Answered

  1. 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

Verified Answer :

syed514

ΔH(reaction) = ΔH(formation of products) - ΔH(formation of reactants) 
ΔH(reaction) = ( 1ΔH(Pb(s)) + 1ΔH(CO2(g)) ) - ( 1ΔH(PbO(s)) + 1ΔH(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.

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.

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