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The first IP lithium is {5.41}{e}{V} and electron gain enthalp

Question

The first IP lithium is $5.41 e V$ and electron gain enthalpy 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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The first IP lithium is

5.41 e V

and electron gain enthalpy of

C l

- 3.61 e V

. Calculate

Δ H

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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Text solutionVerified

Solution:

Δ H /

molecule of

L i^{+}

and

C l^{-} = I F_{1} +

Electron gain enthaply .

= 5.41 - 3.61 = 1.80 e V

∴ L i \to L i^{+} + e, I P_{1} + v e

and

C l + e \to C l^{- 1},

Electron gain enthalpy

= - v e

∴ Δ H / m o l = 1.8 \times 6.023 \times 10^{23} e V

= 1.8 \times 6.023 \times 10^{233} \times 1.602 \times 10^{19} J

= 1.8 \times 6.023 \times 10^{23} \times 1.602 \times 10^{19} \times 10^{- 3} k J

= 173.7 k J

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Question Text	The first IP lithium is $5.41 e V$ and electron gain enthalpy of $C l$ is $- 3.61 e V$ . Calculate $Δ H$ in $K J m o l^{- 1}$ for the reaction: $L i_{g} + C l_{g} \to L i_{g}^{=} + C l_{g}^{-}$ .
Answer Type	Text solution:1
Upvotes	150