Gibbs Free Energy Calculations (College Board AP® Chemistry): Revision Note

Calculate ΔG° for the following reaction at 298 K.

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 × 0.344)

  • ΔG^° = +32.49 kJ mol^-1

  • So, the reaction is thermodynamically unfavorable at 298 K

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

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

• S^° [Ca (s)] = 41.00 J K^-1 mol^-1

• S^° [O₂ (g)] = 205.0 J K^-1 mol^-1

• S^° [CaO (s)] = 40.00 J K^-1 mol^-1

Answer:

• Step 1: Calculate ΔS^° 

  • ΔS^°  = ΣΔS^° _products − ΣΔS^° _reactants

  • ΔS^°  = (2 × ΔS^°  [CaO (s)]) −  (2 × ΔS^° [Ca (s)] + ΔS^° [O₂ (g)])

  • ΔS^°  = (2 × 40.00) − (2 × 41.00 + 205.0)

  • ΔS^°  = −207.0 J K^-1 mol^-1

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

  • ΔS^° = −207.0 J K^-1 mol^-1 ÷ 1000 = −0.207 kJ mol^-1

• Step 3: Calculate ΔG^°

  • ΔG^°= ΔH^° − TΔS^°

  • ΔG^° = −635.5 − (298 × −0.207) 

  • ΔG^°= −573.8 kJ mol^-1

• Since ΔG^° is negative, the reaction is thermodynamically favored