AIIMS AIIMS Solved Paper-2006

  • question_answer
    The enthalpy change \[(\Delta H)\] for the reaction, \[{{N}_{2}}(g)+3{{H}_{2}}(g)\xrightarrow{{}}2N{{H}_{3}}(g)\] is  \[-92.38\text{ }kJ\] at 298 K. The internal energy change \[\Delta U\] at 298 K is:

    A)  \[-\text{ }92.38\text{ }kJ\]                         

    B)  \[-\text{ }87.42\text{ }kJ\]         

    C)         \[-97.34\text{ }kJ\]        

    D)         \[-89.9kJ\]

    Correct Answer: B

    Solution :

    \[\Delta H=\Delta E+P\Delta V\] \[\Delta H=\Delta E+\Delta nRT\] \[\Delta E=\Delta H-\Delta nRT\] \[=-92.38-(-2)(8.314)\times {{10}^{-3}}\times (298)\] \[\Delta E=-92.38+4.895\] \[=-87.48kJ\]


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