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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

Expert verified
a. Since the α form spontaneously converts to the β form and releases energy, the process is exothermic, making ΔH negative. As spontaneous processes increase the total entropy, ΔS is positive. b. The α form has a more ordered crystalline structure as it has a smaller positional probability compared to the β form.

Step by step solution

01

Determine whether the process is exothermic or endothermic

The α form spontaneously converts to the β form. This suggests that the β form is more stable than the α form and the transformation process releases energy. Therefore, the process is exothermic, and we can expect the enthalpy change, ΔH, to be negative.
02

Determine whether the process increases or decreases entropy

Since the transformation process happens spontaneously, the entropy change, ΔS, should be positive. This is because spontaneous processes tend to increase the total entropy of the system and its surroundings. #b. Predicting the more ordered crystalline structure#
03

Analyze the change in positional probability

Because the β form is more stable and the system's entropy increases when it transforms from the α form to the β form, we can assume that the positional probability (how the molecules are randomly organized) is higher in the β form than in the α form.
04

Determine the more ordered crystalline structure

Since the positional probability of the β form is higher than the α form, the α form must have a more ordered crystalline structure. Therefore, the form of phosphorus with the smaller positional probability (more ordered) is the α form.

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Key Concepts

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

Enthalpy Change
Enthalpy change, represented as \(\Delta H\), is an important concept in understanding phase transitions. It reflects whether a process absorbs or releases energy. A negative \(\Delta H\) indicates that the process is exothermic, meaning energy is released, while a positive \(\Delta H\) indicates that the process is endothermic, meaning energy is absorbed. In the case of white phosphorus transitioning from the \(\alpha\) to the \(\beta\) form, the process is exothermic, meaning \(\Delta H\) is negative. This tells us that the transformation releases energy, suggesting that the \(\beta\) form is more stable than the \(\alpha\) form. Stability is often a result of a lower energy state, which is why energy is released during the transition to the \(\beta\) form.
Entropy Change
Entropy, symbolized as \(\Delta S\), measures the degree of disorder or randomness in a system. A positive \(\Delta S\) signifies an increase in disorder, while a negative \(\Delta S\) indicates a decrease.In phase transitions, like the conversion from \(\mathrm{P}_4(\alpha)\) to \(\mathrm{P}_4(\beta)\), a positive \(\Delta S\) is typically associated with spontaneous processes. In this case, the spontaneous transformation suggests that the entropy change is positive. This means that the system becomes more disordered as it shifts to the \(\beta\) form, indicating an increase in molecular randomness.
Crystalline Structure
Crystalline structures are the arrangements of molecules within a solid state, and they can vary in their level of order. Generally, a more ordered crystalline structure implies a specific, repeating pattern of organization.In the context of the exercise, the \(\alpha\) form of phosphorus has a more ordered crystalline structure compared to the \(\beta\) form. This is determined by analyzing the positional probability and entropy changes. A more ordered structure is usually an indication of lower energy and lower entropy, aligning with the smaller positional probability.
Positional Probability
Positional probability relates to the likelihood of molecules arranging in various configurations. A higher positional probability means more possible arrangements of the molecules, indicating a less ordered state.During the transition from \(\mathrm{P}_4(\alpha)\) to \(\mathrm{P}_4(\beta)\), the increase in entropy suggests a higher positional probability in the \(\beta\) form. This form is thus less ordered, as there are more ways to arrange the molecules, contributing to the overall increase in disorder. As a result, the \(\beta\) form, with its higher positional probability, is less ordered than the \(\alpha\) form, making \(\alpha\) phosphorus the more crystalline ordered form.

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Most popular questions from this chapter

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}\) Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

When most biological enzymes are heated, they lose their catalytic activity. This process is called denaturing. The change original enzyme \(\rightarrow\) new form that occurs on heating is endothermic and spontaneous. Is the structure of the original enzyme or its new form more ordered (has the smaller positional probability)? Explain.

Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. $$\begin{array}{lll} {\text { }} & {\text { Boiling Point (1 atm)}} & { \Delta H_{\mathrm{vap}} } \\ \hline {\text { Methane }} & \quad\quad\quad {112 \mathrm{K}} & {8.20 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Hexane }} & \quad\quad\quad {342 \mathrm{K}} & {28.9 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Compare the molar volume of gaseous methane at 112 \(\mathrm{K}\) with that of gaseous hexane at 342 \(\mathrm{K}\) . How do the differences in molar volume affect the values of \(\Delta S_{\mathrm{vap}}\) for these liquids?

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

Consider the reaction $$\begin{array}{l}{2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)} \\\ {\text { sof } \Delta H \text { and } \Delta S}\end{array}$$ a. Predict the signs of \(\Delta H\) and \(\Delta S .\) b. Would the reaction be more spontaneous at high or low temperatures?

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