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24 July, 08:03

Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no (g) + cl2 (g) → 2nocl (g) substance: no (g) cl2 (g) nocl (g) ∆h°f (kj/mol) : 90.29 0 51.71 ∆g°f (kj/mol) : 86.60 0 66.07 s° (j/k·mol) : 210.65 223.0 261.6 what is the value of ∆g° fo this reaction at 550 k?

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  1. 24 July, 10:49
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    - 10.555 kJ/mol.

    Explanation:

    ∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

    Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

    ∆H°rxn is the standard enthalpy change of the reaction (J/mol).

    T is the temperature of the reaction (K).

    ∆S°rxn is the standard entorpy change of the reaction (J/mol. K).

    Calculating ∆H°rxn:

    ∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

    ∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) = (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

    Calculating ∆S°rxn:

    ∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants

    ∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) = (2 x 261.6 J/mol. K) - (1 x 223.0 J/mol. K) - (2 x 210.65 J/mol. K) = - 121.1 J/mol. K. = - 0.1211 kJ/mol. K.

    Calculating ∆G°rxn:

    ∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

    ∴ ∆G°rxn = ∆H°rxn - T∆S°rxn = ( - 77.16 kJ/mol) - (550 K) ( - 0.1211 kJ/mol. K) = - 10.555 kJ/mol.
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