When a solid ionic compound dissolves in water, the giant ionic lattice breaks apart and the separated ions interact with water molecules. To understand why some substances dissolve easily while others do not, we must examine the enthalpy changes that occur during this process.
🔑 Key Principle
The dissolving of an ionic lattice is a two-step process: first, the lattice is broken down into gaseous ions (lattice dissociation, which is highly endothermic), and second, the gaseous ions are surrounded by water molecules (hydration, which is highly exothermic).
Important Enthalpy Definitions
The enthalpy change when 1 mole of an ionic solid is completely dissolved in water under standard conditions to form an infinitely dilute solution. It can be endothermic or exothermic:
\( \text{NaCl}(s) \xrightarrow{\text{H}_2\text{O}} \text{Na}^+(aq) + \text{Cl}^-(aq) \quad \Delta H_{\text{sol}}^\theta = +4\text{ kJ mol}^{-1} \)
The enthalpy change when 1 mole of gaseous ions dissolves in water to form 1 mole of aqueous ions under standard conditions. This process is always exothermic because attractions are formed between the ions and polar water molecules:
\( \text{Na}^+(g) \xrightarrow{\text{aq}} \text{Na}^+(aq) \quad \Delta H_{\text{hyd}}^\theta = -406\text{ kJ mol}^{-1} \)
\( \text{Cl}^-(g) \xrightarrow{\text{aq}} \text{Cl}^-(aq) \quad \Delta H_{\text{hyd}}^\theta = -364\text{ kJ mol}^{-1} \)
Ensure you use correct state symbols in these equations. For hydration, the reactants are gaseous ions, and the products are aqueous. For solution, the reactant is the solid lattice, and the products are aqueous ions. Writing the incorrect state symbols can cost you full marks in written questions.
The Dissolving Enthalpy Cycle
Hess's law allows us to construct an enthalpy cycle to calculate the enthalpy of solution. We connect the solid ionic lattice to the aqueous ions by passing through a gaseous ion intermediate state:
From the cycle above, we can write the equation:
Or if using the lattice enthalpy of formation (which is the negative of the lattice dissociation enthalpy):
\( \Delta H_{\text{L-form}}^\theta(\text{NaCl}) = -788\text{ kJ mol}^{-1} \)
\( \Delta H_{\text{hyd}}^\theta(\text{Na}^+) = -406\text{ kJ mol}^{-1} \)
\( \Delta H_{\text{hyd}}^\theta(\text{Cl}^-) = -364\text{ kJ mol}^{-1} \)
Step 1: Set up the cycle equation. We know that:
\( \Delta H_{\text{sol}}^\theta = -\Delta H_{\text{L-form}}^\theta + \Delta H_{\text{hyd}}^\theta(\text{Na}^+) + \Delta H_{\text{hyd}}^\theta(\text{Cl}^-) \)
Note that we change the sign of the lattice formation enthalpy to get the lattice dissociation enthalpy (\( +788\text{ kJ mol}^{-1} \)).
Step 2: Substitute the values:
\( \Delta H_{\text{sol}}^\theta = -(-788) + (-406) + (-364) \)
\( \Delta H_{\text{sol}}^\theta = 788 - 770 = +18\text{ kJ mol}^{-1} \)
The standard enthalpy of solution for sodium chloride is \( +18\text{ kJ mol}^{-1} \).
\( \Delta H_{\text{sol}}^\theta(\text{MgCl}_2) = -155\text{ kJ mol}^{-1} \)
\( \Delta H_{\text{L-diss}}^\theta(\text{MgCl}_2) = +2526\text{ kJ mol}^{-1} \)
\( \Delta H_{\text{hyd}}^\theta(\text{Cl}^-) = -364\text{ kJ mol}^{-1} \)
Step 1: Express the cycle equation for \( \text{MgCl}_2 \). Note that there are 2 moles of chloride ions per mole of magnesium chloride:
\( \Delta H_{\text{sol}}^\theta = \Delta H_{\text{L-diss}}^\theta + \Delta H_{\text{hyd}}^\theta(\text{Mg}^{2+}) + 2\Delta H_{\text{hyd}}^\theta(\text{Cl}^-) \)
Step 2: Rearrange the equation to solve for \( \Delta H_{\text{hyd}}^\theta(\text{Mg}^{2+}) \):
\( \Delta H_{\text{hyd}}^\theta(\text{Mg}^{2+}) = \Delta H_{\text{sol}}^\theta - \Delta H_{\text{L-diss}}^\theta - 2\Delta H_{\text{hyd}}^\theta(\text{Cl}^-) \)
Step 3: Substitute the known values:
\( \Delta H_{\text{hyd}}^\theta(\text{Mg}^{2+}) = -155 - 2526 - 2(-364) \)
\( \Delta H_{\text{hyd}}^\theta(\text{Mg}^{2+}) = -155 - 2526 + 728 \)
\( \Delta H_{\text{hyd}}^\theta(\text{Mg}^{2+}) = -1953\text{ kJ mol}^{-1} \)
The standard enthalpy of hydration of magnesium ions is \( -1953\text{ kJ mol}^{-1} \).
Factors Affecting Lattice and Hydration Enthalpies
The values of both lattice enthalpy and hydration enthalpy depend on the strength of electrostatic attractions. Two key factors determine this strength: ionic radius and ionic charge.
1. Ionic Radius (Size)
As the ionic radius increases:
- The distance between the centres of opposite charges increases.
- The electrostatic attractions become weaker.
- Lattice Enthalpy: Becomes less exothermic (less energy is released when forming the lattice).
- Hydration Enthalpy: Becomes less exothermic (water molecules are less strongly attracted to the larger ion, which has a lower charge density).
2. Ionic Charge
As the charge on the ions increases:
- The electrostatic attractions between oppositely charged ions in the lattice, and between ions and polar water molecules, become much stronger.
- Lattice Enthalpy: Becomes significantly more exothermic.
- Hydration Enthalpy: Becomes significantly more exothermic due to the increased attraction to the water dipoles.
Comparing Sodium and Magnesium Ions
The magnesium ion (\( \text{Mg}^{2+} \)) has a higher charge (+2 vs +1) and a smaller ionic radius (\( 0.072\text{ nm} \) vs \( 0.095\text{ nm} \)) than the sodium ion (\( \text{Na}^+ \)). Therefore, \( \text{Mg}^{2+} \) has a much higher charge density.
This explains why the hydration enthalpy of \( \text{Mg}^{2+} \) (\( -1953\text{ kJ mol}^{-1} \)) is far more exothermic than that of \( \text{Na}^+ \) (\( -406\text{ kJ mol}^{-1} \)).
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