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Formation of a solution from two components can be considered as : (i) Pure solvent \(\rightarrow\) separated solvent molecules, \(\Delta H_{1}\) (ii) Pure solute \(\rightarrow\) separated solute molecules, \(\Delta H_{2}\) (iii) separated solvent and solute molecules \(\rightarrow\) solution, \(\Delta H_{3}\) Solution so formed will be ideal if : (a) \(\Delta H_{\text {Soln }}=\Delta H_{1}+\Delta H_{2}+\Delta H_{3}\) (b) \(\Delta H_{\text {Soln }}=\Delta H_{1}+\Delta H_{2}-\Delta H_{3}\) (c) \(\Delta H_{\text {Soln }}=\Delta H_{1}-\Delta H_{2}-\Delta H_{3}\) (d) \(\Delta H_{\text {Soln }}=\Delta H_{3}-\Delta H_{1}-\Delta H_{2}\)

Short Answer

Expert verified
The correct formula for the heat of forming an ideal solution is \( \bigtriangleup H_{soln} = \bigtriangleup H_{1} + \bigtriangleup H_{2} - \bigtriangleup H_{3} \) (option b).

Step by step solution

01

Understanding Heat Changes

When forming a solution, there are heat changes associated with breaking up the pure solvent into separated solvent molecules (\( \bigtriangleup H_1 \) ), breaking up the pure solute into separated solute molecules (\( \bigtriangleup H_2 \) ), and mixing the separated solvent and solute molecules to form a solution (\( \bigtriangleup H_3 \) ). An ideal solution does not release or absorb heat on formation; therefore, the heat absorbed or released in later steps should cancel the effects of earlier steps.
02

Analyzing each option

Option (a) implies that the heat of the solution is simply the sum of all three changes. This would not yield an ideal solution unless all three changes were zero. Option (c) involves subtracting the heat of both solute and solvent separation from the heat of mixing, which doesn't fit with the concept of ideal solution formation. Option (d) reverses the sign of the heat of solute and solvent separation which is also incorrect. The correct relationship would balance out the absorption and release of heat to result in no net heat change for an ideal solution.
03

Identifying the Correct Formula

For an ideal solution, the heat absorbed in separating the solute and solvent molecules (\( \bigtriangleup H_1 \) and \( \bigtriangleup H_2 \) ) should be equal to the heat released when they mix to form a solution (\( \bigtriangleup H_3 \) ). This means the heat changes due to separation should cancel the heat change due to mixing, leading to \( \bigtriangleup H_{soln} = \bigtriangleup H_{3} - (\bigtriangleup H_{1} + \bigtriangleup H_{2}) \) as the correct answer.

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Key Concepts

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

Enthalpy change
Understanding the concept of enthalpy change is crucial when studying the formation of solutions. Enthalpy change, denoted by \( \Delta H \), represents the heat exchanged in a process at constant pressure. It can be though of as the 'heat content' of a system. In the context of dissolution, three types of enthalpy changes are considered: \( \Delta H_{1} \) for separating solvent molecules, \( \Delta H_{2} \) for the separation of solute molecules, and \( \Delta H_{3} \) for mixing solvent and solute to form a solution.

For an ideal solution, the heat absorbed or released through these stages should be balanced, leading to no net enthalpy change. That's to say, the heat required to break apart solute and solvent (\( \Delta H_{1} \) and \( \Delta H_{2} \) respectively) is compensated by the heat released when they form the solution (\( \Delta H_{3} \) ). Therefore, the enthalpy of the ideal solution (\( \Delta H_{\text{Soln}} \) ) is the difference between the heat of mixing and the sum of heats required to separate the solute and solvent. It's a beautiful dance of energy, balancing precisely to maintain equilibrium.
Solution chemistry
In solution chemistry, understanding the processes involved in solution formation is essential. When a solute dissolves in a solvent, intermolecular forces come into play. Initially, these forces exist within the pure solute and solvent. When a solution forms, these forces need to be broken down, which requires energy, before new interactions between solute and solvent molecules can be established.

In an ideal solution, the interactions between solute molecules, solvent molecules, and solute-solvent mixtures are similar in strength. Therefore, the energy required to separate molecules (endothermic process) is offset by the energy released when new interactions are formed (exothermic process). This unique balance results in no net energy change, highlighting why an ideal solution maintains a perfect enthalpy balance. It's a clear example of how energy conservation plays out on a molecular level.
Physical Chemistry for JEE
For students preparing for competitive exams like the Joint Entrance Examination (JEE), comprehending physical chemistry topics such as thermodynamics and solution chemistry is fundamental. These principles underpin various real-world applications, from industrial processes to environmental science.

The topic of ideal solution formation directly links to thermodynamics, enthalpy changes, and solution chemistry – essential areas in the JEE syllabus. By mastering the principles that govern how and why solutions form, with emphasis on enthalpic considerations, students can tackle complex problems with confidence. Practicing such problems doesn't just prepare you for exams; it also sharpens your analytical skills, enabling a deeper understanding of how particles interact at a fundamental level.
Thermodynamics
Thermodynamics is a branch of physics and physical chemistry that deals with the relationships between various forms of energy and heat involved in physical and chemical processes. It's centered around the principles that determine the direction in which processes such as the formation of solutions will occur. For an ideal solution where no net enthalpy change is observed, the first law of thermodynamics — conservation of energy — is elegantly demonstrated.

The formation of an ideal solution is akin to a thermodynamic balance beam, with the energy inputs and outputs canceling each other out. This no-net-change in enthalpy implies that the solution is forming under thermodynamic perfection, where the disorder (or entropy) of the system increases without the need for additional energy input or output, aligning beautifully with the second law of thermodynamics which dictates spontaneous increase in entropy.

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Most popular questions from this chapter

A solution containing \(4.0 \mathrm{~g}\) of PVC in 2 litre of dioxane (industrial solvent) was found to have an osmotic pressure \(3.0 \times 10^{-4}\) atm at \(27^{\circ} \mathrm{C}\). The molar mass of the polymer \((\mathrm{g} / \mathrm{mol})\) will be : (a) \(1.6 \times 10^{4}\) (b) \(1.6 \times 10^{5}\) (c) \(1.6 \times 10^{3}\) (d) \(1.6 \times 10^{2}\)

An aqueous solution of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) having a concentration of \(34.2\) gram/litre has an osmotic pressure of \(2.38\) atmospheres at \(17^{\circ} \mathrm{C}\). For an aqueous solution of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) to be isotonic with this solution, its concentration should be : (a) \(34.2\) gram per litre (b) \(17.1\) gram per litre (c) \(18.0\) gram per litre (d) \(36.0\) gram per litre

Equal weight of a solute are dissolved in equal weight of two solvents \(A\) and \(B\) and formed very dilute solution. The relative lowering of vapour pressure for the solution \(B\) has twice the relative lowering of vapour pressure for the solution \(A\). If \(M_{A}\) and \(M_{B}\) are the molecular weights of solvents \(A\) and \(B\) respectively, then: (a) \(M_{A}=M_{B}\) (b) \(M_{B}=2 \times M_{A}\) (c) \(M_{A}=4 M_{B}\) (d) \(M_{A}=2 M_{B}\)

Phenol associates in benzene to a certain extent in dimerisation reaction. A solution containing \(0.02 \mathrm{~kg}\) of phenol in \(1.0 \mathrm{~kg}\) of benzene has its freezing point depressed \(0.69 \mathrm{~K}\). Hence, degree of association of phenol dimerized will be : \(\left[K_{f}\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]\) (a) \(0.63\) (b) \(0.73\) (c) \(0.83\) (d) \(0.93\)

The osmotic pressures of equimolar solutions of urea, \(\mathrm{BaCl}_{2}\) and \(\mathrm{AlCl}_{3}\) will be in the order : (a) \(\mathrm{AlCl}_{3}>\mathrm{BaCl}_{2}>\) urea (b) \(\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}>\) urea (c) urea \(>\mathrm{BaCl}_{2}>\mathrm{AlCl}_{3}\) (d) \(\mathrm{BaCl}_{2}>\) urea \(>\mathrm{AlCl}_{3}\)

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