Final answer:
The entropy change for the reaction Pb₂+₂(aq) + 2Cl⁻₂(aq) → PbCl₂(s) is predicted to be negative due to the reduction in the dispersal of matter. Calculating the enthalpy for PbCl₂(s) + Cl₂(g) → PbCl₄(ℓ) requires the use of Hess's Law and the given enthalpy changes for related reactions.
Step-by-step explanation:
To predict the sign of the entropy change (ΔS°) for the reaction Pb₂+₂(aq) + 2Cl⁻₂(aq) → PbCl₂(s), we consider the state and quantity of reactants and products. In this reaction, two aqueous ions combine to form a solid compound. Generally, this leads to a decrease in the dispersal of matter and degrees of freedom, resulting in a negative entropy change.
For the unknown enthalpy change in the reaction PbCl₂(s) + Cl₂(g) → PbCl₄(ℓ), we can use Hess's Law to calculate the enthalpy. Since we have the enthalpies for the reactions involving the formation of PbCl₂ and PbCl₄ from elemental Pb and Cl₂, we can infer the enthalpy for the missing reaction by rearranging and combining the equations provided.