Answer:
-1435 kJ
Step-by-step explanation:
The Gibbs free energy, ΔG is an indication the amount of (useful) work that can be obtained from a chemical reaction involving reacting species and it is negative for spontaneous reaction
ΔG° = ΔH° - TΔS
ΔG = ΔG° + RT㏑Q
ΔG = G (reactants) - G(products)
Given that the combustion reaction of glucose is as follows;
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
The above reaction has a ΔG° value of -2870 kJ
Therefore, the free energy for half mole of glucose = -2870 kJ/2 = -1435 kJ