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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    The standard enthalpy of formation of \[N{{H}_{3}}\]is\[-46.0\text{ }kJ/mol\]. If the enthalpy of formation of \[{{H}_{2}}\] from its atoms is \[-436\text{ }kJ/mol\] and that of \[{{N}_{2}}\] is\[-712\text{ }kJ/mol\], the average bond enthalpy of \[N-H\] bonding is:

    A) \[-1102\text{ }kJ/mol\]               

    B) \[-964\text{ }kJ/mol\]

    C) \[+352\text{ }kJ/mol\]               

    D) \[+1056\text{ }kJ/mol\]

    Correct Answer: B

    Solution :

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


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