Enthalpy of Atomization - Definition, Heat of Atomization and FAQs

The energy needed to break a substance into individual atoms is known as the Enthalpy of atomisation. Under standard conditions, the amount of energy required to turn one mole of substance into a separate atom into the gas phase. Energy of atomisation is represented in kilojoules per mole. Enthalpy of atomisation represents the strength of the metallic bond for metals and the strength of the covalent bond for covalent compounds. The enthalpy of atomisation is just like the energy needed to separate a puzzle piece until every piece is divided. Consider puzzle pieces as atoms and the puzzle as a compound. There is a hidden force that holds atoms in salt or atoms in an iron rod, and the energy of atomisation represents the energy required to separate these atoms. Enthalpy is a thermodynamic term that talks about the amount of heat released or absorbed in a process. Therefore, the heat of atomization is always positive.

In this article, students will learn about the enthalpy of atomization of diatomic and polyatomic molecules, enthalpy of atomization of d block, enthalpy of vaporization, enthalpy of sublimation, enthalpy of transition, etc.

Enthalpy of Atomization Definition

• Before we understand about enthalpy of atomization it is important to understand the quantity 'Enthalpy' $\Delta \mathrm{H}$ and its origin. Atomization meaning can be inferred from the term which is to convert into atoms.

• Chemical reactions performed in the lab proceed at a constant pressure i.e., atmospheric pressure.

• A thermodynamic quantity known as enthalpy was introduced to study reactions occurring at constant pressure because internal energy $\Delta U$ (thermodynamic quantity introduced from the first law) was only meant for reactions that were occurring at constant volume.

• Every reaction is associated with either absorption of energy or a release of energy. Therefore, enthalpy $\Delta H$ is a quantity related to heat change.

• The Enthalpy change for the reaction -Heat change can occur for a variety of chemical reactions such as combustion, atomization, hydration, solution, neutralization, phase transitions such as vaporization, fusion, etc.

• Enthalpy of atomization is the change in enthalpy when a mole of a substance is converted to its atoms in the gaseous state by breaking the bonds of this substance.

The Heat of Atomization

• Atomization means to convert into atoms.

It is the heat change in breaking bonds of one mole of substance into its atoms in a gaseous state at standard conditions ( 298 K and 1 bar). Enthalpy of atomization is represented as $\Delta \mathrm{aH}$.

• Diatomic molecules-

Consider the following example-

$\mathrm{H}_2(\mathrm{~g}) \xrightarrow{\text { heat, } \Delta H=+435.94 \mathrm{~kJ} / \mathrm{mol}} 2 \mathrm{H}(\mathrm{g})$

Dihydrogen is a diatomic molecule and the energy supplied will be utilized in breaking its bond to produce its individual atoms in a gaseous state. Therefore, the Enthalpy of atomization is always a positive quantity.

• The heat of atomization, in a case of H₂ can also be termed as Heat Dissociation enthalpy.

In this case enthalpy of atomization is the same as that of bond dissociation enthalpy. Bond dissociation enthalpy is the enthalpy change for a mole of substance to break its covalent bonds into its atoms in a gaseous state.

• For All the diatomic molecules Ex- Cl₂, O₂ their enthalpy of atomization will be the same as bond dissociation energy.

• Polyatomic molecules-

For polyatomic molecules, the above is not true. The bond dissociation energy is not as same as that of enthalpy change of atomization.

For eg-

Consider the molecule of methane.

$\mathrm{CH}_4(\mathrm{~g}) \xrightarrow{\text { heat }} \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})$

$\Delta H$=$1648 \mathrm{~kJ} / \mathrm{mol}$

Despite having the same C-H bond length and energy the energy required to break the C-H bond is different for all the bonds. Here the enthalpy change for the reaction 4H is equal to 1665 kjmol-1

• In such a case, we use mean bond enthalpy. Mean bond dissociation energy or bond enthalpy is the average or mean of bond dissociation enthalpies required to break a particular bond. The energy of atomization in the case of H₂ can also be termed as Heat Dissociation enthalpy.

Bond enthalpy is different for different compounds.

$\mathrm{CH}_4(\mathrm{~g}) \xrightarrow{\text { heat }} \mathrm{CH}_3(\mathrm{~g})+\mathrm{H}(\mathrm{g})$, $\Delta H$=+ 427KJ/mol

$\mathrm{CH}_3(\mathrm{~g}) \xrightarrow{\text { heat }} \mathrm{CH}_2(\mathrm{~g})+\mathrm{H}(\mathrm{g})$, $\Delta H$ = +437 KJ/mol

$\mathrm{CH}_2(\mathrm{~g}) \xrightarrow{\text { heat }} \mathrm{CH}(\mathrm{g})+\mathrm{H}(\mathrm{g})$, $\Delta H$ = +459KJ/mol

$\mathrm{CH}(\mathrm{g}) \xrightarrow{\text { heat }} \mathrm{C}(\mathrm{g})+\mathrm{H}(\mathrm{g})$, $\Delta H$ = +347KJ/mol

• Bond dissociation enthalpy or heat of atomization for common molecules like Cl₂ is 242.5 kJ/mol and I₂ is 15.1 kJ/mol

Enthalpy of atomization of d block-

• d- block elements are known to have higher boiling as well as higher melting points. An element with a higher melting point has higher metallic bonding energy. Metallic bonding energy depends on the Enthalpy of atomization.

• The more the number of unpaired electrons in the d-orbital greater is the energy of atomization. The enthalpy of atomization of transition elements increases as the unpaired electrons increase. With the increase in unpaired electrons, the interatomic interactions also increase.

• The observed trend is that of an increase in unpaired electrons when moving from left to right in a period. Melting points decrease from the second half of the transition series because use pairing of electrons takes place.

For example-

Iron has a melting point of 1808K and cobalt has a melting point of 1768K. Iron has a higher enthalpy of atomization than copper because of the number of electrons present in its d- orbitals.

Electronic configuration of Iron- 3d⁶4s²

Electronic configuration of Cobalt- 3d⁷4s²

For iron- 3d_xy² 3d_yz¹3d_zx¹ (3d_x²-y²)¹ (3d_z²)¹ the total number of electrons in iron are four.

For Cobalt- 3d_xy² 3d_yz²3d_zx¹ (3d_x²-y²)1 (3d_z²)¹

The total number of unpaired electrons is three.

It is because of the higher number of unpaired electrons in Iron the enthalpy of atomization is high.

Phase transitions- Transformation of states of matter into one another requires heat because of the difference in intermolecular forces in liquid, gaseous, and solid.

Standard Enthalpy of fusion is heat change for 1 mole of a solid substance to convert into liquid at constant temperature (melting point). It is denoted by $\Delta f u s \mathrm{H}$. For example- ice has $\Delta f u s \mathrm{H}=6.0 \mathrm{kjmol}-1$

It is always a positive quantity.

Standard enthalpy of vaporization:

The standard enthalpy of vaporization is a type of heat change during the phase transition of a liquid to gaseous. Phase transitions are also accompanied by a change in heat.

Enthalpy of vaporization is the heat absorbed to form vapors for one mole of a liquid at constant temperature (boiling point) under standard conditions (1 bar pressure).

Enthalpy of vaporization is denoted by $\Delta$ vaph

For example-

Nitrogen has a heat of vaporization is $5.39 \mathrm{kJmol}-1$
NaCl has the heat of vaporization $170.0 \mathrm{kJmol}-1$

Enthalpies of vaporization also indicate the magnitude of intermolecular forces. The greater the value of enthalpy of vaporization greater the attractive forces. Example- Acetone has dipole-dipole interactions which are relatively weaker therefore, it requires less heat to form vapors of its 1 mole as compared to water.

Enthalpy of sublimation- It is the heat absorbed by one mole of a solid substance to convert to its gaseous state directly at a constant temperature and constant pressure (1 bar). The enthalpy of sublimation is denoted by $\Delta$ subH.

Example- Heat of sublimation for dry ice is 25.2 kJmol-1

Enthalpy of transition

• There are enthalpies of a few reactions which cannot be calculated directly. So those enthalpies can be determined indirectly from available data on other kinds of enthalpies. Allotropic changes from rhombic sulfur to monoclinic sulfur, graphite to Diamond can be determined using Hess’s law.

• The enthalpy of transition is such enthalpy that cannot be determined directly. The heat of transition for allotropic changes of elements can be calculated from the enthalpy of combustion data.

  For example-
  C(diamond) C(graphite)
  Sor this allotropic change combustion of carbon in diamond form and combustion of carbon in graphite can be subtracted to obtain the value for its enthalpy of transition.

Some Solved Examples

Example 1: The heat of Atomisation of $\mathrm{PH}_3(\mathrm{~g})$ and $\mathrm{P}_2 \mathrm{H}_4(\mathrm{~g})$ are 954 $\mathrm{KJ} \mathrm{mol}{ }^{-1}$ and $1488 \mathrm{KJ} \mathrm{mol}^{-1}$ respectively. The P - P bond energy in $\mathrm{KJ} \mathrm{mol}{ }^{-1}$ is -
1) (correct) 216
2) 428
3) 318
4) 1272

Solution

$3 \mathrm{P}-\mathrm{H}$ bond $=954 \mathrm{KJ} / \mathrm{mol}$
$1 \mathrm{P}-\mathrm{H}$ bond $=318 \mathrm{KJ} / \mathrm{mol}$
$4 \mathrm{P}-\mathrm{H}$ bond $+1-\mathrm{P}-\mathrm{P}$ bond $=1488 \mathrm{KJ} / \mathrm{mol}$
$1 \mathrm{P}-\mathrm{P}$ bond $=(1488-4 * 318)=216 \mathrm{KJ} / \mathrm{mol}$

Hence, the answer is ( $216 \mathrm{KJ} / \mathrm{mol}$ ).

Example 2:

$\begin{aligned} & \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2,(\mathrm{~g}) ; \quad \Delta \mathrm{H}=-94.3 \mathrm{kcal} / \mathrm{mol} \\ & \mathrm{CO}(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) ; \quad \Delta H=-67.4 \mathrm{kcal} / \mathrm{mol} \\ & \mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{O}(\mathrm{g}) ; \quad \Delta H=117.4 \mathrm{kcal} / \mathrm{mol} \\ & \mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})+\mathrm{O}(\mathrm{g}) ; \quad \Delta H=230.6 \mathrm{kcal} / \mathrm{mol} \\ & \text { Calculate } \Delta H \text { for } C(\mathrm{~s}) \rightarrow C(\mathrm{~g}) \text { in } \mathrm{kcal} / \mathrm{mol}\end{aligned}$

1) 171
2) 154
3) 117
4) (correct) 145

Solution
$C(s) \rightarrow C(g)$ can be obtained as,

$
\Delta H=\Delta H_1-\Delta H_2-\frac{1}{2} \Delta H_3+\Delta H_4=145
$

Hence, the answer is the option (4).

Example 3: Calculate $\mathrm{A}-\mathrm{X}$ bond enthalpy in $\mathrm{KJ} /$ mole
Given :
$\Delta f H(A X 3, g)=153 \mathrm{~kJ} / \mathrm{mole} \Delta \mathrm{Hf}(\mathrm{X}, \mathrm{g})=61 \mathrm{~kJ} /$ mole $\Delta$ Hatomisation $(\mathrm{A}, \mathrm{S})=1$
$57 \mathrm{~kJ} /$ mole
1) (correct) 62.33
2) 53.33
3) 41.33
4) 37.66

Solution

$\begin{aligned} & A(s)+\frac{3}{2} X_2(g) \longrightarrow A X_3(g) \\ & 153=(157+3 \times 61)-[B \cdot E \cdot(A-X) \times 3] \\ & B \cdot E \cdot(A-X)=\frac{-187}{-3}=62.33 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$