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Electron Affinity

The electron ...

Question

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

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Solution

Given,
$C l (g) + e^{-} \to C l^{-} (g) + 37 e V$
Energy released when $1$ mole $(35.5 g)$ of chlorine atoms change completely into $C l^{-} (g) = 37 \times 96.49 k J = 3570.13 k J / m o l$
Energy released when $3.55 g$ of chlorine atoms change completely into $C l^{-} (g) = (3570.13 / 35.5) \times 3.55 = 357.013 k J$

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+ 37 e V

k J

3.55 g

C l

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

3.7

1

C l^{-}

1

= 23.06

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3.7 e V

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2 g

{C l}^{-}

(1 e V = 23.06 k c a l

{m o l}^{- 1})

3.7 e V

2 g

{C l}^{-}

1 e V = 23.06 k c a l

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C l_{(g)}^{-}

C l_{(g)}

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