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Chapter 17: Problem 95

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{atm},\) 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 $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 are in equilibrium can be found using the Clapeyron equation, \( \Delta G = \Delta H - T \Delta S = 0 \). We are given ∆H = -743.1 J/mol, and ∆S = -17.0 J/(K⋅mol). Rearrange the equation to get T: \( T = \frac{\Delta H}{\Delta S} \). Substitute the given values and solve for T: \( T = \frac{-743.1 \, \mathrm{J/mol}}{-17.0 \, \mathrm{J/(K \cdot mol)}} = 43.7 \, \mathrm{K} \). The equilibrium temperature for solids I and II is 43.7 K.

Step by step solution

01

Identify the Clapeyron Equation

In this case, since we're dealing with phase transitions, the equilibrium condition in terms of temperature can be found using the Clapeyron equation, which is: \( \Delta G = \Delta H - T \Delta S = 0 \) where, ∆G is the Gibbs free energy change, ∆H is the enthalpy change, T is the temperature, and ∆S is the entropy change. Since equilibrium implies that ∆G = 0, we will rearrange the above equation to solve for T.
02

Rearrange the Clapeyron Equation to Solve for T

Rearrange the Clapeyron equation to get T: \( T = \frac{\Delta H}{\Delta S} \) We are given ∆H = -743.1 J/mol, and ∆S = -17.0 J/(K⋅mol). Now we can substitute these values into the equation and solve for T.
03

Substitute the Given Values and Solve for T

Now substitute the given ∆H and ∆S values into the equation: \( T = \frac{-743.1 \, \mathrm{J/mol}}{-17.0 \, \mathrm{J/(K \cdot mol)}} \) Solve for T: \( T = 43.7 \, \mathrm{K} \)
04

State the Equilibrium Temperature for Solids I and II

The temperature at which solids I and II are in equilibrium is found to be 43.7 K.

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Most popular questions from this chapter

For a liquid, which would you expect to be larger, $\Delta S_{\text { fusion }}\( or \)\Delta S_{\text { evaporation }} ?$ Why?

Using Appendix 4 and the following data, determine \(S^{\circ}\) for $\mathrm{Fe}(\mathrm{CO})_{5}(g) . $\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=?$ $\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=107 \mathrm{J} / \mathrm{K}$ $\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) \quad \Delta S^{\circ}=-677 \mathrm{J} / \mathrm{K}$

What information can be determined from \(\Delta G\) for a reaction? Does one get the same information from \(\Delta G^{\circ},\) the standard free energy change? \(\Delta G^{\circ}\) allows determination of the equilibrium constant \(K\) for a reaction. How? How can one estimate the value of \(K\) at temperatures other than \(25^{\circ} \mathrm{C}\) for a reaction? How can one estimate the temperature where \(K=1\) for a reaction? Do all reactions have a specific temperature where \(K=1 ?\)

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Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C},\) the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).
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