ΔG° at 500. K ΔG°= -197.5kJ/mol
The expression for the standard change in free energy is:
ΔG° = ΔH°-TΔS°
Where,
ΔG° is the change in the Gibbs free energy.
T is the absolute temperature. (T in kelvins)
ΔH° is the enthalpy change of the reaction.
ΔS° is the change in entropy.
We have,
at T = 500 K
Because ΔG° = ΔH°-TΔS°
at 500K, ΔG° = -197.5kJ/mol.
ΔG°= -402kJ/mol- (500K) (-0.409kJ/mol.K)
ΔG°= -197.5kJ/mol
hence at 500 K ΔG°= -197.5kJ/mol
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