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Calculate ΔH for
2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
given the following reactions:
1/2N2 (g) + 1/2O2 (g) ⟶ NO (g) Δ H = 90.3 kJ
NO (g) + 1/2Cl2 (g) ⟶ NOCl (g) Δ H = −38.6 kJ