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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).

Short Answer

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a. The sign of ΔH is negative and the sign of ΔS is negative for the phase transition of phosphorus. b. The β form of phosphorus has a more ordered crystalline structure or has a smaller positional probability.

Step by step solution

01

Predict the sign of ΔH for the phase transition of phosphorus

Since the α form of phosphorus spontaneously converts to the β form below -76.9°C, this indicates that the β form is more stable at lower temperatures. Therefore, this process is exothermic, which means that heat is released and the enthalpy change (ΔH) is negative. So, we can predict that the sign of ΔH is negative.
02

Predict the sign of ΔS for the phase transition of phosphorus

Entropy (ΔS) is a measure of the disorder in the system. As the α form spontaneously converts into the β form at lower temperatures, we can predict that the arrangement of P4 molecules in the β form is more ordered than the α form. A spontaneous increase in order implies that the entropy change (ΔS) is negative. So, we can predict that the sign of ΔS is negative.
03

Predict which form of phosphorus has a more ordered crystalline structure

Since the entropy (ΔS) change is negative, this means that the system becomes more ordered as the α form converts to the β form. More ordered crystalline structure implies a lower positional probability. Therefore, we can predict that the β form of phosphorus has a more ordered crystalline structure or has a smaller positional probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy
Enthalpy is a measure of the total energy of a system, including internal energy and energy required to make space for it by displacing its surroundings. It's an important concept when understanding reactions, particularly those involving phase transitions. The change in enthalpy, \( \Delta H \), indicates whether a reaction absorbs or releases heat. - **Exothermic Reactions**: These release heat, resulting in a negative \( \Delta H \). Due to the system losing energy to its surroundings, the enthalpy decreases.- **Endothermic Reactions**: These absorb heat, leading to a positive \( \Delta H \). The system captures energy, increasing its enthalpy.For the transition from the \( \alpha \) to the \( \beta \) form of phosphorus, the negative change in enthalpy means the process is exothermic. Thus, the \( \beta \) form of phosphorus is more stable and energetically favorable under the given conditions, resulting in the release of heat as the \( \alpha \) form converts.
Entropy
Entropy, represented by the symbol \( \Delta S \), refers to the degree of randomness or disorder in a system. In chemical processes, a change in entropy indicates how the arrangement of molecules has changed in terms of order.- **Increased Order**: When a system becomes more ordered, entropy decreases, showing a negative \( \Delta S \).- **Increased Disorder**: Conversely, chaos in the system raises entropy, resulting in a positive \( \Delta S \).During the transformation of phosphorus from the \( \alpha \) form to the \( \beta \) form, the change in entropy is negative. This implies the \( \beta \) structure is more organized compared to the \( \alpha \) form. Despite spontaneous processes frequently leading to increased disorder, in this case, under specific conditions, increased order is more favorable, which is highlighted by the \( \beta \) form's decreased entropy.
Phase transition
Phase transitions are changes of matter from one state to another, such as solid to liquid, liquid to gas, and vice versa. They occur when external conditions like temperature and pressure change. The study of phase transitions often involves enthalpy and entropy to describe the energy and order changes.- **Typical Transitions**: These include melting, freezing, sublimation, and condensation.- **Solid-Solid Transition**: As seen in the phosphorus example, where both phases are solid, but they differ in molecular arrangement.In the studied transition between the \( \alpha \) and \( \beta \) forms of phosphorus, the phenomenon is a solid-solid phase transition. Even though both forms consist of \( \text{P}_4 \) molecules, their differing arrangements signify a change in molecular configuration and energy states. This transition reflects key thermodynamic principles — including the release of heat (negative \( \Delta H \)) and a move towards a more ordered structure (negative \( \Delta S \)). Phase transitions are crucial for understanding many natural and industrial processes, illustrating the ever-present interplay between energy changes and molecular order.

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Most popular questions from this chapter

Given the following data: $$\begin{array}{lr}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta G^{\circ}=-6399 \mathrm{~kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{array}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

Consider the following reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{298}=0.090\) For \(\mathrm{Cl}_{2} \mathrm{O}(g)\), \(\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{~kJ} / \mathrm{mol}\) \(S^{\circ}=266.1 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=\) \(-R T \ln (K)\) b. Use bond energy values (Table 8.4) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and \(b\) to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\). e. Estimate the value of \(K\) at \(500 . \mathrm{K}\). f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when \(P_{\mathrm{H}_{2} \mathrm{O}}=18\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=2.0\) torr, and \(P_{\mathrm{HOCl}}=0.10\) torr.

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), cal- culate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mol} \mathrm{~A}(\mathrm{~g})\) at \(1.00\) atm and \(1.00 \mathrm{~mol} \mathrm{~B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) c. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) d. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{NaCl}(l)\)

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