• # 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 KC)  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]. LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

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