Enthalpy change scales with reaction extent based on limiting reactant

Steps:

• The given ΔH = +66 kJ is for 1 mol N2 + 2 mol O2 → 2 NO2.
• With 2 mol N2 and 2 mol O2 available, O2 is limiting and supports reaction of 1 mol N2.
• The amounts used match exactly one mole of the reaction as written.
• Thus, ΔH = +66 kJ for the process.

Why C is correct:

• Enthalpy is an extensive property; here, the reaction extent is 1 (stoichiometric match), so ΔH equals the given value per the reaction equation.

Why the others are wrong:

• A: +16 kJ – No stoichiometric basis for dividing by 4.
• B: +33 kJ – No basis for halving the value.
• D: +132 kJ – Assumes full reaction of 2 mol N2, ignoring O2 limitation.

Final answer: C