Calculate \(\Delta S\) for following process: $$ \underset{\text { at } 100 \mathrm{~K}}{X(s)} \longrightarrow \underset{\text { at } 200 \mathrm{~K}}{X(l)} $$ Given : Melting point of \(X_{(s)}=100 \mathrm{~K} ; \Delta H_{\text {Fusion }}=20 \mathrm{~kJ} / \mathrm{mol} ; C_{p, m}(X, l)=10 \mathrm{~J} / \mathrm{mol} \mathrm{K}\) (a) \(26.93 \mathrm{~J} / \mathrm{K}\) (b) \(206.93 \mathrm{~J} / \mathrm{K}\) (c) \(203 \mathrm{~J} / \mathrm{K}\) (d) \(206.93 \mathrm{~kJ} / \mathrm{K}\)

Short Answer

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\(206.93 \text{ J/K}\)

Step by step solution

01

Calculate Entropy Change Due to Fusion

Use the formula for entropy change during phase change, \(\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_{fusion}}\). Given the enthalpy of fusion, \(\Delta H_{fusion} = 20 kJ/mol\), and the melting point, \(T_{fusion} = 100 K\), calculate \(\Delta S_{fusion}\).
02

Calculate Entropy Change Due to Temperature Increase

Use the formula for entropy change due to temperature increase for a substance at constant pressure, \(\Delta S = \int_{T_{1}}^{T_{2}} \frac{C_{p,m}}{T} dT\). Given the molar heat capacity, \(C_{p,m}(X,l) = 10 J/mol K\), integrate between the initial temperature, \(T_{1} = 100 K\), and final temperature, \(T_{2} = 200 K\).
03

Calculate Total Entropy Change

Sum the entropy changes calculated in steps 1 and 2 to get the total entropy change for the process, \(\Delta S_{total} = \Delta S_{fusion} + \Delta S_{temperature}\).
04

Convert Units if Necessary

Make sure the entropy changes from steps 1 and 2 are in the same units before adding them. If one is in kJ and the other in J, convert one so both are in J/K.

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Key Concepts

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

Physical Chemistry
Physical chemistry is a branch of chemistry focused on understanding the physical properties of molecules, the forces that act upon them, and the energy changes associated with chemical reactions. It bridges the gap between the microscopic world and the observable macroscopic world. Entropy, represented by the symbol \( S \), is a fundamental concept in physical chemistry that measures the disorder or randomness of a system. The calculation of entropy changes, like the melting of a solid into a liquid, involves understanding how heat and temperature affect the state of matter at a molecular level.

In the exercise, we addressed entropy through a process that involves both a phase change and a temperature increase, both of which are key concepts in physical chemistry. These concepts are essential for predicting the behavior of substances in various conditions and for designing processes in both industrial and laboratory settings.
Phase Transition
A phase transition is a transformation of a substance from one state of matter to another, such as from solid to liquid or from liquid to gas. Each phase has distinct physical properties and requires energy changes for a substance to transition between them. The enthalpy of fusion is associated with the melting of a solid to a liquid, which occurs at a specific temperature called the melting point.

During the phase transition of substance X in our exercise, energy is added to overcome the forces holding the solid together, allowing it to become a liquid at its melting point. This energy addition results in an entropy change, as the molecules in the solid structure gain freedom of movement in the liquid phase, increasing disorder.
Enthalpy of Fusion
The enthalpy of fusion, often symbolized by \( \Delta H_{fusion} \), is the heat required to change a substance from solid to liquid at its melting point without changing its temperature. This is an intrinsic property of the material and is expressed in units of energy per amount of substance, such as joules per mole. It's vital for calculating the entropy change during melting.
For our exercise, the given enthalpy of fusion for substance X is \( 20 \text{ kJ/mol} \). This value is used to calculate the entropy change due to the fusion part of the process. Understanding the enthalpy of fusion is crucial when studying energy requirements for phase transitions in scenarios ranging from industrial freezing and melting to natural phenomena like the melting of ice.
Molar Heat Capacity
Molar heat capacity \( C_{p,m} \) is a physical property that indicates how much heat energy is needed to raise the temperature of one mole of a substance by one degree Kelvin at constant pressure. In the context of the given problem, we consider the molar heat capacity of liquid X to calculate the entropy change associated with heating the liquid from its melting point to a higher temperature.

With the molar heat capacity provided (\( 10 \text{ J/mol K} \)), we can integrate over the temperature range to find the change in entropy due to the temperature increase. This concept also helps in understanding the heat flow in chemical processes, which is important for tasks like designing thermal systems or evaluating reaction conditions.
Temperature Increase
Temperature increase in a substance corresponds to the rise in the average kinetic energy of its molecules. This process involves transferring heat to the substance, leading to changes in its physical properties, and can affect the entropy of the system. In our example, substance X is heated from its melting point (100 K) to a higher temperature (200 K), causing the liquid molecules to move more vigorously.

This increased molecular motion contributes to a higher degree of disorder and results in a positive change in the system's entropy. The computation of this entropy change involves integrating the molar heat capacity of the substance over the temperature range, as demonstrated in the provided solution steps. Temperature increase is a key factor in many scientific and engineering contexts, from weather patterns to the efficiency of engines and refrigerators.

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Most popular questions from this chapter

Mechanical work is specially important in systems that contain. (a) gas-liquid (b) liquid-liquid (c) solid-solid (d) amalgam

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One mole of an ideal gas undergoes a change of state \((2.0 \mathrm{~atm}, 3.0 \mathrm{~L})\) to \((2.0 \mathrm{~atm}, 7.0 \mathrm{~L})\) with a change in internal energy \((\Delta U)=30 \mathrm{~L}-\mathrm{atm} .\) The change in enthalpy \((\Delta H)\) of the process in L-atm : (a) 22 (b) 38 (c) 25 (d) None of thése

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