# Building Born-Haber Cycles ## Born-Haber Cycles - Construction - A **Born-Haber cycle** is a specific application of Hess's Law for ionic compounds - Enables us to calculate lattice enthalpy which cannot be found by experiment - The basic principle is to construct a diagram where **energy increases going up the diagram**

In Born-Haber cycles, endothermic reactions go UP and exothermic reactions go DOWN

### The Basic Principle - The cycle shows all the steps needed to turn atoms into gaseous ions and from gaseous ions into the ionic lattice - The alternative route to the ionic lattice begins from the enthalpy of formation of the elements in their standard states ## Drawing the Cycle for Sodium Chloride ### Step 1: Start with the Elements - Draw the elements with their state symbols about a third of the way up the diagram - This is shown as the left hand side of the equation for the process indicated - The location is marked by drawing a horizontal bar or line which represents the starting energy level For sodium chloride: $\text{Na (s)} + \frac{1}{2}\text{Cl}_2\text{ (g)}$ ### Step 2: Creating the Gaseous Atoms - Next, create the gaseous atoms - This is a two-step process of first creating the gaseous atoms and then turning them into ions - Creating gaseous atoms is a bond breaking process, so arrows must be drawn **upwards** - It doesn't matter whether you start with sodium or chlorine **For sodium:** - The enthalpy of atomisation of sodium is: $$\text{Na (s)} \rightarrow \text{Na (g)} \quad \Delta H_{\text{at}}^{\theta} = +108 \text{ kJ mol}^{-1}$$ **For chlorine:** - The enthalpy of atomisation of chlorine is: $$\frac{1}{2}\text{Cl}_2\text{ (g)} \rightarrow \text{Cl (g)} \quad \Delta H_{\text{at}}^{\theta} = +121 \text{ kJ mol}^{-1}$$ - Show the products of the process on the horizontal lines and the energy value against a vertical arrow connecting the energy levels ### Step 3: Creating the Gaseous Ions **For the sodium ion:** - The sodium ion loses an electron, so this energy change is the first ionisation energy for sodium: $$\text{Na (g)} \rightarrow \text{Na}^+\text{ (g)} + e^- \quad \Delta H_{\text{ie}}^{\theta} = +500 \text{ kJ mol}^{-1}$$ - The change is endothermic so the direction continues **upwards** **For the chlorine ion:** - The chlorine atom gains an electron, so this is electron affinity: $$\text{Cl (g)} + e^- \rightarrow \text{Cl}^-\text{ (g)} \quad \Delta H_{\text{ea}}^{\theta} = -364 \text{ kJ mol}^{-1}$$ - The exothermic change means this is **downwards** - The change is displaced to the right to make the diagram easier to read

Ionisation energy is always endothermic (upwards); electron affinity is typically exothermic (downwards)

### Step 4: Completing the Cycle **Adding enthalpy of formation:** - The enthalpy of formation of sodium chloride is added at the bottom of the diagram: $$\text{Na (s)} + \frac{1}{2}\text{Cl}_2\text{ (g)} \rightarrow \text{NaCl (s)} \quad \Delta H_{\text{f}}^{\theta} = -411 \text{ kJ mol}^{-1}$$ - This is an exothermic change for sodium chloride so the arrow points **downwards** - Enthalpy of formation can be exothermic or endothermic, so you may need to show it above the elements (and displaced to the right) for an endothermic change **Adding lattice enthalpy:** - The final change is lattice enthalpy, which is usually shown as a formation - For sodium chloride the equation is: $$\text{Na}^+\text{ (g)} + \text{Cl}^-\text{ (g)} \rightarrow \text{NaCl (s)} \quad \Delta H_{\text{latt}}^{\theta}$$ - This completes the cycle

The cycle has two routes: the indirect route (via gaseous ions) and the direct route (enthalpy of formation)

### Complete Born-Haber Cycle Structure The completed cycle shows: - Elements at the bottom - Gaseous atoms in the middle - Gaseous ions at the top - The ionic solid connects from elements (direct route) and from gaseous ions (indirect route) ## Worked Example: Potassium Chloride **Construct a Born-Haber Cycle for potassium chloride (KCl)** | Step | Equation | Enthalpy Change | |------|----------|----------------| | Convert K(s) atoms into K(g) atoms | $\text{K(s)} \longrightarrow \text{K(g)}$ | $\Delta H_{\text{at}}^{\theta}$ | | Convert K(g) atoms into K⁺(g) ions | $\text{K(g)} \longrightarrow \text{K}^+\text{(g)} + e^-$ | $IE_1$ | | Convert Cl₂(g) molecules into Cl(g) atoms | $\frac{1}{2}\text{Cl}_2\text{(g)} \longrightarrow \text{Cl(g)}$ | $\Delta H_{\text{at}}^{\theta}$ | | Convert Cl(g) atoms into Cl⁻(g) ions | $\text{Cl(g)} + e^- \longrightarrow \text{Cl}^-\text{(g)}$ | $EA_1$ | | Add up all values to get $\Delta H_{\text{f}}^{\theta}$ | | $\Delta H_{\text{f}}^{\theta}$ | | Apply Hess's Law to find $\Delta H_{\text{latt}}^{\theta}$ | | $\Delta H_{\text{latt}}^{\theta}$ | **Key relationships:** - The direct and indirect route start at the same point on the cycle and end at the same point on the cycle - By Hess's Law: $\Delta H_{\text{f}}^{\theta} = \Delta H_{\text{at}}^{\theta}(\text{K}) + \Delta H_{\text{at}}^{\theta}(\text{Cl}) + IE_1 + EA_1 + \Delta H_{\text{latt}}^{\theta}$ - Rearranging: $\Delta H_{\text{latt}}^{\theta} = \Delta H_{\text{f}}^{\theta} - (EA_1) - (IE_1) - \Delta H_{\text{at}}^{\theta}(\text{Cl}) - \Delta H_{\text{at}}^{\theta}(\text{K})$

A downward arrow means exothermic; an upward arrow means endothermic

## Worked Example: Magnesium Oxide **Construct a Born-Haber Cycle for magnesium oxide (MgO)** For compounds with ions having charges greater than ±1, you need additional steps: | Step | Equation | Enthalpy Change | |------|----------|----------------| | Convert Mg(s) atoms into Mg(g) atoms | $\text{Mg(s)} \longrightarrow \text{Mg(g)}$ | $\Delta H_{\text{at}}^{\theta}$ | | Convert Mg(g) atoms into Mg⁺(g) ions | $\text{Mg(g)} \longrightarrow \text{Mg}^+\text{(g)} + e^-$ | $IE_1$ | | Convert Mg⁺(g) ions into Mg²⁺(g) ions | $\text{Mg}^+\text{(g)} \longrightarrow \text{Mg}^{2+}\text{(g)} + e^-$ | $IE_2$ | | Convert O₂(g) molecules into O(g) atoms | $\frac{1}{2}\text{O}_2\text{(g)} \longrightarrow \text{O(g)}$ | $\Delta H_{\text{at}}^{\theta}$ | | Convert O(g) atoms into O⁻(g) ions | $\text{O(g)} + e^- \longrightarrow \text{O}^-\text{(g)}$ | $EA_1$ | | Convert O⁻(g) ions into O²⁻(g) ions | $\text{O}^-\text{(g)} + e^- \longrightarrow \text{O}^{2-}\text{(g)}$ | $EA_2$ | | Add up all values to get $\Delta H_{\text{f}}^{\theta}$ | | $\Delta H_{\text{f}}^{\theta}$ | | Apply Hess's Law to find $\Delta H_{\text{latt}}^{\theta}$ | | $\Delta H_{\text{latt}}^{\theta}$ | **Important points for MgO:** - Magnesium requires **two ionisation energies** (to form Mg²⁺) - Oxygen requires **two electron affinities** (to form O²⁻) - The second electron affinity (EA₂) is typically **endothermic** because you're adding an electron to an already negative ion

Second electron affinity values are endothermic (positive) as work must be done against repulsion

**Applying Hess's Law:** $$\Delta H_{\text{f}}^{\theta} = \Delta H_{\text{at}}^{\theta}(\text{Mg}) + \Delta H_{\text{at}}^{\theta}(\text{O}) + IE_1 + IE_2 + EA_1 + EA_2 + \Delta H_{\text{latt}}^{\theta}$$ Rearranging: $$\Delta H_{\text{latt}}^{\theta} = \Delta H_{\text{f}}^{\theta} - (EA_1) - (EA_2) - (IE_1) - (IE_2) - \Delta H_{\text{at}}^{\theta}(\text{Mg}) - \Delta H_{\text{at}}^{\theta}(\text{O})$$ ## Key Points to Remember - Always start with elements in their standard states about one-third up the diagram - Bond breaking processes (atomisation, ionisation) go **upwards** (endothermic) - Electron affinity usually goes **downwards** (exothermic), but second electron affinities go **upwards** - Enthalpy of formation connects elements to ionic solid - can be up or down - Lattice enthalpy connects gaseous ions to ionic solid - always **downwards** (when shown as formation) - The cycle is now complete and can be used to calculate unknown enthalpy values using Hess's Law - For ions with charges >±1, include multiple ionisation energies or electron affinities - Label all arrows with the appropriate enthalpy change and value