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Multiple Choice
Given the reaction Pb (s) + CO2 (g) → PbO (s) + CO (g) with ΔH° = +131.4 kJ, and the standard enthalpies of formation ΔHf° of CO2(g) = -393.5 kJ/mol and CO(g) = -110.5 kJ/mol, what is the standard enthalpy of formation for PbO(s)?
A
-262.6 kJ/mol
B
-362.6 kJ/mol
C
-212.6 kJ/mol
D
-312.6 kJ/mol
Verified step by step guidance
1
Identify the given reaction: Pb (s) + CO2 (g) → PbO (s) + CO (g) and note that the standard enthalpy change ΔH° for this reaction is +131.4 kJ.
Recall that the standard enthalpy change of a reaction (ΔH°) is calculated using the formula: ΔH° = ΣΔHf°(products) - ΣΔHf°(reactants).
List the known standard enthalpies of formation: ΔHf° of CO2(g) = -393.5 kJ/mol and ΔHf° of CO(g) = -110.5 kJ/mol. The ΔHf° of Pb(s) is 0 kJ/mol because it is an element in its standard state.
Set up the equation using the enthalpy of formation values: ΔH° = [ΔHf°(PbO) + ΔHf°(CO)] - [ΔHf°(Pb) + ΔHf°(CO2)]. Substitute the known values into the equation: 131.4 kJ = [ΔHf°(PbO) - 110.5 kJ/mol] - [0 kJ/mol - 393.5 kJ/mol].
Solve the equation for ΔHf°(PbO): Rearrange the equation to find ΔHf°(PbO) and calculate its value.