Using the provided table, determine the enthalpy for the reaction 2 NH₃ (g) + 3 N₂O (g)→ 4 N₂ (g) + 3 H₂O (l)

Question

Substance ∆H_f° (kJ/mol)

NH₃ (g) -46.2

N₂O _(g) 81.6

N₂ (g) 0

H₂O (l) -285.8

Answer 1

ΔH° = ΣnH_f^º (products) - ΣmH_f^º (reactants); n and m are total moles

ΔH° = (3 mol H₂O × -285.8 kJ/mol H₂O) - [ (2 mol NH₃ × -46.2 kJ/mol NH₃) + (3 mol N₂O × 81.6 kJ/mol N₂O)]

= 3(-285.8 kJ) - 2(-46.2 kJ) - 3(81.6 kJ) = -857.4 +92.4 - 244.8 = -1009.8 kJ

ΔH° = -1009.8 kJ