Given the reaction: N_2O(g) + NO_2(g) → 3 NO(g), what is the standard enthalpy change (ΔH°) for this reaction? (Standard enthalpies of formation: ΔH_f°[N_2O(g)] = 82 kJ/mol, ΔH_f°[NO_2(g)] = 34 kJ/mol, ΔH_f°[NO(g)] = 90 kJ/mol)

• A. ΔH° = -184 kJ
• B. ΔH° = 184 kJ
• C. ΔH° = 164 kJ
• D. ΔH° = -164 kJ

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{N_2O(g) + NO_2(g) \rightarrow 3\ NO(g)}\).
2. Recall the formula for the standard enthalpy change of a reaction using standard enthalpies of formation: \(\Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})\).
3. Identify the standard enthalpies of formation for each species from the problem: \(\Delta H_f^\circ[\mathrm{N_2O(g)}] = 82\ \mathrm{kJ/mol}\), \(\Delta H_f^\circ[\mathrm{NO_2(g)}] = 34\ \mathrm{kJ/mol}\), and \(\Delta H_f^\circ[\mathrm{NO(g)}] = 90\ \mathrm{kJ/mol}\).
4. Calculate the total enthalpy of the products by multiplying the enthalpy of formation of NO by its coefficient: \$3 \times 90\ \mathrm{kJ/mol}$, and sum these values.
5. Calculate the total enthalpy of the reactants by summing the enthalpies of formation of \(\mathrm{N_2O}\) and \(\mathrm{NO_2}\), then subtract the reactants' total from the products' total to find \(\Delta H^\circ\).