Question

2. Find the experimental Lattice energy of aluminum oxide using a Born-Haber cycle using the following information: AH (aluminum oxide) = -1676 kJ/mol IE, (aluminum) = 577.6 kJ/mol IE, (aluminum) =1816.7 kJ/mol IE, (aluminum) = 2744.8 kJ/mol AH® (aluminum atom, g) = 329.7 kJ/mol AHⓇEAI (oxygen) = -200.4 kJ/mol AHⓇEAT (oxygen) = 780.0 kJ/mol AH® (oxygen atom, g) = 249.2 kJ/mol Write each of the appropriate balanced chemical equations (with physical state) and assign the appropriate enthalpy to each. Be sure the cycle is balanced to give the lattice energy equation, either as forming aluminum oxide or breaking aluminum oxide apart.

Answer 1

2Al(s) + 3/2O2(g) ----------> Al2O3 (s)      $\Delta$ Hf   = -1676KJ/mole

sublimation energy

2Al(s) -------> 2Al(g)                                   $\Delta$ H1   = 2*329.7   = 659.4KJ/mole

ionization energy of Al

2Al(g) -----------> 2Al^+ (g) + 2e^-              $\Delta$H2 = 2*577.6       = 1155.2KJ/mole

2Al^+(g) -----------> 2Al^2+ (g) + 2e^-           $\Delta$ H3   = 2*1816.7   = 3633.4KJ/mole

2Al^2+(g) -----------> 2Al^3+ (g) + 2e^-           $\Delta$ H4 = 2*2744.8 = 5489.6KJ/mole

bond energy of oxygen

3/2O2(g) --------------> 3O(g)                        $\Delta$ H5 = 3/2*498.4   = 747.6KJ/mole

electron affinity

3O(g) + 3e^- ------------> 3O^- (g)                $\Delta$ H6   = 3*-200.4   = -601.2KJ/mole

3O^-(g) + 3e^- ------------> 3O^2- (g)              $\Delta$ H7   = 3*780 = 2340KJ/mole

lattice energy of Al2O3

2Al^3+ (g) + 3O^2- --------------> Al2O3(s)      $\Delta$ H8 =

$\Delta$Hf = $\Delta$ H1 + $\Delta$ H2 + $\Delta$ H3 + $\Delta$ H4 + $\Delta$ H5 + $\Delta$ H6 + $\Delta$ H7 + $\Delta$ H8

-1676 = 659.4 + 1155.2 + 3633.4 + 5489.6 +747.6 -601.2 +2340+ $\Delta$ H8

$\Delta$H8   = -15100KJ/mole >>>answer