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26/09/2021 11:16 am
Topic starter
Given
Reaction Energy Change (in kJ)
Li(s)→ Li(g) 161
Li(g) → Li+(g) 520
1/2F2(g) → F(g) 77
F(g) + e- → F-(g) (Electron gain enthalpy)
Li+(g) + F-(g) → LiF(s) - 1047
Li(s) + 1/2F2(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
1 Answer
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26/09/2021 11:21 am
Correct answer: (c) -328 kJ mol-1
Explanation:
Applying Hess’s Law
ΔfH° = ΔsubH + 1/2ΔdissH + I.E. + E.A + ΔlatticeH
-617 = 161 + 520 + 77 + E.A. + (-1047)
E.A. = –617 + 289 = –328 kJ mol-1
∴ electron affinity of fluorine
= –328 kJ mol-1