2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    The standard enthalpy of formation of \[N{{H}_{3}}\] is -46.0 kJ/mol. If the enthalpy of formation of \[{{H}_{2}}\] from its atoms is -436 kJ/mol and that of \[{{N}_{2}}\] is - 712 kJ/mol, the average bond enthalpy of N - H bond in \[N{{H}_{3}}\] is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) -1102 kJ/mol                       

    B)  - 964 kJ/mol

    C) + 352 kJ/mol      

    D) + 1056 kJ/mol

    Correct Answer: B

    Solution :

    Given\[\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}N{{H}_{3}};\] \[\Delta {{H}_{f}}=-46.0kJ\text{/}mol\] \[H+H{{H}_{2}};\Delta {{H}_{f}}=-436kJ/mol\] \[N+N{{N}_{2}};\Delta {{H}_{f}}=-712kJ/mol\] \[\Delta {{H}_{f}}\left( N{{H}_{3}} \right)=\frac{1}{2}\Delta {{H}_{N-N}}+\frac{3}{2}\Delta {{H}_{H-H}}-\Delta {{H}_{N-F}}\] \[-46=\frac{1}{2}\left( -712 \right)+\frac{3}{2}\left( -436 \right)-\Delta {{H}_{N-F}}\] On calculation\[\Delta {{H}_{N-F}}=-964kJ/mol\]


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