Consider the following reaction: 2NO (g) + O2 (g) →2NO2 (g) Estimate ΔG∘ for this reaction at each of the following temperatures and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the give temperature range.) 718k

-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.