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Question

Chemistry Question on Hybridisation

The electron affinity of chlorine is 3.7eV.13.7\, eV. \,1 gram of chlorine is completely converted to ClCl^- ion in a gaseous state. (1eV=23.06kcalmol1)(1\, eV = 23.06\, kcal\, mol^{-1}). Energy released in the process is

A

4.8kcal4.8 \,kcal

B

7.2kcal7.2 \,kcal

C

8.2kcal8.2 \,kcal

D

2.4kcal2.4 \,kcal

Answer

2.4kcal2.4 \,kcal

Explanation

Solution

Number of moles =135.5=\frac{1}{35.5} Given, 1eV=23.06kcalmol11 eV = 23.06\, kcal\, mol^{-1} 3.7eV=3.7×23.06kcalmol13.7 eV = 3.7 \times 23.06\, kcal\, mol^{-1} i.e. 1 mole realease energy =3.7×23.06kcal=3.7\times23.06\,kcal \therefore Energy released =135.5×3.7×23.06kcal=2.4kcal=\frac{1}{35.5}\times3.7\times23.06\,kcal=2.4\,kcal