If the enthalpy of formation and enthalpy of solution of \(\mathrm{HCl}(g\) ) are \(-92.3 \mathrm{~kJ} / \mathrm{mol}\) and \(-75.14\) \(\mathrm{kJ} / \mathrm{mol}\) respectively then find enthalpy of formation of \(\mathrm{Cl}^{-}(a q):\) (a) \(-17.16 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-167.44 \mathrm{~kJ} / \mathrm{mol}\) (c) \(17.16 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

Short Answer

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-167.44 kJ/mol

Step by step solution

01

Understand the Enthalpy Relationships

The enthalpy of formation of HCl(g) is the heat change associated with the formation of one mole of HCl(g) from its elements in their standard states. The enthalpy of solution of HCl(g) is the heat change when one mole of gaseous HCl dissolves in sufficient water to give an infinitely dilute solution. The enthalpy of formation of Cl^-(aq) will be the sum of these two values since first HCl(g) is formed and then it is dissolved to form Cl^-(aq).
02

Write the Enthalpy of Formation for Cl^-(aq)

To find the enthalpy of formation of Cl^-(aq), we need to add the enthalpy of formation of HCl(g) and the enthalpy of solution of HCl(g).
03

Calculate the Enthalpy of Formation of Cl^-(aq)

The enthalpy of formation of Cl^-(aq) can be calculated using the formula: \[ \Delta H_{\text{formation, Cl}^-(aq)} = \Delta H_{\text{formation, HCl}(g)} + \Delta H_{\text{solution, HCl}(g)} \] where \[ \Delta H_{\text{formation, HCl}(g)} = -92.3 \text{ kJ/mol} \] and \[ \Delta H_{\text{solution, HCl}(g)} = -75.14 \text{ kJ/mol} \]. Therefore, \[ \Delta H_{\text{formation, Cl}^-(aq)} = (-92.3 \text{ kJ/mol}) + (-75.14 \text{ kJ/mol}) \].
04

Compute the Final Answer

After performing the addition: \[ \Delta H_{\text{formation, Cl}^-(aq)} = (-92.3 - 75.14) \text{ kJ/mol} \], \[ \Delta H_{\text{formation, Cl}^-(aq)} = -167.44 \text{ kJ/mol} \]. This corresponds to option (b).

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Key Concepts

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

Understanding Physical Chemistry
Physical chemistry bridges the gap between physics and chemistry, delving into the behavior of matter at a molecular and atomic level, and how chemical reactions occur. It applies principles of physics to understand the physical properties of molecules, the forces that act upon them, and the energy changes that accompany chemical reactions.

In the context of the enthalpy of formation and enthalpy of solution, physical chemistry helps us comprehend how energy is transferred during the formation of chemical compounds and the dissolving process in solvents, respectively. It's this deep dive into energy interactions that allows students to solve problems related to thermodynamics and kinetics within the scope of chemistry.

For students preparing for competitive exams like the JEE, a solid understanding of physical chemistry concepts is essential, as they apply to a wide array of scientific processes and applications.
Exploring Enthalpy of Solution
The enthalpy of solution, or the heat of solution, is the heat change associated with the dissolution of a substance in a solvent. When a solute dissolves in a solvent, bonds between solute particles break and new interactions between the solute and solvent are formed.

The overall heat change can be either endothermic or exothermic. An endothermic reaction absorbs heat from the surroundings, resulting in a positive enthalpy change. Conversely, an exothermic reaction releases heat, leading to a negative enthalpy change.

Understanding the enthalpy of solution is vital in predicting the solubility of substances and designing processes where temperature control is crucial. For example, when solving the provided exercise, recognizing that the enthalpy of solution for HCl(g) is negative indicates that the process is exothermic, releasing energy to the surroundings.
Chemical Thermodynamics
Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It deals with the principles of energy conversion and the directionality of chemical processes.

The first law, concerning the conservation of energy, informs us that the total energy of a system and its surroundings is constant. It's this principle that underpins calculations relating to the enthalpy changes in chemical reactions. The second law introduces the concept of entropy, a measure of disorder that predicts the feasibility of a reaction. And the third law provides a reference point for the determination of absolute entropies of substances.

Students often encounter chemical thermodynamics problems when they need to calculate the spontaneity of a reaction, the energy required for a reaction, or energy changes during phase transitions.
JEE Physical Chemistry Problems
Students preparing for the Joint Entrance Examination (JEE), one of the most competitive engineering entrance exams in India, must have a strong grasp on the physical chemistry syllabus. JEE physical chemistry problems involve numerical calculations, conceptual questions, and application-based scenarios that test students' understanding of chemical thermodynamics, kinetics, equilibrium, and other areas.

Solving such problems often requires an integrated approach that combines theoretical knowledge with practical application. It is vital to understand concepts such as the enthalpy of formation and dissolution, as seen in the exercise example provided, since these are frequent topics in JEE physical chemistry sections. A strategic approach of systematic learning and frequent practice with a variety of problems, would equip students with the necessary skills to excel in this portion of the exam.

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Most popular questions from this chapter

Out of molar entropy (I), specific volume (II), heat capacity (III), volume (IV), extensive properties are : (a) I, II (b) I, II, IV (c) II, III (d) III, IV

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