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. 2.06 K
B. 31.6 K
C. 43.7 K
D. 53.4 K

Answer

To find the temperature at which solids I and II are in equilibrium, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

At equilibrium, \(\Delta G = 0\). Therefore, we can set up the equation:

\[ 0 = \Delta H - T \Delta S \]

Rearranging this gives us:

\[ T = \frac{\Delta H}{\Delta S} \]

Given values:
- \(\Delta H = -743.1 \, \text{J/mol}\)
- \(\Delta S = -17.0 \, \text{J/(K mol)}\)

Now, substituting these values into the equation:

\[ T = \frac{-743.1 \, \text{J/mol}}{-17.0 \, \text{J/(K mol)}} \]

Calculating the temperature:

\[ T = \frac{743.1}{17.0} \]

\[ T \approx 43.7 \, \text{K} \]

Thus, the temperature at which solids I and II are in equilibrium is approximately **43.7 K**.