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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 33

  • question_answer
    The heat of formation of \[N{{H}_{3}}\left( g \right)\] is \[-46kJ\text{ }mo{{l}^{-1}}.\] The \[\Delta H\] (in kJ) of the reaction is: \[2N{{H}_{3}}\left( g \right)\xrightarrow{{}}{{N}_{2}}\left( g \right)+3{{H}_{2}}\left( g \right)\]

    A) 46        

    B)        -46   

    C) 92        

    D)        -92

    Correct Answer: C

    Solution :

    [c] \[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\to N{{H}_{3}}(g)\] \[\Delta {{H}_{r}}=-46\,kJ/mol\] \[2N{{H}_{3}}(g)\to {{N}_{2}}(g)+3{{H}_{2}}(g)\] \[\Delta {{H}_{r}}=-2\Delta {{H}_{r}}\] \[=-2(-46)=92kJ\]          


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