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Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) c. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) d. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{NaCl}(l)\)

Short Answer

Expert verified
a. \(\Delta S^{\circ} < 0\) b. \(\Delta S^{\circ} < 0\) c. \(\Delta S^{\circ} > 0\) d. \(\Delta S^{\circ} > 0\)

Step by step solution

01

Reaction a:

In this reaction, a solid sodium \(\mathrm{(Na)}\) reacts with gaseous chlorine \(\frac{1}{2}\mathrm{Cl}_2\) to form solid sodium chloride \(\mathrm{(NaCl)}\). The products have a lower entropy than the reactants, because a gas is involved in the reactants and both products are solids. Therefore, the sign of \(\Delta S^{\circ}\) is negative.
02

Reaction b:

In this reaction, gaseous nitrogen \(\mathrm{(N_{2})}\) reacts with gaseous hydrogen \(\mathrm{(3H_{2})}\) to produce gaseous ammonia \(\mathrm{(2NH_{3})}\). We have 4 moles of gas in the reactants and 2 moles of gas in the products. The decrease in the number of gas moles results in a decrease in entropy. Therefore, the sign of \(\Delta S^{\circ}\) is negative.
03

Reaction c:

In this reaction, solid sodium chloride \(\mathrm{(NaCl)}\) dissociates into aqueous ions \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\). The dissociation of the solid into separate ions in solution increases the entropy as there is an increase in the number of particles and their freedom of movement. Therefore, the sign of \(\Delta S^{\circ}\) is positive.
04

Reaction d:

In this reaction, solid sodium chloride \(\mathrm{(NaCl)}\) changes its state to a liquid \(\mathrm{(NaCl(l))}\). A change from a solid to a liquid results in higher entropy, as the particles have more freedom to move and rearrange in the liquid state compared to the solid state. Therefore, the sign of \(\Delta S^{\circ}\) is positive.

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy, heat, and work. It examines how energy is transferred and transformed in physical and chemical processes.
When we talk about thermodynamics in chemistry, we often refer to concepts like enthalpy, entropy, and Gibbs free energy.
  • Energy Transformation: Energy can be converted from one form to another, like potential energy into kinetic energy.
  • First Law of Thermodynamics: Energy cannot be created or destroyed in an isolated system.
  • Entropy: Often symbolized by "S," entropy is a measure of disorder or randomness in a system.
The second law of thermodynamics tells us that the total entropy of an isolated system can never decrease over time.
This is why processes such as mixing or dissolving tend to increase entropy.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products, changing the chemical structure and energy content.
These reactions often involve the breaking and forming of chemical bonds.
  • Endothermic Reactions: Absorb energy, resulting in a positive change in enthalpy ( ΔH > 0 ).
  • Exothermic Reactions: Release energy, resulting in a negative change in enthalpy ( ΔH < 0 ).
Catalysts can increase the rate of a reaction without being consumed by lowering the activation energy.
During chemical reactions, we consider the state of reactants and products to predict changes in entropy.
State of Matter
The state of matter refers to the physical form in which a substance exists: solid, liquid, or gas.
Each state has distinct characteristics based on particle arrangement and movement.
  • Solids: Particles are tightly packed with little movement, resulting in low entropy.
  • Liquids: Particles have more freedom to move, increasing entropy.
  • Gases: Particles are far apart and move freely, possessing the highest entropy.
Transitions between these states, such as melting and boiling, involve changes in entropy.
For example, when a solid turns into a liquid, the entropy increases because particles move more freely.
Entropy Change
Entropy change ( ΔS ) measures how the disorder or randomness in a system changes during a process.
For reactions:
  • Negative ΔS: Indicates a decrease in entropy, meaning the system becomes more ordered.
  • Positive ΔS: Indicates an increase in entropy, meaning the system becomes more disordered.
For the reactions given in the original exercise: 1. The formation of NaCl from Na and Cl yields a negative ΔS because gas turns into a solid. 2. The production of NH₃ decreases entropy due to fewer moles of gas in the products. 3. The dissociation of NaCl into ions results in positive ΔS due to increased particle freedom. 4. The melting of NaCl results in positive ΔS as solids transition to liquids, increasing disorder.

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Most popular questions from this chapter

As \(\mathrm{O}_{2}(I)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). At what temperature are solids I and II in equilibrium?

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for each of the three steps in the Ostwald process (see Appendix 4 ). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(C_{\text {graphite }}(s)\) or \(C_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\). a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(S_{\text {rhcmbic }} \longrightarrow\) \(\mathrm{S}_{\text {monoclinic }}\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

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