Use the following data to calculate a value for the Xe-F bond enthalpy in XeF4

Use the following data to calculate a value for the Xe-F bond enthalpy in XeF4
Xe(g) + 2F2(g) -> XeF4(g) delta H = -252 kJ/mol
F2(g) -> 2F(g) delta H = +158 kJ/mol

Is this a hess's law question or a mean enthalpy question (bonds broken - bonds formed). Not sure how to do this, any help is greatly appreciated thanks!

Bit of both ...

At the end of the day all enthalpy questions boil down to Hess' law ... which is just another way of stating the law of conservation of energy.

For this question you first have to write out an equation for the average bond enthalpy of the Xe-F bond.

XeF₄(g) --> Xe(g) + 4F(g) ........... ΔH = 4 x bond enthalpy Xe-F

Now use the equations given to construct a cycle with this equation.

Thanks, so is this enthalpy of formation right? So it will be the sum of enthalpy change of the products - the sum of the enthalpy change of the reactants?

i got +142 kj/mol

Would you please be able to explain how? I'm still not sure I understand this. Thanks

Check this out ...

Check this out ...

Why did we make the third equation??
I am so confused, thank you!

The third equation is the definition of bond enthalpy (x 4 in this case)

Bond enthalpy = "the energy required to break 1 mole of bonds in gaseous molecules measured over a range of similar structures"

Xe-F does not exist as a molecule so:

XeF₄(g) --> Xe(g) + 4F(g) is equal to four times the Xe-F bond enthalpy term.

Why did we make the third equation??
I am so confused, thank you!

Check this out ...

I know this is an old thread, so may not get a response... but why do the arrows point down? Surely they should point up?