Chemical Thermodynamics Question 205
Question: Given
Reaction Energy Change (in KJ) $ Li( s )\to Li( g ) $ 161 $ Li( 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 LiF(s) $ -1047 $ Li(s)+\frac{1}{2}F_2(g)\to LiF(s) $ -617 Based on data provided, the value of electron gain enthalpy of fluorine would be:
Options:
A) $ -300,kJ,mo{{l}^{-1}} $
B) $ -350kJ,mo{{l}^{-1}} $
C) $ -328kJmo{{l}^{-1}} $
D) $ -228kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Applying Hess’s Law $ {\Delta_{f}}{{H}^{{}^\circ }}={\Delta_{sub}}H+\frac{1}{2}\Delta dissH+I.E.+E.A+{\Delta_{lattice}}H $
$ -617=161+520+77+E.A.+( -1047 ) $
$ E.A.=-617+289=-328kJmo{{l}^{-1}} $
$ \therefore $ electron affinity of fluorine $ =-328kJ,mo{{l}^{-1}} $
BETA
