IB Chemistry R1.4 R1.4.2

Gibbs energy (ΔG)

The single master equation that determines whether a chemical reaction is thermodynamically spontaneous.

Reactivity 1.4 HL Extension ⏱️ ~5 min revision

The Spontaneity Paradox

The Second Law of Thermodynamics states that the total entropy of the universe must always increase for any process to occur spontaneously. But this creates a problem: how can we measure the entropy change of the entire universe for a reaction in a beaker?

In the 19th century, Josiah Willard Gibbs resolved this by creating a single equation that focuses entirely on the system, wrapping the universe's entropy requirement into one measurable value.

IB Definition

Gibbs Free Energy Change (\(\Delta G\))

Gibbs energy change (\(\Delta G\)): The thermodynamic quantity that combines enthalpy, entropy, and absolute temperature to determine the maximum non-expansion work obtainable from a closed system at constant pressure. It serves as the ultimate criterion for reaction spontaneity.

The Gibbs Equation

The Master Equation

The Gibbs-Helmholtz Equation

\( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \)

Symbol Meaning Units
\( \Delta G^\circ \) Standard Gibbs energy change kJ mol⁻¹
\( \Delta H^\circ \) Standard enthalpy change kJ mol⁻¹
\( T \) Absolute temperature K (Kelvin)
\( \Delta S^\circ \) Standard entropy change J K⁻¹ mol⁻¹ (convert to kJ!)
Examiner Trap

Critical Unit Matching Trap (\(\Delta H\) in kJ vs \(\Delta S\) in J)

Critical Unit Trap: \(\Delta H\) is provided in \(\text{kJ mol}^{-1}\), while \(\Delta S\) is in \(\text{J K}^{-1}\text{mol}^{-1}\). You must convert \(\Delta S\) to kiloJoules (divide by 1000) or convert \(\Delta H\) to Joules before substituting into \(\Delta G = \Delta H - T\Delta S\).

Spontaneity Rules

Spontaneity Criteria

Interpreting the Sign of \(\Delta G\)

  • \(\Delta G < 0\) (negative): Reaction is spontaneous in the forward direction.
  • \(\Delta G > 0\) (positive): Reaction is non-spontaneous in the forward direction (reverse reaction is spontaneous).
  • \(\Delta G = 0\): System is at dynamic equilibrium (\(Q = K\)).
Examiner Trap

Spontaneous Does Not Mean Fast (Thermodynamics vs Kinetics)

A reaction with \(\Delta G < 0\) is thermodynamically spontaneous, but \(\Delta G\) provides zero information about reaction rate. A spontaneous reaction with a large activation energy (\(E_{\text{a}}\)) will proceed at an imperceptible rate under ambient conditions (e.g. diamond converting to graphite has \(\Delta G < 0\), but is kinetically stable for billions of years).

Worked Example: Calculating ΔG°

Worked Example

Calculating \(\Delta G^\circ\) for Ammonia Synthesis at 298 K

For the Haber process: \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})\)

Given: \(\Delta H^\circ = -92.2\text{ kJ mol}^{-1}\) and \(\Delta S^\circ = -198.7\text{ J K}^{-1}\text{mol}^{-1}\).
Calculate \(\Delta G^\circ\) at \(298\text{ K}\) and determine if the reaction is spontaneous.

Step 1: Convert \(\Delta S^\circ\) to \(\text{kJ K}^{-1}\text{mol}^{-1}\)
\(\Delta S^\circ = \dfrac{-198.7}{1000} = -0.1987\text{ kJ K}^{-1}\text{mol}^{-1}\)

Step 2: Substitute into Gibbs equation
\(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\)
\(\Delta G^\circ = -92.2 - (298 \times -0.1987) = -92.2 - (-59.2) = \mathbf{-33.0\text{ kJ mol}^{-1}}\)

Conclusion: Because \(\Delta G^\circ\) is negative (\(-33.0\text{ kJ mol}^{-1}\)), the synthesis of ammonia is thermodynamically spontaneous at 298 K.

The Four Scenarios

The sign of ΔG depends on the interplay between ΔH and TΔS. There are four possible combinations:

ΔH ΔS ΔG Spontaneous?
− (exo) + (increase) Always − Always
− (exo) − (decrease) Depends on T At low T
+ (endo) + (increase) Depends on T At high T
+ (endo) − (decrease) Always + Never
Crossover Calculation

Finding the Temperature Boundary for Spontaneity (\(\Delta G = 0\))

At dynamic equilibrium (\(\Delta G = 0\)), the temperature at which a reaction shifts between spontaneous and non-spontaneous is given by setting \(\Delta H^\circ - T\Delta S^\circ = 0\):

\( T_{\text{crossover}} = \dfrac{\Delta H^\circ}{\Delta S^\circ} \)

For the Haber process: \(T = \dfrac{-92.2}{-0.1987} = 464\text{ K} = 191\text{ }^\circ\text{C}\). Above 464 K, the \(-T\Delta S\) penalty exceeds \(\Delta H\), and \(\Delta G\) becomes positive.

AQA GCSE & IB Chemistry

Study this topic on the go

Get active recall flashcards, notes, and topic quizzes in ChemEasy, or build your revision schedule with ChemPlan IB.

See our apps