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Chapter 19: Problem 51

The standard entropies at \(298 \mathrm{~K}\) for certain of the group \(4 \mathrm{~A}\) elements are as follows: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) \(\mathrm{Si}(s)=18.81 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) and \(\operatorname{Sn}(s)=\) \(51.818 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) All but \(\mathrm{Sn}\) have the diamond structure. How do you account for the trend in the \(S^{\circ}\) values?

Short Answer

Expert verified
The trend in the standard entropy values of group 4A elements can be accounted for by considering the increasing atomic mass and structural differences between the elements. The diamond structure of carbon (C), silicon (Si), and germanium (Ge) gives them lower entropy values due to its stability and rigidity, while the non-diamond structure of tin (Sn) allows for more configurational freedom, resulting in a higher entropy. Additionally, the increase in atomic mass leads to more atoms and electrons in a sample, contributing to the rise in entropy down the group.

Step by step solution

01

Identify Trends and Properties of Elements in Group 4A

Group 4A elements include carbon (C), silicon (Si), germanium (Ge), tin (Sn), and lead (Pb). The given values of standard entropies at 298 K indicate that the entropy values increase down the group from carbon to tin. Let's organize this information in a table. | Element | Standard Entropy (J/mol·K) | Structure | |---------|---------------------------|-----------| | C | 2.43 | Diamond | | Si | 18.81 | Diamond | | Ge | 31.09 | Diamond | | Sn | 51.818 | Non-Diamond | From the table, note that all elements except tin (Sn) have the diamond structure.
02

Explain The Effect of Atomic Mass

As we go down the group 4A, the atomic mass of the elements increases. This increase in atomic mass means more atoms in a given sample, which subsequently results in a higher number of microstates and greater entropy. Additionally, elements with larger atomic masses also have more electrons, and these electrons are distributed over a larger volume. This can lead to a higher number of possible electronic configurations, further contributing to the increase in entropy as we go down the group from carbon (C) to tin (Sn).
03

Explain The Effect of Crystal Structure

The diamond structure has a very stable and rigid lattice that is held together by strong covalent bonds. This leads to a more ordered structure with limited vibrational freedom, which typically results in lower entropy values. On the other hand, tin (Sn) does not have the diamond structure, meaning that it might have a less ordered and more flexible lattice, leading to higher entropy.
04

Relate Atomic Mass and Crystal Structure to the Entropy Trend

The combination of increasing atomic mass and the structural differences between the elements in group 4A results in the observed trend of increasing entropy from carbon (C) to tin (Sn). The diamond structure of carbon (C), silicon (Si), and germanium (Ge) gives them lower entropy values, whereas the non-diamond structure of tin (Sn) allows for more configurational freedom, resulting in a higher entropy value. In conclusion, the trends in the standard entropy values of group 4A elements can be accounted for by considering the increasing atomic mass and structural differences between the elements, mainly the presence of the diamond structure in all elements except tin (Sn).

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Most popular questions from this chapter

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

(a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of \(\Delta S_{\text {surr }}\) ?

(a) For a process that occurs at constant temperature, express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b) For a certain process that occurs at constant \(T\) and \(P,\) the value of \(\Delta G\) is positive. What can you conclude? (c) What is the relationship between \(\Delta G\) for a process and the rate at which it occurs?

Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Molten gold solidifies. (b) Gaseous \(\mathrm{Cl}_{2}\) dissociates in the stratosphere to form gaseous \(\mathrm{Cl}\) atoms. (c) Gaseous CO reacts with gaseous \(\mathrm{H}_{2}\) to form liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\). (d) Calcium phosphate precipitates upon mixing \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\)
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