For the reaction H₂O₂(g)$\rarr$ H₂O(g)+ ¹/₂O₂(g), $\Delta$H° = -106 kJ/mol and $\Delta$S° = 58 J/K·mol at 25°C.Calculate $\Delta$G° for this reaction at this temperature.

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The Answer of For the reaction H₂O₂(g)$\rarr$ H₂O(g)+ ¹/₂O₂(g), $\Delta$H° = -106 kJ/mol and $\Delta$S° = 58 J/K·mol at 25°C.Calculate $\Delta$G° for this reaction at this temperature.

For the Reaction H2O2(g) →\rarr→ H2O(g)+ 1/2O2(g) Δ\DeltaΔ H° = -106 KJ/mol And

For the Reaction H₂O₂(g) $\rarr$ H₂O(g)+ ¹/₂O₂(g) $\Delta$ H° = -106 KJ/mol And