Question

Use the thermochemical data that follow to calculate the value of ΔH° for the reaction: 4 B 2 A + C + 2 F A + 2B E ΔH° = 80 kJ C + 2D 2E ΔH° = -25 kJ F + 2A D ΔH° = 45 kJ -100 kJ 225 kJ -225 kJ 95 kJ 60 kJ Using the information below, calculate ΔHf° for PbO(s) PbO(s) + CO(g) → Pb(s) + CO₂(g) ΔH° = –131.4 kJ ΔHf° for CO₂(g) = –393.5 kJ/mol ΔHf° for CO(g) = –110.5 kJ/mol +252.1 kJ/mol –372.6 kJ/mol +283.0 kJ/mol –151.6 kJ/mol –283.0 kJ/mol

Answer 1

The enthalpy of formation (ΔHf°) for PbO(s) is calculated to be 151.6 kJ/mol using Hess's Law and the provided thermochemical data for the reaction and enthalpies of formation for CO(g) and CO₂(g).

Step-by-step explanation:

To calculate the enthalpy change (ΔH°) for the reaction PbO(s) + CO(g) → Pb(s) + CO₂(g), we need to know the enthalpies of formation (ΔHf°) for each substance involved in the reaction. The given ΔH° for the reaction is –131.4 kJ, and the enthalpies of formation for CO₂(g) and CO(g) are –393.5 kJ/mol and –110.5 kJ/mol respectively.

The enthalpy of formation for Pb(s) is assumed to be 0 kJ/mol because it is in its standard state. To find the enthalpy of formation for PbO(s), we apply Hess's Law. The equation ΔHf°(PbO) = ΔH° + ΔHf°(CO) - ΔHf°(CO₂) is used, where ΔH° is the enthalpy change for the overall reaction.

Inserting the known values into the equation, we get:

• ΔHf°(PbO) = (–131.4 kJ) + (–110.5 kJ) - (–393.5 kJ)

• ΔHf°(PbO) = –131.4 kJ - 110.5 kJ + 393.5 kJ

• ΔHf°(PbO) = 151.6 kJ/mol