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DaniilM [7]
3 years ago
14

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

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
8 0

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

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⇒ PH2(g) = 1.0079 bar - 0.03167 bar = 0.97623 bar = 0.9635 atm

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⇒ molar mass M(s) =   (0.595 g) / (9.1082 E-3 mol)

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