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) A)3.77 × 10³ kJ B)-521 kJ C)-3.77 × 10³ kJ D)-639 kJ E)-769 kJ

Question

Quizplus · Accepted Answer

To find ΔH° for the given reaction, we need to combine the two given reactions. The first reaction is already in the desired form. For the second reaction, we need to reverse it and multiply by 2 to match the stoichiometry of the target reaction. This gives us ΔH° = 2 × 130 kJ = 260 kJ. Adding this to the ΔH° of the first reaction (-509.2 kJ), we get -509.2 kJ + 260 kJ = -769.2 kJ, which rounds to -769 kJ.

Given: Pb(s)+ PbO2(s)+ 2H2SO4(l)→ 2PbSO4(s)+ 2H2O(l); ΔH° = -509.2 KJ

Given: Pb(s)+ PbO₂(s)+ 2H₂SO₄(l)→ 2PbSO₄(s)+ 2H₂O(l); ΔH° = -509.2 KJ