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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2009

  • question_answer
    The enthalpy of formation of \[N{{H}_{3}}\] is\[-46\,\,kJ\,\,mo{{l}^{-1}}\]. The enthalpy change for the reaction                 \[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g)\] is

    A)  +184 kJ                               

    B)  +23 kJ

    C)  +92 kJ                                  

    D)  +46 kJ

    Correct Answer: C

    Solution :

    \[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g)\] \[\Delta {{H}_{r}}=-\](\[2\times \] enthalpy of formation of \[N{{H}_{3}}\])                 \[=-(2\times -46)=92\,kJ\]


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