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The enthalpy of solution of sodium chloride is \(4 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and its enthalpy of bydration of ions is \(-784\) kJ mol \(^{-1}\). What will be the lattice enthalpy of sodium chloride? (a) \(+780 \mathrm{k} J \mathrm{~mol}^{-1}\) (b) \(+394 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(+788 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(+398 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Short Answer

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+780 kJ/mol

Step by step solution

01

Understand the Enthalpy of Solution

The enthalpy of solution is the total heat effect when one mole of solute forms an infinitely dilute solution with the solvent at constant pressure. It is given as 4 kJ/mol for sodium chloride.
02

Understand the Enthalpy of Hydration

The enthalpy of hydration is the heat released when one mole of gaseous ions dissolves in enough solvent to form an infinitely dilute solution. It is given as -784 kJ/mol for sodium chloride.
03

Formula for Lattice Enthalpy

The lattice enthalpy can be calculated through Hess's law, using the enthalpy of solution and the enthalpy of hydration. The lattice enthalpy is the sum of the enthalpy of solution and the enthalpy of hydration: Lattice Enthalpy = Enthalpy of Solution + Enthalpy of Hydration.
04

Calculate the Lattice Enthalpy

Substituting the given values in the formula provides us with the lattice enthalpy: Lattice Enthalpy = 4 kJ/mol + (-784 kJ/mol) = -780 kJ/mol (negative sign indicates that energy is released). Since lattice enthalpy is usually expressed as a positive value referring to the energy required to separate the ions, the correct answer is +780 kJ/mol.

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Key Concepts

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

Enthalpy of Solution
The enthalpy of solution quantifies the heat change that occurs when one mole of a solute is dissolved in a solvent, forming an infinitely dilute solution, all while maintaining constant pressure. It's a key concept in thermochemistry, the branch of chemistry concerning energy changes during chemical processes. In our specific case regarding sodium chloride (NaCl), the process absorbs a small amount of energy, indicated by the enthalpy of solution given as 4 kJ/mol. This value can be either positive or negative, signifying endothermic (heat absorbed) or exothermic (heat released) processes, respectively. In this scenario, sodium chloride's dissolution is slightly endothermic.

Understanding the enthalpy of solution is crucial for predicting the solubility of substances as well as their interactions with solvents. This knowledge can help to design better industrial processes that require precise solubility control, such as in pharmaceuticals and agricultural products.
Enthalpy of Hydration
The enthalpy of hydration is another pivotal concept in thermochemistry that describes the heat released when gaseous ions are hydrated, or surrounded by solvent molecules, to form an infinitely dilute solution. For sodium chloride, the enthalpy of hydration is given as an exothermic value, -784 kJ/mol. This means that a substantial amount of energy is released when NaCl's ions, initially in a gaseous state, become surrounded by water molecules.

The large negative number indicates a strong attraction between sodium and chloride ions to water, leading to a significant energy release upon hydration. This process is crucial for understanding how ionic substances interact with water, influencing everything from biochemistry within living organisms to the functioning of batteries. Hydration enthalpies are highly important for industries that rely on the solvation of ions, including water treatment and the manufacturing of synthetics.
Hess's Law
Hess's law is a principle that asserts the total enthalpy change during a chemical process is the same regardless of the number of steps the process is carried out in. This law is indispensable because it allows us to calculate the enthalpy change of reactions that are difficult to study directly, such as the lattice enthalpy of ionic compounds. By using Hess's law, we can determine the lattice enthalpy by combining the enthalpy of solution and the enthalpy of hydration.

In the given problem, Hess's law is used to compute the lattice enthalpy for sodium chloride. It's a straightforward method: simply add the enthalpy of solution (4 kJ/mol) to the enthalpy of hydration (-784 kJ/mol) to find the lattice enthalpy. This law is a staple in chemical thermodynamics and offers a reliable way to understand and calculate energy changes in various chemical systems.
Thermochemistry
Thermochemistry is the study of energy and heat associated with chemical reactions and physical transformations. It involves understanding concepts like enthalpy, entropy, and Gibbs free energy. The cases of enthalpy of solution and enthalpy of hydration discussed herein are integral parts of thermochemistry. Thermochemistry provides insights into reaction spontaneity, energy efficiency, and the stability of substances.

Thermochemical principles are applied widely, from engineering environmentally-friendly processes to predicting the behavior of chemical species in various conditions. Knowledge of thermochemistry is essential for professionals in fields such as environmental science, engineering, and materials science, allowing them to harness and control the energy exchanges in chemical processes for practical applications.

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Most popular questions from this chapter

For a reaction, \(\mathrm{CaCO}_{3(\mathrm{~s})} \rightarrow \mathrm{CaO}_{(0)}+\mathrm{CO}_{2(g)}\) \(\Delta_{j} H^{\circ}(\mathrm{CaO})=-635.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) \(\Delta_{f} H^{\circ}\left(\mathrm{CO}_{2}\right)=-393.5 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) and \(\Delta_{f} H^{\circ}\left(\mathrm{CaCO}_{3}\right)=-1206.9 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}\) Which of the following is a correct statement? (a) A large amount of heat is evolved during the decomposition of \(\mathrm{CaCO}_{3}\). (b) Decomposition of \(\mathrm{CaCO}_{3}\) is an endothermic process and heat is provided for decomposition. (c) The amount of heat evolved cannot be calculated from the data provided. (d) \(\Delta_{r} H^{\circ}=\Sigma \Delta_{f} H^{\circ}\) (reactants) \(-\Sigma \Delta_{f} H^{\circ}\) (products)

\(\Delta U=q+w\), is mathematical expression for (a) first law of thermodynamics (b) second law of thermodynamics (c) third law of thermodynamics (d) zeroth law of thermodynamics.

At \(373 \mathrm{~K}\), steam and water are in equilibrium and \(\Delta H=40.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\). What will be \(\Delta S\) for conversion of water into steam? \(\mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(g)}\) (a) \(109.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(31^{-1} \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(21.93 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(326 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

What will be the work done when one mole of a gas expands isothermally from \(15 \mathrm{~L}\) to \(50 \mathrm{~L}\) against a constant pressure of 1 atm at \(25^{\circ} \mathrm{C}\) ? (a) \(-3542 \mathrm{cal}\) (b) \(-843.3 \mathrm{cal}\) (c) \(-718 \mathrm{cl}\) (d) \(-60.23 \mathrm{cal}\)

Hess's law is applicable for the determination of heat of (a) transition (b) formation (c) reaction (d) all of these.

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