Question

Determine the enthalpy of the reaction: S (s) + 3/2 O2 (g)  SO3 (g), given the following reactions. SO2 (g)  S(s) + O2 (g) ΔH = +296.0 kJ 2 SO3 (g)  2 SO2 (g) + O2 (g) ΔH = +198.2 kJ

Answer 1

-196 kJ

Step-by-step explanation:

By the Hess' Law, the enthalpy of a global reaction is the sum of the enthalpies of the steps reactions. If the reaction is multiplied by a constant, the value of the enthalpy must be multiplied by the same constant, and if the reaction is inverted, the signal of the enthalpy must be inverted too.

2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ

S(s) + O₂(g) → SO₂(g) ΔH = -297 kJ (inverted and multiplied by 2)

2S(s) + 3O₂(g) → 2SO₃(g) ΔH = -790 kJ

2SO₂(g) → 2S(s) + 2O₂(g) ΔH = +594 kJ

-------------------------------------------------------------

2S(s) + 3O₂(g) + 2SO₂(g) → 2SO₃(g) + 2S(s) + 2O₂(g)

Simplifing the compounds that are in both sides (bolded):

2SO₂(g) + O₂(g) → 2SO₃(g) ΔH = -790 + 594 = -196 kJ