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Chapter 17: Problem 87

As \(\mathrm{O}_{2}(I)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). 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 \(-17.0 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). At what temperature are solids I and II in equilibrium?

Short Answer

Expert verified
The temperature at which solids I and II of O₂ are in equilibrium is \(43.7 \mathrm{~K}\), which is found using the Gibbs free energy change formula and the given enthalpy and entropy change values for the phase transition.

Step by step solution

01

Write the Gibbs free energy change formula

Gibbs free energy change (ΔG) is related to enthalpy change (ΔH) and entropy change (ΔS) by the following equation: ΔG = ΔH - TΔS At equilibrium, the Gibbs free energy change (ΔG) for the phase transition is zero, so we have 0 = ΔH - TΔS
02

Rearrange the formula to solve for temperature

We need to find the temperature (T) at which the two solids are in equilibrium, so we rearrange the equation to solve for T: T = ΔH / ΔS
03

Plug in the given values for ΔH and ΔS

We are given that the enthalpy change (ΔH) for the phase transition is -743.1 J/mol and the entropy change (ΔS) is -17.0 J/(K⋅mol). Substituting these values into the formula, we get: T = (-743.1 J/mol) / (-17.0 J/(K⋅mol))
04

Calculate the temperature

Now, calculate the temperature where the two phases are in equilibrium: T = 43.7 K Therefore, at a temperature of \(43.7 \mathrm{~K}\), solids I and II of O₂ are in equilibrium.

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Most popular questions from this chapter

Consider the following reaction at \(800 . \mathrm{K}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm} .\) Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ}$$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{~g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{~g} / \mathrm{cm}^{3}\) ). The elevated boiling point of the solution was \(355.4 \mathrm{~K}\). What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{~kJ} / \mathrm{mol}\), the entropy of vaporization is \(95.95\) \(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\), and the boiling-point elevation constant is \(2.5 \mathrm{~K} \cdot \mathrm{kg} / \mathrm{mol}\).

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?
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