NEET Sample Paper NEET Sample Test Paper-14

  • question_answer As\[{{O}_{2}}(l)\]is cooled at 1 atm pressure, it freezes to form solid I at 54.5 K. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that for the phase transition solid I to solid II, \[\Delta H=-743.1\,J\,mo{{l}^{-1}}\]and \[\Delta S=-17.0\,J{{K}^{-1}}\,mo{{l}^{-1}}.\]At what temperature are solids I and II in equilibrium?

    A)  2.06 K                        

    B)  31.6 K

    C)  43.7 K                        

    D)  53.4 K

    Correct Answer: C

    Solution :

    Since, \[\Delta G=\Delta H-T\Delta S.\] At equilibrium \[\Delta G=0\] \[\therefore \]    \[\Delta H=T\Delta S\] Given \[\Delta H=-743.1\,\text{J/mol}\text{.}\] \[\Delta S=-17.0\,J\,{{K}^{-1}}\,mo{{l}^{-1}}\] \[\therefore \]    \[\frac{-743.1}{-17.0}=T\] \[T=43.7\,K\] At this temperature solid I and II are in equilibrium. Hence, the correct option is [c].

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