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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2009

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

    A) \[+184kJ\]       

    B) \[+23\text{ }kJ\]

    C) \[+92kJ\]        

    D) \[+46kJ\]

    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 \[N{{H}_{3}}\]) \[=-(2\times -46)=93kJ\]]


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