Gibbs Free Energy Calculations (College Board AP® Chemistry)

Worked Example

Calculate ΔG° for the following reaction at 298K.

2NaHCO₃ (s) → Na₂CO₃ (s) + H₂O (l) + CO₂ (g)

• ΔH° = +135 kJ mol^-1        

• ΔS° = +344 J K^-1 mol^-1

Answer:

• Step 1: Convert the entropy value to kilojoules

  • ΔS°= +344 J K^-1 mol^-1 ÷ 1000 = +0.344 kJ K^-1 mol^-1 

• Step 2: Substitute the terms into the Gibbs Equation

  • ΔG^° = ΔH^°- TΔS^°

  • ΔG^° = +135 - (298 x 0.344)

  • ΔG^° = +32.49 kJ mol^-1

  • So, the reaction is thermodynamically unfavourable at 298K

Worked Example

Determine whether the following reaction is thermodynamically favored at 298 K. 

2Ca (s) + O₂ (g) → 2CaO (s)         ΔH = –635.5 kJ mol^-1

• S^o [Ca (s)] = 41.00 J K^-1 mol^-1

• S^o [O₂ (g)] = 205.0 J K^-1 mol^-1

• S^o [CaO (s)] = 40.00 J K^-1 mol^-1

Answer:

• Step1: Calculate ΔS^o 

  • ΔS^o  = ΣΔS^o _products – ΣΔS^o _reactants

  • ΔS^o  = (2 x ΔS^o  [CaO (s)]) –  (2 x ΔS^o [Ca (s)] + ΔS^o [O₂ (g)])

  • ΔS^o  = (2 x 40.00) – (2 x 41.00 + 205.0)

  • ΔS^o  = -207.0 J K^-1 mol^-1

• Step 2:Convert ΔS^o to kJ K^-1 mol^-1

  • ΔS^o = –207.0 J K^-1 mol^-1 ÷ 1000 = –0.207 kJ mol^-1

• Step 3: Calculate ΔG^o

  • ΔG^o= ΔH^o – TΔS^o

  • ΔG^o = –635.5 – (298 x -0.207) 

  • ΔG^o= –573.8 kJ mol^-1

• Since ΔG^o is negative, the reaction is thermodynamically favoured