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Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightleftharpoons \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \rightleftharpoons \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Short Answer

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(a) Positive; (b) Negative; (c) Positive; (d) Positive.

Step by step solution

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01

Identify Phase Changes and Count Molecules - Reaction (a)

For the reaction \( \mathrm{Zn}(s)+2 \mathrm{HCl}(aq) \rightarrow \mathrm{ZnCl}_2(aq)+\mathrm{H}_2(g) \), observe the phases and number of molecules. Reactants include a solid and an aqueous solution, producing an aqueous solution and a gas. The production of a gas generally leads to increased entropy due to the greater freedom of movement of gas-phase particles compared to solids and liquids. Hence, the entropy change is positive.
02

Identify Phase Changes and Count Molecules - Reaction (b)

For the reaction \( \mathrm{O}(g)+\mathrm{O}(g) \rightarrow \mathrm{O}_2(g) \), two gas molecules combine to form one gas molecule. The decrease in the number of gas molecules typically results in a decrease in entropy, as fewer molecules represent a lower number of microstates. Therefore, entropy change is negative.
03

Identify Phase Changes and Count Molecules - Reaction (c)

In the reaction \( \mathrm{NH}_4 \mathrm{NO}_3(s) \rightarrow \mathrm{N}_2 \mathrm{O}(g)+2 \mathrm{H}_2 \mathrm{O}(g) \), a solid reactant is transformed into gaseous products. Moving from a solid to multiple gas molecules greatly increases entropy. This is due to the transition from a highly ordered solid state to a less ordered gaseous state with more microstates, making the entropy change positive.
04

Identify Phase Changes and Count Molecules - Reaction (d)

In the reaction \( 2 \mathrm{H}_2 \mathrm{O}_2(l) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(l)+\mathrm{O}_2(g) \), liquid reactants are transformed into both liquid and gas products. The formation of a gas generally increases entropy, as gases have more possible arrangements of particles compared to liquids, leading to higher entropy. Thus, the entropy change is positive.

Key Concepts

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

Phase Changes
Phase changes are transformations from one state of matter to another. They fundamentally change the way particles in a substance are arranged. During phase changes, the degree of order in a system alters: moving from a solid to a liquid, or a liquid to a gas, increases disorder, thereby increasing entropy. Similarly, transitioning from a gas to a liquid or a liquid to a solid reduces disorder, resulting in a decrease in entropy.

Consider water turning into steam. This is a phase change from liquid to gas. Here, the molecules gain energy and move more freely, occupying more space and having more possible positions, increasing entropy.
Microstates
In the realm of thermodynamics, microstates refer to the different possible arrangements of molecules in a system. Entropy is deeply connected to the number of microstates. More microstates mean higher entropy, as the system's particles have more ways to be arranged.

For instance, let's take a look at a pack of playing cards. If we just shuffle them, the number of ways we can arrange them is vast. The greater the number of arrangements, the higher the entropy. Similarly, a gas with more dispersed molecules has more microstates than a tightly packed solid, contributing to its higher entropy.
Gaseous State
Gases represent the state of matter with the highest degree of freedom. In a gaseous state, particles are spread apart and move freely. This contributes to a higher entropy compared to solids and liquids.

When a reaction involves the formation of gases from solids or liquids, like the evolution of \(H_2\) gas in a chemical reaction, it typically results in a positive change in entropy. This is because gas molecules occupy a larger volume, allowing for more random arrangements, i.e., more microstates.
  • Gases are highly compressible due to the large distances between molecules.
  • Each molecule in the gas is independent of others, moving randomly and in various directions.
This significant freedom of movement and increased volume capability lead to a greater number of microstates, and thus, higher entropy.
Solid State
The solid state is characterized by a fixed, rigid structure where particles are tightly packed together. This strong degree of order equates to low entropy. Solids have fewer microstates available because their tightly bound particles are limited in how they can be rearranged.

In the reaction where \(NH_4NO_3(s)\) changes into gaseous products like \(N_2O(g)\) and \(H_2O(g)\), the shift from solid to gas greatly increases potential microstates. The structure of solids restricts particle movement, significantly reducing entropy as compared to gases, where particles are free to move.

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Most popular questions from this chapter

How does the entropy of a system change for each of the following processes? (a) A solid melts. (b) A liquid freezes. (c) A liquid boils. (d) A vapor is converted to a solid. (e) A vapor condenses to a liquid. (f) A solid sublimes. (g) A solid dissolves in water.

The following reaction represents the removal of ozone in the stratosphere: $$ 2 \mathrm{O}_{3}(g) \rightleftarrows 3 \mathrm{O}_{2}(g) $$ Calculate the equilibrium constant \(\left(K_{P}\right)\) for this reaction. In view of the magnitude of the equilibrium constant, explain why this reaction is not considered a major cause of ozone depletion in the absence of humanmade pollutants such as the nitrogen oxides and CFCs. Assume the temperature of the stratosphere is \(-30^{\circ} \mathrm{C}\) and \(\Delta G_{\mathrm{i}}^{\circ}\) is temperature independent.

From the values of \(\Delta H\) and \(\Delta S\), predict which of the following reactions would be spontaneous at \(25^{\circ} \mathrm{C}:\) reaction A: \(\Delta H=10.5 \mathrm{~kJ} / \mathrm{mol}, \Delta S=30 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} ;\) reaction B: \(\Delta H=1.8 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-113 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). If either of the reactions is nonspontaneous at \(25^{\circ} \mathrm{C},\) at what temperature might it become spontaneous?

Calculate \(\Delta S_{\text {sys }}\) for (a) the isothermal compression of 0.0050 mol of an ideal gas from \(112 \mathrm{~mL}\) to \(52.5 \mathrm{~mL}\) (b) the isothermal compression of \(0.015 \mathrm{~mol}\) of an ideal gas from \(225 \mathrm{~mL}\) to \(22.5 \mathrm{~mL},\) and \((\mathrm{c})\) the isothermal expansion of \(22.1 \mathrm{~mol}\) of an ideal gas from 122 L to \(275 \mathrm{~L}\).

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l),(b) C_{2} H_{5} O H(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (Hint: Which molecule can hydrogen- (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g),\) (e) \(\mathrm{O}_{2}(g)\) bond?), (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\), or \(\mathrm{O}_{3}(g),(\mathrm{f}) \mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g) .\)

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