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Atomic Size and Electronegativity

Given :Reacti...

Question

Given :

Reaction Energy charge (in kJ)
$L i (s) ⟶ L i (g)$ $161$
$L i (g) ⟶ {L i}^{+} (g)$ $520$
$\frac{1}{2} F_{2} (g) ⟶ F (g)$ $77$
$F (g) + e^{-} ⟶ F^{-} (g)$ (Electron gain enthalpy)
${L i}^{+} (g) + F^{-} (g) ⟶ L i F (s)$ $- 1047$
$L i (s) + \frac{1}{2} F_{2} (g) ⟶ L i F (s)$ $- 617$

Based on data provided, the value of electron gain enthalpy of fluorine would be

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is C $- 328 k J {m o l}^{- 1}$

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Similar questions

Q. What is the standard enthalpy change for the formation of

L i F_{(s)}

? Refer to the table given below.

Process	Reaction	$Δ H^{\circ} (k J / m o l)$
Sublimation	$L i_{(s)} \to L i_{(g)}$	$155.2$
Dissociation	$\frac{1}{2} F_{2 (g)} \to F_{(g)}$	$75.3$
Ionization	$L i_{(g)} \to L i_{(g)}^{+} + e^{-}$	$520$
Electron affinity	$F_{(g)} + e^{-} \to F_{(g)}^{-}$	$- 328$
Formation of $LiF(s)	$L i_{(g)}^{+} + F_{(g)}^{-} \to L i F_{(s)}$	$- 1017$

Q. Calculate the lattice formation enthalpy in

{kJ mol}^{- 1}

L i F

given that
Enthalpy of sublimation of lithium =

155.3 {kJ mol}^{- 1}

Dissociation enthalpy of half mole of

F_{2} = 75.3 kJ

Ionization enthalpy of lithium

= 520 {kJ mol}^{- 1}

Electron gain enthalpy of 1 mol of

F (g) = - 333 kJ Δ_{f} H_{overall} = - 594 {kJ mol}^{- 1}

Q. The first I.E. of

L i

is 5.41 eV and first electron gain enthalpy of

C l

is - 3.61 eV. 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)}^{-}

Q. Calculate the lattice enthalpy(nearest integer value) in

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

L i F

, given that enthalpy of,

(i) sublimation of lithium is

155.3 k J

{m o l}^{- 1}

;

(ii) dissociation of

1 / 2

F_{2}

75.3 k J

;

(iii) ionization enthalpy of lithium is

520 k J

{m o l}^{- 1}

;

(iv) electron gain enthalpy of 1 mole of

F (g)

- 333 k J

;

Δ_{f} H

- 594 k J

{m o l}^{- 1}

Q. Calculate the lattice enthalpy in

k J m o l^{- 1}

L i F

given that
Sublimation of lithium is

155.3 k J m o l^{- 1}

Dissociation of half mole of

F_{2} = 75.3 k J

Ionization enthalpy of lithium

= 520 k J m o l^{- 1}

Electron gain enthalpy of 1 mol of

F (g) = - 333 k J Δ_{f} H_{o v e r a l l} = - 594 k J m o l^{- 1}

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