Ch 2 · Enthalpies of solution and hydration · AS & A Level Chemistry 9701

Topic 2 of 4 in Chemical energetics — AS & A Level Chemistry 9701 — 45 lessons.

Introduction

When an ionic solid dissolves in water, its lattice breaks apart and each ion becomes surrounded by water molecules. The overall enthalpy change of this process, the enthalpy of solution, can be calculated using an energy cycle that links it to two quantities from earlier lessons: lattice energy and a new quantity, enthalpy of hydration.

Key ideas — Defining enthalpy of hydration and enthalpy of solution

The standard enthalpy change of hydration, Δ H_hyd^ominus, is the enthalpy change when one mole of a specified gaseous ion is dissolved in water to form one mole of aqueous ions under standard conditions, e.g. Na^+(g) arrow Na^+(aq). It is always exothermic, since ion–dipole attractions form between the ion and surrounding water molecules. The standard enthalpy change of solution, Δ H_sol^ominus, is the enthalpy change when one mole of an ionic solid dissolves in water to form an infinitely dilute aqueous solution under standard conditions, e.g. NaCl(s) arrow Na^+(aq) + Cl^-(aq). Unlike hydration, Δ H_sol^ominus can be either exothermic or endothermic.

🧑‍🏫 Teacher's explanation

Imagine you are trying to move into a new house. The 'lattice energy' is like the energy required to break down an old, sturdy building into individual bricks. The 'enthalpy of hydration' is like the energy released when those bricks are surrounded by a cozy, protective layer of insulation (water molecules).

📖 Definitions

🔍 Worked example — Standard

The enthalpy of solution for KNO_3 is +20.0 kJ mol^-1. The enthalpy of lattice formation is -750 kJ mol^-1. Calculate the total enthalpy of hydration for the ions K^+ and NO_3^-.

🔍 Worked example — Exam-style

Explain why the enthalpy of hydration of Mg^2+ is more exothermic than the enthalpy of hydration of Na^+.

⚠️ Common errors

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🎯 Exam question

Key ideas — An energy cycle linking solution, lattice energy and hydration

Dissolving can be imagined as two steps: first breaking the lattice apart into gaseous ions (the reverse of lattice energy, so -Δ H_latt^ominus), then hydrating each mole of gaseous ion (summing the relevant Δ H_hyd^ominus values). By Hess's Law, the direct route (dissolving) equals the sum of this indirect route:

🧑‍🏫 Teacher's explanation

Imagine you are trying to move a group of people from a crowded, organized bus (the crystal lattice) into a large, open swimming pool (the water/solvent). To get them out of the bus, you have to spend energy to break them apart from their seats. Once they are out, they feel much more relaxed and 'comfortable' when they jump into the pool because they can move around freely. This 'feeling of comfort' is like the energy released when ions bond with water molecules.

📖 Definitions

🔍 Worked example — Easy

Calculate the enthalpy change of solution (Δ H_sol^ominus) for NaCl given that the lattice enthalpy (Δ H_latt^ominus) is -785 kJ mol^-1 and the hydration enthalpies (Δ H_hyd^ominus) are Na^+ = -406 kJ mol^-1 and Cl^- = -364 kJ mol^-1.

🔍 Worked example — Standard

Calculate the enthalpy change of solution for MgCl_2 given Δ H_latt^ominus = -2930 kJ mol^-1, Δ H_hyd^ominus(Mg^2+) = -1920 kJ mol^-1, and Δ H_hyd^ominus(Cl^-) = -364 kJ mol^-1.

🔍 Worked example — Exam-style

A student finds that the enthalpy of solution for KNO_3 is +36 kJ mol^-1. If the lattice enthalpy is -750 kJ mol^-1, calculate the sum of the hydration enthalpies for K^+ and NO_3^-.

⚠️ Common errors

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🎯 Exam question

Key ideas — Calculations using the cycle

Any one of Δ H_sol^ominus, Δ H_latt^ominus or the hydration enthalpies can be found if the other two are known, by rearranging the equation above. Because Δ H_sol^ominus is usually a small difference between two much larger quantities (Δ H_latt^ominus and the sum of hydration enthalpies, both typically several hundred mathrmkJ\ mol^-1), it can take either sign depending on which term is slightly larger in magnitude, which is why some salts dissolve exothermically and others endothermically.

🧑‍🏫 Teacher's explanation

Imagine you are trying to move a heavy piece of furniture from a box into a room. The 'lattice energy' is the energy required to break the furniture out of its tight packaging. The 'hydration enthalpy' is the energy released when the furniture settles comfortably into its new spot in the room. The 'enthalpy of solution' is the net energy change of this whole process.

📖 Definitions

🔍 Worked example — Exam-style

A student finds that the enthalpy of solution for NH_4Cl is +17 kJ mol^-1. The lattice enthalpy is -767 kJ mol^-1. Calculate the sum of the hydration enthalpies for NH_4^+ and Cl^-.

⚠️ Common errors

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🎯 Exam question

Key ideas — Effect of ionic charge and radius on hydration enthalpy

Hydration enthalpy arises from the electrostatic attraction between an ion and the permanent dipoles of surrounding water molecules. A higher ionic charge gives a more exothermic (larger magnitude) hydration enthalpy, since a more highly charged ion attracts the partially charged oxygen (for cations) or hydrogen (for anions) atoms of water molecules more strongly. A smaller ionic radius also gives a more exothermic hydration enthalpy, since it allows water molecules to approach closer to the centre of charge, increasing the attraction. Consequently, small, highly charged ions such as mathrmMg^2+ or mathrmAl^3+ have far more exothermic hydration enthalpies than large, singly charged ions such as Cs^+.

🧑‍🏫 Teacher's explanation

Imagine you are holding a magnet. If you have a tiny, powerful neodymium magnet (high charge, small radius), it will pull metal objects toward it very forcefully. If you have a large, weak refrigerator magnet (low charge, large radius), the pull is much weaker. Hydration works exactly like this. Water molecules are polar; they have a positive end and a negative end. When an ion is placed in water, these water molecules are 'pulled' toward the ion.

Ch 2 · Enthalpies of solution and hydration · AS & A Level Chemistry 9701

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