Estimate the average `S-F` bond enthalpy in `SF_(6)`. The values of standard enthalpy of formation of `SF_(6)(g),S(g)` and `F(g)` are :-`1100, 274` and `80 kJ mol^(-1)` respectively.

Answer

`309 kJ mol^(-1)`

Explanation

To calculate the average `S-F` bonds enthalpy, we determine the enthalpy change for the following reaction.
`S(g)" "6F(g) rarrSF(g)`
`DeltaH_(f)^(0)275 kJ mol^(-1)+80 kJ mol^(-1)1100 kJ mol^(_1)`
`Delta H_(f)^(0)=DeltaH_(f)^(0)(SF_(6),g)-[DeltaH_(f)^(0)(S,g)+6DeltaH_(f)^(0)(F,g)]`
`[-1100-(274+6xx80)]kJ mol^(-1)`
Hence `DeltaH_((S-F))=(-DeltaH_(f)^(0)(SF_(6),g))/(6)=(1854 kJ mol^(-1))/(6)=309 kJ mol^(-1)`