Enthalpy Changes in Solution Revision Notes for OCR A-Level Chemistry A

Enthalpy Changes in Solution

Introduction

When ionic compounds dissolve in water, there is an overall energy change associated with the process. Understanding these energy changes helps explain why some ionic compounds dissolve easily whilst others do not, and whether the dissolution process absorbs or releases heat.

Dissolving ionic compounds

Water molecules possess a remarkable ability to break apart the giant ionic lattice structure and overcome the strong electrostatic forces between oppositely-charged ions. This property makes water an excellent solvent for many ionic compounds.

When an ionic salt dissolves in water:

The regular arrangement of ions in the solid lattice is disrupted
Water molecules surround and separate the individual ions
The ions become dispersed throughout the solution as aqueous ions

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The ability of water to dissolve ionic compounds stems from its polar nature - the partial charges on water molecules interact with the charged ions in the lattice, gradually pulling them away from their fixed positions.

For example, when sodium chloride dissolves:

$\text{Na}^+\text{Cl}^-\text{(s)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$

In the solid lattice, $\text{Na}^+$ and $\text{Cl}^-$ ions are held together by strong electrostatic attractions. In the aqueous solution, these ions are separate but now surrounded by water molecules.

Water molecules interact with the ions through their partial charges. The oxygen atom in water carries a partial negative charge ( $\delta^-$ ), which is attracted to positive ions. The hydrogen atoms carry partial positive charges ( $\delta^+$ ), which are attracted to negative ions.

Standard enthalpy change of solution

The standard enthalpy change of solution ( $\Delta_{\text{sol}}H^\circ$ ) is defined as the enthalpy change that occurs when one mole of a solute dissolves completely in a solvent under standard conditions. When the solvent is water, the ions from the ionic lattice become surrounded by water molecules to form aqueous ions.

The enthalpy change of solution can be either:

Exothermic (negative $\Delta_{\text{sol}}H^\circ$ ) - heat is released to the surroundings, temperature increases
Endothermic (positive $\Delta_{\text{sol}}H^\circ$ ) - heat is absorbed from the surroundings, temperature decreases

For sodium chloride dissolving in water:

$\text{Na}^+\text{Cl}^-\text{(s)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \qquad \Delta_{\text{sol}}H^\circ = +4 \text{ kJ mol}^{-1}$

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The positive value indicates this process is endothermic - the solution gets cooler as sodium chloride dissolves. This is why adding salt to ice can lower the temperature further, as the dissolving process absorbs heat from the surroundings.

Experimental determination of enthalpy change of solution

You can determine the enthalpy change of solution directly in the laboratory using simple calorimetry equipment. The method involves measuring the temperature change when a known mass of ionic compound dissolves in a known volume of water.

Procedure for potassium chloride

Weigh out a sample of $\text{KCl(s)}$ accurately
Using a measuring cylinder, transfer a known volume of distilled water (e.g. 25.0 cm³) into a plastic cup in a beaker
Measure the initial temperature of the water to the nearest 0.5°C
Quickly add all the $\text{KCl(s)}$ to the water and stir with the thermometer
Record the lowest temperature reached (for endothermic) or highest temperature (for exothermic) to the nearest 0.5°C

Calculation steps

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Exemplo Trabalhado: Determining the enthalpy change of solution for KCl

Step 1: Calculate the energy change $q$ in the solution using:

$q = mc\Delta T$

where:

$m$ = mass of solution (NOT just water)
$c$ = specific heat capacity of solution (≈ 4.18 J g⁻¹ K⁻¹)
$\Delta T$ = temperature change (in °C or K)

For the potassium chloride example:

Density of water = 1.00 g cm⁻³
Therefore, 25.0 cm³ of water has mass = 25.0 g
Mass of solution = mass of water + mass of KCl = 25.0 + 3.73 = 28.73 g

$q = 28.73 \times 4.18 \times 6.5 = 780.594 \text{ J} = 0.781 \text{ kJ}$

Step 2: Calculate the amount in moles of compound that dissolved.

$n(\text{KCl}) = \frac{m}{M} = \frac{3.73}{74.6} = 0.0500 \text{ mol}$

Step 3: Calculate $\Delta_{\text{sol}}H$ in kJ mol⁻¹.

The equation shows: $\text{KCl(s)} \rightarrow \text{K}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$

In the experiment, 0.0500 mol KCl absorbed 0.781 kJ of energy from the solution (temperature decreased).

Therefore, 1 mol KCl would absorb:

$\frac{0.781}{0.0500} = 15.6 \text{ kJ}$

$\Delta_{\text{sol}}H = +15.6 \text{ kJ mol}^{-1}$

The positive sign indicates the process is endothermic.

Important considerations

chatImportant

Mass in calculations: When using $q = mc\Delta T$ , the mass $m$ must be the mass of the solution (water + dissolved compound), not just the mass of water. The thermometer measures the temperature of the entire solution. Using only the water mass would introduce a significant error - in the example above, this would give 13% error.

chatImportant

Sign conventions: Take care with signs. If the temperature decreases, the dissolution is endothermic and $\Delta_{\text{sol}}H$ is positive. If the temperature increases, the dissolution is exothermic and $\Delta_{\text{sol}}H$ is negative.

The dissolving process

When a solid ionic compound dissolves in water, two distinct processes occur:

The ionic lattice breaks apart to form separate gaseous ions
Water molecules surround the gaseous ions to form hydrated aqueous ions

Each process involves an enthalpy change.

Stage 1: Breaking the lattice

Breaking up the ionic lattice requires energy input to overcome the strong electrostatic attractions between oppositely-charged ions. This process is endothermic.

For sodium chloride:

$\text{NaCl(s)} \rightarrow \text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)} \qquad \Delta H = +788 \text{ kJ mol}^{-1}$

infoNote

This is the opposite of lattice formation. The lattice enthalpy (lattice formation) would be:

$\text{Na}^+\text{(g)} + \text{Cl}^-\text{(g)} \rightarrow \text{NaCl(s)} \qquad \Delta_{\text{LE}}H^\circ = -788 \text{ kJ mol}^{-1}$

Breaking the lattice is therefore the negative of lattice enthalpy (endothermic process).

Stage 2: Hydration of ions

The separate gaseous ions then interact with polar water molecules to form hydrated aqueous ions. This process releases energy and is exothermic. The energy change is called the enthalpy change of hydration.

The enthalpy change of hydration ( $\Delta_{\text{hyd}}H^\circ$ ) is the enthalpy change that occurs when gaseous ions dissolve in water to form one mole of aqueous ions.

For sodium and chloride ions:

$\text{Na}^+\text{(g)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} \qquad \Delta_{\text{hyd}}H^\circ = -406 \text{ kJ mol}^{-1}$

$\text{Cl}^-\text{(g)} + \text{aq} \rightarrow \text{Cl}^-\text{(aq)} \qquad \Delta_{\text{hyd}}H^\circ = -378 \text{ kJ mol}^{-1}$

Both hydration processes are exothermic (negative values).

Energy cycles and Hess's law

We can link the enthalpy change of solution, lattice enthalpy, and hydration enthalpies using an energy cycle and Hess's law.

The energy cycle shows two routes from ionic lattice to aqueous ions:

Route 1 (direct): Ionic lattice dissolves directly to form aqueous ions - this is $\Delta_{\text{sol}}H$
Route 2 (indirect): Ionic lattice breaks into gaseous ions (lattice breaking), then gaseous ions hydrate to form aqueous ions (hydration)

According to Hess's law, the total enthalpy change is the same regardless of route:

$\text{Route 1 = Route 2}$

Therefore:

$\Delta_{\text{sol}}H^\circ = \Delta_{\text{LE}}H^\circ + \Delta_{\text{hyd}}H^\circ(\text{cation}) + \Delta_{\text{hyd}}H^\circ(\text{anion})$

Or rearranging:

$\Delta_{\text{LE}}H^\circ + \Delta_{\text{sol}}H^\circ = \Delta_{\text{hyd}}H^\circ(\text{cation}) + \Delta_{\text{hyd}}H^\circ(\text{anion})$

infoNote

This relationship allows us to calculate any one of these values if we know the other three. It's particularly useful because lattice enthalpies and hydration enthalpies cannot be measured directly but can be calculated from measurable quantities.

Whether dissolution is exothermic or endothermic

The enthalpy change of solution can be exothermic or endothermic depending on the relative magnitudes of:

The lattice enthalpy (breaking the lattice - endothermic)
The sum of hydration enthalpies (forming aqueous ions - exothermic)

infoNote

If the energy released during hydration is greater than the energy required to break the lattice, dissolution is exothermic overall. If less energy is released during hydration than is required to break the lattice, dissolution is endothermic overall.

Worked example: Calculating lattice enthalpy of sodium chloride

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Exemplo Trabalhado: Calculating the lattice enthalpy of sodium chloride

Given information:

$\text{Na}^+\text{(g)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} \qquad \Delta_{\text{hyd}}H^\circ = -406 \text{ kJ mol}^{-1}$

$\text{Cl}^-\text{(g)} + \text{aq} \rightarrow \text{Cl}^-\text{(aq)} \qquad \Delta_{\text{hyd}}H^\circ = -378 \text{ kJ mol}^{-1}$

$\text{Na}^+\text{Cl}^-\text{(s)} + \text{aq} \rightarrow \text{Na}^+\text{(aq)} + \text{Cl}^-\text{(aq)} \qquad \Delta_{\text{sol}}H^\circ = +4 \text{ kJ mol}^{-1}$

Calculate the lattice enthalpy of sodium chloride.

Step 1: Construct the energy cycle

The energy cycle must show:

Gaseous ions at the top
Ionic lattice at the bottom left
Aqueous ions at the bottom right

Label the arrows:

Arrow A: Lattice enthalpy (gaseous ions → ionic lattice) - we want to find this
Arrow B: Hydration of Na⁺
Arrow C: Hydration of Cl⁻
Arrow D: Enthalpy change of solution

Step 2: Apply Hess's law

Identify the two routes:

Route 1: A + D
Route 2: B + C

Using Hess's law: A + D = B + C

$\Delta_{\text{LE}}H^\circ(\text{NaCl}) + \Delta_{\text{sol}}H^\circ(\text{NaCl}) = \Delta_{\text{hyd}}H^\circ(\text{Na}^+) + \Delta_{\text{hyd}}H^\circ(\text{Cl}^-)$

Substitute the values:

$\Delta_{\text{LE}}H^\circ(\text{NaCl}) + (+4) = (-406) + (-378)$

$\Delta_{\text{LE}}H^\circ(\text{NaCl}) + (+4) = -784$

$\Delta_{\text{LE}}H^\circ(\text{NaCl}) = -784 - 4 = -788 \text{ kJ mol}^{-1}$

The lattice enthalpy of sodium chloride is -788 kJ mol⁻¹.

chatImportant

Important study tip: Between each horizontal energy level in the cycle, only one chemical species changes state or form, and all other species remain balanced. This helps ensure your energy cycle is correctly constructed.

Worked example: Calculating enthalpy change of hydration

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Exemplo Trabalhado: Calculating the enthalpy change of hydration of Ca²⁺

Given information:

Table

Calculate the enthalpy change of hydration of $\text{Ca}^{2+}$ ions.

Solution

This example differs from the previous one in two important ways:

Two $\text{Br}^-$ ions are involved, so we must multiply the hydration enthalpy of $\text{Br}^-$ by 2
The enthalpy change of solution of calcium bromide is exothermic (negative), which affects the positioning in the energy cycle

Energy cycle construction

The cycle shows:

Gaseous ions $\text{Ca}^{2+}\text{(g)} + 2\text{Br}^-\text{(g)}$ at the top
Ionic lattice $\text{CaBr}_2\text{(s)}$ at the middle left
Aqueous ions $\text{Ca}^{2+}\text{(aq)} + 2\text{Br}^-\text{(aq)}$ at the bottom

Applying Hess's law

Two routes from gaseous ions to aqueous ions:

Route 1: A + D
Route 2: B + C

Using Hess's law: A + D = B + C

$\Delta_{\text{LE}}H^\circ(\text{CaBr}_2) + \Delta_{\text{sol}}H^\circ(\text{CaBr}_2) = \Delta_{\text{hyd}}H^\circ(\text{Ca}^{2+}) + 2 \times \Delta_{\text{hyd}}H^\circ(\text{Br}^-)$

Substitute the values:

$(-2176) + (-99) = \Delta_{\text{hyd}}H^\circ(\text{Ca}^{2+}) + (2 \times -348)$

$-2275 = \Delta_{\text{hyd}}H^\circ(\text{Ca}^{2+}) - 696$

$\Delta_{\text{hyd}}H^\circ(\text{Ca}^{2+}) = -2275 + 696 = -1579 \text{ kJ mol}^{-1}$

The enthalpy change of hydration of $\text{Ca}^{2+}$ ions is -1579 kJ mol⁻¹.

Application: Cold packs

Cold packs used for sports injuries demonstrate an endothermic dissolution process. These packs contain two separate compartments - one with a chemical (commonly ammonium nitrate, $\text{NH}_4\text{NO}_3$ ), and the other with water. When the pack is squeezed, the seal breaks and the chemical dissolves in the water.

infoNote

Since the dissolution of ammonium nitrate is endothermic, it absorbs heat energy from the surroundings (the injured area), causing a cooling effect. This reduces swelling and provides pain relief.

The dissolving process involves:

Breaking the ionic lattice (endothermic - requires energy)
Hydrating the ions (exothermic - releases energy)

For the cold pack to work, the energy required to break the lattice must be greater than the energy released during hydration, making the overall process endothermic.

Remember!

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Key Points to Remember:

Enthalpy change of solution ( $\Delta_{\text{sol}}H^\circ$ ) is the enthalpy change when one mole of solute dissolves in a solvent to form aqueous ions. It can be exothermic or endothermic.
Dissolving occurs in two stages: (1) Breaking the ionic lattice into gaseous ions (endothermic), and (2) Hydrating the gaseous ions to form aqueous ions (exothermic).
Enthalpy change of hydration ( $\Delta_{\text{hyd}}H^\circ$ ) is the enthalpy change when gaseous ions dissolve in water to form one mole of aqueous ions. This is always exothermic.
Energy cycles and Hess's law connect solution enthalpy, lattice enthalpy, and hydration enthalpies:

$\Delta_{\text{LE}}H^\circ + \Delta_{\text{sol}}H^\circ = \Delta_{\text{hyd}}H^\circ(\text{cation}) + \Delta_{\text{hyd}}H^\circ(\text{anion})$
Experimental determination uses $q = mc\Delta T$ where $m$ is the mass of the entire solution (not just water), and careful attention to signs is essential.
Multiple ions: When more than one of a particular ion is present (e.g., $\text{CaBr}_2$ has two $\text{Br}^-$ ions), multiply that ion's hydration enthalpy by the appropriate factor in calculations.

Enthalpy Changes in Solution (OCR A-Level Chemistry A): Revision Notes