ΔH° = -769.2 kJ is required.
Explanation:
Given equations are:
Pb(s) + PbO₂(s) + 2 H₂SO₄(l) → 2 PbSO₄(s) + 2H₂O(l); ΔH° = –509.2 kJ ------1
SO₃(g) + H₂O(l) → H₂SO₄(l); ΔH° = –130. kJ ----2
From the above 2 equations, by adding or subtracting or multiplying or dividing the required amount to get the final equation by means of Hess's law.
Multiplying eq. 2 by 2 we will get,
2 SO₃(g) + 2 H₂O(l) → 2 H₂SO₄(l) ; ΔH° = -260 kJ ----3
Adding it to eq. 1 we will get,
Pb(s) + PbO₂(s) + 2 H₂SO₄(l) → 2 PbSO₄(s) + 2H₂O(l); ΔH° = –509.2 kJ ------1
2 H₂SO₄ and 2 H₂O gets cancelled since they are on opposite sides, and the ΔH° values are added to get the ΔH° value of the required equation as,
Pb(s) + PbO₂(s) + 2 SO₃(g) → 2 PbSO₄(s)
ΔH° = -509.2 - 260 = -769.2 kJ
