The enthalpy of decomposition of NO2Cl is -114 kJ. Use the following data to calculate the heat of formation of NO2Cl from N2, O2, and Cl2:
NO2Cl (g) → NO2 (g) + 1/2Cl2(g) ΔH°rxn = -114 kJ 1/2
N2 (g) + O2 (g) → NO2 (g) ΔH°f = 33.2 kJ
1/2 N2 (g) + O2 (g) + 1/2 Cl2 (g) → NO2Cl (g) ΔH°f = ?