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Manipal Engineering Manipal Engineering Solved Paper-2013

  • question_answer
    Calculate second electron affinity of oxygen for the process, \[{{O}^{-}}(g)+{{e}^{-}}(g)\to {{O}^{2-}}(g)\] by using the following data
    (i) Heat of sublimation of \[Mg(s)=+147.7\,\,kJ\,\,mo{{l}^{-1}}\]
    (ii) lionisation energy of\[Mg(g)\]to form \[M{{g}^{2+}}(g)=2189.0\,\,kJ\,\,mo{{l}^{-1}}\]
    (iii) Bond dissociation energy for \[{{O}_{2}}=498.4\,\,kJ\,\,mo{{l}^{-1}}\]
    (iv) First electron affinity of \[O(g)=-141.0\,\,kJ\,\,mo{{l}^{-1}}\]
    (v) Heat formation of \[MgO=-601.7\,\,kJ\,\,mol-1\]
    (vi) Lattice energy of \[MgO=-3791.0\,\,kJ\,\,mo{{l}^{-1}}\]

    A) \[235.6\,\,kJ\,\,mo{{l}^{-1}}\]

    B) \[468.7\,\,kJ\,\,mo{{l}^{-1}}\]

    C) \[544.4\,\,kJ\,\,mo{{l}^{-1}}\]

    D) \[744.4\,\,kJ\,\,mo{{l}^{-1}}\]

    Correct Answer: D

    Solution :

    Stepwise formation of \[MgO\] involves
    Step Reaction \[\Delta H\]in\[KJ\,\,mo{{l}^{-1}}\]
    (i) \[Mg(s)\to Mg(g)\] \[147.7\]
    (ii) \[Mg(g)\to M{{g}^{2+}}+2{{e}^{-}}\] \[2189.0\]
    (iii) \[\frac{1}{2}{{O}_{2}}(g)\to O(g)\] \[498.4/2\]
    (iv) \[O(g)+1{{e}^{-}}\to {{O}^{-}}(g)\] \[-141.0\]
    (v) \[{{O}^{-}}(g)t{{e}^{-}}\to {{O}^{2-}}(g)\] \[q\]
    (vi) \[M{{g}^{2+}}(g)+{{O}^{2-}}(g)\to MgO(s)\] \[-3791.0\]
    \[Mg(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}MgO(s)\Delta H=-1346.1+q\] \[Mg(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}MgO(s)\Delta {{H}_{2}}=-601.7\] By Born Herber cycle (based on Hess law) \[\Delta {{H}_{1}}=\Delta {{H}_{2}}\] \[-1346.1+q=601.7\] \[q=744.4\,\,J\,\,mo{{l}^{-1}}\]


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