Calculate the∆H_rxn for

Given the following reactions:

• 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 enthalpy change (∆Hrxn) for this reaction.
2. List the given reactions and their enthalpy changes: 1) Cl₂O(g) + F₂O(g) → 2 ClF(g) + O₂(g), ∆H = -167.4 kJ; 2) 4 ClF₃(g) + 4 O₂(g) → 2 Cl₂O(g) + 6 F₂O(g), ∆H = 682.8 kJ; 3) 2 F₂(g) + O₂(g) → 2 F₂O(g), ∆H = -181.7 kJ.
3. 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 need to manipulate the given reactions to derive the target reaction.
4. Reverse the second reaction to get Cl₂O(g) + 3 F₂O(g) → 2 ClF₃(g) + 2 O₂(g), and change the sign of ∆H to -682.8 kJ. This will help in forming ClF₃ on the product side.
5. Combine the manipulated reactions to cancel out intermediates and obtain the target reaction. Adjust stoichiometry as needed, ensuring that the sum of the enthalpy changes gives the ∆Hrxn for the target reaction.