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Chapter 16: Problem 85

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

Short Answer

Expert verified
The temperature at which solids I and II of \(\mathrm{O}_{2}(l)\) are in equilibrium is 43.7 K. This means that at this temperature, both solid phases can coexist without further change.

Step by step solution

01

Write the equation for Gibbs free energy change for the phase transition

We know that \(\Delta G = \Delta H - T\Delta S\), where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, T is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.
02

Set \(\Delta G\) to 0 and plug in given values

Since the solids are in equilibrium, we can set \(\Delta G = 0\). Plug in the given values for \(\Delta H\) and \(\Delta S\): 0 = -743.1 \(\mathrm{J/mol}\) - T(-17.0 \(\mathrm{J/K\cdot mol}\))
03

Solve for the equilibrium temperature T

Rearrange the equation to solve for T: T(-17.0 \(\mathrm{J/K\cdot mol}\)) = -743.1 \(\mathrm{J/mol}\) Now, divide both sides by -17.0 \(\mathrm{J/K\cdot mol}\): T = \(\frac{-743.1 \mathrm{J/mol}}{-17.0 \mathrm{J/K\cdot mol}}\) T ≈ 43.7 K
04

State the answer and the meaning in context

The temperature at which solids I and II of \(\mathrm{O}_{2}(l)\) are in equilibrium is 43.7 K. This means that at this temperature, both solid phases can coexist without further change.

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Most popular questions from this chapter

The equilibrium constant for a certain reaction increases by a factor of 6.67 when the temperature is increased from \(300.0 \mathrm{K}\) to \(350.0 \mathrm{K}\). Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: $$\begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6} \text { at } 600 . \mathrm{K}\\\ &2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37} \mathrm{at} 600 . \mathrm{K} \end{aligned}$$

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