Predict the sign of \(\Delta S_{\text { sys }}\) for each of the following processes: (a) Molten gold solidifies. (b) Gaseous \(C l_{2}\) dissociates in the stratosphere to form gaseous Cl atoms. (c) Gaseous CO reacts with gaseous \(\mathrm{H}_{2}\) to form liquid methanol, \(\mathrm{CH}_{3} \mathrm{OH} .(\mathbf{d})\) 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) \(\Delta S_\text{sys}<0\), as the system decreases disorder during the phase change from liquid to solid. (b) \(\Delta S_\text{sys}>0\), as the dissociation of \(Cl_2\) molecules to Cl atoms increases disorder. (c) \(\Delta S_\text{sys}<0\), as the conversion from gaseous to liquid molecules decreases disorder. (d) \(\Delta S_\text{sys}<0\), as the precipitation of calcium phosphate from a solution decreases disorder.

Step by step solution

01

(a) Molten gold solidifies

Initially, the gold is in a molten (liquid) state, and the final state of gold is a solid. Therefore, we have a phase change from liquid to solid. The process of a liquid becoming a solid decreases the disorder in the system, as particles in a solid have less freedom of motion than particles in a liquid. This means that the entropy decreases, and \(\Delta S_\text{sys}<0\).
02

(b) Gaseous \(Cl_2\) dissociates to form gaseous Cl atoms

Initially, the system contains gaseous \(Cl_2\) molecules, and the final state has gaseous Cl atoms. The dissociation of \(Cl_2\) molecules into individual Cl atoms increases the number of gas particles in the system, which increases the disorder. Therefore, the entropy increases, and \(\Delta S_\text{sys}>0\).
03

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

Initially, the system contains gaseous CO and H2, and the final state has liquid methanol (\(CH_3OH\)). The process involves converting gas molecules to liquid molecules, which reduces the freedom of motion for these particles. This decrease in disorder results in a decrease in entropy. Therefore, \(\Delta S_\text{sys}<0\).
04

(d) Calcium phosphate precipitates upon mixing

Initially, the system contains aqueous \(Ca(NO_3)_2\) and \((NH_4)_3PO_4\). As these compounds mix, solid calcium phosphate (\(Ca_3(PO_4)_2\)) precipitates. The process of a solute precipitating from a solution decreases the disorder in the system, as solid particles have less freedom of motion than particles in a solution. Therefore, the entropy decreases, and \(\Delta S_\text{sys}<0\).

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Key Concepts

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

Thermodynamics in Chemistry
Thermodynamics is a fundamental branch of physical chemistry that deals with the relationships between energy, heat, work, and the changes these cause in the chemical composition of substances. One of the core concepts of thermodynamics is entropy, symbolized as 'S', which is a measure of the disorder or randomness in a system. The second law of thermodynamics states that entropy in an isolated system will tend to increase over time.

In the context of chemistry, this law helps predict the direction in which a chemical process will naturally proceed. A chemical reaction or physical change will favor the direction that increases the system's overall entropy, making entropy an essential indicator of reaction spontaneity.
Phase Changes and Entropy
Phase changes, such as melting, freezing, vaporization, condensation, and sublimation, are physical changes that involve a transition between different states of matter. During a phase change, the entropy of a system changes significantly because the molecular arrangement and degree of movement change drastically.

When a substance melts or vaporizes, it goes from an ordered phase (solid or liquid) to a more disordered phase (liquid or gas), resulting in an increase in entropy (\texttt{\(\Delta S_{sys}>0\)}). Conversely, when a substance freezes or condenses, the disorder decreases as the molecules become more ordered and fixed, which leads to a decrease in entropy (\texttt{\(\Delta S_{sys}<0\)}). These changes are predictable through the concept of entropy.
Chemical Reactions and Entropy
The principle of entropy applies not only to phase changes but also to chemical reactions. Chemical reactions involve the breaking and forming of chemical bonds, leading to changes in the molecular structure and, consequently, the entropy of the system. A reaction resulting in more complex molecules from simpler ones, or in molecules with more motion possibilities like gases, will generally increase the entropy.

It's also noteworthy that reactions that produce more moles of gas from fewer moles of gas or from liquids or solids usually lead to an increase in entropy. This links back to the notion that gases have more freedom of movement than liquids and solids, contributing to a more disordered system.
Entropy Changes in Reactions
The change in entropy, denoted as \texttt{\(\Delta S_{sys}\)}, is a key parameter to determine when discussing chemical reactions. When we predict the sign of \texttt{\(\Delta S_{sys}\)} for a given process, we consider whether the process induces more disorder or less within the system.

For example, reactions that result in a solid from either a liquid or a gaseous state will typically lead to a decrease in entropy (\texttt{\(\Delta S_{sys}<0\)}) due to the increased order and rigidity in the solid state. Similarly, reactions that create fewer or simpler molecules from many or more complex ones will also reduce the system's entropy. Understanding these entropy changes assists chemists in predicting the behavior of chemical reactions under various conditions.

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Most popular questions from this chapter

A system goes from state 1 to state 2 and back to state \(1 .\) (a) Is \(\Delta E\) the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state 1\(?(\mathbf{c})\) Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state 1\(?\)

(a) Using data in Appendix \(C,\) estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. (b) Use a reference source, such as Web Elements (www. webelements.com), to find the experimental melting and boiling points of \(I_{2}\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)?

(a) What sign for \(\Delta S\) do you expect when the volume of 0.200 mol of an ideal gas at \(27^{\circ} \mathrm{Cis}\) increased isothermally from an initial volume of 10.0 \(\mathrm{L} ?(\mathbf{b})\) If the final volume is 18.5 \(\mathrm{L}\) , calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

(a) For a process that occurs at constant temperature, does the change in Gibbs free energy depend on changes in the enthalpy and entropy of the system? (b) For a certain process that occurs at constant \(T\) and \(P\) , the value of \(\Delta G\) is positive. Is the process spontaneous? (c) If \(\Delta G\) for a process is large, is the rate at which it occurs fast?

Indicate whether each statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increases during a reversible process, the entropy change of the surroundings must decrease by the same amount. (c) In a certain spontaneous process the system undergoes an entropy change of \(4.2 \mathrm{J} / \mathrm{K} ;\) therefore, the entropy change of the surroundings must be \(-4.2 \mathrm{J} / \mathrm{K}\)

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