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The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /\) \(\mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\) a n d \(\quad \mathrm{Sn}(s)=51.818 \mathrm{~J} /\) mol-K. All but \(S\) n have the same (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 for the group 14 elements at 298 K can be attributed to the increase in atomic mass as we move down the group. This increasing atomic mass increases the number of vibrational modes, which in turn increases the entropy values of the elements.

Step by step solution

01

Recall the meaning of entropy

Entropy is a thermodynamic property that describes the measure of disorder in a system. In this case, it refers to the disorder of the elements in their solid-state lattice structures. A higher entropy value indicates a more disordered state.
02

Identify a possible factor

Since all the given elements have the same diamond structure, we must identify some other factors which can lead to the trend observed in entropy values. One possible factor for this trend is the increasing atomic mass of the elements in question.
03

Connect atomic mass to the entropy trend

As we go down the group 14 elements from carbon to silicon to germanium to tin, there is an increase in atomic mass. This increase in atomic mass causes an increase in the number of vibrational modes available in each element. A higher number of vibrational modes contribute to a higher entropy value.
04

Consider the fact that Sn does not have a diamond structure

Although Sn does not have a diamond structure, its increased entropy value can still be explained by the fact that it has an even higher atomic mass compared to the other elements. This increased atomic mass effectively contributes to the higher entropy value observed for Sn.
05

Conclusion

The trend in the standard entropy values for the group 14 elements at 298 K can be attributed to the increase in atomic mass as we move down the group. This increasing atomic mass increases the number of vibrational modes, which in turn increases the entropy values of the elements.

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Key Concepts

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

Thermodynamic Property
In thermodynamics, entropy is a pivotal property that helps us understand the level of disorder or randomness in a system. It offers insights into how energy spreads in materials and can increase as temperature rises.
In a material, especially solid states like crystals, entropy is related to how atoms are arranged or vibrate. When we talk about standard entropy, we're referring to the entropy value of a substance under standard conditions, typically at a temperature of 298 K and a pressure of 1 bar.
Entropy increases when the structure of a material allows for more molecular or atomic movements. For the group 14 elements, we observe variations in entropy that relate to both the structures they form and their intrinsic properties. Understanding entropy is key to predicting how materials will interact energetically and how they can potentially change states.
Atomic Mass
Atomic mass is the weight of an atom, largely determined by the total number of protons and neutrons in its nucleus. As you move down the periodic table, atoms usually contain more protons and neutrons, leading to an increase in atomic mass.
For elements in group 14, from carbon to silicon to germanium and tin, we observe a gradual increase in atomic mass. This progressive increase is not just a number; it impacts the physical properties—including entropy.
Greater atomic mass means the element has heavier atoms, which can affect how atoms vibrate and interact with each other within a lattice. In essence, the mass affects the number and types of vibrational modes present, influencing the overall entropy.
Vibrational Modes
Atoms within a solid are not static; they vibrate. These atomic movements are termed vibrational modes, crucial in determining a material's entropy. As the atomic mass of an element increases, the vibrational modes can become more varied and complex.
The more vibrational modes present, the greater potential for disorder, thus enhancing entropy. Heavier atoms mean that there might be additional ways to distribute energy among these various modes.
In the context of entropy in group 14 elements, these modes explain why entropy increases from carbon to tin. Recognizing these vibrational modes helps us understand why, even though certain elements like tin don't fit the diamond structure, they still exhibit higher entropy due to their more complex vibrational nature.

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Most popular questions from this chapter

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2}\). (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)+2 \mathrm{CO}_{2}(g) \end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

For each of the following processes, indicate whether the signs of \(\Delta S\) and \(\Delta H\) are expected to be positive, negative, or about zero. (a) A solid sublimes. (b) The temperature of a sample of \(\mathrm{Co}(s)\) is lowered from \(60^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C} .\) (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\).

(a) For each of the following reactions, predict the sign of \(\Delta H^{*}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\) at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{FeO}(s)\) (ii) \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (iii) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (iv) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CaO}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a}\). (b) By using the value of \(K_{a}\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} \mathrm{M}\), \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} \mathrm{M},\) and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?\)

Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H,\) we can know absolute values of \(S .(\mathbf{b})\) If you heat a gas such as \(\mathrm{CO}_{2}\), you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

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