Question

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?

Answer 1

Answer:

- 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.