Answer:
-8.953 kJ/mol
It's a spontaneous reaction.
Explanation:
The variation of the free energy at standard conditions (ΔG°) can be calculated with the change in the enthalpy at standard conditions (ΔH°) and the variation the entropy at standard conditions (ΔS°):
ΔG° = ΔH° - TΔS°
For the reaction, ΔH° and ΔS° can be calculated by the values of the enthalpy of formation (Hf) and the entropy (S) of the reactants and products, which can be found in a thermodynamics table.
NO(g): Hf = 90.25 kJ/mol; S = 0.2108 kJ/K.mol
O₂(g): Hf = 0; S = 0.2051 kJ/K.mol
NO₂(g): Hf = 33.18 kJ/mol; S = 0.2401 kJ/K.mol
ΔH° = ∑n*Hf, products - ∑n*Hf, reactants (n is the number of moles)
ΔH° = (2*33.18) - (2*90.25)
ΔH° = -114.14 kJ/mol
ΔS° = ∑n*S, products - ∑n*S, reactants
ΔS° = (2*0.2401) - (0.2051 + 2*0.2108)
ΔS° = -0.1465 kJ/K.mol
ΔG° = -114.14 - (-0.1465)*718
ΔG° = -8.953 kJ/mol
ΔG° indicates if the reaction is spontaneous or nonspontaneous. If ΔG° < 0 the reaction is spontaneous. So, the reaction given is spontaneous.
