), (l), (g), or (aq).

What “at a given temperature” means operationally

To apply the syllabus skill precisely:

• Use the temperature provided (not automatically 298 K).

• Convert that temperature to K.

• Compute the entropy contribution as $T\Delta S^\circ$ with consistent units.

• Subtract that contribution from $\Delta H^\circ$ to obtain $\Delta G^\circ$ for that temperature under standard conditions.