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)+(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)`

`(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+(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+(1)/(2)* 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