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Standard XII

Chemistry

Electron Gain Enthalpy

The electron ...

Question

The electron affinity of chlorine is $3.7$ eV. $1$ gram of chlorine is completely converted to $C l^{-}$ ion in a gaseous state. ( $1$ eV $= 23.06$ kcal $m o l^{- 1}$ ). Energy released in the process is:

4.8

kcal

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7.2

kcal

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8.2

kcal

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2.4

kcal

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Solution

The correct option is D $2.4$ kcal
Given $u (g) + e^{-} ⟶ u^{-} (g) E_{a} = 3.7 e v$

For $1 g m \to e n e r g y = - 3.7 \times 23.06 \times \frac{1}{35.5} = - 2.4 K c a l$

Energy released $= 2.4$ Kcal

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Similar questions

Q. The electron affinity of chlorine is

3.7 e V

. How much energy (nearest integer value) in

k c a l

is released when

2 g

of chlorine is completely converted to

{C l}^{-}

ion in a gaseous state?

(1 e V = 23.06 k c a l

{m o l}^{- 1})

Q. The electron gain enthalpy of chlorine is

3.7 e V

. How much energy in kcal is released when

2 g

of chlorine is completely converted to

{C l}^{-}

ions in the gaseous state?
(

1 e V = 23.06 k c a l

)

Q. Calculate the electronegativity of chlorine from bond energy of

C l - F

bond

(61 k c a l m o l^{- 1})

F - F (38 k c a l m o l^{- 1})

and

C l - C l

bond

(58 k c a l m o l^{- 1})

and electronegativity of fluorine

4.0 e V

Q. The electron affinity of chlorine is

+ 37 e V

. How much energy in

k J

is released when

3.55 g

of chlorine is converted completely into

C l

ion in the gaseous state .

(1 e V = 96.49 k J / m o l e)

Q. Calculate the electronegativity (eV) of chlorine from the bond energy of

C l - F

bond (61 kcal mol

^{- 1}

F - F

bond (

38

kcal mol

^{- 1}

) and

C l - C l

bond (

58

kcal mol

^{- 1}

) and electronegativity of fluorine (

4.0

eV).
Based on Pauling's scale :

(E N)_{F} - (E N)_{C l} = k \sqrt{Δ} = 0.208 \sqrt{Δ}

where,

Δ

is resonance energy and k is the conversion factor (which is

0.208

for converting kcal into eV).

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The electron affinity of chlorine is 3.7 eV. 1 gram of chlorine is completely converted to Cl− ion in a gaseous state. (1 eV=23.06 kcal mol−1). Energy released in the process is:

The correct option is D 2.4 kcalGiven u(g)+e−⟶u−(g)Ea=3.7 evFor 1gm→energy=−3.7×23.06×135.5=−2.4 KcalEnergy released =2.4 Kcal

The electron affinity of chlorine is $3.7$ eV. $1$ gram of chlorine is completely converted to $C l^{-}$ ion in a gaseous state. ( $1$ eV $= 23.06$ kcal $m o l^{- 1}$ ). Energy released in the process is:

The correct option is D $2.4$ kcal
Given $u (g) + e^{-} ⟶ u^{-} (g) E_{a} = 3.7 e v$

For $1 g m \to e n e r g y = - 3.7 \times 23.06 \times \frac{1}{35.5} = - 2.4 K c a l$

Energy released $= 2.4$ Kcal