Consider the following: li(s) (g) n lii(s) 222 kj. lii(s) has a lattice energy of 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 kj/mol. use these data to determine the heat of sublimation of li(s

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

Explanation :  

The formation of lithium iodide is,

= enthalpy of formation of lithium iodide

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

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

= sublimation energy of lithium

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

= ionization energy of lithium

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

= dissociation energy of iodine

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

= electron affinity energy of iodine

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

= lattice energy of lithium iodide

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

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

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

hypothesising or postulating