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.

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

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