IB CHEMISTRY

HL Only
Reactivity 1.4.2

Gibbs Free Energy (ΔG\Delta G)

Combining Enthalpy and Entropy to find the "Free Energy" available to do work.

ΔGθ=ΔHθ−TΔSθ\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta
ΔH\Delta H (Enthalpy)

Heat energy change. Usually in kJ.

ΔS\Delta S (Entropy)

Disorder change. Usually in J (Needs converting!).

TT (Temperature)

Must be in Kelvin (K).

ΔG\Delta G (Free Energy)

Measure of spontaneity.

Deep Think Concept

The Unit Trap

This calculation will be WRONG if you don't convert units.

  • ΔH\Delta H is typically given in kJ mol−1kJ \: mol^{-1} (e.g., -50).
  • ΔS\Delta S is typically given in J K−1 mol−1J \: K^{-1} \: mol^{-1} (e.g., +100).

You MUST divide ΔS\Delta S by 1000 to get it into kJkJ.

What does the result mean?

ΔG<0\Delta G < 0 (Negative)Reaction is SPONTANEOUS
ΔG>0\Delta G > 0 (Positive)Reaction is NON-SPONTANEOUS
ΔG=0\Delta G = 0 (Zero)System is at EQUILIBRIUM

Putting it into Practice

Calculating Gibbs Free Energy

Paper 2 Style

For a reaction at 298 K:

  • ΔH=−92 kJ mol−1\Delta H = -92 \: kJ \: mol^{-1}
  • ΔS=−198 J K−1 mol−1\Delta S = -198 \: J \: K^{-1} \: mol^{-1}

Calculate ΔG\Delta G and determine if the reaction is spontaneous.

Practice: Spontaneity Logic

[2 Marks]

A reaction is endothermic (ΔH>0\Delta H > 0) but has a positive entropy change (ΔS>0\Delta S > 0).

Explain how the temperature affects the spontaneity of this reaction.

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