For NaCl, the lattice formation enthalpy is -787 kJ mol -1. Thanks. NaCl(s) -> Na(s) + 1/2 Cl2(g) H = 411kJ Na(s) - … For NaCl, the lattice dissociation enthalpy is +787 kJ mol -1.
In an exam paper it says that the lattice energy of NaCl is more exothermic than MgCl as the Na+ cation has a smaller ionic radius than Mg+. The lattice energy of an ionic solid is a measure of the strength of bonds in that ionic compound. The lattice energy is the reaction: NaCl(s) = Na+(g) + Cl-(g) Using Hess's Law this reaction can be made from the given reactions. Here are the resources given: What is the reasoning for for Na+ having a smaller ionic radius even though Mg+ has a higher proton number? You should talk about "lattice formation enthalpy" if you want to talk about the amount of energy released when a lattice is formed from its scattered gaseous ions. That immediately removes any possibility of confusion.