Answer:
ΔHr = -275 kj
Explanation:
It is possible to obtain the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid using the reactions:
(1) H₂SO₄(l) → SO₃(g) + H₂O (l) ΔH=+113kJ
(2) Pb(s) + PbO₂(s) + 2SO₃(g) → 2PbSO₄(s) ΔH=−775kJ
If you sum (1) + ¹/₂(2) you will obtain:
H₂SO₄(l) + ¹/₂Pb(s) + ¹/₂PbO₂(s) → PbSO₄(s) + H₂O(l)
Using Hess's law, the net change in enthalpy for this reaction could be obtained as:
ΔHr = ΔH(1) + ¹/₂ΔH(2)
ΔHr = +113kJ + ¹/₂ -775kJ
ΔHr = -275 kJ
