Question

The lead-acid storage battery is the oldest rechargeable battery in existence. It was invented in 1859 by French physician Gaston Plante and still retains application today, more than 150 years later. There are two reactions that take place during discharge of the lead-acid storage battery. In one step, sulfuric acid decomposes to form sulfur trioxide and water: H2SO4 (l) → SO3 (g) + H2O (l) ΔH=+113.kJ In another step, lead, lead(IV) oxide, and sulfur trioxide react to form lead(II) sulfate: Pb (s) + PbO2 (s) + 2SO3 (g) → 2PbSO4 (s) ΔH=−775.kJ Calculate the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid from these reactions. Round your answer to the nearest kJ .

Answer 1

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

I hope it helps!