Enthalpies of reactions!

I dont know whether I get this or not so I basically want my work verfied :P
So I need to work out the enthalpy for the reaction of propane decomposited to cyclopropane and hydrogen.
If the standard heat of combustion for propane is -2220,
and the standard heat of combustion for cyclopropane & hydrogen is -1966 and -286.
Would the enthalpy of reaction be -4472 kjmol-1?
It just seems way too high :s

There is no such word as 'decomposited'

the answer's wrong ...

So reaction you want is:

C3H8 ---> C3H6 + H2

you have:

C3H8 + 5O2 ---> 3CO2 + 4H2O = -2220
C3H6 + 4.5O2 ----> 3CO2 + 3H2O = -1966
H2 + 0.5O2 ----> H2O = -286

Use Hess's law

ermm..can you possibly explain it in the context of hess's law :P We've only just started it and im slightly confused :L

Hess lets you go from A to B via any intermediate phase (eg Combustion products)

A --------------> combustion products <-------------------- B

To go from A to combustion products is the enthalpy

BUT to go from combustion products to B is the REVERSE of the combustion enthalpy

so you add the combution enthalpy of the reactants to the reverse of the combustion enthalpy of the products

does that help?

Thankyou, I get the basics now (:
However, how do you work it out with the reversion combustion? Would you just change the negative to a positive?
I now got an answer of 32kjmol-1??