Using the provided table and the equation below, determine the heat of formation for PbS. 2 PbS (s) + 3 O₂ (g) → 2 SO₂ (g) + 2 PbO (s) ∆H° = -828.4 kJ/mol

Question

Substance ∆H_f° (kJ/mol)

O₂ (g) 0

SO_{2 (g)} -296.9

PbO (s) -217.3

Answer 1

2PbS(s) + 3O₂(g) ===> 2SO₂(g) + 2PbO(s) ... ∆Hº = -828.4 kJ/mol

∆H_{reaction =} ∑n∆H_products - ∑n∆H_reactants ... equation 1

∆H_reaction = -828.4 kJ/mol

∑n∆H_products = (2 x -296.9) + (2 x -217.3) = -593.8 + -434.6 = -1028.4 kJ

∑n∆H_reactants = 2 x ∆H_PbS + O = 2∆H_PbS which we will call 2x

Using equation 1 and solving for x:

-828.4 = -1028.4 - 2x

-2x = -828.4 + 1028.4

-2x = +200

2x = -200

x = -100 kJ/mol