JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer Given                 Reaction                                Energy Change (in KJ)                 \[\operatorname{L}\operatorname{i}(\operatorname{s})\to \operatorname{Li}(g)\]                                           161                 \[\operatorname{L}\operatorname{i}(g)\to {{\operatorname{Li}}^{+}}(g)\]                             520                 \[\frac{1}{2}{{\operatorname{F}}_{2}}(g)\to \operatorname{F}(g)\]                          77\[{{\operatorname{F}}_{2}}(g)+{{e}^{-}}\to {{\operatorname{F}}^{-}}(g)\]               (Electron gain                                                                                 enthalpy)                 \[\operatorname{L}{{\operatorname{i}}^{+}}(g)+{{\operatorname{F}}^{-}}(g)\to Li\,\,\operatorname{F}(s)\]   -1047                 \[\operatorname{L}\operatorname{i}(\operatorname{s})+\frac{1}{2}{{\operatorname{F}}_{2}}(g)\to \operatorname{Li}\operatorname{F}(s)\]       -617                 Based on data provided, the value of electron gain enthalpy of fluorine would be:     JEE Main  Online Paper (Held On 22 April 2013)

    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 \[=-328\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]


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