Given the following reactions 2s (s) + 3o2 (g) → 2so3 (g) δh = -790 kj s (s) + o2 (g) → so2 (g) δh = -297 kj the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 2so2 (g) + o2 (g) → 2so3 (g) is kj

-196 kJ

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

b. examine and describe cells