Calculating Standard Enthalpy of Formation?

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?

Calculating Standard Enthalpy of Formation?
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.