Question

15. 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

A) –151.6 kJ/mol
B) –283.0 kJ/mol
C) +283.0 kJ/mol
D) –372.6 kJ/mol
E) +252.1 kJ/mol

Answer 1

Answer: The correct answer is Option A.

Step-by-step explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as
`\Delta H^o`

The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H^o_f(product)]-\sum [n* \Delta H^o_f(reactant)]`

For the given chemical reaction:

`PbO(s)+CO(g)\rightarrow Pb(s)+CO_2(g);\Delta H^o=-131.4kJ`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(1* \Delta H^o_f_((Pb(s))))+(1* \Delta H^o_f_((CO_2(g))))]-[(1* \Delta H^o_f_((PbO(s))))+(1* \Delta H^o_f_((CO(g))))]`

We are given:

`\Delta H^o_f_((CO_2(g)))=-393.5kJ/mol\\\Delta H^o_f_((CO(g)))=-110.5kJ/mol\\\Delta H^o_f_((Pb(s)))=0kJ/mol\\\Delta H^o_(rxn)=-131.4kJ`

Putting values in above equation, we get:

`-131.4=[(1* \Delta H^o_f_((0)))+(1* (-393.5))]-[(1* \Delta H^o_f_((PbO(s))))+(1* (-110.5))]\\\\\Delta H^o_f_((PbO(s)))=-151.6kJ/mol`

Hence, the correct answer is Option A.

Answer 2

ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants)
ΔH(reaction) = ( 1*ΔH(Pb(s)) + 1*ΔH(CO2(g)) ) - ( 1*ΔH(PbO(s)) + 1*ΔH(CO(g)) )
ΔH(reaction) = ( 0 + -393.5 ) - ( ΔH(PbO(s)) + -110.5 )
ΔH(reaction) = -283 - ΔH(PbO(s))
-131.4 = -283 -ΔH(PbO(s))
ΔH(PbO(s)) = -151.6 kJ

So, the best answer is A.