Determine the enthalpy for the reaction s+0_2 --> so_2

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

the answer would be b i believe