Thermochemistry of ammonia decomposition:\nGiven ΔH_f°(NH3, g) = −46 kJ per kg·mole, find ΔH° for 2 NH3(g) → N2(g) + 3 H2(g).

Introduction / Context:
Calculating reaction enthalpy from standard heats of formation is a staple skill in energy balances and reactor design. This problem reverses ammonia formation to evaluate the decomposition enthalpy.

Given Data / Assumptions:

• Standard heat of formation: ΔH_f°(NH3, g) = −46 kJ per kg·mole.
• Reaction: 2 NH3(g) → N2(g) + 3 H2(g).
• ΔH_f°(N2, g) = 0; ΔH_f°(H2, g) = 0 (elements in standard state).

Concept / Approach:
Use Hess’s law: ΔH°_rxn = Σ ν_p ΔH_f°(products) − Σ ν_r ΔH_f°(reactants). For elemental products, heats of formation are zero. The reaction is the reverse of ammonia formation; reversing a reaction changes the sign of ΔH and scaling by stoichiometric coefficients scales ΔH proportionally.

Step-by-Step Solution:
ΔH°_rxn = [1·0 + 3·0] − [2·(−46)] kJ per kg·mole of reaction as written.ΔH°_rxn = 0 − (−92) = +92 kJ per kg·mole.Therefore, decomposition is endothermic by 92 kJ per kg·mole (of reaction).

Verification / Alternative check:
Formation: N2 + 3 H2 → 2 NH3 has ΔH° = −92 kJ per kg·mole (twice −46). Reversing gives +92 kJ per kg·mole, consistent.

Why Other Options Are Wrong:
46 and −23 use wrong scaling/sign; −92 is the formation value, not decomposition; 0 conflicts with the known exothermic formation of NH3.

Common Pitfalls:
Forgetting to multiply by stoichiometric coefficient 2; sign reversal when flipping the reaction.

Final Answer:
92