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 .

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

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

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the accuracy of the data.