Question

Rutle (TIO2) (Iv) (8) [25] Question 2 2 (a) () Using a Bom-Haber cycle, calculate the lattice enthalpy of LIO2, given the following data AH°(kJ/mol) +161 Sublimation of Li(s) +520 lonization of LI(g) +498 Dissociation of O2(g) -141 Electron affinity of O(g) +865 Electron affinity of O (g) -598 Enthalpy of formation of LIO2 TURN OVE

Answer 1

The formation of Li2O takes place as -

2 Li(s) + 1/2 O2(g) ------------> Li2O(s).

by using Born habber cycle ,

It actually happens in many steps which are given below.

step 1 in the first step, Li changes form solid to gas ,while the sublimation energy is given, 161 KJ, since because Li has 2 mole , therefore we multiply 2 by each term in the given equation ,and also the sublimation energy. Such as.

2 [ Li(s) ---------> Li(g) ] , sublimation energy = 2*161= 322 KJ.

step2. . In this step, Li(g) ionises to Li+ , the equation is given.

2 [ Li(g) -------------> Li+ + e- ] , ionization energy = 2*520 kJ = 1040 KJ.

step 3. in this step, dissociation of O2 takes place, we multiply by 1/2 the whole equation for sake of convenience to calculate.

1/2 [ O2 ---------> 2 O ] , Dissociation energy = 1/2*498=249 KJ.

step 4. In this step, the oxygen gets ionise by accepting an electron.

O(g) + e- -------> O- , electron affinity of oxygen = -141 kJ.

step 5. in This step, O- accepts an electron and becomes O2-

O-(g) --------->> O2- (g) , electron affinity of O- = +865 kJ.

step 6. Two Li+ cations adds O2- , to form Li2O(s) ,

2Li+ + O2- ----------> Li2O(s) , lattice enthalpy = ? ,To be find.

NOW, , IF WE SUM UP ALL THE SIX STEP EQUATION, WE HAVE ONLY ,

2Li(s) + 1/2 O2(g) --------> Li2O (s) , enthalpy of formation is given= -554 KJ.

applying Hess law of summation of energies, we get,

(322+ 1040+ 249 -141+865 + ∆H ) KJ= -554 KJ

2335KJ + ∆H = -554 KJ

∆H = -2335- 554 KJ =- 2879 KJ.

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