The `DeltaH_(f)^(0)(KF,s)` is `-563 kJ mol^(-1)`. The ionization enthalpy of `K(g)` is `419 kJ mol^(-1)`. and the enthalpy of sublimation of potassium is `88 kJ mol^(-1)`. The electron affinity of `F(g)` is `322 kJ mol^(-1)` and `F-F` bond enthalpy is `158 kJ mol^(_1)`. Calculate the lattice enthalpy of `KF(s)`.
The given data are as follows:
(i) `K(s)+1//2F_(2)(g)rarrKF(s)" "DeltaH_(f)^(0)= -563 kJ mol^(-1)`
(ii) `K(g)rarrK^(+)(g)+e^(-)" "Delta_("Ioniz")^(0)=419 kJ mol^(-1)`
(iii) `K(s)rarrK(g)" "DeltaH_("sub")^(0)=88kJ mol^(-1)`
(iv) `F(g)+e^(-)rarrF^(-)(g)" "DeltaH_(eg)^(0)= -322 kJ mol^(-1)`
(v) `F_(2)(g)rarr2F(s)" "DeltaH_("diss")^(0)= 158 kJ mol^(-1)`
(vi) `K^(+)(g)+F^(-)(g)rarrKF(s)" " DeltaH_(L)^(0)=?`

Answer

`-827 kJ mol^(-1)`

Explanation

Equation (vi) can be generated by the following manipulations.
Eq.(i)-Eq.(ii)-Eq.(iii)-Eq.(iv)`-1//2` E.(v)
Carrying out the corresponding manipulations on `Delta H_(L)^(0)`, we get
`DeltaH_(L)^(0)=(-563-419-88+322-79)kJ mol^(-1)`
`= -827 kJ//"mole"`