Given: 4 NO₂(g)+ O₂(g)→ 2 N₂O₅(g)ΔH° = -110.2 kJ Find ΔH° for N₂O₅(g)→ 2 NO₂(g)+ 1/2 O₂(g).

Name: Given: 4 NO 2 (g)+ O 2 (g)→ 2 N 2 O 5 (g)ΔH° = -110.2 kJ Find ΔH° for N 2 O 5 (g)→ 2 NO 2 (g)+ 1/2 O 2 (g).
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Question

Quizplus · Accepted Answer

The given reaction is the reverse of the target reaction. Reversing a reaction changes the sign of ΔH°. Therefore, ΔH° for the reverse reaction is +110.2 kJ for 2 moles of N₂O₅. For 1 mole, it is 55.1 kJ.

Given: 4 NO2(g)+ O2(g)→ 2 N2O5(g)ΔH° = -110

Given: 4 NO₂(g)+ O₂(g)→ 2 N₂O₅(g)ΔH° = -110