Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The lattice energy* of NaI is \(-686 \mathrm{~kJ} / \mathrm{mol}\), and the enthalpy of hydration is \(-694 \mathrm{~kJ} / \mathrm{mol}\). Calculate the enthalpy of solution per mole of solid NaI. Describe the process to which this enthalpy change applies.

Short Answer

Expert verified
The enthalpy of solution for NaI can be calculated by adding the given lattice energy and enthalpy of hydration: \(-686 \mathrm{~kJ/mol} + (-694 \mathrm{~kJ/mol}) = -1380 \mathrm{~kJ/mol}\). This value represents the energy change when one mole of solid NaI dissolves in water, indicating an exothermic process as energy is released during dissolution.

Step by step solution

01

Understand the terms

The lattice energy is the energy required to break one mole of a solid ionic compound into gaseous ions. The enthalpy of hydration is the energy released when one mole of gaseous ions is solvated by water molecules.
02

Write the given values

The lattice energy of NaI is given as -686 kJ/mol and the enthalpy of hydration of NaI is given as -694 kJ/mol.
03

Calculate the enthalpy of solution

The enthalpy of solution is the sum of the lattice energy and the enthalpy of hydration. Therefore, we can calculate the enthalpy of solution using the given values: Enthalpy of solution = Lattice energy + Enthalpy of hydration Enthalpy of solution = (-686 kJ/mol) + (-694 kJ/mol)
04

Find the answer

Adding the values together, we get: Enthalpy of solution = -1380 kJ/mol
05

Describe the process

The enthalpy change of -1380 kJ/mol applies to the process of dissolving one mole of solid NaI in water. This value indicates that the dissolution of NaI in water is an exothermic process, as energy is released during the process.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lattice Energy
The lattice energy is a critical concept in understanding ionic compounds and their interactions. Imagine a crystal structure where ions are neatly arranged and bound together in a repeating pattern. The lattice energy is the total amount of energy it takes to break apart this robust structure into individual gaseous ions. For a compound like sodium iodide (NaI), this energy is quite significant due to the strong electrostatic forces between the positively charged sodium ions and the negatively charged iodide ions.

The process of breaking these bonds is always endothermic, meaning it requires the absorption of energy. Students can think of lattice energy as a measure of an ionic compound's stability: the higher the lattice energy, the more stable the compound and the more energy it would take to disassemble the crystal lattice.
Enthalpy of Hydration
When these gaseous ions enticed by lattice energy encounter water, a new type of energy change comes into play called the enthalpy of hydration. This is the energy change associated with the solvation process, where water molecules surround and stabilize the ions. In the enthalpy of hydration, we observe the reverse of lattice energy: energy is released as water molecules, owing to their polarity, are attracted to and interact with ions. The negative sign of the enthalpy of hydration value indicates that it's an exothermic process.

For example, the hydration of Na+ and I- ions from sodium iodide (NaI) results in a substantial energy release. This process is critical to understand as it explains how ionic compounds dissolve in water and also greatly influences the solubility of different substances in aqueous solutions.
Exothermic Process
An exothermic process is a reaction or change that releases heat into its surroundings. The 'exo' prefix means 'outward', and thermic pertains to 'heat'. In chemistry, this is a fundamental concept that students must discern, as it underpins the behavior of substances during reactions.

When ionic compounds like NaI dissolve in water, the overall process is exothermic. The sum of the endothermic lattice energy and the exothermic enthalpy of hydration results in net energy release. This release of energy typically manifests as an increase in the temperature of the solution, a tangible effect students can observe during laboratory experiments. Understanding exothermic reactions is also important in various real-world applications, such as in the design of heat packs or even within industrial chemical processes.
Ionic Compounds
Ionic compounds are substances composed of positively charged ions (cations) and negatively charged ions (anions) that are held together by potent electrostatic forces known as ionic bonds. In a classic example, sodium iodide (NaI) comprises sodium ions (Na+) and iodide ions (I-). The formation of ionic compounds often results from the transfer of electrons from one atom to another, leading to a full outer shell for each ion – a stable electron configuration.

Students must appreciate the unique properties of ionic compounds: high melting points, solid at room temperature, and the ability to dissolve in water due to their ionic nature. The dissolution process involves the disruption of the ionic lattice and interaction with water molecules, key factors in understanding the enthalpy changes discussed in lattice energy and the enthalpy of hydration.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A solution is prepared by mixing \(0.0300 \mathrm{~mol} \mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(0.0500\) \(\mathrm{mol} \mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at \(25^{\circ} \mathrm{C}\). At \(25^{\circ} \mathrm{C}\), the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and \(11.4\) torr, respectively.

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) is a monoprotic acid that ionizes only partially in aqueous solutions. A \(0.10 M\) formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of \(0.10 M\) formic acid.

Using the following information, identify the strong electrolyte whose general formula is $$ \mathrm{M}_{x}(\mathrm{~A})_{y} \cdot z \mathrm{H}_{2} \mathrm{O} $$ Ignore the effect of interionic attractions in the solution. a. \(\mathrm{A}^{n-}\) is a common oxyanion. When \(30.0 \mathrm{mg}\) of the anhydrous sodium salt containing this oxyanion \(\left(\mathrm{Na}_{n} \mathrm{~A}\right.\), where \(n=1,2\), or 3 ) is reduced, \(15.26 \mathrm{~mL}\) of \(0.02313 M\) reducing agent is required to react completely with the \(\mathrm{Na}_{n}\) A present. Assume a \(1: 1\) mole ratio in the reaction. b. The cation is derived from a silvery white metal that is relatively expensive. The metal itself crystallizes in a body-centered cubic unit cell and has an atomic radius of \(198.4 \mathrm{pm}\). The solid, pure metal has a density of \(5.243 \mathrm{~g} / \mathrm{cm}^{3}\). The oxidation number of \(\mathrm{M}\) in the strong electrolyte in question is \(+3\). c. When \(33.45 \mathrm{mg}\) of the compound is present (dissolved) in \(10.0 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\), the solution has an osmotic pressure of 558 torr.

A solution of phosphoric acid was made by dissolving \(10.0 \mathrm{~g}\) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) in \(100.0 \mathrm{~mL}\) water. The resulting volume was \(104 \mathrm{~mL}\) Calculate the density, mole fraction, molarity, and molality of the solution. Assume water has a density of \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\).

You and your friend are each drinking cola from separate 2 - \(\mathrm{L}\) bottles. Both colas are equally carbonated. You are able to drink \(1 \mathrm{~L}\) of cola, but your friend can drink only about half a liter. You each close the bottles and place them in the refrigerator. The next day when you each go to get the colas, whose will be more carbonated and why?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free