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The lattice energy of solid \(\mathrm{NaCl}\) is 180 kcal/mol. The dissolution of the solid in water, in the form of ions is endothermic to the extent of \(1 \mathrm{kcal} / \mathrm{mol}\). If the solvation energies of \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions are in the ratio \(6: 5\), what is the enthalpy of hydration of sodium ion? (a) \(-85.6 \mathrm{kcal} / \mathrm{mol}\) (b) \(-97.6 \mathrm{kcal} / \mathrm{mol}\) (c) \(82.6 \mathrm{kcal} / \mathrm{mol}\) (d) \(100 \mathrm{kcal} / \mathrm{mol}\)

Short Answer

Expert verified
-97.6 kcal/mol

Step by step solution

01

Understand the Lattice Energy and Dissolution Process

Lattice energy is the energy released when ions bond to form a solid. The given lattice energy for NaCl is 180 kcal/mol. The dissolution process of NaCl in water is endothermic, and it absorbs 1 kcal/mol. During dissolution, energy is absorbed to break the ionic bonds (endothermic), and energy is released when water molecules surround the ions (exothermic).
02

Calculate Total Hydration Energy of Both Ions

Since dissolution is endothermic by 1 kcal/mol, but the ionic solid had released 180 kcal/mol upon formation, the total energy released during hydration of Na+ and Cl- ions is the sum of lattice energy and dissolution energy. Total hydration energy = Lattice energy + Dissolution energy = 180 kcal/mol + 1 kcal/mol = 181 kcal/mol.
03

Calculate Hydration Energy of Sodium and Chloride Ions

The given ratio of solvation energies of Na+ to Cl- is 6:5, which means the hydration energy for Na+ is 6x and for Cl- is 5x. Therefore, the sum of their energies (6x + 5x) is equal to the total hydration energy. Hence, 11x = 181 kcal/mol. Solve for x to find the individual hydration energies.
04

Determine the Enthalpy of Hydration for Sodium Ion

By solving the equation 11x = 181 kcal/mol, we find that x = 181 kcal/mol / 11. Once we have x, we can calculate the hydration energy for Na+, which is 6 times x. The enthalpy of hydration for the sodium ion is thus 6x.

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Key Concepts

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

Lattice Energy
Lattice energy plays a critical role in understanding the thermodynamics of ionic compounds such as salt (NaCl). By definition, lattice energy is the amount of energy released when oppositely charged ions in the gaseous state come together to form a solid ionic compound. In simpler terms, think of it as the 'glue' that holds the ions together in the crystal lattice structure. A high lattice energy indicates that the compound is very stable and it takes a considerable amount of energy to break those ionic bonds during dissolution.

For instance, in our exercise, the lattice energy for NaCl is given as 180 kcal/mol. This high value signifies that sodium chloride's ionic bonds are quite strong. Consequently, when NaCl dissolves in water, energy is required to overcome this lattice energy. Understanding this concept is essential as it serves as a basis for calculating other thermodynamic quantities, such as the enthalpy of hydration of individual ions.

Factors Influencing Lattice Energy

Several factors affect lattice energy, including the size of the ions and the charges they carry. The smaller and more highly charged the ions, the greater the lattice energy as they can get closer and exert stronger electrostatic attractions to one another.
Solvation Energy
When a solute dissolves in a solvent, solvation occurs, in which solvent molecules surround each solute particle, stabilizing it in solution. The energy change associated with this process is known as solvation energy. In the case of ionic compounds like NaCl dissolving in water, the solvation energy is specific to hydration, where water is the solvent, and therefore it is often referred to as hydration energy.

Solvation energy can be exothermic or endothermic depending on whether the process releases or absorbs energy respectively. In our givens, the ratio of solvation energies for Na+ and Cl- ions is 6:5. This ratio indicates that different ions have varying affinities for the solvent and their individual enthalpies of hydration can be calculated based on this ratio.

Implications of Solvation Energy

The solvation process significantly affects the solubility of compounds. The hydration energy can help overcome the lattice energy of an ionic solid, leading to its dissolution. The degree to which solvation energy can compensate for the lattice energy directly influences the solubility of that substance in a particular solvent.
Dissolution Process
Understanding the dissolution process is fundamental to predicting the behavior of solids in liquids. The dissolution process can be viewed as a two-step mechanism: first, the solid's lattice structure is broken down (an endothermic process requiring energy); second, the individual ions are stabilized by the solvent in a process known as solvation or hydration (often exothermic as it releases energy).

In the exercise, we're informed that the dissolution of NaCl is endothermic to the extent of 1 kcal/mol – this is the energy required to separate the ionic lattice. However, the overall process of dissolution can be either endothermic or exothermic, depending on the balance between the lattice energy and the solvation energy. To find the enthalpy of hydration for sodium ions, we analyze the total energy changes including this dissolution process.

Practical Importance

This balance influences the solubility of an ionic compound in a solvent. In industrial applications and daily life, understanding and manipulating the dissolution process enables us to control the concentration and rate at which a compound will dissolve, which is crucial in fields such as pharmaceuticals, cooking, and chemical manufacturing.

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Most popular questions from this chapter

The molar heat capacities of \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are in the ratio \(1: 2: 3 .\) The enthalpy change for the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) at temperature \(T_{1}\) is \(\Delta H_{1} .\) Assuming that the heat capacities do not change with temperature, the enthalpy change, \(\Delta H_{2}\), at temperature, \(T_{2}\left(T_{2}>T_{1}\right)\) will be (a) greater than \(\Delta H_{1}\) (b) equal to \(\Delta H_{1}\) (c) less than \(\Delta H_{1}\) (d) greater or less than \(\Delta H_{1}\), depending on the values of \(T_{2}\) and \(T_{1}\).

Which of the following salts shall cause more cooling when one mole of the salt is dissolved in the same amount of water? (Integral heat of solution at \(298 \mathrm{~K}\) is given for each solute.) (a) \(\mathrm{KNO}_{3} ; \Delta H=35.4 \mathrm{~kJ} / \mathrm{mol}\) (b) \(\mathrm{NaCl} ; \Delta H=5.35 \mathrm{~kJ} / \mathrm{mol}\) (c) \(\mathrm{KOH} ; \Delta H=-55.6 \mathrm{~kJ} / \mathrm{mol}\) (d) \(\mathrm{HBr} ; \Delta H=-83.3 \mathrm{~kJ} / \mathrm{mol}\)

The enthalpy of formation of ammonia gas is \(-46.0 \mathrm{~kJ} / \mathrm{mol}\). The enthalpy change for the reaction: \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\) is (a) \(46.0 \mathrm{~kJ}\) (b) \(92.0 \mathrm{~kJ}\) (c) \(23.0 \mathrm{~kJ}\) (d) \(-92.0 \mathrm{~kJ}\)

Calculate the standard free energy of the reaction at \(27^{\circ} \mathrm{C}\) for the combustion of methane using the given data: \(\mathrm{CH}_{4}(\mathrm{~g})\) \(+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) $$\begin{array}{lcccc} \hline \text { Species } & \mathbf{C H}_{4}(\mathrm{~g}) & \mathrm{O}_{2}(\mathrm{~g}) & \mathrm{CO}_{2}(\mathrm{~g}) & \mathbf{H}_{2} \mathrm{O}(\mathrm{l}) \\ \hline \Delta_{\mathrm{f}} \boldsymbol{H}^{\circ} /(\mathrm{kJ} & -74.5 & 0 & -393.5 & -286.0 \\ \left.\mathrm{~mol}^{-1}\right) & & & & \\ \boldsymbol{S}^{\circ} /\left(\mathrm{JK}^{-1}\right. & 186 & 205 & 212 & 70 \\\ \left.\mathbf{m o l}^{-1}\right) & & & & \\ \hline \end{array}$$ (a) \(-891.0 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-240 \mathrm{~kJ} / \mathrm{mol}\) (c) \(-819 \mathrm{~kJ} / \mathrm{mol}\) (d) \(-963 \mathrm{~kJ} / \mathrm{mol}\)

A geyser, operating on LPG (liquefied petroleum gas) heats water flowing at the rate of \(3.0\) litres per minutes, from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\). If the heat of combustion of LPG is \(40,000 \mathrm{~J} / \mathrm{g}\), how much fuel, in \(\mathrm{g}\), is consumed per minute? (Specific heat capacity of water is \(4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}\) ) (a) \(15.25\) (b) \(15.50\) (c) \(15.75\) (d) \(16.00\)

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