Question

15. Using the information below, calculate ΔHf° for PbO(s)

Answer 1

Answer: The correct answer is Option A.

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:  

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

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:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Pb(s))})+(1\times \Delta H^o_f_{(CO_2(g))})]-[(1\times \Delta H^o_f_{(PbO(s))})+(1\times \Delta H^o_f_{(CO(g))})]$

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:

$-131.4=[(1\times \Delta H^o_f_{(0)})+(1\times (-393.5))]-[(1\times \Delta H^o_f_{(PbO(s))})+(1\times (-110.5))]\\\\\Delta H^o_f_{(PbO(s))}=-151.6kJ/mol$

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.