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Chapter 54: Problem 5

Why do the entropies of atom combination generally become more negative as the number of atoms in the molecule increases?

Short Answer

Expert verified
The negative entropy of atom combination is due to the ordered process of forming a molecule from individual atoms, leading to a decrease in the overall entropy of the system (represented by a negative value). As the number of atoms in the molecule increases, there are more atoms going from an individual state to an ordered state in the molecule, which leads to a greater decrease in entropy and hence makes the entropy value more negative.

Step by step solution

01

Understand the Concept of Entropy

Entropy is a measure of the disorder or randomness of a system. In thermodynamics, it is used to determine the energy dispersal in a system. In an isolated system, entropy tends to increase. This is based on the second law of thermodynamics.
02

Effect of increasing the number of atoms

Entropy is dependent on the number of available states a system can occupy. For systems with larger number of atoms, there are more possible arrangements and positions for the atoms, leading to a higher number of possible states and thus a higher entropy.
03

Explain negative entropy

Entropy values can be either positive, negative or zero. For the entropy change during the formation of a molecule from individual atoms (combination entropy), the expected entropy will be negative. This is because the process of forming a molecule from individual atoms is generally an ordered process, leading to a decrease in entropy (which is represented by a negative value). As the number of atoms increases, the combination entropy gets more negative because more individual atoms become ordered when forming a molecule.

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Most popular questions from this chapter

Complete the diagram below, similar to that in Model 2, to depict \(\Delta S T\) for the reaction: $$ \mathrm{A}_{2} \mathrm{X}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightrightarrows 2 \mathrm{XB}(\mathrm{g})+\mathrm{A}_{2}(\mathrm{~g}) $$ using \(\Delta S_{\mathrm{ac}}^{\circ}\left(\mathrm{A}_{2} \mathrm{X}_{2}\right), \Delta S_{\mathrm{ac}}^{\circ}\left(\mathrm{B}_{2}\right), \Delta S_{\mathrm{ac}}^{\circ}(\mathrm{XB}), \Delta S_{\mathrm{ac}}^{\infty}\left(\mathrm{A}_{2}\right)-\) do not use numerical values.

Why is \(\Delta S_{\mathrm{ac}}^{\circ}\) of \(\mathrm{N}(\mathrm{g})=0 ?\)

Do you expect \(\Delta S\) for the following reaction to be positive or negative? Explain. $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

Do you expect \(\Delta S\) for the following reaction to be positive or negative? Explain. $$ \mathrm{N}(\mathrm{g})+2 \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{NO}_{2}(\mathrm{~g}) $$
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