Question

Find the ΔG° for the combustion of half a mole of glucose C6H12O6 in kJ

Answer 1

Answer:

-1435 kJ

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