How is lattice energy estimated using a Born-Haber cycle? Estimating lattice energy using the Born-Haber cycle has been discussed in Ionic Solids. For a quick review, the following is an example that illustrate the estimate of the energy of crystallization of NaCl. Hsub of Na = 108 kJ/mol (Heat of sublimation) D of Cl2 = 244 (Bond dissociation energy) IP of Na(g) = 496 (Ionization potential or energy) EA of Cl(g) = -349 (Electron affinity of Cl) Hf of NaCl = -411 (Enthalpy of formation) The Born-Haber cycle to evaluate Elattice is shown below: -----------Na+ + Cl(g)-------- | | |-349 |496+244/2 ↓ | Na+(g) + Cl-(g) | | Na(g) + 0.5Cl2(g) | | |108 | | |Ecryst= -788 Na(s) + 0.5Cl2(l) | | | |-411 | ↓ ↓ -------------- NaCl(s) -------------- Ecryst = -411-(108+496+244/2)-(-349) kJ/mol = -788 kJ/mol. Discussion The value calculated for U depends on the data used. Data from various sources differ slightly, and so is the result. The lattice energies for NaCl most often quoted in other texts is about 765 kJ/mol. Compare with the method shown below Na(s) + 0.5 Cl2(l) → NaCl(s) - 411 Hf Na(g) → Na(s) - 108 -Hsub Na+(g) + e → Na(g) - 496 -IP Cl(g) → 0.5 Cl2(g) - 0.5 * 244 -0.5*D Cl-(g) → Cl(g) + 2 e 349 -EA Add all the above equations leading to Na+(g) + Cl-(g) → NaCl(s) -788 kJ/mol = Ecryst The bond between ions of opposite charge is strongest when the ions are small.The lattice energies for the alkali metal halides is therefore largest for LiF and smallest for CsI, as shown in the table below. Lattice Energies of Alkali Metals Halides (kJ/mol) F- Cl- Br- I- Li+ 1036 853 807 757 Na+ 923 787 747 704 K+ 821 715 682 649 Rb+ 785 689 660 630 Cs+ 740 659 631 604 The ionic bond should also become stronger as the charge on the ions becomes larger. The data in the table below show that the lattice energies for salts of the OH- and O2- ions increase rapidly as the charge on the ion becomes larger. Lattice Energies of Salts of the OH- and O2- Ions (kJ/mol) OH- O2- Na+ 900 2481 Mg2+ 3006 3791 Al3+ 5627 15,916 Here is an additional explanation Why Does Sodium Form NaCl? Sodium reacts with chlorine to form Na+ ions and Cl- ions in spite of the fact that the first ionization energy of sodium is larger than the electron affinity of chlorine. To explain this, we need to divide the reaction between sodium and chlorine into a number of hypothetical steps for which we know the amount of energy given off or absorbed. The starting materials for this reaction are solid sodium metal and chlorine molecules in the gas phase, and the product of the reaction is solid sodium chloride. 2 Na(s) + Cl2(g) 2 NaCl(s) Let's imagine that the reaction takes place by the following sequence of steps.• A mole of sodium is converted from the solid to a gas. As might be expected, this reaction is endothermic. Na(s) Na(g) Ho = 107.3 kJ • An electron is then removed from each sodium atom to form a mole of Na+ ions. The energy consumed in this reaction is equal to the first ionization energy of sodium. Na(g) Na+(g) Ho = 495.8 kJ • A mole of chlorine atoms is formed by breaking the bonds in one-half a mole of chlorine molecules. Like the previous steps, this is an endothermic reaction. 1/2 Cl2(g) Cl(g) Ho = 121.7 kJ • An electron is then added to each chlorine atom to form a Cl- ion. This is the first exothermic step in this process, and the energy released is equal to the electron affinity of chlorine. Cl(g) + e- Cl-(g) Ho = -348.8 kJ • The isolated Na+ and Cl- ions in the gas phase then come together to form solid NaCl. This is a strongly exothermic reaction, for which Ho is equal to the lattice energy of NaCl. Na+(g) + Cl-(g) NaCl(s) Ho = -787.3 kJ If we consider just the first four steps in this reaction, Hess's law suggests that it 5.16 takes 376.0 kJ/mol to form Na+ and Cl- ions from sodium metal and chlorine gas. Na(s) + 1/2 Cl2(g) Na+(g) + Cl-(g) Ho = 376.0 kJ/mol When we include the last step in the calculation, the lattice energy of NaCl is large enough to compensate for all of the steps in this reaction that consume energy, as shown in the figure below. Na(s) + 1/2 Cl2(g) NaCl(s) Ho = -411.3 kJ/molThe primary driving force behind this reaction is therefore the force of attraction between the Na+ and Cl- ions formed in the reaction, not the affinity of a chlorine atom for electrons. Why does the reaction stop at NaCl? Why doesn't it keep going to form NaCl2 or NaCl3? The lattice energy would increase as the charge on the sodium atom increased from Na+ to Na2+ or Na3+. But to form an Na2+ ion, we have to remove a second electron from the sodium atom, and the second ionization energy of sodium (4562.4 kJ/mol) is almost 10 times as large as the first ionization energy. The increase in the lattice energy that would result from forming an Na2+ ion can't begin to compensate for the energy needed to break into the filled-shell configuration of the Na+ ion to remove a second electron. The reaction between sodium and chlorine therefore stops at