Lattice Energy Calculations:

 

Lattice Energy (LE) is defined for the reaction of gas-phase ions combining to form the ionic solid.

For Example:

                          Li+(g)     +  F-(g)    →  LiF(s)                    ΔHLE  =  ???

                          Mg2+(g)  +  O2-(g)  →  MgO(s)                 ΔHLE  =  ???

 

ΔHLE  can be calculated using Hess’s Law and tabulated values.

For example:

             (6)         Li+(g)   +  F-(g)    →  LiF(s)                      ΔHLE  =  ???

Using:

             (1)         Li(s)                    →  Li(g)                          ΔHVAP               =  +161 kJ/mol

             (2)         ½F2(g)                →  F(g)                            ½BE(F-F)            =  +79.5 kJ/mol

             (3)         Li(g)                    →  Li+(g) + e_                 IE                       =  +520 kJ/mol

             (4)         F(g)     + e_         →  F_(g)                          EA                     =  -328 kJ/mol

             (5)         Li(s)    + ½F2(g)  →  LiF(s)                        ΔHf                    =  -617 kJ/mol

 

Hess’s Law states that the enthalpy of a reaction is sum of the enthalpies of individual reactions that add to give the overall reaction.   Here:

 

flip        (1)         Li(g)                    →  Li(s)                          -ΔHVAP             =  -161 kJ/mol

flip        (2)         F(g)                     →  ½F2(g)                       -½BE(F-F)          =  -79.5 kJ/mol

flip        (3)         Li+(g) + e_          →  Li(g)                          -IE                    =  -520 kJ/mol

flip        (4)         F_(g)                    →  F(g)     + e_                -EA                   =  +328 kJ/mol

keep      (5)         Li(s)    + ½F2(g)  →  LiF(s)                       ΔHf                    =  -617 kJ/mol

             (6)         Li+(g)   +  F-(g)    →  LiF(s)                      ΔHLE                  =  -1050 kJ/mol

 

Or:

             (6)         Mg2+(g)   +  O2-(g)    →  MgO(s)              ΔHLE  =  ???

Using:

             (1)         Mg(s)                 →  Mg(g)                        ΔHVAP               =  +148 kJ/mol

             (2)         ½O2(g)               →  O(g)                           ½BE(O-O)           =  +249 kJ/mol

             (3)         Mg(g)                 →  Mg+(g) + 2e_            IE1 + IE2           =  +2188 kJ/mol !!!

             (4)         O(g)     + 2e_      →  O2-(g)                        EA1 + EA2        =  +737 kJ/mol   !!!

             (5)         Mg(g) + ½O2(g)  →  MgO(s)                     ΔHf                    =  -601 kJ/mol

 

Hess’s Law here:

 

flip        (1)         Mg(g)                  →  Mg(s)                        -ΔHVAP             =  -148 kJ/mol

flip        (2)         O(g)                    →  ½O2(g)                      -½BE(O-O)         =  -249 kJ/mol

flip        (3)         Mg2+(g) + 2e_     →  Mg(g)                        -IE1-IE2           =  -2188 kJ/mol

flip        (4)         O2-(g)                  →  O(g)   + 2e_               -EA1-EA2        =  -737 kJ/mol

keep      (5)         Mgs)   + ½O2(g)→  MgO(s)                     ΔHf                    =  -601 kJ/mol

             (6)         Mg2+(g) + O2-(g) →  MgO(s)                    ΔHLE                  =  -3932 kJ/mol

 

Note also that the magnitude of the lattice energy is proportional to the size of the opposite charges and inversely proportional to the distance between the ions (larger opposite charges and/or smaller ions generally yield more negative lattice energy).  This should make sense.