Born Haber Cycle

Lattice enthalpy Δlattice Hө is the enthalpy change when one mole of ionic compound dissociates into its constituent atoms in the form of gaseous ions. According to this definition lattice enthalpy is always positive which means it will be endothermic process.

            MX(s) ⟶ M⁺(g) + X^-(g)            Δ_lattice H^ө = “+”

But you can also say that lattice energy is the energy which is released when gaseous ions bind together to form ionic compound. By this definition lattice enthalpy is always negative which means it will be exothermic process.

            M⁺(g) + X^-(g) ⟶ MX(s)            Δ_lattice H^ө = “-”

In general terms we can say that lattice energy is the energy difference between separate gaseous ions and ionic solid. In the previous post of thermodynamics you have seen that there is no direct way to calculate Δ_lattice H^ө experimentally. Two German scientists Max Born and Fritz Haber developed an indirect way to calculate the lattice energy.

They applied Hess’s law of constant heat summation in the enthalpy diagram. They designed a cycle of reactions which has 5 steps.

Step 1: Formation of NaCl from its constituent elements in their standard stage.

            Na(g) + Cl(g)  ⟶ NaCl(s)         ...........Δ_f H^ө = -411.2 kJ mol^-1

Step 2: Sublimation of Na(s) to Na(g) or you can also call it atomization

            Na(s) ⟶ Na(g)             ...........Δ_a H^ө = 108.4 kJ mol^-1

Step 3: Atomization of Cl₂ to Cl(g). This process involves the dissociation of Cl-Cl bond, so it will be equal to the bond dissociation enthalpy of Cl-Cl.

            Cl₂(g) ⟶ Cl(g)              ...........Δ_Cl-Cl H^ө = 242 kJ mol^-1

For the formation of NaCl we need only one Cl and the value given abov is for Cl₂, so we will take half of its value (242/2) = 121 kJ mol^-1.

Step 4: (a) Ionization of Na(g) to Na⁺

                    Na(g) ⟶ Na⁺(g) + e^-                 ...........Δ_ionization H^ө = 496 kJ mol^-1

(b) Ionization of Cl(g) to Cl^-(g). In this process Cl gains an electron to get converted to Cl^-(g). So we need to calculate the electron gain enthalpy.

               Cl(g) + e^- ⟶ Cl^-(g)        ...........Δ_eg H^ө = -348.6 kJ mol^-1

Step 5: Na⁺ and Cl^- ions come closer and arrange themselves in lattice to form NaCl.

            Na⁺(g) + Cl^-(g) ⟶ NaCl(s)         ...........Δ_lattice H^ө = -? kJ mol^-1

Cycle starts from the formation of NaCl from its constituent elements and ends by the union of Na⁺ and Cl^- to form NaCl. We know that the enthalpy change in a cyclic process is zero.

0 = Δ_f H^ө + Δ_a H^ө+ ½ Δ_Cl-Cl H^ө + Δ_ionization H^ө + Δ_egH^ө + Δ_lattice H^ө

Now let's put the sign of all enthalpies. Δ_eg H^өis always negative since energy is required to add an electron to the neutral atom. Δ_lattice H^ө also is negative because energy is released when two ions come together to form ionic solid.

0 = Δ_f H^ө + Δ_a H^ө+ ½ Δ_Cl-Cl H^ө + Δ_ionization H^ө - Δ_egH^ө - Δ_lattice H^ө

Enthalpy of formation, bond dissociation, ionization and electron gain enthalpy can be measured experimentally. As we want to calculate lattice enthalpy, we will transfer it to the left side of the equation.  

Δ_lattice H^ө = Δ_f H^ө+ Δ_a H^ө + ½ Δ_Cl-Cl H^ө + Δ_ionizationH^ө - Δ_eg H^ө

Now let's put in the values:

Δ_lattice H^ө = 411.2 kJ mol^-1+ 108.4 kJ mol^-1 + 121 kJ mol^-1 + 496 kJ mol^-1- 348.6 kJ mol^-1

Δ_lattice H^ө = + 788 kJ mol^-1

Born Haber Cycle  

That means 788 kJ mol^-1 energy is required to break one mole of NaCl into Na⁺ and Cl^- ions. Lattice energy defines the stability of ionic solids. The high value of lattice energy is the reason why ionic solids have such high melting and boiling point.

As you have seen that ionization enthalpy (IE) and electron gain enthalpy (EG) play an important role in determining the value of lattice energy and we can compare the lattice energy of given ionic compounds by observing the trends of IE and EG. Let’s try:

Compare the lattice energy of NaCl and MgCl₂

IE of Mg  > Na so the lattice energy of MgCl₂(2326 kJ mol^-1) will be higher than NaCl (788 kJ mol^-1).

Compare the lattice energy of NaCl and LiCl.

Li has higher IE than Na, so LiCl has higher lattice energy (860 kJ mol^-1) than NaCl (788 kJ mol^-1).

Compare the lattice energy of MgCl₂ and MgO

Electron gain enthalpy: first Eg of O is lesser than Cl, but the second Eg will be much higher as it becomes difficult to add one more electron to the negatively charged ion.

So the lattice energy of MgO (3795 kJ mol^-1) will be higher than that of MgCl₂ (2326 kJ mol^-1).

Now you have understood the first law of thermodynamics. You have learnt about internal energy and enthalpy. But we need one more thermodynamic property to understand the spontaneity of a reaction. You may think that if a reaction doesn’t require heat to start and is exothermic, it must be spontaneous. But it doesn’t happen in every reaction. In the next post of thermodynamics we will discuss this in detail.

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