# 6.5 BORN-HABER CYCLE The **Born-Haber cycle** is a specific application of Hess's Law used to determine the lattice energies of binary ionic compounds (like $M^+X^-$). ## Lattice Energy The **lattice energy** ($\Delta H_{lattice}$) is the change in energy that occurs when one mole of an ionic solid is formed from its separated gaseous ions. It is a measure of the strength of the ionic bonds in a crystal lattice. $$ \mathrm{M}^{+}_{(g)} + \mathrm{X}^{-}_{(g)} \longrightarrow \mathrm{MX}_{(s)} $$ Lattice energy cannot be measured directly in an experiment. However, it can be calculated indirectly using the Born-Haber cycle, which breaks down the formation of an ionic compound into a series of well-defined steps. ## Constructing the Born-Haber Cycle for NaCl We can illustrate the cycle using sodium chloride ($NaCl$) as an example. The standard enthalpy of formation ($\Delta H_f^{\circ}$) for $NaCl$ is $-411 \, kJ\,mol^{-1}$. The cycle connects this value to other measurable enthalpy changes. The overall formation reaction is: $$ \mathrm{Na}_{(s)} + \frac{1}{2} \mathrm{Cl}_{2(g)} \longrightarrow \mathrm{NaCl}_{(s)} \quad \Delta H_f^{\circ} = -411 \, kJ\,mol^{-1} $$ This process can be broken down into five distinct steps: **Step I: Sublimation of Sodium** Solid sodium is converted into gaseous sodium atoms. $$ \mathrm{Na}_{(s)} \longrightarrow \mathrm{Na}_{(g)} \quad \Delta H_s^{\circ} = +108 \, kJ\,mol^{-1} $$ **Step II: Ionization of Sodium** Gaseous sodium atoms lose an electron to form gaseous sodium ions. This is the first ionization energy ($IE_1$) of sodium. $$ \mathrm{Na}_{(g)} \longrightarrow \mathrm{Na}^{+}_{(g)} + 1e^{-} \quad \Delta H_{IE}^{\circ} = +496 \, kJ\,mol^{-1} $$ **Step III: Dissociation (Atomization) of Chlorine** The covalent bond in chlorine molecules is broken to form gaseous chlorine atoms. We only need one mole of Cl atoms, so we start with half a mole of $Cl_2$. $$ \frac{1}{2} \mathrm{Cl}_{2(g)} \longrightarrow \mathrm{Cl}_{(g)} \quad \Delta H_{at}^{\circ} = +121 \, kJ\,mol^{-1} $$ **Step IV: Formation of Chloride Ion** Gaseous chlorine atoms gain an electron to form gaseous chloride ions. This is the electron affinity ($EA$) of chlorine. $$ \mathrm{Cl}_{(g)} + 1e^{-} \longrightarrow \mathrm{Cl}^{-}_{(g)} \quad \Delta H_{EA}^{\circ} = -349 \, kJ\,mol^{-1} $$ **Step V: Formation of the Ionic Lattice** Gaseous sodium and chloride ions combine to form one mole of solid sodium chloride. This is the lattice energy ($\Delta H_{l}^{\circ}$) that we want to calculate. $$ \mathrm{Na}^{+}_{(g)} + \mathrm{Cl}^{-}_{(g)} \longrightarrow \mathrm{NaCl}_{(s)} \quad \Delta H_{l}^{\circ} = ? $$ ## Calculation of Lattice Energy for NaCl According to Hess's Law, the total enthalpy change of the direct route (enthalpy of formation) must equal the sum of the enthalpy changes of the indirect route (the five steps). $$ \Delta H_f^{\circ} = \Delta H_s^{\circ} + \Delta H_{IE}^{\circ} + \Delta H_{at}^{\circ} + \Delta H_{EA}^{\circ} + \Delta H_l^{\circ} $$ 1. **Given Values:** - $\Delta H_f^{\circ} = -411 \, kJ\,mol^{-1}$ - $\Delta H_s^{\circ} = +108 \, kJ\,mol^{-1}$ - $\Delta H_{IE}^{\circ} = +496 \, kJ\,mol^{-1}$ - $\Delta H_{at}^{\circ} = +121 \, kJ\,mol^{-1}$ - $\Delta H_{EA}^{\circ} = -349 \, kJ\,mol^{-1}$ 2. **Apply the Formula:** $$ -411 = (+108) + (+496) + (+121) + (-349) + \Delta H_l^{\circ} $$ 3. **Solve for Lattice Energy ($\Delta H_l^{\circ}$):** $$ -411 = (725) + (-349) + \Delta H_l^{\circ} $$ $$ -411 = 376 + \Delta H_l^{\circ} $$ $$ \Delta H_l^{\circ} = -411 - 376 $$ $$ \Delta H_l^{\circ} = -787 \, kJ\,mol^{-1} $$ Thus, the lattice energy of NaCl is **-787 kJ/mol**. The negative sign indicates that energy is released when the lattice is formed. ## Concept Assessment Exercise 6.5 **Q:** Draw a complete Born-Haber Cycle for the formation of $MgO(s)$ and calculate its lattice energy from the following data. | Enthalpy Change | Value | | :--- | :--- | | Standard enthalpy of formation of MgO | $-602 \, kJ\,mol^{-1}$ | | Standard enthalpy of sublimation of Mg | $+150 \, kJ\,mol^{-1}$ | | Total Ionization energy of Mg to $Mg^{2+}$ | $+2180 \, kJ\,mol^{-1}$ | | Standard enthalpy of atomization of $O_2$ ($1/2 O_2 \rightarrow O$) | $+247 \, kJ\,mol^{-1}$ | | Total Electron affinity of O to $O^{2-}$ (EA₁ + EA₂) | $(-141) + (+878) = +737 \, kJ\,mol^{-1}$ | **A:** The steps are: 1. $Mg_{(s)} \rightarrow Mg_{(g)}$ 2. $Mg_{(g)} \rightarrow Mg^{2+}_{(g)} + 2e^-$ 3. $\frac{1}{2} O_{2(g)} \rightarrow O_{(g)}$ 4. $O_{(g)} + 2e^- \rightarrow O^{2-}_{(g)}$ 5. $Mg^{2+}_{(g)} + O^{2-}_{(g)} \rightarrow MgO_{(s)}$ Using Hess's Law: $$ \Delta H_f^{\circ} = \Delta H_{sub} + \Delta H_{IE} + \Delta H_{at} + \Delta H_{EA} + \Delta H_{lattice} $$ $$ -602 = (+150) + (+2180) + (+247) + (+737) + \Delta H_{lattice} $$ $$ -602 = 3314 + \Delta H_{lattice} $$ $$ \Delta H_{lattice} = -602 - 3314 = -3916 \, kJ\,mol^{-1} $$ ## Factors Affecting Lattice Energy Lattice energy is influenced by the **charge density** of the ions involved. $$ \text{Charge Density} = \frac{\text{Ionic Charge}}{\text{Ionic Radius}} $$ Two main factors determine the magnitude of lattice energy: 1. ***Ionic Charge***: As the magnitude of the ionic charges increases, the electrostatic attraction between ions becomes stronger, leading to a higher (more exothermic) lattice energy. - *Example*: The lattice energy of $MgO$ ($Mg^{2+}$ and $O^{2-}$) is significantly higher than that of $NaCl$ ($Na^+$ and $Cl^-$) because of the greater charges. 2. ***Ionic Radius***: As the ionic radius increases, the distance between the centers of the ions increases. This weakens the electrostatic attraction, leading to a lower (less exothermic) lattice energy. - *Example*: Both $LiF$ and $KF$ have ions with +1 and -1 charges. However, the $K^+$ ion is larger than the $Li^+$ ion. Therefore, $LiF$ has a higher lattice energy than $KF$. ## Concept Assessment Exercise 6.6 **Q:** Which of the following compounds is expected to have higher lattice energy? Why? a. LiF or NaF b. NaCl or $MgCl_2$ **A:** a. **LiF** has a higher lattice energy. Both have a -1 fluoride ion, but the $Li^+$ ion is smaller than the $Na^+$ ion, resulting in a stronger attraction. b. **$MgCl_2$** has a higher lattice energy. The magnesium ion has a +2 charge ($Mg^{2+}$), which creates a much stronger electrostatic attraction with the chloride ions compared to the +1 charge of the sodium ion ($Na^+$) in NaCl. ## Role of Hydration in the Dissolving Process When an ionic solute dissolves in a solvent like water, two main energy changes occur: 1. **Breaking the Lattice**: Energy is required to overcome the ionic attractions and separate the ions from the crystal lattice. This is an *endothermic* process, equal in magnitude but opposite in sign to the lattice energy ($- \Delta H_{lattice}$). 2. **Hydration of Ions**: Energy is released when the separated gaseous ions are surrounded and stabilized by solvent molecules (water). This is an *exothermic* process known as the **enthalpy of hydration** ($\Delta H_{hyd}$). The overall energy change, the **enthalpy of solution** ($\Delta H_{solution}$), is the sum of these two steps. $$ \Delta H_{solution} = (-\Delta H_{lattice}) + \Delta H_{hydration} $$ - If $\Delta H_{solution}$ is **negative (exothermic)**, the solution gets warmer. This occurs when the hydration energy released is greater than the energy required to break the lattice. - Example: $NaOH_{(s)} + xH_2O_{(l)} \rightarrow Na^{+}_{(aq)} + OH^{-}_{(aq)} \quad \Delta H_{soln} = -44.5 \, kJ\,mol^{-1}$ - If $\Delta H_{solution}$ is **positive (endothermic)**, the solution gets colder. This occurs when the energy required to break the lattice is greater than the hydration energy released. - Example: $NH_4NO_{3(s)} + xH_2O_{(l)} \rightarrow NH_{4(aq)}^{+} + NO_{3(aq)}^{-} \quad \Delta H_{solution} = +25.7 \, kJ\,mol^{-1}$ | Substance | Enthalpy of Solution ($kJ\,mol^{-1}$) | | :--- | :--- | | LiCl | -37.0 | | NaCl | +2.98 | | KCl | +17.2 | | KI | +20.3 | | $NH_4NO_3$ | +25.7 | | $AlCl_3$ | -321.0 | ## Concept Assessment Exercise 6.7 **Q:** Calculate the enthalpy of solution for NaCl from the following data: - $\Delta H_{lattice} = -786 \, kJ/mol$ - $\Delta H_{hydration} = -783 \, kJ/mol$ **A:** 1. **Energy to break the lattice:** $-\Delta H_{lattice} = -(-786) = +786 \, kJ/mol$ 2. **Energy released by hydration:** $\Delta H_{hydration} = -783 \, kJ/mol$ 3. **Calculate enthalpy of solution:** $$ \Delta H_{solution} = (-\Delta H_{lattice}) + \Delta H_{hydration} $$ $$ \Delta H_{solution} = (+786) + (-783) = +3 \, kJ/mol $$ This small positive value indicates that dissolving NaCl in water is slightly endothermic, which is consistent with experimental data. ## Factors Affecting Enthalpy of Hydration Similar to lattice energy, the enthalpy of hydration is influenced by the **charge density** of the ion. $$ \text{Charge Density} = \frac{\text{Ionic Charge}}{\text{Ionic Radius}} $$ 1. ***Ionic Charge***: Ions with a higher charge create a stronger ion-dipole attraction with water molecules, resulting in a more exothermic (more negative) enthalpy of hydration. - *Example*: The hydration enthalpy of $Mg^{2+}$ ions is much more exothermic than that of $Na^+$ ions. 2. ***Ionic Radius***: Smaller ions have a higher charge density, allowing water molecules to approach more closely and form stronger ion-dipole interactions. This leads to a more exothermic enthalpy of hydration. - *Example*: The $Mg^{2+}$ ion is smaller than the $Ca^{2+}$ ion. Therefore, the enthalpy of hydration for $MgSO_4$ is more exothermic than for $CaSO_4$. ## Concept Assessment Exercise 6.8 **Q:** Which has a higher (more exothermic) enthalpy of hydration? Justify your answer. a. LiF or NaF b. $KNO_3$ or $NaNO_3$ **A:** a. **LiF**. The cations ($Li^+$ vs $Na^+$) are the difference. Since $Li^+$ is smaller than $Na^+$, it has a higher charge density and thus a more exothermic enthalpy of hydration. b. **$NaNO_3$**. The anions are the same ($NO_3^-$). The $Na^+$ ion is smaller than the $K^+$ ion, giving it a higher charge density and a more exothermic enthalpy of hydration.