A) \[-300\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]
B) \[-350\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]
C) \[-328\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]
D) \[-228\operatorname{k}\operatorname{J}{{\operatorname{mol}}^{-1}}\]
Correct Answer: A
Solution :
Applying Hess?s Law \[{{\Delta }_{f}}{{H}^{o}}={{\Delta }_{sub}}H+\frac{1}{2}{{\Delta }_{diss}}H+I.E.+E.A+{{\Delta }_{lattice}}H\]\[-617=161+520+77+E.A+(-1047)\] \[E.A.=-617+289=-328kJ\,mo{{l}^{-1}}\] \[\therefore \] electron affinity of fluorine \[=-328\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]You need to login to perform this action.
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