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Determine the residual molar entropies for molecular crystals of the following: a. \(^{35} \mathrm{Cl}^{37} \mathrm{Cl}\) b. \(\mathrm{CFCl}_{3}\) c. \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) d. \(\mathrm{CO}_{2}\)

Short Answer

Expert verified
The residual molar entropies for the molecular crystals are: a. \(^{35}Cl^{37}Cl\) : \(S_R = 11.74 \: \mathrm{J/(mol \cdot K)}\) b. \(CFCl_{3}\) : \(S_R = 0 \: \mathrm{J/(mol \cdot K)}\) c. \(CF_{2}Cl_{2}\) : \(S_R = 0 \: \mathrm{J/(mol \cdot K)}\) d. \(CO_{2}\) : \(S_R = 0 \: \mathrm{J/(mol \cdot K)}\)

Step by step solution

01

a. ^{35}Cl^{37}Cl

Since there are two isotopes of chlorine, ^{35}Cl and ^{37}Cl, the possible combinations are ^{35}Cl^{35}Cl, ^{37}Cl^{37}Cl, and ^{35}Cl^{37}Cl. The natural abundance of ^{35}Cl is 75.77% and of ^{37}Cl is 24.23%. The mole fractions for each combination are: - p(^{35}Cl^{35}Cl) = 0.7577 × 0.7577 = 0.5734 - p(^{35}Cl^{37}Cl) = 0.7577 × 0.2423 = 0.1836 - p(^{37}Cl^{37}Cl) = 0.2423 × 0.2423 = 0.0587
02

b. CFCl₃

There are no isotopes in this molecule, so each position has only one possibility and mole fraction equals to 1.
03

c. CFâ‚‚Clâ‚‚

In this case, we have two positions with different possibilities: one for F and the other for Cl. The mole fractions for each position are: - p(F) = 1 (only one isotopic possibility) - p(Cl) = 1 (only one positional possibility)
04

d. COâ‚‚

In this case, we have two positions with different possibilities: one for C and the other for O. The mole fractions for each position are: - p(C) = 1 (only one isotopic possibility) - p(O) = 1 (only one positional possibility) ##Step 2: Calculate the residual molar entropies## Now, we will use the formula \(S_R = -R \sum_{i = 1}^n p_i \ln p_i\) to determine the residual molar entropies for each compound.
05

a. ^{35}Cl^{37}Cl

For this molecule, we have: \(S_R = -R [p(^{35}Cl^{35}Cl) \ln p(^{35}Cl^{35}Cl) + p(^{35}Cl^{37}Cl) \ln p(^{35}Cl^{37}Cl) + p(^{37}Cl^{37}Cl) \ln p(^{37}Cl^{37}Cl)]\) \(S_R = -8.314 [0.5734\ln(0.5734) + 0.1836\ln(0.1836) + 0.0587\ln(0.0587)]\) \(S_R = 11.74 \: \mathrm{J/(mol \cdot K)}\)
06

b. CFCl₃

For this molecule, the residual molar entropy will be zero since there is only one possibility in each position and mole fractions are 1.
07

c. CFâ‚‚Clâ‚‚

For this molecule, the residual molar entropy will also be zero since there is only one possibility in each position and mole fractions are 1.
08

d. COâ‚‚

For this molecule also, the residual molar entropy will be zero since there is only one possibility in each position and mole fractions are 1. In conclusion: a. ^{35}Cl^{37}Cl : \(S_R = 11.74 \: \mathrm{J/(mol \cdot K)}\) b. CFCl₃ : \(S_R = 0 \: \mathrm{J/(mol \cdot K)}\) c. CF₂Cl₂ : \(S_R = 0 \: \mathrm{J/(mol \cdot K)}\) d. CO₂ : \(S_R = 0 \: \mathrm{J/(mol \cdot K)}\)

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Key Concepts

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

Molecular Crystals
Molecular crystals are a type of crystalline solid where the individual units are molecules. They have unique properties because the constituents are held together by relatively weak intermolecular forces, such as Van der Waals interactions, rather than strong ionic or covalent bonds found in other types of crystals. This structure has a significant influence on a property known as residual molar entropy, which can be thought of as the remaining entropy of a substance as it approaches absolute zero temperature. Intuitively, it quantifies the degree of disorder even at a state where a perfect crystal is thought to have no entropy (as proposed by the third law of thermodynamics).

However, in cases such as isotopic molecular crystals, some positional disorder can remain because the different isotopes create distinguishable arrangements. This is what gives rise to residual molar entropy in such molecular crystals. For example, in a crystal composed of isotopic chlorine ([](^{35}Cl^{37}Cl)), there can be different arrangements of the isotopes within the crystal lattice, contributing to its residual molar entropy.
Mole Fractions
Mole fractions are a way to express the concentration of a component in a mixture, defined as the number of moles of that component divided by the total number of moles of all components. In the context of isotopic molecular crystals, mole fractions help determine the probability of finding a particular isotopic combination in the crystal lattice.

For instance, when calculating the residual molar entropy for a molecule like ([](^{35}Cl^{37}Cl)), we first find the natural abundances of each isotope and then calculate the mole fraction of each possible isotopic pair. These mole fractions are essential in determining the entropy since they represent the probability of each configuration occurring within the crystal structure, which directly enters into the entropy calculation as seen in the step by step solution provided.
Isotopic Abundance
Isotopic abundance refers to the naturally occurring distribution of isotopes of a particular element. It's important in calculating the residual molar entropy of substances that have isotopes because these variations lead to different potential configurations within a molecular crystal. The isotopic abundance of an element like chlorine has significant implications for the residual molar entropy calculation since the various isotopic combinations will occur in the crystal according to their abundances in nature.

The exercise earlier provided showed us how to use isotopic abundances of chlorine to compute mole fractions, which are then used to calculate the entropy of the crystal. Understanding isotopic abundance is crucial because it leads to the realization that even at very low temperatures, not all positional order is lost due to the presence of isotopes, and thus, there is a residual entropy associated with these materials.
Enthalpy Calculation
Enthalpy calculation is a critical part of thermochemistry, fundamentally linked to residual molar entropy but serving a different aspect of understanding chemical processes. While residual molar entropy deals with the randomness or disorder in a system at a given temperature, enthalpy represents the heat content of the system. It is commonly used to calculate the heat absorbed or released during a reaction or phase change at constant pressure.

Even though enthalpy does not directly contribute to the residual molar entropy calculation, it's important to highlight the relationship between enthalpy, entropy, and temperature, which together determine the spontaneity of processes through the Gibbs free energy. Educator emphasis is often on entropy and enthalpy differences (∆S and ∆H) to explore these thermodynamic concepts further in other chemistry problems.

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Most popular questions from this chapter

Consider the following isotope exchange reaction: \\[ \mathrm{DCl}(g)+\operatorname{HBr}(g) \rightleftharpoons \operatorname{DBr}(g)+\mathrm{HCl}(g) \\] The amount of each species at equilibrium can be measured using proton and deuterium NMR (see Journal of Chemical Education \(73[1996]: 99) .\) Using the following spectroscopic information, determine \(K_{p}\) for this reaction at \(298 \mathrm{K}\). For this reaction, \(\Delta \varepsilon=41 \mathrm{cm}^{-1}\) equal to the difference in zero-point energies between products versus reactants, and the groundstate electronic degeneracy is zero for all species. $$\begin{array}{lccc} & M\left(\mathrm{g} \mathrm{mol}^{-1}\right) & B\left(\mathrm{cm}^{-1}\right) & \widetilde{\nu}\left(\mathrm{cm}^{-1}\right) \\ \hline \mathrm{H}^{35} \mathrm{Cl} & 35.98 & 10.59 & 2991 \\ \mathrm{D}^{35} \mathrm{Cl} & 36.98 & 5.447 & 2145 \\ \mathrm{H}^{81} \mathrm{Br} & 81.92 & 8.465 & 2649 \\ \mathrm{D}^{81} \mathrm{Br} & 82.93 & 4.246 & 1885 \end{array}$$

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The molecule NO has a ground electronic level that is doubly degenerate, and a first excited level at \(121.1 \mathrm{cm}^{-1}\) that is also twofold degenerate. Determine the contribution of electronic degrees of freedom to the standard molar entropy of NO. Compare your result to \(R \ln (4)\). What is the significance of this comparison?

Consider an ensemble of units in which the first excited electronic state at energy \(\varepsilon_{1}\) is \(m_{1}\) -fold degenerate, and the energy of the ground state is \(m_{o}\) -fold degenerate with energy \(\varepsilon_{0}\) a. Demonstrate that if \(\varepsilon_{0}=0,\) the expression for the electronic partition function is \\[ q_{E}=m_{o}\left(1+\frac{m_{1}}{m_{o}} e^{-\varepsilon_{1} / k T}\right) \\]\ b. Determine the expression for the internal energy \(U\) of an ensemble of \(N\) such units. What is the limiting value of \(U\) as the temperature approaches zero and infinity?

What is the contribution to the internal energy from translations for an ideal monatomic gas confined to move on a surface? What is the expected contribution from the equipartition theorem?

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