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The enthalpies(∆H), entropies (∆S) and temperaturesof three biochemical reactions are shown in the table. The Gibbs free energy equation is :∆G=∆H-T∆S.
What is the change in free energy of the oxidation of glucose? (Express your answer in kj/mol to the nearest tenths place.)
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Answer:
ΔG∘=−1486 kJ/mol
T=3032 K
Explanation:
ΔG∘=ΔH∘−TΔS∘
Under standard conditionsT=298K
∴ΔG∘=−1648.4×103−(298×−543.7)
ΔG∘=−1486 kJ/mol
At equilibriumΔG=0
∴0=ΔH∘−TΔS∘
assuming the enthalpy and entropychanges don't change much with temperature we can use the standard values.
∴T=ΔH∘ΔS∘
T=−1648.4×103−543.7 K
T=3032 K
Strictly speaking, that is not the equation for rusting. This requires both air and water and the product, rust, is hydrated iron(III) oxideFe2O3.xH2O.
Please follow
ΔG∘=−1486 kJ/mol
T=3032 K
Explanation:
ΔG∘=ΔH∘−TΔS∘
Under standard conditionsT=298K
∴ΔG∘=−1648.4×103−(298×−543.7)
ΔG∘=−1486 kJ/mol
At equilibriumΔG=0
∴0=ΔH∘−TΔS∘
assuming the enthalpy and entropychanges don't change much with temperature we can use the standard values.
∴T=ΔH∘ΔS∘
T=−1648.4×103−543.7 K
T=3032 K
Strictly speaking, that is not the equation for rusting. This requires both air and water and the product, rust, is hydrated iron(III) oxideFe2O3.xH2O.
Please follow
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