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Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Molten gold solidifies. (b) Gaseous \(\mathrm{Cl}_{2}\) dissociates in the stratosphere to form gaseous \(\mathrm{Cl}\) atoms. (c) Gaseous CO reacts with gaseous \(\mathrm{H}_{2}\) to form liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH}\). (d) Calcium phosphate precipitates upon mixing \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\left(\mathrm{NH}_{4}\right)_{3} \mathrm{PO}_{4}(a q)\)

Short Answer

Expert verified
(a) ΔS_sys < 0 (negative) since molten gold solidifies and becomes more ordered. (b) ΔS_sys > 0 (positive) due to the increase in the number of gas particles, increasing disorder. (c) ΔS_sys < 0 (negative) as the gaseous reactants combine to form a liquid, decreasing disorder. (d) ΔS_sys < 0 (negative) because precipitate formation causes the system to become more ordered.

Step by step solution

01

(a) Molten gold solidifies

In this process, molten gold (which is a liquid) is transformed into solid gold. The liquid state of matter has more disorder than the solid state since the particles in a liquid have more kinetic energy and a higher degree of freedom. Therefore, as the gold solidifies, the disorder of the system is decreasing. Consequently, the entropy change for this process, ΔS_sys, is expected to be negative.
02

(b) Gaseous Cl2 dissociates in the stratosphere to form gaseous Cl atoms.

For this process, one mole of gaseous Cl2 molecules dissociate into two moles of gaseous Cl atoms. The increase in the number of gas particles results in an increase in disorder of the system. The entropy change ΔS_sys is therefore positive.
03

(c) Gaseous CO reacts with gaseous H2 to form liquid methanol, CH3OH.

Through this reaction, one mole of gaseous CO and two moles of gaseous H₂ combine to form one mole of liquid methanol. This results in a decrease in the number of gas particles and a change in state from gas to liquid, accompanied by a decrease in disorder. Consequently, the entropy change ΔS_sys for this process is negative.
04

(d) Calcium phosphate precipitates upon mixing Ca(NO3)2(aq) and (NH4)3PO4(aq)

When calcium nitrate and ammonium phosphate solutions are mixed, a precipitate of calcium phosphate forms. The process involves changing states from the more disordered aqueous ions to a more organized solid precipitate. Thus, the entropy change ΔS_sys for this process is negative as the system becomes more ordered.

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Key Concepts

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

Second Law of Thermodynamics
The Second Law of Thermodynamics tells us about the natural tendency of systems to go towards disorder. This law states that in an isolated system, the entropy, which we can think of as a measure of disorder or randomness, will tend to increase over time. Entropy is denoted by S and can help us predict how energy and matter will behave in different conditions.

Another key aspect of this law is that energy transfer processes will always result in a net increase in the total entropy of the universe. In other words, processes generally tend to move towards more disordered states. This is crucial when considering whether a process will spontaneously occur or not.

For example, when molten gold solidifies, this process is moving towards a more ordered state as the atoms become fixed in place. This means the entropy of the system, ext{ΔS}_{ ext{sys}}, is negative. It's one of the exceptions where order actually increases.
Phase Changes
Phase changes are transitions between different states of matter: solid, liquid, and gas. These changes are fascinating as they prominently involve entropy changes.

For instance, when a liquid solidifies, like in the case of molten gold becoming a solid, the particles lose energy and become more ordered. This results in a decrease in entropy, ext{ΔS}_{ ext{sys}}, which is why it has a negative sign.

Conversely, when a substance goes from a solid to a liquid, or a liquid to a gas, the particles gain more freedom and move more energetically, which increases disorder. Here, the entropy increases, and ext{ΔS}_{ ext{sys}} becomes positive. You can see this when water boils into steam or ice melts to become water.
Chemical Reactions
Chemical reactions often involve changes in entropy. When we predict the sign of ext{ΔS}_{ ext{sys}} for a reaction, we analyze changes in the number and state of the molecules.

Take for example the reaction where gaseous CO and H₂ convert to liquid methanol. Initially, we have three gaseous molecules, but they form a single liquid molecule, reducing the gas molecules' randomness. Therefore, ext{ΔS}_{ ext{sys}} is negative.

Alternatively, when gaseous ext{Cl}_2 splits into two separate gaseous ext{Cl} atoms, the number of particles and disorder increases, leading to a positive ext{ΔS}_{ ext{sys}}. Watching these shifts can help us understand how reactions proceed and specify which conditions might favor them.
Disorder and Order in Chemistry
Disorder and order are central to understanding entropy in chemistry. A more disordered system has higher entropy, while an ordered system shows lower entropy.

This can be observed in processes like precipitation, where ions from a solution form a solid. For instance, when calcium phosphate precipitates, the initially dispersed ions become organized into a solid lattice. The system becomes more ordered, and ext{ΔS}_{ ext{sys}} is negative.

In general, processes that move towards order, like solidification or precipitation, tend to decrease entropy. On the other hand, changing from solid or liquid to gas moves towards disorder, increasing entropy. Understanding these tendencies helps us anticipate and explain the behavior of chemical systems.

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Most popular questions from this chapter

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

Isomers are molecules that have the same chemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. [Section 19.4\(]\)

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of \(\mathrm{SO}_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix \(\mathrm{C}\). (a) What is the equilibrium constant for the reaction at \(298 \mathrm{~K}\) ? (b) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2} ?\) (c) If \(P_{\mathrm{SO}_{2}}=P_{\mathrm{H}_{2} \mathrm{~S}}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances: $$ \begin{array}{lrr} \text { Substance } & \Delta \boldsymbol{H}_{f}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l}) & \Delta \mathbf{G}_{f}^{\circ}(\mathbf{k J} / \mathbf{m o l}) \\ \hline \mathrm{AgNO}_{3}(s) & -124.4 & -33.4 \\ \mathrm{AgNO}_{3}(a q) & -101.7 & -34.2 \\ \mathrm{MgSO}_{4}(s) & -1283.7 & -1169.6 \\ \mathrm{MgSO}_{4}(a q) & -1374.8 & -1198.4 \end{array} $$ (a) Write the formation reaction for \(\mathrm{AgNO}_{3}(s) .\) Based on this reaction, do you expect the entropy of the system to increase or decrease upon the formation of \(\mathrm{AgNO}_{3}(s) ?\) (b) Use \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) of \(\mathrm{AgNO}_{3}(s)\) to determine the entropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolving \(\mathrm{AgNO}_{3}\) in water an exothermic or endothermic process? What about dissolving \(\mathrm{MgSO}_{4}\) in water? (d) For both \(\mathrm{AgNO}_{3}\) and \(\mathrm{MgSO}_{4},\) use the data to calculate the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with reference to material presented in this chapter and in the "A Closer Look" box on page 540 .

Using the data in Appendix \(C\) and given the pressures listed, calculate \(\Delta G^{\circ}\) for each of the following reactions: $$ \begin{array}{l} \text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \\ \quad P_{\mathrm{N}_{2}}=2.6 \mathrm{~atm}, P_{\mathrm{H}_{2}}=5.9 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=1.2 \mathrm{~atm} \\ \text { (b) } 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=P_{\mathrm{NO}_{2}}=5.0 \times 10^{-2} \mathrm{~atm} \\ \quad P_{\mathrm{N}_{2}}=0.5 \mathrm{~atm}, P_{\mathrm{H}_{2} \mathrm{O}}=0.3 \mathrm{~atm} \\ \text { (c) } \mathrm{N}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \\ \quad P_{\mathrm{N}_{2} \mathrm{H}_{4}}=0.5 \mathrm{~atm}, P_{\mathrm{N}_{2}}=1.5 \mathrm{~atm}, P_{\mathrm{H}_{2}}=2.5 \mathrm{~atm} \end{array} $$

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