Question

Use the Born-Haber cycles to determine the solvation energy for these two compounds: NaCl MgCl₂. Solvation energy is deltaH for the process.

ions gas phase --> ions aqueous solution.

(Make the ions in aq solution the starting & ending point for the cylce.)

MgCl₂ pulls water from your skin in a painful manner where NaCl(s) in your hand has no effecfts.

Calculate the relevant enthalpychange for MgCl₂ and NaCl that can be used to explain the statement above.

Answer 1

From the Born-Haber cycle, we can write as shown below.

For NaCl: The enthalpy of formation of NaCl (∆H_f) = +S + I + (1/2)D -E + ∆H_lattice

Where, +S = sublimation energy for the conversion of Na(s) to Na(g); +I = Ionization energy for the conversion of Na(g) to Na⁺(g), +D = dissociation energy for the conversion of Cl₂(g) to 2Cl(g); -E = electron affinity for the conversion of Cl(g) to Cl^-(g), ∆H_lattice = lattice enthalpy for the conversion fo Na⁺(g) and Cl^-(g) to NaCl(s)

By substituting the all the standard values, you will get ∆H_lattice = -787 KJ/mol

The standard value for ∆H_hydration of NaCl = -769 KJ/mol

Now, ∆H_hydration = ∆H_lattice + ∆H_solvation

i.e. ∆H_solvation for NaCl = -(-787) + (-769) = +18 KJ/mol

For MgCl2: The enthalpy of formation of MgCl₂ (∆H_f) = +S + I₁ + I₂ + D -2*E + ∆H_lattice

By substituting the all the standard values, you will get ∆H_lattice = -2526 KJ/mol

The standard value for ∆H_hydration of MgCl₂ = -2647 KJ/mol

Now, ∆H_hydration = ∆H_lattice + ∆H_solvation

i.e. ∆H_solvation for MgCl₂ = -(-2526) + (-2647) = -121 KJ/mol