The question is incomplete, the complete question is;
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:
-275 kJ
Step-by-step explanation:
We could use the two reactions outlined below to obtain the enthalpy change for the formation of lead sulphate following the Hess law of constant heat summation.
H₂SO₄(l) → SO₃(g) + H₂O (l) ΔH=+113kJ Reaction 1
Pb(s) + PbO₂(s) + 2SO₃(g) → 2PbSO₄(s) ΔH=−775kJ Reaction 2
The sum of reaction 1 and half of reaction 2 yields:
H₂SO₄(l) + ¹/₂Pb(s) + ¹/₂PbO₂(s) → PbSO₄(s) + H₂O(l)
Applying Hess law of constant heat summation;
ΔHreaction = ΔH(1) + ¹/₂ΔH(2)
ΔHreaction = +113kJ + ¹/₂ (-775kJ)
ΔHreaction = -275 kJ