Respuesta :
Answer : The heat of sublimation of Li(s) is 179.5 kJ/mole
Explanation :
The formation of lithium iodide is,
[tex]Li^{1+}(g)+\frac{1}{2}I_2(g)\overset{\Delta H_L}\rightarrow LiI(s)[/tex]
[tex]\Delta H_f^o[/tex] = enthalpy of formation of lithium iodide
The steps involved in the born-Haber cycle for the formation of [tex]LiI[/tex]:
(1) Conversion of solid lithium into gaseous lithium atoms.
[tex]Li(s)\overset{\Delta H_s}\rightarrow Li(g)[/tex]
[tex]\Delta H_s[/tex] = sublimation energy of lithium
(2) Conversion of gaseous lithium atoms into gaseous lithium ions.
[tex]Li(g)\overset{\Delta H_I}\rightarrow Li^{+1}(g)[/tex]
[tex]\Delta H_I[/tex] = ionization energy of lithium
(3) Conversion of molecular gaseous iodine into gaseous iodine atoms.
[tex]I_2(g)\overset{\Delta H_D}\rightarrow 2I(g)[/tex]
[tex]\frac{1}{2}I_2(g)\overset{\Delta H_D}\rightarrow I(g)[/tex]
[tex]\Delta H_D[/tex] = dissociation energy of iodine
(4) Conversion of gaseous iodine atoms into gaseous iodine ions.
[tex]I(g)\overset{\Delta H_E}\rightarrow I^-(g)[/tex]
[tex]\Delta H_E[/tex] = electron affinity energy of iodine
(5) Conversion of gaseous cations and gaseous anion into solid lithium iodide.
[tex]Li^{1+}(g)+I^-(g)\overset{\Delta H_L}\rightarrow LiI(s)[/tex]
[tex]\Delta H_L[/tex] = lattice energy of lithium iodide
To calculate the overall energy from the born-Haber cycle, the equation used will be:
[tex]\Delta H_f^o=\Delta H_s+\Delta H_I+\frac{1}{2}\Delta H_D+\Delta H_E+\Delta H_L[/tex]
Now put all the given values in this equation, we get:
[tex]-273kJ/mole=\Delta H_s+520kJ/mole+\frac{1}{2}\times 151kJ/mole+(-295kJ/mole)+(-753kJ/mole)[/tex]
[tex]\Delta H_s=179.5kJ/mole[/tex]
Therefore, the heat of sublimation of Li(s) is 179.5 kJ/mole
