Question

The first, second and third ionization energies of $Al$ are $578$,$1817$ and$2745$ $kJ\;mo{l^{ - 1}}$ respectively. Calculate the energy required to convert all the atoms of$Al$ to $A{l^{3 + }}$ present in $270$mg of $Al$ vapors.

Answer 1

Ionization energy is the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule. It is quantitatively expressed as
$A\left( g \right) + energy \to {{\text{A}}^ + }\left( g \right) + {e^ - }$
Where $A$is any isolated atom or molecule, ${{\text{A}}^ + }$ is resultant ion when an electron is removed from original atom or molecules and ${e^ - }$is removed electron.
The ionization energy of a chemical element is expressed in joules or electron volts. The ionization energy is often reported as the amount of energy (in joules) required to ionize the number of atoms or molecules present in one mole.

Complete answer:
First ionization energy of $Al$corresponds to:
$Al \to A{l^ + } + {e^ - }$ $I.{E_1} = 578{\text{ }}kJ{\text{ }}mo{l^{ - 1}}$
Second ionization energy of $Al$corresponds to:
$A{l^ + } \to A{l^{2 + }} + {e^ - }$ $I.{E_2} = 1817{\text{ }}kJ{\text{ }}mo{l^{ - 1}}$
Third ionization energy of $Al$corresponds to:
$A{l^{2 + }} \to A{l^{3 + }} + {e^ - }$ $I.{E_3} = 2745{\text{ }}kJ{\text{ }}mo{l^{ - 1}}$
Adding all three equations we get:
$Al \to A{l^{3 + }} + 3{e^ - }$ $I.E = I.{E_1} + I.{E_2} + I.{E_3}$
$I.E = I.{E_1} + I.{E_2} + I.{E_3}$
$I.E = (578 + 1817 + 2745)kJ{\text{ }}mo{l^{ - 1}}$
$I.E = 5140{\text{ }}kJ{\text{ }}mo{l^{ - 1}}$
Now, consider the case for $270$mg of $Al$ vapors.
$Number{\text{ }}of{\text{ }}moles,{\text{ }}n = \dfrac{{given{\text{ }}weight}}{{molar{\text{ }}mass}}$
$n = \dfrac{{270 \times {{10}^{ - 3}}}}{{27}} = 0.01mol$
$I.E = 5140{\text{ }}kJ{\text{ }}mo{l^{ - 1}} \times 0.01mol$
$I.E = 51.40{\text{ }}kJ$
Hence, $51.40{\text{ }}kJ$is the energy required to convert all the atoms of$Al$ to $A{l^{3 + }}$ present in $270$mg of $Al$ vapors.

ADDITIONAL INFORMATION:
Ionization energy is generally an endothermic process. As a rule, the closer the outermost electrons to the nucleus of the atom, the higher is the atom ionization energy.
Comparison of ionization energies of atoms in the periodic table reveals two periodic trends which follow the rules of Coulombic attraction:
Ionization energy generally increases from left to right within a given period.
Ionization energy generally decreases from top to bottom in a given group.

Note:
Generally, the ${(n + 1)^{th}}$ ionization energy of a particular element is larger than the nth ionization energy. When the next ionization energy involves removing an electron from the same electron shell, the increase in ionization energy is primarily due to the increased net charge of the ion from which the electron is being removed. Electrons removed from more highly charged ions experience greater forces of electrostatic attraction; thus, their removal requires more energy. In addition, when the next ionization energy involves removing an electron from a lower electron shell, the greatly decreased distance between the nucleus and the electron also increases both the electrostatic force and the distance over which that force must be overcome to remove the electron. Both of these factors further increase the ionization energy.