Enthalpy of Atomization - Definition, Heat of Atomization and FAQs

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 utilised 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 the 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, O_2, their enthalpy of atomization will be the same as the bond dissociation energy.

• Polyatomic molecules-

For polyatomic molecules, the above is not true. The bond dissociation energy is not as same as that of the 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 kJ/mol

• In such a case, we use the 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 the 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 the 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 greater 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 increases. 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 of 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 is 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 Vaporisation

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

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

Enthalpy of vaporisation is denoted by $\Delta$H_vap

For example-

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

Enthalpies of vaporisation also indicate the magnitude of intermolecular forces. The greater the value of the enthalpy of vaporisation greater the attractive forces. Example- Acetone has dipole-dipole interactions, which are relatively weaker; therefore, it requires less heat to form vapours of 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 an 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 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

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

Question 4: Which of the following has the lowest enthalpy of atomisation?

1) Fe

2) Co

3) Ni

4) (correct) Cu

Solution:

The enthalpy of atomisation values of the given elements (in $\mathbf{k J m o l}^{-}$)are listed below:
$\mathrm{Fe}: 416 \mathrm{Co}: 425 \mathrm{Ni}: 430 \mathrm{Cu}: 339$
Hence, the answer is option (4).

Practice more questions from the link given below

For more questions to practice, the following MCQs will help in the preparation for competitive examinations:

Read More: