2. Solved Papers
  3. JCECE Medical

JCECE Medical JCECE Medical Solved Paper-2008

  • question_answer
    If at 298 K the bond energies of \[C-H,C-C,C=C\]and \[H-H\]bonds are respectively 414, 347,  615 and \[435\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}},\] the value of enthalpy change for the reaction \[{{H}_{2}}C=C{{H}_{2}}(g)+\xrightarrow{{}}{{H}_{3}}C-C{{H}_{3}}(g)\]at \[298\,K\]will be

    A) \[+250\,kJ\]

    B)  \[-250\,kJ\]

    C)  \[+125\,kJ\]

    D)  \[-125\,kJ\]

    Correct Answer: D

    Solution :

     \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}\xrightarrow{{}}C{{H}_{3}}-C{{H}_{3}}\] \[\Delta \Eta ={{(BE)}_{\text{reactants}}}-{{(BE)}_{\text{products}}}\] \[=4{{(BE)}_{C-H}}+{{(BE)}_{C=C}}+{{(BE)}_{H-H}}\] \[-[6{{(BE)}_{C-H}}+{{(BE)}_{C-C}}]\] \[=-125\,kJ\]


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