Found on the IChO, and it seemed very interesting to me:
Calculate the Si–Si bond dissociation enthalpy of Si
2H
6 from the following informations:
Bond dissociation enthalpy for H–H = 436 kJ/mol
Bond dissociation enthalpy for Si–H = 304 kJ/mol
ΔfH [Si
(g)] = 450 kJ/mol
ΔfH [Si
2H
6(g)] = 80.3 kJ/mol
2Si
(g)+3H-H(g)
H
3Si-SiH
3(g)I have two expresions to calculate ΔH:
ΔH=ΔfH [Si
2H
6(g)] - 2ΔfH [Si
(g)]
ΔH=3Bond dissociation enthalpy(H-H) - 6Bond dissociation enthalpy(Si-H) - Bond dissociation enthalpy(Si-Si)
And I got that the bond dissociation enthalpy(Si-Si)=303.7kJ/mol, but I found somewhere on the net that Si-Si=around 50. Did I calculate this wrong or not?