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The direct conversion of \(\mathrm{A}\) to \(\mathrm{B}\) is difficult, hence it is carried out by the following path: Given \(\Delta \mathrm{S}(\mathrm{A} \longrightarrow \mathrm{C})=50 \mathrm{e} . \mathrm{u} .\) \(\Delta \mathrm{S}(\mathrm{C} \longrightarrow \mathrm{D})=30 \mathrm{e} . \mathrm{u} .\) \(\Delta \mathrm{S}(\mathrm{B} \longrightarrow \mathrm{D})=20 \mathrm{e} . \mathrm{u} .\) where e.u. is entropy unit then \(\Delta \mathrm{S}(\mathrm{A} \longrightarrow \mathrm{B})\) is: (a) \(+100\) e.u. (b) \(+60\) e.u. (c) \(-100\) e.u. (d) \(-60\) e.u.

Short Answer

Expert verified
(b) +60 e.u.

Step by step solution

01

Analyze the statement

We are given the entropy changes for converting A to C, C to D, and B to D. We need to find the entropy change for converting A directly to B.
02

Use Hess's Law

According to Hess's Law, the total entropy change for a reaction that occurs in multiple steps is the sum of the entropy changes for each individual step. Therefore, we can write the equation:\[\Delta S(\text{A} \to \text{B}) = \Delta S(\text{A} \to \text{C}) + \Delta S(\text{C} \to \text{D}) - \Delta S(\text{B} \to \text{D})\]
03

Plug in the values

Substitute the given values into the equation:\[\Delta S(\text{A} \to \text{B}) = 50\, \text{e.u.} + 30\, \text{e.u.} - 20\, \text{e.u.}\]
04

Calculate the result

Add the values together:\[\Delta S(\text{A} \to \text{B}) = 50 + 30 - 20 = 60 \, \text{e.u.}\]
05

Determine the sign and option

The calculated \( \Delta S \) is positive, so the correct answer is (b) \(+60\) e.u.

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Key Concepts

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

Entropy and Its Changes
Entropy is a fundamental concept in thermodynamics related to the degree of randomness or disorder in a system. It helps us understand how energy disperses in a physical system. When we say there is an entropy change, we are essentially talking about the shift in disorder from an initial state to a final state. For reactions, it's important to know that the total entropy change is the sum of the entropy changes in all steps involved.

In the context of the exercise provided, we use these entropy changes to better understand the process from A to B through multiple steps, and how these steps affect the overall disorder of the system. Entropy changes are often measured in entropy units (e.u.), which help quantify these changes numerically.
  • Increase in entropy indicates a move towards greater disorder.
  • Decrease in entropy suggests more ordered arrangements.
Understanding how entropy changes in reactions is key to predicting how spontaneous a process is, as processes that lead to an increase in total entropy are more likely to be spontaneous.
Basics of Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. At the core of thermodynamics are four essential laws that govern these relationships and help us predict how systems behave. In essence, it is about understanding how energy is converted and exchanged within physical systems.

In chemical reactions like the one found in our original exercise, applying thermodynamics helps us calculate vital properties such as entropy changes. Hess's Law, for instance, is an application of the first law of thermodynamics, which makes it possible to determine the total energy change by using step processes.
  • First law: Energy cannot be created or destroyed, only converted from one form to another.
  • Second law: Total entropy of an isolated system can only increase over time.
  • Hess's Law: The enthalpy change in a chemical reaction is independent of the pathway taken.
By understanding these principles, students can better grasp how reactions can turn one set of substances into another. Monitoring how energy changes and affects systems over different processes helps us predict reaction feasibility.
The Role of Multi-Step Reactions
Multi-step reactions are essential because not all reactions occur smoothly from reactants to products in one go. Instead, many reactions involve intermediates and transition states, resulting in a series of smaller steps. In our exercise, going from A to B is simplified via the path through C and D.

Hess's Law is particularly useful for multi-step reactions. It tells us that the total change in a state function like entropy is the same whether the reaction is completed in a single step or multiple steps. This is crucial when a direct pathway is unknown or difficult, and alternative pathways can be used to calculate desired changes accurately.
  • Each step contributes to the overall change in the process.
  • By adding up the changes from each step, we can find the total change for the reaction.
Understanding and calculating changes in multi-step reactions require recognizing each individual step's contribution and compiling them towards managing bigger, complex reactions with confidence.

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Most popular questions from this chapter

For the reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+3 \mathrm{D}(\mathrm{g})\) The value of \(\Delta \mathrm{H}\) at \(27^{\circ} \mathrm{C}\) is \(19.0 \mathrm{kcal}\). The value of \(\Delta \mathrm{E}\) for the reaction would be (given \(\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(20.8 \mathrm{kcal}\) (b) \(19.8 \mathrm{kcal}\) (c) \(18.8 \mathrm{kcal}\) (d) \(17.8 \mathrm{kcal}\)

The Gibbs free energy is defined as: (a) \(\mathrm{G}=\mathrm{H}-\mathrm{T} \cdot \mathrm{S}\) (b) \(\mathrm{G}=\mathrm{H}+\mathrm{T} . \mathrm{S}\) (c) \(\mathrm{G}=\mathrm{E}-\mathrm{T} . \mathrm{S}\) (d) \(\mathrm{G}=\mathrm{E}+\mathrm{T} \cdot \mathrm{S}\)

\(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \longrightarrow \mathrm{HS}(\mathrm{g})+\mathrm{H}(\mathrm{g}), \Delta \mathrm{H}^{\circ}=\mathrm{x}_{1}\), \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\right]=\mathrm{x}_{2}, \Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{s}}[\mathrm{H}(\mathrm{g})]=\mathrm{x}_{3}\) hence, \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}(\mathrm{HS})\) is: (a) \(x_{1}+x_{2}-x_{3}\) (b) \(\mathrm{x}_{3}-\mathrm{x}_{1}-\mathrm{x}_{2}\) (c) \(\mathrm{x}_{1}-\mathrm{x}_{2}-\mathrm{x}_{3}\) (d) \(x_{3}-x_{1}+x_{2}\)

Which of the following reaction defines \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\) ? (a) \(\mathrm{C}\) (diamond) \(+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) (b) \(\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{HF}(\mathrm{g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) (d) \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\)

If the value of \(\Delta \mathrm{H}\) in a reaction is positive, then the reaction is called: (a) Exothermic (b) Endothermic (c) Polymorphic (d) Polytropic

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