Born–Haber Cycle in the energy states of the Chemistry

It is a series of steps (chemical processes) used to calculate the lattice energy of ionic solids, which is difficult to determine experimentally. You can think of BH cycle as a special case of Hesse’s law, which states that the overall energy change in a chemical process can be calculated by breaking down the process into several steps and adding the energy change from each step.

To understand the BH cycle fully let us define the meaning of lattice energy first.

It is the energy released  when a metal ion in its gaseous state combines with a nonmetal anion in its gaseous state to form an ionic solid. The magnitude of the lattice energy relates to how stable the ionic solid is. A larger value indicates a more stable ionic compound.

The energy released in the following processes is called is called lattice energy:

Li+(g) + F-(g)	->	LiF(s)	LE = -1047kJ
Mg+(g) + O2-(g)	->	MgO(s)	LE = -3916kJ

where k is a constant, Q₁ and Q₂ are the charges on the cation and anion respectively, and r is the inter-nuclear distance.

Test yourself: (answers found at end of page)

1. Explain why the LE for LiF is much less than for MgO ?
2. Explain why LE of Na₂S is -2203 kJ and Cs₂S is -1850 kJ?

Lattice Energy Calculations via Born-Haber cycle

Example 1: Li(s) + ¹/₂F₂(g) -> LiF(s)           -617kJ

This overall chemical process can be broken down to many intermediate steps:

1.	Li(s)	->	Li(g)	energy of sublimation	+161 kJ	endothermic
2.	Li(g)	->	Li+(g)	ionization energy	+520 kJ	endothermic
3.	1/2F2(g)	->	F(g)	energy of dissociation	+77 kJ	endothermic
4.	F(g)	->	F-(g)	electron affinity	-328 kJ	exothermic
5.	Li+(g) + F-(g)	->	LiF(s)	Lattice Energy (LE)	??

If we add 1through 5, after cancelling the intermediates,
Li⁺(s) + ¹/₂F₂(g) -> LiF(s)     [+161+520+77+(-328)+LE] kJ = -617 kJ
So, LE = -1047 kJ

As you can see, the lattice energy is hugely exothermic so offsets the other endothermic processes including energy of sublimation, ionization energy of Li(g) and the bond dissociation energy of F2(g). The final overall process of formation of LiF from lithium (s) and F2(g) is exothermic: energetically favorable.

Test yourself: (answers found at end of page)

3. From your knowledge of ionic radii and the charge on the cations and anions in the following compounds, arrange the compounds in the order of increasing lattice energy.
MgSe,       MgTe,       K2Se,       K2Te

Example 2: Lets calculate the lattice energy for KCl from the following literature values via the Born-Haber cycle.

Ionization energy for K	419.0 kJ/mole
Electron affinity for Cl	-349 kJ/mole
Bond energy for Cl2	239 kJ/mole
Heat of sublimation for K	64 kJ/mole
ΔH0f for KCl	437 kJ/mole
LE	?

Test yourself: (answers found at bottom of page)

4. Through the BH cycle calculate its lattice energy from the following energy data.

Energy of sublimation for K(s):	77.08 kJ/mole
Ionization energy for K(g):	418.6 kJ/mole
Bond dissociation energy for Br2(g)	193 kJ/mole
Electron affinity for bromine:	-324.6 kJ/mole
Heat of vaporization of Br2:	29.96 kJ/mole
Heat of formation for KBr(s)	-393.8 kJ/mole

Answers to Test Yourself questions:

1. , For LiF the product of Q1 and Q2 is -1(+1 x -1) whereas for MgO the product of Q1 and Q2 is -4(+2 x -2), this makes the numerator much larger for LE for MgO compared to LiF.
2. IN this case the product of Q1 and Q2 is the same for Na2S and Cs2S, so the numerator for LE is the same for both, but the Cs being much larger than Na means the inter-nuclear distance, r, for Cs2S is much larger, making the magnitude for LE for Cs2S much smaller. (LE is inversely proportional to the r and directly proportional to the Q1 and Q2).
3. Based on the formula for LE and considering the charges (Q1 and Q2) and inter-nuclear distance (r), the trend for LE for the compounds will be:
CaSe >> CaTe >> Na2Se >> Na2Te
4. BH cycle for the formation of KBr:

1.	K(s) + 1/2Br2(l)	->	K(g) + 1/2Br2(l)	77.08 kJ
2.	K(g) + 1/2Br2(l)	->	K(g) + 1/2Br2(g)	29.96/2 kJ/(1/2 mole)
3.	K(g) + 1/2Br2(g)	->	K+(g) + 1/2Br2(g)	418.6 kJ
4.	K+(g) + 1/2Br2(g)	->	K+ + Br(g)	193/2 kJ
5.	K+ + Br(g)	->	K+(g) + 1/2Br-(g)	-324.6 kJ
6.	K+(g) + 1/2Br-(g)	->	KBr(s)	??

Add 1 through 6

K(s) + 1/2Br2(l) -> KBr(s)       -393.8 kJ

77.08+14.98+418.6+96.9+(-324.6)+LE = -393.8 kJ; LE = 283.6 + LE = -393.4
LE = -677.4 kJ

Hess’s Law is probably one of the favorite parts of general chemistry, particularly in high school or first level undergraduate courses, when it is finally understood. In classroom practice, Hess’s Law becomes a game or puzzle that a student attempts to solve after being given a set of clues in which to answer a question. However, before delving into Hess’s Law, something should be said about state functions, energy, and then enthalpy.

State Functions

Referring to the Merriam Webster dictionary [1], the definition of state is given as “a condition or stage in the physical being of something,” and of function is “a variable (as a quality, trait, or measurement) that depends on and varies with another.” Taking these together, a state function is a property or feature of an object which varies depending upon some other property of the object. This may seem like a rather mealy-mouth description for a function of state, but it is best understood by examining the total energy in a system and through the laws of thermodynamics.

Consider, perhaps, the most famous law of nature: conservation of energy. This is often applied in both chemistry and physics courses, particularly in thermodynamics for the former. The law is sometimes referred to as the first law of thermodynamics. It turns out that given a closed or isolated system, the total energy of that system is a function of state. We can see an example of this with mechanical potential energy.

Say our system consists of a floor, a table, and a book, with the book on the floor (you will not be apart of this system, only these objects). Now, when the book is just lying on the floor, it has some energy associated with it. Usually, we would say the book has 0 Joules of energy, defining the ground as having a height of 0 meters. Moving the book to the table top and leaving it resting on the table top gives the book some amount of potential energy (PE = mass*gravity*height). It does not matter if you take that book and run around the room with it, waving it above your head before setting it on the table, OR just lifting it straight up to the table. In the end, the book started on the floor and it ended on the table, what happened in between does not matter. Thus, the potential energy for the book is a state function.

Enthalpy

This brings us to enthalpy (H). Enthalpy is another state function. It is often used in physics to define the amount of total energy that would be needed to sort of “poof” of object into existence. That is, H = U + PV, where U is the total internal energy of the object/system, P is the pressure, and V is the volume. From a physics viewpoint, H consists of the total internal energy that an object contains U and the work (PV) needed to create space for that object to exist. In terms of chemistry, enthalpy is often used to determine if a process is endothermic or exothermic. However, it also gives us a means of using Hess’s Law to determine how a series of reactions will take place. Hess’s Law will take advantage of the fact that enthalpy is a state function, so it does not matter if by starting at point A to get to point Z we go straight from A to Z or go from A to B to C to ... to Y to Z.

Let’s use an example in order to illustrate Hess’s Law and how to use the feature that enthalpy is a function of state. I will extract the “clues” for the problem from NIST. The question will be: What is the change in entropy for the reaction when oxygen is mixed with sulfur to create sulfur trioxide? Your clues are presented in the following table:

Molecule	standard enthalpy (kJ/mol)
O (g)	249.18
O2 (g)	0
S (g)	276.98
SO (g)	5.01
SO2 (g)	-296.84
SO3 (g)	-395.77

Sometimes you will get the clues as reactions as well. But we will actually build the reactions ourselves. I will tell you the answer up front, then show you how we could have obtained the answer in many ways, thanks to Hess’s Law. It is -1420.29 kJ/mol (exothermic reaction).

The first way we could have developed our equation was to write the balanced equation:

3O + S <--> SO3

Since enthalpy is a state function, all that matters is the initial and final states. The initial state of the system is the reactants (3O + S). The final state is the products (SO3). Since we want to know the change in enthalpy, we always calculate a “change-in” by subtracting the initial from the final.

H(SO3) - [3*H(O) + H(S)] = -395.77 - [3*249.18 + 276.98]
= -395.77 - [1024.52]
= -395.77 - 1024.52
= -1420.29

3O + S <--> SO3 with dH = -1420.29 kJ/mol

The problem has been solved (Note: notice that I multiplied by 3. Tables like the one above given the energy per mole usually, so you need to have a balanced equation first, and then make sure to multiply by the correct coefficient). However, Hess’s Law tells us that we can use any path to get to the final answer. What if instead we went the route below:

2O + S <--> SO2
SO2 + O <--> SO3

Working out the math for the first step gives:

H(SO2) - [2*H(O) + H(S)] = -296.84 - [2*249.18 + 276.98]
= -296.84 - [775.34]
= -296.84 - 775.34
= -1027.18

And then the second step:

H(SO3) - [H(O) + H(SO2)] = -395.77 - [249.18 -296.84]
= -395.77 - [-47.66]
= -395.77 +47.66
= -348.11

Adding the two steps together: -1027.18-348.11 = -1420.29 kJ/mol (exothermic process)

Hess’s Law has been used to show that it does not matter which path I take to get from the first equation to the last equation. I could have even taken a step back and first made O2(g) and proceeded from there, the total change in enthalpy would be the same. As a final example of the process, a figure is presented which shows various steps that could have led to the conclusion of the example reaction. Note: If you ever want to use an equation in the opposite direction, just switch the reaction and then change the sign of the enthalpy. For example,

3O + S <--> SO3 with dH = -1420.29 kJ/mol
and
SO3 <--> 3 O + S with dH = 1420.29 kJ/mol

shows that forming sulfur trioxide from oxygen and sulfur (synthesis) is an exothermic reaction, but then to separate the sulfur trioxide back into its parts (decomposition) would be an endothermic reaction and require the energy to be put back into the system (to break the bonds, etc).