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Chapter 52: Problem 13

Complete the diagram below, similar to that in Model 2, to depict \(\Delta S^{\circ}\) for the reaction: $$ \mathrm{A}_{2} \mathrm{X}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) \rightleftarrows 2 \mathrm{XB}(\mathrm{g})+\mathrm{A}_{2}(\mathrm{~g}) $$

Short Answer

Expert verified
For this reaction, \(\Delta S^{\circ}\) increases, which means it is positive. The depiction will start with \(S_{initial}\) at a certain level, rise to \(S_{final}\) representing the increase in molecules as the reaction occurs, and label this difference as \(\Delta S_{sys}\).

Step by step solution

01

Identify Molecule count

The initial side (reactant side) of the reaction has 2 molecules (A2X2 and B2) while the final side (product side) of the reaction has 3 molecules (2XB and A2). The product side has more molecules than does the reactant side.
02

Draw and Label the Diagram

We can represent the entropy change through a diagram with a baseline representing the initial entropy, a sloping line increasing upwards to represent the increase in entropy (as more dispersal of energy and particles occurs), and a final line representing the final entropy level. Label the initial entropy as \(S_{initial}\), the final entropy as \(S_{final}\), and the change in entropy as \(\Delta S_{sys} = S_{final} - S_{initial}\) . This change happens to be positive as the number of molecules increases from reactants to products.
03

Indicate Entropy Change

Indicate that the \(\Delta S^{\circ}\) is positive using an '+' symbol or mentioning that the entropy increases. This is because the system shifts from less entropy (fewer molecules and less spreading out) to higher entropy (more molecules and more spreading out) when the reaction occurs.

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Most popular questions from this chapter

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

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 your reasoning. $$ \mathrm{N}(\mathrm{g})+2 \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{NO}_{2}(\mathrm{~g}) $$

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