Using the relevant \(S^{\circ}\) values listed in Appendix G, calculate \(\Delta S^{\circ}\) for the following changes: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{N}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)\)

Short Answer

Expert verified
Calculate the \(\Delta S^{\circ}\) for each reaction using the standard entropy values from the Appendix, making sure to account for the stoichiometry in each reaction.

Step by step solution

01

Determine \(\Delta S^{\circ}\) for Reaction (a)

To find the change in standard entropy \(\Delta S^{\circ}\) for reaction (a), subtract the sum of standard entropy values \(S^{\circ}\) of reactants from the sum of \(S^{\circ}\) of products. Use \(S^{\circ}\) values for \(\mathrm{N}_{2}\), \(\mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\): \[\Delta S^{\circ}_{(a)} = [2S^{\circ}(\mathrm{NH}_{3}(g))] - [S^{\circ}(\mathrm{N}_{2}(g)) + 3S^{\circ}(\mathrm{H}_{2}(g))]\]
02

Determine \(\Delta S^{\circ}\) for Reaction (b)

Similarly, for reaction (b), subtract the sum of \(S^{\circ}\) of reactants from the \(S^{\circ}\) of the product. Use \(S^{\circ}\) values for \(\mathrm{N}_{2}\), \(\mathrm{O}_{2}\), and \(\mathrm{N}_{2} \mathrm{O}_{5}\): \[\Delta S^{\circ}_{(b)} = S^{\circ}(\mathrm{N}_{2} \mathrm{O}_{5}(g)) - [S^{\circ}(\mathrm{N}_{2}(g)) + \frac{5}{2}S^{\circ}(\mathrm{O}_{2}(g))]\]

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Key Concepts

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

Entropy
Entropy is a crucial concept in thermodynamics, often represented with the symbol 'S'. It's a measure of the disorder or randomness of particles in a system. The higher the entropy, the greater the disorder and the higher the number of possible states for a system. Imagine a room filled with children holding balloons; if they're all sitting quietly, the room has low entropy. Once they start running around randomly, the entropy increases.

In the context of chemical reactions, the standard entropy change, denoted as \(\Delta S^\circ\), indicates how the entropy of the system changes as the reaction proceeds from reactants to products. If \(\Delta S^\circ > 0\), the products have greater disorder than the reactants, signaling an increase in entropy during the reaction. Conversely, if \(\Delta S^\circ < 0\), the reaction leads to a decrease in entropy.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and the forms of energy transformation in a system. The laws of thermodynamics govern how and why energy is transferred within the physical world. Particularly in chemistry, these laws help us understand how energy changes during chemical reactions and how these reactions can do work.

For example, the second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. Applied to chemical reactions, this law implies that, while individual steps of a reaction may decrease entropy, the overall reaction must increase the entropy of the universe, which includes the system and surrounding environment.
Chemical Reactions
Chemical reactions involve the rearrangement of atoms to transform reactants into products. These reactions are driven by various factors, including enthalpy changes, entropy changes, and the resulting Gibb's free energy change. It's the interplay of these factors that determines whether a reaction is spontaneous or non-spontaneous.

During a reaction, bonds between atoms are broken and formed, which requires or releases energy, leading to enthalpy changes. Meanwhile, entropy indicates how the energy distribution among particles alters, impacting the system's disorderliness. Understanding how these factors interact provides insight into the behavior of chemical systems and the likelihood of a reaction occurring under given conditions.
Gibbs Free Energy
Gibbs free energy, symbolized as 'G', is a thermodynamic quantity representing the maximum reversible work that may be performed by a thermodynamic system at constant temperature and pressure. It combines enthalpy (H), temperature (T), and entropy (S) in a single equation: \( G = H - TS \). The change in Gibbs free energy, \(\Delta G\), during a process is a crucial indicator of a reaction's spontaneity.

If \(\Delta G < 0\), the reaction is exergonic and will proceed spontaneously. Conversely, if \(\Delta G > 0\), the reaction is endergonic and non-spontaneous without additional energy input. A \(\Delta G = 0\) indicates the system is in equilibrium. By calculating the standard entropy change \(\Delta S^\circ\) in a reaction, one can assess the entropy contribution to \(\Delta G\) and better predict reaction behavior.

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Most popular questions from this chapter

Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and \(25^{\circ} \mathrm{C}\). Identify each as either spontaneous or nonspontaneous at these conditions. (a) \(\mathrm{C}(s, \text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (b) \(\mathrm{O}_{2}(g)+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)\) (c) \(2 \mathrm{Cu}(s)+\mathrm{S}(g) \longrightarrow \mathrm{Cu}_{2} \mathrm{S}(s)\) (d) \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)\) (e) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)\) (f) \(\operatorname{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CaSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Indicate whether the following processes are spontaneous or nonspontaneous. (a) Liquid water freezing at a temperature below its freezing point (b) Liquid water freezing at a temperature above its freezing point (c) The combustion of gasoline (d) A ball thrown into the air (e) A raindrop falling to the ground (f) Iron rusting in a moist atmosphere

From the following information, determine \(\Delta S^{\circ}\) for the following: \(\mathrm{N}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g) \quad \Delta S^{\circ}=?\) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta S^{\circ}=24.8 \mathrm{J} / \mathrm{K}\) \(\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g) \quad \Delta S^{\circ}=115.0 \mathrm{J} / \mathrm{K}\) \(\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \quad \Delta S^{\circ}=117.0 \mathrm{J} / \mathrm{K}\)

Use the standard entropy data in Appendix G to determine the change in entropy for each of the following reactions. All the processes occur at the standard conditions and \(25^{\circ} \mathrm{C}\). (a) \(\mathrm{MnO}_{2}(s) \longrightarrow \mathrm{Mn}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(l) \longrightarrow 2 \mathrm{HBr}(g)\) (c) \(\operatorname{Cu}(s)+S(g) \rightarrow \operatorname{CuS}(s)\) (d) \(2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{C}(s, \text { graphite })+2 \mathrm{H}_{2} \mathrm{O}(g)\) (f) \(\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g)\)

Explain what happens as a reaction starts with \(\Delta G<0\) (negative) and reaches the point where \(\Delta G=0\).

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