/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 The standard entropies at \(298 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /\) mol-K, Ge \((s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\) a n d \(\quad \operatorname{Sn}(s)=51.818 \mathrm{~J} /\) mol-K. All but Sn have the same (diamond) structure. How do you account for the trend in the \(S^{\circ}\) values?

Short Answer

Expert verified
The trend in entropy increases down the group due to larger atomic size and Sn's unique structure.

Step by step solution

01

Define Standard Entropy

Standard entropy, denoted as \( S^{\circ} \), is a measure of the disorder or randomness of a substance in the standard state (1 atm pressure and 298 K). It is usually given in units of J/mol-K.
02

Review Provided Data

The given standard entropies at 298 K are: \( \mathrm{C}(s, \text{diamond})=2.43 \mathrm{~J/mol-K} \), \( \mathrm{Si}(s)=18.81 \mathrm{~J/mol-K} \), \( \text{Ge}(s)=31.09 \mathrm{~J/mol-K} \), and \( \mathrm{Sn}(s)=51.818 \mathrm{~J/mol-K} \).
03

Consider Structural Differences

Identify that all elements except Sn have a diamond-like structure. Sn has a different, more metallic structure, which increases its entropy due to greater atomic vibrations and less ordered arrangement.
04

Analyze Trends in Entropy Values

Notice that the entropy increases as you move down the group from C to Sn. This is due to the increase in atomic size and corresponding increase in the number of accessible microstates, leading to higher entropy values.
05

Draw Conclusion about Entropy Trend

The trend is accounted for by the increase in atomic size from C to Sn and the difference in structural arrangement, with Sn having a higher entropy due to its unique structure compared to the other group 14 elements.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Entropy Trends
Entropy, symbolized as \( S^{\circ} \), reflects the amount of disorder or randomness in a substance. As elements move down the periodic table, there tends to be an increase in entropy. This is evident in Group 14 elements, from carbon to tin.

Whats drives this increase in entropy? Primarily, as the atomic size grows, so does the number of possible arrangements, or microstates, the atoms can have. This increase in arrangements results in higher entropy values. For instance, tin (Sn) has a higher standard entropy than carbon (C) because its atoms can occupy more positions due to a larger atomic size.
  • Entropy is influenced by the structure of the substance.
  • Structural differences result in variations in entropy values due to atomic arrangement possibilities.
To sum up, entropy trends in Group 14 elements are mainly influenced by atomic size and structural arrangement.
Group 14 Elements
Group 14 elements include carbon (C), silicon (Si), germanium (Ge), and tin (Sn). These elements are known to form the backbone of many compounds and are fundamental in various materials. They display interesting trends in their standard entropies.

One key factor in these trends is their crystal structures. With the exception of tin, these elements share a diamond-like structure at room temperature, which is highly ordered. Tin, however, has a more metallic and less ordered structure, which allows for higher entropy. Here's a breakdown:
  • Carbon (C), in diamond form, is very ordered, resulting in a much lower entropy value.
  • Silicon (Si) and germanium (Ge), while having the same structure, have progressively higher entropies due to their increasing atomic sizes.
  • Tin (Sn) diverges from the diamond structure, contributing to its higher entropy.
Understanding these elements and their characteristics helps to understand the trends in entropy and their applications in materials science.
Atomic Structure and Entropy
The atomic structure is closely tied to a substance's entropy. In simple terms, the more ways atoms can be arranged in a material, the higher the entropy.

Atoms in elements like carbon, silicon, and germanium are bound in a tightly ordered diamond structure. This restricts their relative positions, resulting in lower entropy. However, as we progress from carbon to tin, we notice:
  • Each element, with increasing atomic number, shows more available microstates, leading to higher entropy.
  • Sn's unique structure breaks away from the diamond form, resulting in even more possible configurations.
These variations illustrate how atomic makeup and structure can govern the thermodynamic properties, like entropy, of elements and compounds.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following processes are spontaneous and which are nonspontaneous: (a) mixing of water and ethanol, \((\mathbf{b})\) dissolution of sugar in a cup of hot coffee, (c) formation of oxygen atoms from \(\mathrm{O}_{2}\) molecules at STP, (d) rusting of iron, (e) formation of glucose from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) at \(\mathrm{STP} ?\)

Indicate whether each statement is true or false. (a) A reaction that is spontaneous in one direction will be nonspontaneous in the reverse direction under the same reaction conditions. (b) All spontaneous processes are fast. (c) Most spontaneous processes are reversible. (d) An isothermal process is one in which the system loses no heat. (e) The maximum amount of work can be accomplished by an irreversible process rather than a reversible one.

For a certain chemical reaction, \(\Delta H^{\circ}=-40.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=-150.0 \mathrm{~J} / \mathrm{K} .(\mathbf{a})\) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

(a) For a process that occurs at constant temperature, does the change in Gibbs free energy depend on 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. Is the process spontaneous? (c) If \(\Delta G\) for a process is large, is the rate at which it occurs fast?

(a) What sign for \(\Delta S\) do you expect when the volume of \(0.200 \mathrm{~mol}\) of an ideal gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from an initial volume of \(10.0 \mathrm{~L} ?(\mathbf{b})\) If the final volume is \(18.5 \mathrm{~L},\) calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.