[Johnny 2375]

Can't seem to figure out this one could you help me please thanks.

Calculate the average C-C bond enthalpy in benzene (C6H6) given the following data:

C(s) → C(g)
enthalpy change = +715 kJmol-1

H2(g) → 2 H(g)
enthalpy change = +436 kJmol-1

C6H6(l) → C6H6(g)
enthalpy change = +31 kJmol-1

enthalpy of formation (C6H6(l)) = +49 kJmol-1

E (C-H) = +413 kJmol-1

formulated this equation:  6C + 3H2 → C6H6   

Then I tried using hess's law

((6(413)+31+6x)) - ((6(715)+6(436)) = 49

6x - 4397 = 49

6x = 4446
x= 741
The actual answer is +507 but I do not know why

[mjc123]

Have you actually drawn your Hess's law cycle explicitly? That might make things clearer.

Have you got your signs right in your expression for the heat of formation?

How many H-H bonds do you break?