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Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(C\). In each case explain the sign of \(\Delta S^{\circ}\). (a) \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(g) \longrightarrow \mathrm{MgCl}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)\)

Short Answer

Expert verified
(a) Δ³§Â° = 99 ´³/³¾´Ç±ô·°­, positive sign indicates an increase in disorder. (b) Δ³§Â° = -148 ´³/³¾´Ç±ô·°­, negative sign indicates a decrease in disorder. (c) Δ³§Â° = -208 ´³/³¾´Ç±ô·°­, negative sign indicates a decrease in disorder. (d) Δ³§Â° = -220 ´³/³¾´Ç±ô·°­, negative sign indicates a decrease in disorder.

Step by step solution

01

(a) Calculate Δ³§Â° for the reaction Nâ‚‚Hâ‚„(g) + Hâ‚‚(g) → 2 NH₃(g)

First, find the standard molar entropies of all species in Appendix C: ³§Â°(Nâ‚‚Hâ‚„(g)) = 156 ´³/³¾´Ç±ô·°­ ³§Â°(Hâ‚‚(g)) = 131 ´³/³¾´Ç±ô·°­ ³§Â°(NH₃(g)) = 193 ´³/³¾´Ç±ô·°­ Now, use the entropy change equation: Δ³§Â° = Σ³§Â°(products) - Σ³§Â°(reactants) Δ³§Â° = 2 × 193 - (156 + 131) Δ³§Â° = 386 - 287 Δ³§Â° = 99 ´³/³¾´Ç±ô·°­ The sign of Δ³§Â° is positive, which means that the disorder (or randomness) in the system increases during this reaction.
02

(b) Calculate Δ³§Â° for the reaction K(s) + Oâ‚‚(g) → KOâ‚‚(s)

Find the standard molar entropies of all species in Appendix C: ³§Â°(K(s)) = 64 ´³/³¾´Ç±ô·°­ ³§Â°(Oâ‚‚(g)) = 205 ´³/³¾´Ç±ô·°­ ³§Â°(KOâ‚‚(s)) = 121 ´³/³¾´Ç±ô·°­ Use the entropy change equation: Δ³§Â° = 121 - (64 + 205) Δ³§Â° = -148 ´³/³¾´Ç±ô·°­ The sign of Δ³§Â° is negative, which means that the disorder (or randomness) in the system decreases during this reaction.
03

(c) Calculate Δ³§Â° for the reaction Mg(OH)â‚‚(s) + 2 HCl(g) → MgClâ‚‚(s) + 2 Hâ‚‚O(l)

Find the standard molar entropies of all species in Appendix C: ³§Â°(Mg(OH)â‚‚(s)) = 63 ´³/³¾´Ç±ô·°­ ³§Â°(HCl(g)) = 187 ´³/³¾´Ç±ô·°­ ³§Â°(MgClâ‚‚(s)) = 89 ´³/³¾´Ç±ô·°­ ³§Â°(Hâ‚‚O(l)) = 70 ´³/³¾´Ç±ô·°­ Use the entropy change equation: Δ³§Â° = (89 + 2 × 70) - (63 + 2 × 187) Δ³§Â° = 229 - 437 Δ³§Â° = -208 ´³/³¾´Ç±ô·°­ The sign of Δ³§Â° is negative, which means that the disorder (or randomness) in the system decreases during this reaction.
04

(d) Calculate Δ³§Â° for the reaction CO(g) + 2 Hâ‚‚(g) → CH₃OH(g)

Find the standard molar entropies of all species in Appendix C: ³§Â°(CO(g)) = 198 ´³/³¾´Ç±ô·°­ ³§Â°(Hâ‚‚(g)) = 131 ´³/³¾´Ç±ô·°­ ³§Â°(CH₃OH(g)) = 240 ´³/³¾´Ç±ô·°­ Use the entropy change equation: Δ³§Â° = 240 - (198 + 2 × 131) Δ³§Â° = 240 - 460 Δ³§Â° = -220 ´³/³¾´Ç±ô·°­ The sign of Δ³§Â° is negative, which means that the disorder (or randomness) in the system decreases during this reaction.

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Key Concepts

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

Standard Molar Entropy
The concept of standard molar entropy, denoted as ³§Â°, is a fundamental thermodynamic quantity that provides insight into the level of randomness or disorder in a system on a per mole basis. It is measured in joules per mole kelvin (´³/³¾´Ç±ô·°­). Every substance has a characteristic ³§Â° value under standard conditions, which are typically 1 bar of pressure and a specified temperature, commonly 298.15 K (25°C).

When examining a chemical reaction, it's essential to understand that the total entropy of a system plus its surroundings always increases for a spontaneous process. The change in standard molar entropy for a reaction, denoted as ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej (iej iej iej iej). This calculation is possible thanks to the tabulated entropy values found in reference materials such as Appendix C in many textbooks.

In essence, standard molar entropies allow us to quantify how much energy per mole at a given temperature is 'dispersed' or 'spread out'. This knowledge plays a pivotal role in understanding and predicting the direction and spontaneity of chemical reactions.

Entropy and State Change

Entropy is directly related to the physical state of a substance. Gases typically have higher entropies than liquids, which in turn have higher entropies than solids. This is due to the increased freedom of movement for molecules in the gas state compared to those in the liquid or solid states. For example, in a reaction where gas molecules form from a solid or a liquid, we would generally expect an increase in entropy.
Thermodynamic Entropy
When discussing thermodynamic entropy, we're addressing a broader concept that encompasses not just the entropy of a system at the molar level, but the total entropy change throughout a process. Thermodynamic entropy is a state function, which means its value depends only on the current state of the system, not on how the system arrived at that state.

Entropy can also be viewed as a measure of energy dispersal at a specific temperature. An increase in entropy (ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej (iej iej iej iej). This is often associated with changes in heat, where heat absorbed by a system at a given temperature increases the randomness of particles. In contrast, releasing heat generally means a decrease in the system's entropy.

One of the cornerstones of thermodynamics is the second law, which states that the entropy of the universe is always increasing. The implications of this law are significant for understanding chemical reactions, as a reaction is only spontaneous if it leads to an increase in the total entropy of the universe. Practically, this means considering both the system's change in entropy and its surroundings' when determining the spontaneity of a process.
Chemical Reaction Spontaneity
Understanding chemical reaction spontaneity is crucial for predicting whether a reaction will occur without external intervention. Spontaneity in this context does not imply speed—rather, it refers to the natural tendency of a process to progress in one direction under given conditions. At the heart of this tendency is entropy, more specifically, the change in entropy for the system and its surroundings.

A spontaneous reaction is one that results in an overall increase in entropy of the universe (ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej iej (iej iej iej iej), as oppose to non-spontaneous or

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Most popular questions from this chapter

About \(86 \%\) of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio \(\left(T_{\text {high }}-T_{\text {low }}\right) / T_{\text {high }}\), where \(T_{\text {high }}\) is the temperature of the heat going into the engine and \(T_{\text {low }}\) is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of \(700 \mathrm{~K}\) and an exit temperature of \(288 \mathrm{~K} ?\) (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near \(100 \%\) efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure \(5.10\) that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(\mathrm{~s})\) to \(\mathrm{I}_{2}(\mathrm{~g})\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as WebElements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2}\). (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

(a) Give two examples of endothermic processes that are spontaneous. (b) Give an example of a process that is spontaneous at one temperature but nonspontaneous at a different temperature.

Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Gaseous Ar is liquefied at \(80 \mathrm{~K}\). (b) Gaseous \(\mathrm{N}_{2} \mathrm{O}_{4}\) dissociates to form gaseous \(\mathrm{NO}_{2}\). (c) Solid potassium reacts with gaseous \(\mathrm{O}_{2}\) to form solid potassium superoxide, \(\mathrm{KO}_{2}\). (d) Lead bromide precipitates upon mixing \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)\) and \(\mathrm{KBr}(a q)\)

Use Appendix \(C\) to compare the standard entropies at \(25^{\circ} \mathrm{C}\) for the following pairs of substances: (a) \(\mathrm{Sc}(s)\) and \(\mathrm{Sc}(g) ;\) (b) \(\mathrm{NH}_{3}(g)\) and \(\mathrm{NH}_{3}(a q) ;\) (c) \(1 \mathrm{~mol} \mathrm{P}_{4}(g)\) and \(2 \mathrm{~mol}\) \(\mathrm{P}_{2}(\mathrm{~g}) ;\) (d) C(graphite) and \(\mathrm{C}\) (diamond). In each case explain the difference in the entropy values.

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