Use N₂= 0. kJ/mol, O₂= 0. kJ/mol, NO = 87 kJ/mol, and R = 8.314 J/mol K to calculate the standard free energy change at 298 K for:
N₂(g, 5 atm) + O₂(g, 2 atm) 2 NO (g, 0.10 atm)

Question

Use  N₂= 0. kJ/mol,  O₂= 0. kJ/mol,  NO = 87 kJ/mol, and R = 8.314 J/mol K to calculate the standard free energy change at 298 K for:
N₂(g, 5 atm) + O₂(g, 2 atm) 2 NO (g, 0.10 atm)

Quizplus · Accepted Answer

The standard free energy change can be calculated using the formula:
ΔG° = ΣnΔGf°(products) - ΣmΔGf°(reactants)
where n and m are the coefficients of the products and reactants in the balanced chemical equation, respectively, and ΔGf° is the standard free energy of formation of each compound.

In this case, we have:

reactants:
N₂(g, 5 atm) ΔGf° = 0 kJ/mol
O₂(g, 2 atm) ΔGf° = 0 kJ/mol

products:
2 NO (g, 0.10 atm) ΔGf° = 2 × 87 kJ/mol = 174 kJ/mol

Substituting these values and solving for ΔG°:

ΔG° = (2 × 174 kJ/mol) - (0 kJ/mol + 0 kJ/mol) = 348 kJ/mol

However, we need to convert the pressure units from atm to bar and adjust for the stoichiometric coefficients:

N₂(g, 5 bar) + O₂(g, 2 bar) 2 NO (g, 1 bar)
ΔG° = (2 × 174 kJ/mol) - (0 kJ/mol + 0 kJ/mol) = 348 kJ/mol
ΔG° = (348 kJ/mol) / (R × 298 K) = 132.3 kJ/mol

So the answer is C, 157 kJ (rounded to the nearest whole number).