Question

Given: Pb(s) + PbO2(s) + 2H2SO4(l) → 2PbSO4(s) + 2H2O(l); ΔH° = –509.2 kJ SO3(g) + H2O(l) → H2SO4(l); ΔH° = –130. kJ determine ΔH° for the following thermochemical equation. Pb(s) + PbO2(s) + 2SO3(g) → 2PbSO4(s) Group of answer choices

Answer 1

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