Question

As O₂ (I) is cooled at 1atm, it freezes at 54K to form solid I. At a lower temperature, Solid I, rearranges to solid II, which has a different crystal structure. Thermal measurements show that $\mathbf{∆}\mathbf{H}$for the$\mathbf{I}\mathbf{\to }\mathbf{II}$phase transition is -743.1 J/Kmol and $\mathbf{∆}\mathbf{S}$ and for the same transition is -17.0JK^-1mol^-1. At what temperature are Solids I and II in equilibrium?

Short answer

Answer:

The temperature at which solids I and II will be in equilibrium is 43.7K

Step-by-step solution

Definition of Equilibrium

A condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs is defined as Equilibrium

Calculation to find the temperature of Solids I and II in equilibrium

$$\begin{gathered} ∆\mathrm{H}°=-743.1\mathrm{J}/\mathrm{Kmol} \\ ∆\mathrm{S}°=-17\mathrm{J}/\mathrm{mol} \end{gathered}$$

The phase transition reaction is $\mathrm{SolidI}\rightleftharpoons \mathrm{SolidII}$

For a reaction at equilibrium,

$\begin{array}{rcl}∆\mathrm{H}& =& \mathrm{T}∆\mathrm{S}\\ \mathrm{T}& =& \frac{∆\mathrm{H}}{∆\mathrm{S}}\\ & =& \frac{-743.1\mathrm{J}/\mathrm{mol}}{-17.0\mathrm{J}/\mathrm{mol}}\\ \mathrm{T}& =& 43.7\mathrm{K}\end{array}$