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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    Given
    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
    Based on data provided, the value of electron gain enthalpy of fluorine would be:

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


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