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BCECE Medical BCECE Medical Solved Papers-2015

  • question_answer
    What is the enthalpy of the disproportionation of \[MgCl\] if the enthalpy of formation of hypothetical \[MgCl\] is \[-125\,kJ/mol\] and the \[MgC{{l}_{2}}\] is \[-642\,kJ/mol\]?

    A)  - 767 kJ/mol

    B)  767 kJ/mol

    C)  -392 kJ/mol

    D)  392 kJ/mol

    Correct Answer: C

    Solution :

    \[Mg(s)+C{{l}_{2}}(g)\xrightarrow{{}}MgC{{l}_{2}}(s)\],                  \[\Delta {{H}_{1}}=-642\,\,kJ/mole\] \[Mg(s)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}MgCl(s)\],                  \[\Delta {{H}_{2}}=-125\,kJ/mol\] \[2MgCl\xrightarrow{_{{}}}MgC{{l}_{2}}+Mg\],                 \[\Delta H=?\]                 \[\Delta H=\Delta {{H}_{1}}-2\Delta {{H}_{2}}\]                 \[=-642-2\times (-125)\] \[=-392\,kJ/mol\]


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