As `O_2` (I) is cooled at 1 atm pressure , it freezes to form solid I at 54.5 K. At a lower temperature , solid rearrange to solid II, which has a different crystal that for the phase transition solid to slid II , `DeltaH=-743.1 Jmol^(-1) and DeltaS=-17.0JK^(-1) mol^(-1)` . At what temperature are solids I and II in equilibrium ?

Options

A. 2.06K
B. 31.5K
C. 43.7 K
D. 53.4 K

Answer

C

Explanation

For solid `I hArr `solid II euilibrium , `DeltaG=0` . Hence,from `DeltaG= DeltaH-T DeltaS ,0=DeltaH - T DeltaS` or `T= ( DeltaH )/( DeltaS)= ( - 743.15 mol^(-1))/( - 17.0 JK^(-1) mol^(-1)) = 43.7 K`