Question

49. The first ionization energy for Li is 5.4 eV and
electron affinity of Cl is 3.61 eV. The AH (in kJ/
mol) for the reaction Lia+CI-Li* +Cr is (if
resulting ions do not c
mbine with each other) (lev
= 1.6 x 10-9 ) :​

Answer 1

we know, ionisation enthalpy is the amount of energy required to remove an electron from isolated gaseous atom.

so, Li(g) ⇔Li+(g) ; ∆H1 = 5.41eV

and electron affinity is the amount of energy released to add an electron from isolated gaseous atom.

so, Cl(g) ⇔Cl-(g) ; ∆H2 = -3.61 eV

∆H for the reaction is :

∆H = ∆H1 + ∆H2

= 5.41eV - 3.61eV

= 1.8 eV

we know, 1eV = 96.4869 kJ/mol

so, ∆H = 1.8 × 96.4869 = 173.67642kJ/mol ≈ 173.7 kJ/mol

Answer 2

ionization Potential are

Li=5.4eV

Electron affinity Cl=3.61eV

Delta H=I.P.-E.A =5.4-3.61=1.80eV=1.80* 1.6* {{10}^{-22}}kJ

Delta H=2.86\times {{10}^{-22}}kJ

Avogadro's number =6.02* {{10}^{23}}

Delta H=2.86* {{10}^{-22}}* 6.02* {{10}^{23}}=170,kJ/mole.