Ammonium chloride when dissolved in water leads to cooling sensation. The dissolution of \(\mathrm{NH}_{4} \mathrm{Cl}\) at constant temperature is accompanied by (a) increase in entropy (b) decrease in entropy (c) no change in entropy (d) no change in enthalpy

Short Answer

Expert verified
The dissolution of H_{4}Cl in water is accompanied by an increase in entropy (a).

Step by step solution

01

Understanding Entropy

Entropy is a measure of the disorder or randomness in a system. When a solid dissolves in water, the ions or molecules that make up the solid become dispersed in the solvent, increasing the disorder of the system.
02

Analyzing Dissolution Process

When solid ammonium chloride (H_{4}Cl) dissolves in water, the solid's structured lattice breaks apart, and the ammonium (H_{4}^+) and chloride (Cl^-) ions become distributed throughout the water. This process results in a greater number of possible microstates for the ions, thus increasing the entropy of the system.
03

Determining the Change in Entropy

Given the dissolution process increases the randomness of the system, the entropy increases. Therefore, the correct answer is an increase in entropy.

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Key Concepts

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

Dissolution of Salts
When we dissolve salts like ammonium chloride (H_{4}Cl) in water, we observe fascinating physical and chemical changes. This process, which often occurs at a constant temperature and pressure, involves breaking the bonds of the salt's crystal lattice structure and dispersing the individual ions throughout the solvent, in this case, water. During dissolution, the solute particles, which were originally in a fixed and orderly arrangement, become randomly distributed within the solvent, which leads to an increase in the system's disorder. This disorder is a physical representation of a thermodynamic quantity known as entropy. Despite the energy required to break the initial bonds of the salt lattice, the overall randomness or entropy of the system increases because the ions are more dispersed and can occupy many more positions than they could in the solid state.Furthermore, the interesting sensation of cooling when ammonium chloride is dissolved is a result of the endothermic reaction: the system absorbs heat from the surroundings, contributing to the sensory experience.
Thermodynamics
Thermodynamics plays a crucial role in understanding the dissolution of salts, by providing insights into the energy changes and the notions of entropy in such processes. In thermodynamics, entropy is a measure of the number of specific ways in which a system may be arranged, often thought of as the degree of disorder or randomness. The Second Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases.In the case of dissolving ammonium chloride in water, we see an application of this law. The initial lattice structure of the salt, which is highly ordered, transitions to a more disordered state as the ions disperse in the water. This change suggests an increase in entropy, aligning with what the Second Law predicts for spontaneous processes. To fully comprehend such an occurrence within the realm of thermodynamics, one must consider both the system and its surroundings, including the heat transfer during the dissolution.
Physical Chemistry Problems
Solving physical chemistry problems like the dissolution of salts calls for a thorough understanding of concepts such as enthalpy, entropy, and the Gibbs free energy. These problems aren't meant to be only academic exercises; they offer practical insights into the behavior of substances in real-world applications, extending from industrial processes to biological systems. In an educational setting, enhancing one's problem-solving skills requires recognizing patterns, applying fundamental principles, and evaluating the implications of each step. For instance, identifying an increase in entropy as per our ammonium chloride example is not merely about pairing facts. It involves grasping the underlying principles of thermodynamic stability and predicting the direction of chemical processes. These problems encourage students to think critically about molecular interactions and the laws governing energy changes, ultimately forming a more comprehensive understanding of our world at the molecular level.

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Most popular questions from this chapter

When an ideal diatomic gas is a heated at a constant pressure, the fraction of heat energy supplied which increase the internal energy of the gas is (a) \(\frac{2}{5}\) (b) \(\frac{3}{5}\) (c) \(\frac{5}{7}\) (d) \(\frac{3}{7}\)

One mole of an ideal gas undergoes the following cyclic process: (i) Isochoric heating from \(\left(P_{1}, V_{1}, T_{1}\right)\) to double temperature. (ii) Isobaric expansion to double volume. (iii) Linear expansion (on \(P-V\) curve) to \(\left(P_{1}, 8 V_{1}\right) .\) (iv) Isobaric compression to initial state. If \(T_{1}=300 \mathrm{~K}\), the magnitude of net work done by the gas in the cyclic process is (a) \(2400 \mathrm{cal}\) (b) \(1200 \mathrm{cal}\) (c) \(4800 \mathrm{cal}\) (d) \(3600 \mathrm{cal}\)

The work involved \((w)\) in an isothermal expansion of \(n\) moles of an ideal gas from an initial pressure of ' \(P\) ' atm to final pressure of 1 atm in number of steps such that in every step the constant external pressure exactly 1 atm less than the initial pressure of gas is maintained, is given as (a) \(-n R T \sum_{i=1}^{i=P-1}\left(\frac{1}{P+1-i}\right)\) (b) \(-n R T \sum_{i=1}^{i=P}\left(\frac{1}{P+1-i}\right)\) (c) \(-n R T \sum_{i=1}^{i=P}\left(\frac{i}{P+1-i}\right)\) (d) \(-n R T \sum_{i=1}^{i=P-1}\left(\frac{i}{P+1-i}\right)\)

Given the following entropy values (in \(\mathrm{J} / \mathrm{K}-\mathrm{mol}\) ) at \(298 \mathrm{~K}\) and 1 atm \(\mathrm{H}_{2}(\mathrm{~g})\) \(=130.6, \mathrm{Cl}_{2}(\mathrm{~g})=223.0\) and \(\mathrm{HCl}(\mathrm{g})=186.7\) The entropy change (in J/K-mol) for the reaction: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})\), is (a) \(+540.3\) (b) \(+727.0\) (c) \(-166.9\) (d) \(+19.8\)

What is the entropy change when \(3.6 \mathrm{~g}\) of liquid water is completely converted into vapours at \(373 \mathrm{~K} ?\) The molar heat of vaporization is \(40.85 \mathrm{~kJ} / \mathrm{mol}\). (a) \(218.9 \mathrm{~J} / \mathrm{K}\) (b) \(2.189 \mathrm{~J} / \mathrm{K}\) (c) \(21.89 \mathrm{~J} / \mathrm{K}\) (d) \(0.2189 \mathrm{~J} / \mathrm{K}\)

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