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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The enthalpy of formation of\[N{{H}_{3}}\]is\[-46\,\,kJ\,\,mo{{l}^{-1}}\]. The enthalpy change for the reaction \[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g)\]is

    A) \[+184\,\,kJ\]                   

    B)        \[+23\,\,kJ\]

    C)        \[+92\,\,kJ\]                      

    D)        \[+46\,\,kJ\]

    Correct Answer: C

    Solution :

    \[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g)\] \[\Delta {{H}_{r}}=-\](\[2\times \]enthalpy of formation of\[N{{H}_{3}}\])          \[=-(2\times -46)=92\,\,kJ\]


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