Homework Problem 8.67

[Tim Foster 2A]

8.67) Use the information in Tables 8.3, 8.6, 8.7 to estimate the enthalpy of formation of each of the following compounds in the liquid state. The standard enthalpy of sublimation of carbon is +717 kj/mol.

b)CH3OH
c)C6H6 (without resonance)
d)C6H6 (with resonance)

For b), I created the formation reaction CH2O+H2=>CH3OH. I calculated the bond enthalpies of the product to be -2059 KJ, and the bond enthalpies of the reactants to be 2(412) + 743 + 436 and a reaction enthalpy of -56, - Hvap for rxn enthalpy -91. The answer is -257 KJ; where am I going wrong?

[Suchita 2I]

Since enthalpy of formation by definition is the enthalpy change associated with the formation of 1 mole of a compound in its standard state from its ELEMENTS in their standard states, the equation you should be using is:
C(s) + 2H2(g) + 1/2O2(g) = CH3OH (l)
However, when calculating the enthalpy change of a reaction using bond enthalpies, all the reactants must be in gaseous state. So two additional steps are required of adding the enthalpy of sublimation of carbon and the enthalpy of condensation of methanol (- enthalpy of vaporization). This gives:
717+{(2*436)+(0.5*496)-(3*412)-360-463}-35.3
717-939-35.3= -257 kJ

[Nicole 1F]

For Part b of this problem, what does "atomize" mean/refer to in the solutions manual? Thanks!

[Suchita 2I]

The enthalpy of atomization basically refers to the enthalpy of sublimation, that is, the enthalpy of fusion+the enthalpy of vaporization. This is necessary because solid carbon needs to be converted to gaseous carbon in order to use bond enthalpies. Hope this helps!