ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ
Calculating ΔGꝋ
ΔGꝋ from ΔHꝋ and ΔSꝋ valuesCalculate the free energy change for the following reaction:
2NaHCO3 (s) → Na2CO3 (s) + H2O (l) + CO2 (g)
ΔHꝋ = +135 kJ mol-1 ΔSꝋ = +344 J K-1 mol-1
Answer
Step 1: Convert the entropy value in kilojoules
ΔSꝋ = +344 J K-1 mol-1 ÷ 1000 = +0.344 kJ K-1 mol-1
Step 2: Substitute the terms into the Gibbs Equation
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ
= +135 - (298 x 0.344)
= +32.49 kJ mol-1
The temperature is 298 K since standard values are quoted in the question
ΔGꝋ from other ΔGꝋ valuesWhat is the standard free energy change, ΔGꝋ, for the following reaction?
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)

Answer
ΔGꝋ = ΣΔGproductsꝋ - ΣΔGreactantsꝋ
ΔGꝋ = [(2 x CO2 ) + (3 x H2O )] - [(C2H5OH) + (3 x O2)]
ΔGꝋ = [(2 x -394 ) + (3 x -229 )] - [-175 + 0]
ΔGꝋ = -1300 kJ mol-1
The idea of free energy is what's 'leftover' to do useful work when you've carried out the reaction. The enthalpy change is the difference between the energy you put in to break the chemical bonds and the energy out when making new bonds; the entropy change is the 'cost' of carrying our the reaction, so free energy is what is you are left with!
转载自savemyexams
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