Question

Calculate ΔG° at 298.0 K for this reaction given that ΔH°rxn = –12.0 kJ and ΔS°rxn = –147 J/K:

2NO(g) + O₂(g) → 2NO₂(g)

A) –1353 kJ
B) 1353 kJ
C) –1153 kJ
D) 1153 kJ
E) None of the above.

Answer 1

By using the standard free energy change equation, ΔG° = ΔH° - TΔS°, and substituting the provided values for ΔH° and ΔS° at 298.0 K, we find that ΔG° for the reaction is 31.806 kJ or 31806 J, which does not match any of the given options, so the answer is E) None of the above.

Step-by-step explanation:

To calculate the standard free energy change (ΔG°) at 298.0 K for the reaction 2NO(g) + O₂(g) → 2NO₂(g) given that the standard enthalpy change (ΔH°) for the reaction is − 12.0 kJ and the standard entropy change (ΔS°) for the reaction is − 147 J/K, we use the standard free energy change equation:

ΔG° = ΔH° - TΔS°

Substituting the provided values:

ΔG° = (-12.0 kJ) - (298.0 K)(-147 J/K × 1 kJ/1000 J)

ΔG° = (-12.0 kJ) - (298.0 K)(-0.147 kJ/K)

ΔG° = (-12.0 kJ) + (43.806 kJ)

ΔG° = 31.806 kJ

ΔG° = 31806 J

Since the result is positive, the reaction is not spontaneous under standard conditions at 298.0 K. The closest option to our calculated value is E) None of the above.