The first IP of lithium is 5-41eV and electron affinity of Cl is -3-61eV- Calculate - H in kJ mol-1 for the reaction- Lig-Clg-Lig- Clg-

First step refers to the image . Δ H=(1.8×1.602×10^ 19×6.023×10^23×10^ 3)=173.6 kJ/mol. ...

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Electron Gain Enthalpy

The first IP ...

Question

The first IP of lithium is $5.41 e V$ and electron affinity of $C l$ is $- 3.61 e V$ . Calculate $Δ H$ in $k J$ ${m o l}^{- 1}$ for the reaction:
${L i}_{(g)} + {C l}_{(g)} \to {L i}_{(g)}^{+} + {C l}_{(g)}^{-}$

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I E

5.41 e V

C l

3.61 e V

Δ H

k J {m o l}^{- 1}

L i (g) + C l (g) ⟶ {L i}^{+} (g) + {C l}^{-} (g)

L i

C l

Δ H

k J m o l^{- 1}

L i_{(g)} + C l_{(g)} \to L i_{(g)}^{+} + C l_{(g)}^{-}

5.4

C l

3.6

Δ H

L i_{(g)} + C l_{(} g) \to L i_{(g)}^{+} + C l_{(g)}^{-}

L i

5.4

C l

3.61

Δ H

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

5.4

3.61

Δ

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

1

= 1.6 \times 10^{- 19}

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