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NEET AIPMT SOLVED PAPER 2001

  • question_answer
                    Enthalpy of \[C{{H}_{4}}+\frac{1}{2}{{O}_{2}}\to C{{H}_{3}}OH\] negative. If enthalpy of combustion of \[C{{H}_{4}}\] and \[C{{H}_{3}}OH\] are x and y respectively. Then which relation is correct?      

    A)                                                                                            x > y      

    B)                 x < y      

    C)                 x = y                      

    D)                 x > y  

    Correct Answer: A

    Solution :

                    \[C{{H}_{4}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}C{{H}_{3}}OH(l)\]                 \[\therefore \Delta H=-[(\Delta H\,of\,combustion\,of\,C{{H}_{3}}OH)\]                 -(\[\Delta H\] of combustion of CH4)]                 \[\begin{align}   & =-[(-y)-(-x)] \\  & =-[-y+x]=y-x \\ \end{align}\]                 \[\therefore \,\,x>y\]


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