Given the following reactions:
2 NOCl(g) → 2 NO(g) + Cl₂(g) Δ_rH° = +75.56 kJ/mol
2 NO(g) + O₂(g) → 2 NO₂(g) Δ_rH° = -113.05 kJ/mol
2 NO₂(g) → N₂O₄(g) Δ_rH° = -58.03 kJ/mol
Compute Δ_rH° of N₂O₄(g) + Cl₂(g) → 2 NOCl(g) + O₂(g) in kJ/mol.

Question

Quizplus · Accepted Answer

First, we need to flip the first reaction to get NOCl(g) on the reactant side. Then, we can add the reactions together to get the overall reaction: 2 NO(g) + O2(g) + 2 Cl2(g) → 2 NOCl(g) + 2 O2(g) The Δ_rH° of this reaction can be found by summing the Δ_rH° values of the individual reactions: Δ_rH° = 2(-113.05 kJ/mol) - 58.03 kJ/mol + 2(75.56 kJ/mol) = -95.52 kJ/mol Therefore, the answer is D.

Given the Following Reactions