Given the following,determine $\Delta$_fG° at 298 K for PbO. Pb(s)+ PbO₂(s)$\to$ 2PbO(s); $\Delta$_rG° = -158.5 kJ/mol-rxn at 298K A) -187.9 kJ/mol B) -375.8 kJ/mol C) 58.8 kJ/mol D) 29.4 kJ/mol E) 117.6 kJ/mol

Question

Quizplus · Accepted Answer

We can use Hess's Law to solve for the standard enthalpy of formation of PbO.

First, we can write the balanced chemical equation for the reaction given:

3Pb(s) + 2PbO(s) → 5PbO(s)

Next, we can use the enthalpy of formation values given in the table to calculate the enthalpy change for the reaction:

ΔH° = ΣnΔH°f(products) - ΣnΔH°f(reactants)
ΔH° = 5(0) - 3ΔH°f(Pb) + 2ΔH°f(PbO)

Simplifying and solving for ΔH°f(PbO), we get:

ΔH°f(PbO) = [3ΔH°f(Pb) - 2ΔH°]/2

Substituting the given values, we get:

ΔH°f(PbO) = [3(0) - 2(-158.5)]/2
ΔH°f(PbO) = 187.9 kJ/mol

Therefore, the standard enthalpy of formation of PbO is -187.9 kJ/mol.

Given the Following,determine Δ\DeltaΔ fG° at 298 K for PbO →\to→ 2PbO(s); Δ\DeltaΔ

Given the Following,determine $\Delta$ _fG° at 298 K for PbO $\to$ 2PbO(s); $\Delta$