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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Hess’s Law of Constant Heat Summation

Russian chemist Germain Henri Hess gave an important law about the enthalpy change. He said that enthalpy change of one compound to the other compound is always same, even if it occurs in one step or more steps.

Enthalpy is a state function that means it doesn’t depend on the path taken by the system. Let’s take an example to understand it better:

A + B ⟶ D      Δ_rH

This reaction may take place into more steps like:

A + B ⟶ C         Δ_rH₁       .....equation 1                 

C ⟶ D             Δ_rH₂       .....equation 2

Then you can get Δ_rH by simply applying mathematical operations by adding equation 1 and 2.

[A + B ⟶ C Δ_rH₁] + [ C ⟶ D Δ_rH₂] = A + B ⟶ D  

  

 So, Δ_rH = [Δ_r H₁ + Δ_rH₂]

Let’s take another example: if you have given two different reactions:

C(graphite) + O₂(g) ⟶ CO₂(g) Δ_r H= -393.5 kJ mol^-1    ........equation 1                 

CO(g) + ½ O₂(g) ⟶ CO₂(g)      Δ_r H= -283 kJ mol^-1           .........equation 2                 

And asked you to calculate the Δ_r H of the following reaction

C(graphite) + ½ O₂(g) ⟶ CO(g)           Δ_r H =?

It’s puzzle time; try to rearrange the above two equation so that you can get the desired equation. We want the carbon and oxygen on the reactant side and CO on the product side. We can take equation 1 as it is given, but we have to reverse the equation 2 to get the CO on the product side. When you reverse the equation 2, sign of Δ_rH also gets reversed. And you get a new equation 3.

CO₂(g) ⟶ CO(g) + ½ O₂(g)     Δr H= +283 kJ mol^-1          .........equation 3                 

Now if we add the equation 1 and 3 we can get the desired equation.

Δ_r H = [Δ_rH₁ -393.5 kJ mol^-1] + [Δ_r H₃ + 283 kJ mol^-1]

Δ_r H = -110.5 kJ mol^-1

So you can state the law of constant heat summation as ‘If a reaction takes place in several steps then its Δ_r H^ө is the sum of standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.’

 Δ_r H^ө= Δ_r H^ө₁ + Δ_r H^ө₂+ Δ_r H^ө₃ + ....Δ_r H^ө_n

Hess’s Law of Constant Heat Summation

Bond Dissociation Enthalpy Δbond H^ө

Now you have learnt to calculate enthalpy of reaction by using standard enthalpy of formation and by using standard enthalpies of intermediate reactions. There is one more way to calculate the enthalpy of reaction. When any reaction occurs, some bonds are broken and some new bonds are formed. Energy is needed to break the bonds and energy is released when new bonds are formed. It means if we calculate the bond enthalpies of reactant and products we can get the enthalpy of reaction.

Δ_r H = Ʃ Δbond H^ө(reactants) - Ʃ Δbond H^ө(products)

Bond enthalpy or bond dissociation enthalpy is the enthalpy change when one mole of covalent bond of a gaseous covalent compound is broken into gaseous products.

H₂(g) ⟶ 2H(g) ....... Δ_H-H H^ө = 435.0 kJ mol^-1

Cl₂(g) ⟶ 2H(g) ....... Δ_H-H H^ө = 435.0 kJ mol^-1

The above examples are of simple diatomic molecules. What happens in polyatomic molecule? Let’s take an example of methane CH₄. It has four C-H bond. C is sp³ hybridised, all C-H bonds are identical and have same energy. But each successive step requires different energy to break C-H bond. Because for each C-H bond there is different environment. In first step there are three more C-H bonds while in second step there are two more C-H bonds and so on.

CH₄(g) ⟶ CH₃(g) + H(g)          ........ Δ_C-HH^ө = 427.0 kJ mol^-1

CH₃(g) ⟶ CH₂(g) + H(g)          ........ Δ_C-HH^ө = 439.0 kJ mol^-1

CH₂(g) ⟶ CH(g) + H(g)           ........ Δ_C-HH^ө = 452.0 kJ mol^-1

CH(g) ⟶ C(g) + H(g)               ........ Δ_C-H H^ө = 347.0 kJ mol^-1

In such cases we take the mean value for the bond dissociation enthalpy. So Δ_C-H H^ө will be:

Δ_C-H H^ө = ¼ (427 + 439 + 452 + 347) kJ mol^-1

Δ_C-H H^ө = 416 kJ mol^-1

We have calculated Δ_C-H H^ө for methane. It can be different for other molecules but it will be nearer to the mean value so we can take it as a reference for any C-H bond.

Lattice enthalpy

Bond dissociation enthalpy is useful for covalent bonds but not for ionic compounds. In ionic compounds, ions are trapped in lattice and we need to overcome the lattice energy to set these ions free. You know that ionic compounds dissociate into its constituent ions only when they get dissolved in a solvent. Let’s try to understand the process of dissolution of any ionic compound in a solvent. This process occurs in two steps, in the first step lattice gets broken and ions are freed, and in the second step solvationoccurs. If the solvent is water then solvation is known as hydration.

MX(s) ⟶ M⁺(g) + X^-(g) ⟶ M⁺(aq) + X^-(aq)

So the enthalpy change for the overall reaction is called the enthalpy of solution Δ_solution H^ө which is the sum of lattice enthalpy Δ_lattice H^ө and solvation enthalpy Δ_hydrationH^ө.

Δ_solution H^ө = Δ_lattice H^ө+ Δ_hydration H^ө

It means if you want to know the lattice enthalpy of NaCl you must know the values of Δ_solution H^ө and Δ_hydrationH^ө.  Two German scientists Max Born and Fritz Haber developed an indirect way to calculate the lattice enthalpy of NaCl. They constructed an enthalpy diagram by applying Hess’s law of constant heat summation. We will learn about it in the next post of Thermodynamics. 

This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.