4NO_2(g) + O_2(g) ⟶ 2N₂O_5(g), Δ_rH = –111 kJ.

If N₂O_5(s) is formed instead of N₂O_5(g) in the above reaction, the Δ_rH value will be:
(given, ΔH of sublimation for N₂O₅ is 54 kJ mol^–1)