[enthalpyburns]

I have a question in my Chemistry textbook I can't seem to solve. 

I know that heat of formation= delta H= sum of bonds broken minus sum of bonds formed.

Can you help? 

 The standard enthalpies of formation for S (g), F (g), SF4 (g), and SF6 (g) are +278.8, +79.0, -775, and -1209 kJ per mole, respectively. What is the energy of the S--F bond WITHOUT using the dissociation energy bond chart. (Step by step proof is needed)

 Thank you!

[magician4]

hint: try to derive a pathway form educts to products related to the particles given to you
use Hess's law on this scheme


regards

Ingo

[enthalpyburns]

Magician, I tried another method by dividing the total SF6 (1209Jk/mol) by 6 and dividing the total SF4 (-776Kj/mol) by 4. I thought that should get the "Average" energy per bond but it did not work.. Am I doing something wrong?  (BTW, I can't seem to find the
S-F BOND by using Hess's Law)

[enthalpyburns]

   Is this right?

Modify the bond energy equation:

From: Energy of bond= sum of D (bonds broken)- sum of D (bonds formed) 

And replace it with the regular enthalpy equation

Heat of formation= sum of products minus sum of reactants

Heat of formation= overall energy needed to break and form the NEW bonds   

((278.8_)+(79.0)6) MINUS (-1209)=

753 MINUS (-1209)= 

1962 (sig figs) / 6= 327 Kj/mol for the S-F Bond


DELTA H is not a state function.

[magician4]

if your proposal is related to SF₆, I would agree


regards

Ingo