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Calculate the standard enthalpy change for the reaction 2HCl(g) + F2(g) → 2HF(ℓ) + Cl2(g) given 4HCl(g) + O2(g) → 2H2O(ℓ) + 2Cl2(g) ΔH° = -202.4 kJ/mol rxn 1/2H2(g) + 1/2F2(g) → HF(ℓ) ΔH° = -600.0 kJ/mol rxn H2(g) + 1/2O2(g) → H2O(ℓ) ΔH° = -285.8 kJ/mol rxn 1. ΔH° = +1116.6 kJ/mol rxn 2. ΔH° = -516.6 kJ/mol rxn 3. ΔH° = +516.6 kJ/mol rxn 4. ΔH° = +1587.2 kJ/mol rxn 5. ΔH° = -1015.4 kJ/mol rxn 6. ΔH° = +1015.4 kJ/mol rxn 7. ΔH° = +1088.2 kJ/mol rxn 8. ΔH° = -1088.2 kJ/mol rxn 9. ΔH° = -1116.6 kJ/mol rxn 10. ΔH° = -1587.2 kJ/mol rxn