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State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

Short Answer

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(a) Negative, (b) Positive, (c) Positive, (d) Negative

Step by step solution

01

Analyze Reaction (a)

The reaction \( \mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) \) involves converting a liquid and a gas into a single gaseous product. Generally, the entropy tends to decrease (negative) when a gas reacts with a liquid to form a single gas because there is a reduction in the total number of gas particles and thus less randomness.
02

Analyze Reaction (b)

The reaction \( 2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) \) involves the decomposition of a solid into a liquid and a gas. This increases disorder because gases have higher entropy than solids. The formation of oxygen gas especially contributes to an increase in randomness, leading to a positive entropy change.
03

Analyze Reaction (c)

In the reaction \( \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \), a diatomic molecule dissociates into two individual atoms. This increases the number of particles and the overall randomness of the system, thus increasing entropy. Therefore, the entropy change is positive.
04

Analyze Reaction (d)

For the reaction \( \mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s) \), a solid and a gas react to form another solid. Typically, the formation of a solid from a gas and a solid leads to a decrease in randomness and results in negative entropy change because gases have higher entropy than solids.

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Key Concepts

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

Chemical Reactions
Chemical reactions involve the transformation of reactants into products. During this process, the structure and arrangement of molecules change, affecting the system's entropy. For instance, consider the combination of phosphorus trichloride and chlorine gas forming phosphorus pentachloride gas. In this reaction, the number of gaseous molecules decreases. This decrease leads to less disorder, indicating a negative entropy change. When predicting entropy changes, it's crucial to observe how the number of molecules and their states (solid, liquid, gas) change. Reactions that increase the total number of gaseous particles typically result in a positive entropy change due to increased disorder.
Thermodynamics
Thermodynamics deals with the principles governing energy transformations in chemical processes. Entropy is a central concept in this field, representing the degree of disorder. Generally, systems tend to move towards greater entropy (disorder) as it aligns with the principle of energy dispersion. Consider the decomposition of mercuric oxide into mercury and oxygen gas. In this process, solid decomposes to produce a gas, increasing the disorder and thus a positive entropy change. Thermodynamics helps us to understand these transformations and predictions about entropy by observing energy distribution within the system. It provides insights into why certain reactions occur spontaneously, guided by the quest to attain maximum entropy.
Disorder and Randomness
Disorder and randomness are integral to understanding entropy. Entropy is a measure of the randomness or disorder within a system. Chemical reactions can either increase or decrease this disorder. Take, for example, the dissociation of diatomic hydrogen into individual hydrogen atoms. This reaction increases randomness because splitting a molecule into two generates more pieces, increasing the potential arrangements of the particles. Therefore, the entropy change is positive. Always assess the particle count and their distribution shifts in reactions to gauge how randomness is affected. Increased molecular motion and breakup typically means higher entropy.
Phase Changes
Phase changes can significantly impact entropy. A substance transitioning from a solid to a liquid or gas generally involves increased entropy. This is because particles in a gas move more freely than in a liquid or solid, contributing to greater disorder. Take for instance the reaction where uranium combines with fluorine gas to form uranium hexafluoride solid. Here, the formation of a solid from a gas and a solid results in decreased entropy, as the system has moved to a more ordered state. Observing such transitions helps in predicting entropy changes as they reflect how the molecular order alters typically with a vast difference between gaseous and solid states.

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Most popular questions from this chapter

The enthalpy change in the denaturation of a certain protein is \(125 \mathrm{~kJ} / \mathrm{mol}\). If the entropy change is \(397 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol},\) calculate the minimum temperature at which the protein would denature spontaneously.

A \(74.6-\mathrm{g}\) ice cube floats in the Arctic Sea. The pressure and temperature of the system and surroundings are at 1 atm and \(0^{\circ} \mathrm{C},\) respectively. Calculate \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}}\) and \(\Delta S_{\text {univ }}\) for the melting of the ice cube. What can you conclude about the nature of the process from the value of \(\Delta S_{\text {univ }}\) ? (The molar heat of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol} .\) )

As an approximation, we can assume that proteins exist either in the native (physiologically functioning) state or the denatured state. The standard molar enthalpy and entropy of the denaturation of a certain protein are \(512 \mathrm{~kJ} / \mathrm{mol}\) and \(1.60 \mathrm{~kJ} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Comment on the signs and magnitudes of these quantities, and calculate the temperature at which the denaturation becomes spontaneous.

The equilibrium constant \(K_{P}\) for the reaction: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ is \(5.62 \times 10^{35}\) at \(25^{\circ} \mathrm{C}\). Calculate \(\Delta G_{i}^{o}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C}\)

The sublimation of carbon dioxide at \(-78^{\circ} \mathrm{C}\) is given by: $$ \mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H_{\mathrm{sub}}=25.2 \mathrm{~kJ} / \mathrm{mol} $$ Calculate \(\Delta S_{\text {sub }}\) when \(84.8 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) sublimes at this temperature.

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