[marcoteran97]

Hi, Im just getting trouble doing this question 1, I dont know centanly what to do there, but my Professor told that I can solve this fist question if I have made de second one.
The second was easy to solve by only solving de DH.
I'm so stuked on this fist question, thanks if you guys can help me by trying to give me an advice to solve this.
1 - The fact that the argon di-fluoride was not prepared suggests that the argon-fluorine bond is very weak. Use a thermodynamic cycle to determine the approximate value of the Ar-F bond enthalpy. List thermodynamic factors that favor and disfavor the formation of argon di-fluoride.
2 - Some compounds have positive enthalpy of formation, with this in mind let's assume that ArF2 (g) present DH0 = 100 kJ mol -1. a) What should be the value of the Ar-F bond energy so that you can form ArF2 (g), knowing that the F-F bond has DH0 = 159 kJ mol -1. b) What SEA binding energy is Ar-F DH0 (ArF2 (g)> 100 kJ mol-1?
a) DH = 100kJ / mol
Ar (g) + F2 (g) -> ArF2 (g)
DH = Σ DH. Bonds broken (reagents) -Σ DH bonds formed (products)
DH = (F-F) - 2 (Ar-F)
100kJ / mol = 159kJ / mol - 2 (Ar-F)
2 (Ar-F) = 59kJ / mol
Air-F = 29,5kJ / mol

b) If DH ArF (g)> 100kJ / mol, then the value of the Ar-F bond energy is lower.
For example:
Suppose it is DH ArF (g) = 120kJ / mol
DH = (F-F) - 2 (Ar-F)
120kJ / mol = 159kJ / mol - 2 (Ar-F)
2 (Ar-F) = 39kJ / mol
Air-F = 19,5kJ / mol
Confirming the above statement, the value of the binding energy was lower.