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Manipal Medical Manipal Medical Solved Paper-2001

  • question_answer
    \[2{{N}_{2(g)}}+6{{H}_{2(g)}}\xrightarrow[{}]{{}}4N{{H}_{3(g)}}\] what is the enthalpy of the above reaction when the enthalpy formation of ammonia is\[-46kJ\text{ }mo{{l}^{-1}}\]:

    A)  184 kJ           

    B)  \[-184kJ\]

    C)  46 kJ         

    D)  \[-46kJ\]

    Correct Answer: B

    Solution :

     The enthalpy formation of ammonia is\[-46\text{ }kJ/\]mol. Therefore, in the reaction \[2{{N}_{2(g)}}+6{{H}_{2(g)}}\xrightarrow[{}]{{}}4N{{H}_{3(g)}}\] four molecules of ammonia are formed \[\therefore \]Enthalpy formation \[=4\times (-)46=-184\text{ }kJ.\]


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