Given... #C(s) +O_2(g) rarr CO_2(g)# #;DeltaH^@=-394*kJ*mol^-1# #(i)# #Pb(s) +1/2O_2(g) rarr PbO(s)# #;DeltaH^@=-217*kJ*mol^-1# #(ii)# #PbO(s) +CO(g) rarr Pb(s)+CO_2(g)# #;DeltaH^@=-66*kJ*mol^-1# #(iii)# How do we find #DeltaH_f^@CO(g)#?

1 Answer
Dec 16, 2017

We add the individual equations algebraically to get...

#C(s) +1/2O_2(g) rarr CO(g)#

Explanation:

And so we gots three thermochemical equations...

#C(s) +O_2(g) rarr CO_2(g)# #;DeltaH^@=-394*kJ*mol^-1# #(i)#

#Pb(s) +1/2O_2(g) rarr PbO(s)# #;DeltaH^@=-217*kJ*mol^-1# #(ii)#

#PbO(s) +CO(g) rarr Pb(s)+CO_2(g)# #;DeltaH^@=-66*kJ*mol^-1# #(iii)#

And so take #-(iii)+(i)-(ii)# to get....

#Pb(s)+CO_2(g)+C(s) +O_2(g)+PbO(s)rarrPbO(s) +CO(g)+CO_2(g)+Pb(s) +1/2O_2(g) #....and we cancel out common reactants/products....

#cancel(Pb(s))+cancel(CO_2(g))+C(s) +O_2(g)+cancel(PbO(s))rarrcancel(PbO(s)) +CO(g)+cancel(CO_2(g))+cancel(Pb(s)) +cancel(1/2O_2(g)) # to give....

#C(s) +1/2O_2(g)rarrCO(g)# which is the equation we wanted....

And to get the #DeltaH^@# reaction for this equation, we take the algebraic sum of the thermodynamic parameters, and OBSERVE the sign conventions....

#-(iii)+(i)-(ii)={-(-66)-394-(-217)}*kJ*mol^-1=-111*kJ*mol^-1#

...so your option is pretty clear.....but this is a lot of work for a multiple choice question; I take it is first year chem?