Calculating Standard Enthalpy of Formation?

Question

The standard enthalpy of combustion for napthalene, C10H8 (s) is -5156.8 kj/mol. Using this data and the standard enthalpies of formation, delta Hf H2O (l) = -285.9 kj/mol ; CO2 (g) = -393.5 kj/mol, calculate the STANDARD ENTHALPY OF FORMATION (was bolded in the text) of C10H8 (s) in kj/mol.

a) 78.2kj b) 935.9kj c) -1065.4kj d) 3619.7kj e) -10235.4kj

i think it's A, but i'm not sure. can someone show me how to solve this?

Answer 1

It looks like you got it. This is how it came out for me:

This is the reaction:
C10H8(s) + 12 O2(g) ® 10 CO2(g) + 4 H2O(l)

and Hc=(Summation of Hf of products)-(Summation of Hf of reactants)
So...

Hc (C10H8) = 10 Hf(CO2) + 4 Hf(H2O) – Hf(C10H8)
-5156.8kJ/mol=(10)(-393.5 kJ/mol)+(4)(-285.9kJ/mol0-Hf (C10H8)
-5156.8kJ/mol=(-3935 kJ/mol)+(-1143.6 kJ/mol)- Hf (C10H8)
-5156.8kJ/mol=(-5078.6)- Hf(C10H8)
-78.2kJ/mol=-Hf (C10H8)
78.2kJ/mol=Hf (C10H8)

I believe that's how it's done, but it's been a while since I've done that kind of problem, so it's probably not a terrible idea to double check with someone else.