Use the following reaction enthalpies to determine the reaction enthalpy for 2 HCl(g) + F2(g) → 2 HF(g) + Cl2(g).
4 HCl(g) + O2(g) → 2 H2O(l) + 2 Cl2(g) ΔHrxn = - 202.4 kJ/mol rxn
1/2 H2(g) + 1/2 F2(g) → HF(g) ΔHrxn = - 271.0 kJ/mol rxn
H2(g) + 1/2 O2(g) → H2O(l) ΔHrxn = - 285.8 kJ/mol rxn
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