Ionisation of energy of F -is 320 kJ mol -1- The electron gain enthalpy of fluorine would beA- -320 kJ mol -1B- -320 kJ mol -1C- 160 kJ mol -1D- -160 kJ mol -1

The correct option is A 320 kJ mol^ 1F^ → F + e; Δ H =320 kJ or F +e → F^ ; Δ H = 320 kJ The second process is the reverse of first process. Hence the e ...

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Standard XII

Chemistry

Electron Gain Enthalpy

Ionisation of...

Question

Ionisation of energy of $F^{-}$ is $320 {kJ mol}^{- 1}$ . The electron gain enthalpy of fluorine would be

- 320 {kJ mol}^{- 1}

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- 160 {kJ mol}^{- 1}

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+ 320 {kJ mol}^{- 1}

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160 {kJ mol}^{- 1}

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Solution

The correct option is A $- 320 {kJ mol}^{- 1}$
$F^{-} \to F + e; Δ H = 320 kJ$ or
$F + e \to F^{-}; Δ H = - 320 kJ$
The second process is the reverse of first process. Hence the enthalpy of the second process is the negative of the first process.

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Electron Gain Enthalpy

Standard XII Chemistry

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Ionisation of energy of F− is 320 kJ mol−1. The electron gain enthalpy of fluorine would be

The correct option is A −320 kJ mol−1F−→F+e; ΔH=320 kJ or F+e→F−; ΔH=−320 kJ The second process is the reverse of first process. Hence the enthalpy of the second process is the negative of the first process.

Ionisation of energy of $F^{-}$ is $320 {kJ mol}^{- 1}$ . The electron gain enthalpy of fluorine would be

The correct option is A $- 320 {kJ mol}^{- 1}$
$F^{-} \to F + e; Δ H = 320 kJ$ or
$F + e \to F^{-}; Δ H = - 320 kJ$
The second process is the reverse of first process. Hence the enthalpy of the second process is the negative of the first process.