Question

<th Reaction <th energy change (in kj) Li(s) → Li(g) 161 Li(g) → Li+(g) 520 (1/2) F2 (g) → F(g) 77 F(g) + e- → F-(g) (Electron gain enthalpy) Li+ (g) + F-(g) → Li F(s) -1047 Li(s) + (1/2) F2(g) → LiF(s) -617 Based on data provided, the value of electron gain enthalpy of fluorine would be : (A) - 300 kJ mol-1 (B) - 350 kJ mol-1 (C) - 328 kJ mol-1 (D) - 228 kJ mol-1

Tardigrade Admin · Accepted Answer

Applying Hess's Law Δf H°=Δ text sub H+(1/2) Δ text diss H+I . E+E . A+Δ text lattice H -617=161+520+77+ E.A. +(-1047) E. A .=-617+289=-328 kJ mol -1 ∴ electron affinity of fluorine =-328 kJ mol -1

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

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

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

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

Applying Hess's Law
$Δ_{f} H^{\circ} = Δ_{sub} H + \frac{1}{2} Δ_{diss} H + I . E + E . A + Δ_{lattice} H$
$- 617 = 161 + 520 + 77 +$ E.A. $+ (- 1047)$
E. $A . = - 617 + 289 = - 328 k J m o l^{- 1}$
$∴$ electron affinity of fluorine $= - 328 k J m o l^{- 1}$


$L i (s) \to L i (g)$	161
$L i (g) \to L i^{+} (g)$	520
$\frac{1}{2} F_{2} (g) \to F (g)$	77
$F (g) + e^{-} \to F - (g)$	(Electron gain enthalpy)
$L i^{+} (g) + F^{-} (g) \to L i F (s)$	-1047
$L i (s) + \frac{1}{2} F_{2} (g) \to L i F (s)$	-617

Q. Li(s)→Li(g) 161 Li(g)→Li+(g) 520 21​F2​(g)→F(g) 77 F(g)+e−→F−(g) (Electron gain enthalpy) Li+(g)+F−(g)→LiF(s) -1047 Li(s)+21​F2​(g)→LiF(s) -617 Based on data provided, the value of electron gain enthalpy of fluorine would be :

−300kJmol−1

−350kJmol−1

−328kJmol−1

−228kJmol−1

Applying Hess's Law Δf​H∘=Δsub ​H+21​Δdiss ​H+I.E+E.A+Δlattice ​H −617=161+520+77+ E.A. +(−1047) E. A.=−617+289=−328kJmol−1 ∴ electron affinity of fluorine =−328kJmol−1

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

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

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

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

Applying Hess's Law
$Δ_{f} H^{\circ} = Δ_{sub} H + \frac{1}{2} Δ_{diss} H + I . E + E . A + Δ_{lattice} H$
$- 617 = 161 + 520 + 77 +$ E.A. $+ (- 1047)$
E. $A . = - 617 + 289 = - 328 k J m o l^{- 1}$
$∴$ electron affinity of fluorine $= - 328 k J m o l^{- 1}$