Using the data `Delta_(f) H^(@) (Nf_(3), g) = - 114 kH mol^(-1)`,
`Delta_(N -= N) H^(@) = 946 kJ mol^(-1)`, and `Delta_(f - f) H^(@) = 158 kJ mol^(-1)`, calculate the average bond enthalpy of `N - F` bond in `NF_(3)`.
Strategy : First write the thermochemical equation corresponding to `Delta_(f) H^(@) (NF_(3), g)`:
`(1)/(2) N_(2) (g) + (3)/(2) F_(2) (g) rarr NF_(3) (g)`
Now define `Delta_(r) H^(@)` of this reaction in terms of bonds made and bonds broken. Notice the `1//2` mol of `N -= N` bonds and `3//2` mol of `F - F` bond are broken, while 3 mole of `N - F` bonds are formed each `NF_(3)` has three `N - F` bonds: we are given the blood
enthalpies of `N -= N` and `F - F` bonds, while the average bond enthalpy of `N -F` bond is not known. Applying Eq. for the reaction, we can calculate this unknown.

Explanation

According to Eq. we can write
`Delta_(r) H^(@) = [(1)/(2) Delta_(N -= N) H^(@) + (3)/(2) Delta_(F - F) H^(@)] - [3 Delta_(N - F) H^(@)]`
`- 114 = [(1)/(2) (946) + (3)/(2) (158)] - [3 Delta_(N - f) H^(@)]`
Rearranging to solve for `Delta_(N - F) H^(@)`, we have
`Delta_(N - F) H^(@) = ((473 + 237) + (144))/(3) = (824)/(3)`
`275 kJ mol^(-1)`