Question

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.

Answer 1

2S (s) + 3O2 (g)  → 2SO3 (g)  ΔH = -790 kJ

S (s) + O2 (g)  → SO2(g) ΔH = -297 kJ

get

2SO2 (g) + O2 (g)  → 2SO3 (g)

invert(2) and multiply by 2

2S (s) + 3O2 (g)  → 2SO3 (g)  ΔH = -790 kJ

2SO2(g)  → 2S (s) + 2O2 (g) ΔH =2*(297) =594 kJ

add both equations

2S (s) + 3O2 (g) + 2SO2(g)  → 2SO3+ 2S (s) + 2O2 Hrxn = -790 + 594

cancel common terms

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