ΔH = -296.7 kJ
We have two equations:
(1) 2S +3O₂ → 2SO₃; ΔH = -791.6 kJ
(2) 2SO₂ + O₂ → 2SO₃; ΔH = -198.2 kJ
From these, we must devise the target equation:
(3) S + O₂ → SO₂; ΔH = ?
__________________
The target equation has 1S on the left, so you divide equation (1) by 2.
When you halve an equation, you halve its ΔH.
(4) S + ³/₂O₂ → SO₃; ΔH = -395.8 kJ
Equation (4) has 1SO₃ on the right, and that is not in the target equation.
You need an equation with 1SO₃ on the left.
Reverse Equation (2) and divide it by 2.
When you reverse an equation, you reverse the sign of its ΔH.
(5) SO₃ → SO₂ + ½O₂; ΔH = +99.1 kJ
Now, you add equations (4) and (5), cancelling species that appear on opposite sides of the reaction arrows.
When you add equations, you add their ΔH values.
_______________________________________
We get the target equation (3):
(4) S + ³/₂O₂ → SO₃; _ΔH = -395.8 kJ
(5) SO₃ → SO₂ + ½O₂; ΔH = +99.1 kJ
(3) S + O₂ → SO₂; __ΔH = -296.7 kJ
