Calculate the ∆H_rxn for
ClF (g) + F₂ (g) → ClF₃ (g) ΔH_rxn = ?
Given the following reactions:
Cl₂O (g) + F₂O (g) → 2 ClF (g) + O₂ (g) ΔH° = −167.4 kJ
4 ClF₃ (g) + 4 O₂ (g) → 2 Cl₂O (g) + 6 F₂O (g) ΔH° = 682.8 kJ
2 F₂ (g) + O₂ (g) → 2 F₂O (g) ΔH_rxn = −181.7 kJ

• A. -177.85 kJ
• B. -355.73 kJ
• C. -258.36 kJ
• D. -487.27 kJ

Answer: (A)

Verified step by step guidance

1. Identify the target reaction: ClF (g) + F2 (g) → ClF3 (g). We need to find the ΔHrxn for this reaction using the given reactions.
2. Use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction. We will manipulate the given reactions to achieve the target reaction.
3. Reverse the second given reaction: 2 Cl2O (g) + 6 F2O (g) → 4 ClF3 (g) + 4 O2 (g). This changes the sign of ΔH° to -682.8 kJ.
4. Multiply the third given reaction by 3/2 to match the stoichiometry of F2 in the target reaction: 3 F2 (g) + 3/2 O2 (g) → 3 F2O (g). This scales the ΔHrxn to -272.55 kJ.
5. Combine the modified reactions to cancel out intermediates and obtain the target reaction. Sum the ΔH values of the modified reactions to find the ΔHrxn for the target reaction.