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AP® Chemistry · Unit 9 · Topic 9.3

Gibbs Free Energy and Thermodynamic Favorability: Revision Notes (Chemistry 9.3)

One-page recap of ΔG°, the two ways to calculate it, the four sign combinations of ΔH° and ΔS°, and the unit and sign slips that cost marks.

Short on time? This page is the recap. For explanations and worked examples, use the full study guide.

Recap

  • ΔG° is the Gibbs free energy change with everything in its standard state: pure solids and liquids, 1 M solutions, gases at 1 bar (about 1 atm).
  • ΔG° < 0 means the process is thermodynamically favoured (older books say “spontaneous”). ΔG° > 0 means the reverse is favoured.
  • ΔG° says which direction is favoured, never how fast.
  • ΔG° combines two drives: releasing energy (ΔH° < 0) and dispersing matter and energy (ΔS° > 0).

Key relationships

Relationship Units and notes Use it to
ΔG° = ΣnΔG°f(products) − ΣnΔG°f(reactants) kJ mol⁻¹; ΔG°f = 0 for elements in standard states find ΔG° from a data table
ΔG° = ΔH° − TΔS° ΔH° in kJ mol⁻¹, ΔS° ÷ 1000 to kJ K⁻¹ mol⁻¹, T in K find ΔG° at any temperature
T = ΔH° ÷ ΔS° same units top and bottom find the changeover temperature (ΔG° = 0)
ΔH° ΔS° Favoured at
− + all T
+ − no T
+ + high T
− − low T

Assumptions

  • ΔH° and ΔS° are treated as constant with temperature, so ΔG° at a new temperature is an estimate.
  • Every species is in its standard state. Different concentrations or pressures change ΔG (later in the unit).

Mistakes to avoid

  1. J and kJ mixed. Divide ΔS° by 1000 first. An answer of tens of thousands of kJ is a warning sign.
  2. °C instead of K. 25 °C is 298 K.
  3. Reactants minus products. It is always products minus reactants.
  4. Coefficients dropped. Multiply each ΔG°f by its coefficient.
  5. “Exothermic, so favoured.” Check ΔS° too: freezing water is exothermic but not favoured above 273 K.
  6. “Favoured, so fast.” Rate is a kinetics question.

Quick self-check

  1. ΔH° = −50 kJ mol⁻¹ and ΔS° = +100 J K⁻¹ mol⁻¹. When is the process favoured? (At all temperatures: both terms make ΔG° negative.)
  2. ΔH° = +30.0 kJ mol⁻¹, ΔS° = +80.0 J K⁻¹ mol⁻¹. Find ΔG° at 500 K. (30.0 − 500 × 0.0800 = −10.0 kJ mol⁻¹, favoured)
  3. ΔH° = +45.0 kJ mol⁻¹, ΔS° = +150 J K⁻¹ mol⁻¹. Above what temperature is it favoured? (45 000 ÷ 150 = 300 K)

Next: practice questions.

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