Enthalpy Change of Atomisation
What Exactly is Atomisation Energy?
Imagine you have a piece of sodium metal. It looks solid and continuous, but it's actually made up of countless sodium atoms all bonded together. Now, imagine you could carefully pluck each atom away from its neighbors, one by one, turning the solid metal into a gas of individual, separate sodium atoms floating around. The energy change that happens during this process is the enthalpy change of atomisation.
It is defined as the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state under standard conditions[1]. The standard state is the most stable physical form of an element at 100 kPa pressure and a specified temperature, usually 298 K (25 °C).
For example, the atomisation of sodium is represented by the equation:
$Na(s) \rightarrow Na(g)$ $\Delta H_{at} = +107 \text{ kJ mol}^{-1}$
This tells us that 107 kJ of energy is needed to convert one mole of solid sodium into one mole of gaseous sodium atoms.
Atomisation for Different Types of Elements
The process and the energy required differ depending on whether the element is a metal, a non-metal, or a noble gas. The structure of the element dictates how many bonds need to be broken.
| Element Type | Standard State | Atomisation Process | Example & $\Delta H_{at}$ (kJ mol$^{-1}$) |
|---|---|---|---|
| Metals (e.g., Sodium) | Giant Metallic Lattice | Breaking strong metallic bonds in the solid lattice. | $Na(s) \rightarrow Na(g)$ +107 |
| Non-Metals (e.g., Chlorine) | Diatomic Molecules ($Cl_2$) | First turning liquid to gas, then breaking the covalent bond in the $Cl_2$ molecule. | $\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$ +121 |
| Noble Gases (e.g., Argon) | Monatomic Gas | No bonds to break, as atoms are already separate. | $Ar(g) \rightarrow Ar(g)$ 0 |
A crucial point from the table is for diatomic molecules like chlorine. The definition is for the formation of one mole of gaseous atoms. Since one mole of $Cl_2$ gas produces two moles of Cl atoms, the equation is written for half a mole of $Cl_2$: $\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$. This ensures the stoichiometry matches the definition.
The Link to Bond Enthalpy
For simple diatomic molecules, the atomisation enthalpy is directly equal to the bond enthalpy of that molecule. Bond enthalpy is the energy required to break one mole of a specific chemical bond in the gaseous state.
For instance, the bond enthalpy for the $H-H$ bond is +436 kJ mol$^{-1}$. This is the same value as the atomisation enthalpy for hydrogen:
$\frac{1}{2}H_2(g) \rightarrow H(g)$ $\Delta H_{at} = +218 \text{ kJ mol}^{-1}$
Wait, that's half the value! Remember the definition: for one mole of atoms. Breaking one mole of $H-H$ bonds ($H_2(g) \rightarrow 2H(g)$) produces two moles of H atoms and has $\Delta H = +436 \text{ kJ mol}^{-1}$. Therefore, to get the atomisation enthalpy for one mole of H atoms, we take half of that value: +218 kJ mol$^{-1}$.
For a molecule like oxygen, $O_2$, the atomisation enthalpy is half the energy required for $O_2(g) \rightarrow 2O(g)$.
A Practical Application: The Born-Haber Cycle
One of the most important uses of atomisation enthalpy is in the Born-Haber cycle[2]. This is an energy cycle that allows us to calculate the lattice energy[3] of an ionic compound or to check the consistency of other thermochemical data, like the enthalpy of formation.
Let's consider the formation of sodium chloride (table salt) from its elements:
$Na(s) + \frac{1}{2}Cl_2(g) \rightarrow NaCl(s)$ $\Delta H_f = -411 \text{ kJ mol}^{-1}$
This is the direct route, and the enthalpy change is the enthalpy of formation. The Born-Haber cycle breaks this down into a series of steps, including atomisation.
1. Atomise Sodium: $Na(s) \rightarrow Na(g)$ $\Delta H_{at}(Na) = +107 \text{ kJ mol}^{-1}$
2. Atomise Chlorine: $\frac{1}{2}Cl_2(g) \rightarrow Cl(g)$ $\Delta H_{at}(Cl) = +121 \text{ kJ mol}^{-1}$
3. Ionise Sodium: $Na(g) \rightarrow Na^+(g) + e^-$ $\Delta H_{I.E.} = +496 \text{ kJ mol}^{-1}$
4. Add Electron to Chlorine (Electron Affinity): $Cl(g) + e^- \rightarrow Cl^-(g)$ $\Delta H_{E.A.} = -349 \text{ kJ mol}^{-1}$
5. Form the Lattice: $Na^+(g) + Cl^-(g) \rightarrow NaCl(s)$ $\Delta H_{lattice} = ?$
According to Hess's Law[4], the sum of these steps must equal the direct enthalpy of formation:
$\Delta H_f = \Delta H_{at}(Na) + \Delta H_{at}(Cl) + \Delta H_{I.E.} + \Delta H_{E.A.} + \Delta H_{lattice}$
We can plug in the values to find the lattice energy: -787 kJ mol$^{-1}$.
This example shows how crucial the atomisation energies are. Without them, we couldn't complete the cycle and understand the stability of the ionic lattice.
Important Questions
Why is the enthalpy change of atomisation always endothermic?
It is always endothermic because energy is always required to break chemical bonds. Whether it's the metallic bonds in a metal or the covalent bonds in a diatomic molecule, overcoming the attractive forces between atoms demands an input of energy from the surroundings.
How does atomisation enthalpy relate to the periodic table?
Atomisation enthalpy shows clear trends across the periodic table. It generally increases across a period due to an increasing number of electrons involved in bonding and a decreasing atomic radius, leading to stronger bonds. It generally decreases down a group because the atomic radius increases, and the outer electrons are farther from the nucleus, resulting in weaker metallic or covalent bonds.
Can we have an atomisation enthalpy for a compound?
The term is specifically defined for elements. For a compound, the equivalent concept is the "enthalpy of atomisation of a compound," which would be the energy required to convert one mole of a compound in its standard state into its constituent gaseous atoms. For example, for water, it would be the energy for $H_2O(l) \rightarrow 2H(g) + O(g)$.
The enthalpy change of atomisation, $\Delta H_{at}$, is more than just a number in a data table. It is a direct window into the strength of the forces that hold elements together. From explaining why tungsten has such a high melting point (very strong metallic bonds, hence high $\Delta H_{at}$) to being an indispensable part of the Born-Haber cycle for understanding ionic solids, this concept bridges the gap between the microscopic world of atoms and bonds and the macroscopic world of measurable energy changes. Grasping this idea is a fundamental step in mastering chemical energetics.
Footnote
[1] Standard Conditions: A set of specific conditions used to allow fair comparison of thermodynamic data: a pressure of 100 kPa and a temperature of 298 K (25 °C).
[2] Born-Haber Cycle: A thermochemical cycle that applies Hess's Law to calculate the lattice energy of an ionic compound by relating it to other measurable quantities like enthalpy of formation, ionisation energy, electron affinity, and atomisation enthalpy.
[3] Lattice Energy: The enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions. It is a measure of the strength of the ionic bonds in the crystal lattice.
[4] Hess's Law: A fundamental law in chemistry which states that the total enthalpy change for a reaction is independent of the pathway taken, depending only on the initial and final states.