Reaction | Energy Change (in KJ) |
\[Li\left( s \right)\to Li\left( g \right)\] | 161 |
\[Li\left( g \right)\to L{{i}^{+}}\left( g \right)\] | 520 |
\[\frac{1}{2}{{F}_{2}}(g)\to F(g)\] | 77 |
\[F\left( g \right)+{{e}^{-}}\to {{F}^{-}}\left( g \right)\] | Electron gain enthalpy |
\[L{{i}^{+}}(g)+{{F}^{-}}(g)\to LiF(s)\] | -1047 |
\[Li(s)+\frac{1}{2}{{F}_{2}}(g)\to LiF(s)\] | -617 |
A) \[-300\,kJ\,mo{{l}^{-1}}\]
B) \[-350kJ\,mo{{l}^{-1}}\]
C) \[-328\text{ }kJ\text{ }mo{{l}^{-1}}\]
D) \[-228\text{ }kJ\text{ }mo{{l}^{-1}}\]
Correct Answer: C
Solution :
[c] Applying Hess's Law \[{{\Delta }_{f}}{{H}^{{}^\circ }}={{\Delta }_{sub}}H+\frac{1}{2}\Delta dissH+I.E.+E.A+{{\Delta }_{lattice}}H\] \[-617=161+520+77+E.A.+\left( -1047 \right)\] \[E.A.=-617+289=-328\text{ }kJ\text{ }mo{{l}^{-1}}\] \[\therefore \] electron affinity of fluorine \[=-328\text{ }kJ\,mo{{l}^{-1}}\]You need to login to perform this action.
You will be redirected in
3 sec