Determine the standard enthalpy change, ΔH298∘, for the reaction
2 N2O5(l) → 4 NO2(g) + O2(g)
in two independent ways and state the numerical difference (in kJ) between the two results.
Route A – Combine the following reactions:
A. 2 NO2(g) + 1⁄2 O2(g) → N2O5(l) ΔH∘ = –56.5 kJ
B. NO(g) + 1⁄2 O2(g) → NO2(g) ΔH∘ = –57.2 kJ
C. N2O5(l) + NO(g) → 3⁄2 N2(g) + O2(g) ΔH∘ = –348.5 kJ
Route B – Use the standard enthalpies of formation
ΔHf∘(N2O5, l) = –364.0 kJ mol–1
ΔHf∘(NO2, g) = +33.2 kJ mol–1
ΔHf∘(N2, g) = 0 kJ mol–1
ΔHf∘(O2, g) = 0 kJ mol–1.
Provide the two calculated ΔH298∘ values and their difference.