Given the following two equations and their respective enthalpies:
S (s) + O₂ (g) →SO₂ (g) D H _rxn= - 296.8 kJ/mol S (s) + ³/₂ O₂ (g) →SO₃ (g) D H _rxn = - 395.6 kJ/mol What is the enthalpy change D H _rxn for the following reaction?
2 SO₃ (g) →2 SO₂ (g) + O₂ (g) D H _rxn = ???

Question

Quizplus · Accepted Answer

The enthalpy change for the reaction 2 SO₃ (g) → 2 SO₂ (g) + O₂ (g) can be calculated by reversing the second reaction and doubling it, then subtracting the first reaction doubled. This gives: 2(-395.6 kJ/mol) - 2(-296.8 kJ/mol) = 197.6 kJ/mol.

Given the Following Two Equations and Their Respective Enthalpies