2. Solved Papers
  3. AIIMS

AIIMS AIIMS Solved Paper-2002

  • question_answer
    The enthalpy of formation of ammonia is \[=46.0\,kJ\,mo{{l}^{-1}}\]. The enthalpy change for following reaction is : \[2N{{H}_{3}}\xrightarrow{{}}{{N}_{2}}+3{{H}_{2}}\]

    A)  42.0kJ                  

    B)         64.0kJ  

    C)         86.0kJ                  

    D)         92.0kJ

    Correct Answer: D

    Solution :

    When one mole of \[N{{H}_{3}}\] is formed from its constituent elements the enthalpy change \[=-46.0kJ\] Therefore when one mole of \[N{{H}_{3}}\] decompose to give its constituent elements enthalpy change \[=-46.0kJ\] \[\Rightarrow \] When 2 mole \[N{{H}_{3}}\] decompose, enthalpy change \[=2\times 46=92.0kJ\]


You need to login to perform this action.
You will be redirected in 3 sec spinner