Working the Numbers
STEP 1: Enthalpy of Li atomization = 161 kJ
STEP 2: 1/2 the bond energy of F2(g)= 0.5(159 kJ) = 79.5 kJ
STEP 3: IE1 for Li(g) = 520 kJ
STEP 4: EA of F(g) = -328 kJ
The enthalpy change for the overall process, ²Hfo, = -617 kJ
Only the lattice energy is unknown, and it is equal to the enthalpy
change of the overall process minus the sum of the above four
steps = -1050 kJ
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