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Electron Gain Enthalpy

The amount of...

Question

The amount of energy released when $10^{12}$ atoms of $C l$ vapours are converted to $C l^{-}$ ions, according to the equation:
$C l_{(g)} + e^{-} \to C l^{-} i s 60 \times 10^{- 10} J$ . The magnitude of
$Δ_{e g} H$ of $C l$ atom in kJ/mol is (take $N_{A} = 6 \times 10^{23}$

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Solution

The amount of energy released when 1 mol $(6 \times 10^{23})$ of Cl are converted to $C l^{-}$ ions according to the above equation is $Δ_{e a} H$ of Cl vapours.
$Δ_{e g} H$ of Cl vapours= $\frac{- 60 \times 10^{- 10} J \times 6 \times 10^{23}}{10^{12}}$
= $- 3600 J / m o l$
= $- 3.60 k J / m o l$

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10^{12}

C l

C l^{-}

C l (g) + e^{-} \to C l^{-} (g)

58 \times 10^{- 10} J

Δ_{e g} H

C l

k J m o l^{- 1}

10^{12}

C l

C l^{-}

C l_{(g)} + e^{-} \to C l_{(g)}^{-} is 58 \times 10^{- 10} J

Δ_{e g} H^{-}

C l

{eV atom}^{- 1}

10^{12}

C l

C l^{-}

C l_{(g)} + e^{-} \to C l_{(g)}^{-} is 58 \times 10^{- 10} J

Δ_{e g} H^{-}

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{eV atom}^{- 1}

C l

C l_{(g)}^{-}

C l_{(g)}

(C l \to C l^{+} + e^{-})

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6.023 \times 10^{23}

(I P f o r C l = 1250 k J

m o l e^{- 1} a n d E A = 350 k J m o l e^{- 1}

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