Thermodynamics of NO and NO2 at 298 K: Given N2 + 2 O2 → 2 NO2 has ΔG° = 100 kJ per reaction (i.e., 50 kJ per mol NO2), and NO + 1/2 O2 → NO2 has ΔG° = -35 kJ per mol NO2, determine the standard free energy of formation of NO (kJ/mol).

Introduction / Context:Standard Gibbs free energies allow us to relate reaction energetics to the free energies of formation (ΔG°f) of species. By combining two reactions that involve NO2 and NO, we can algebraically solve for ΔG°f(NO) at 298 K. This skill is essential for reaction feasibility estimates and building consistent thermodynamic tables.

Given Data / Assumptions:

• N2 + 2 O2 → 2 NO2 has ΔG° = +100 kJ per reaction, i.e., ΔG°f(NO2) = +50 kJ/mol because elements have ΔG°f = 0.
• NO + 1/2 O2 → NO2 has ΔG° = −35 kJ per mol NO2. (Equivalently, 2 NO + O2 → 2 NO2 has ΔG° = −70 kJ per reaction.)
• Standard state is 1 bar and 298 K.

Concept / Approach:Use reaction free energy definitions in terms of formation values: ΔG°rxn = Σ ν ΔG°f(products) − Σ ν ΔG°f(reactants). For NO + 1/2 O2 → NO2, we have ΔG°rxn = ΔG°f(NO2) − ΔG°f(NO) − (1/2)ΔG°f(O2). Since ΔG°f(O2) = 0, this simplifies to ΔG°rxn = ΔG°f(NO2) − ΔG°f(NO). Insert the given numbers and solve for ΔG°f(NO).

Step-by-Step Solution:

Verification / Alternative check:Tabulated values at 298 K typically list ΔG°f(NO) near +86 kJ/mol and ΔG°f(NO2) near +51 kJ/mol, consistent with the rounded problem data and the computed answer of about 85 kJ/mol.

Why Other Options Are Wrong:

• 15, 30 kJ/mol: far below the well-known positive ΔG°f for NO.
• 170 kJ/mol: double-counting per mole stoichiometry.
• 135 kJ/mol: arises if one misinterprets per-reaction vs per-mole data.

Common Pitfalls:Misreading the per-mole basis of ΔG°; using NO + O2 → 2 NO2 directly without halving to the per-mole basis; forgetting ΔG°f for elements is zero.

Final Answer:85