Question

Consider the following: Li(s) ???? 1???? (g) n LiI(s) ????H???? 222 ????292 kJ. LiI(s) has a lattice energy of ????753 kJ/mol. The ionization energy of Li(g) is 520. kJ/mol, the bond en- ergy of I2(g) is 151 kJ/mol, and the electron affinity of I(g) is ????295 kJ/mol. Use these data to determine the heat of sublimation of Li(s

Answer 1

Answer :  The heat of sublimation of Li(s) is 179.5 kJ/mole

Explanation :  

The formation of lithium iodide is,

$Li^{1+}(g)+\frac{1}{2}I_2(g)\overset{\Delta H_L}\rightarrow LiI(s)$

$\Delta H_f^o$ = enthalpy of formation of lithium iodide

The steps involved in the born-Haber cycle for the formation of $LiI$:

(1) Conversion of solid lithium into gaseous lithium atoms.

$Li(s)\overset{\Delta H_s}\rightarrow Li(g)$

$\Delta H_s$ = sublimation energy of lithium

(2) Conversion of gaseous lithium atoms into gaseous lithium ions.

$Li(g)\overset{\Delta H_I}\rightarrow Li^{+1}(g)$

$\Delta H_I$ = ionization energy of lithium

(3) Conversion of molecular gaseous iodine into gaseous iodine atoms.

$I_2(g)\overset{\Delta H_D}\rightarrow 2I(g)$

$\frac{1}{2}I_2(g)\overset{\Delta H_D}\rightarrow I(g)$

$\Delta H_D$ = dissociation energy of iodine

(4) Conversion of gaseous iodine atoms into gaseous iodine ions.

$I(g)\overset{\Delta H_E}\rightarrow I^-(g)$

$\Delta H_E$ = electron affinity energy of iodine

(5) Conversion of gaseous cations and gaseous anion into solid lithium iodide.

$Li^{1+}(g)+I^-(g)\overset{\Delta H_L}\rightarrow LiI(s)$

$\Delta H_L$ = lattice energy of lithium iodide

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\frac{1}{2}\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-273kJ/mole=\Delta H_s+520kJ/mole+\frac{1}{2}\times 151kJ/mole+(-295kJ/mole)+(-753kJ/mole)$

$\Delta H_s=179.5kJ/mole$

Therefore, the heat of sublimation of Li(s) is 179.5 kJ/mole