Question

For Aluminium the values of \(\mathrm{I}, \mathrm{I}_{2}, \mathrm{I}_{3}\) are 578,1817 and \(2745 \mathrm{~kJ} / \mathrm{mole}\) respectively. What will be the energy needed to convert all the atoms of \(\mathrm{Al}\) to \(\mathrm{Al}^{3+}\) present in \(135 \mathrm{mg}\) of \(\mathrm{Al}\) vapours. (a) \(51.4 \mathrm{~kJ}\) (b) \(25.7 \mathrm{~kJ}\) (c) \(77.1 \mathrm{~kJ}\) (d) \(12.85 \mathrm{~kJ}\)

Short answer

The energy needed is 25.7 kJ, corresponding to option (b).

Step-by-step solution

Calculate Moles of Aluminium

First, find the number of moles of Aluminium in 135 mg. The atomic mass of
{Al}
is 27 g/mol.Convert the mass from mg to grams:\[ 135 \text{ mg} = 0.135 \text{ g} \]Now, calculate the number of moles:\[ \text{moles of Al} = \frac{0.135 \text{ g}}{27 \text{ g/mol}} = 0.005 \text{ moles} \]

Calculate Total Ionization Energy

The ionization energies to convert Al to Al\(^{3+}\) are given by: \( \mathrm{I_{1}} = 578 \text{ kJ/mol} \)\( \mathrm{I_{2}} = 1817 \text{ kJ/mol} \)\( \mathrm{I_{3}} = 2745 \text{ kJ/mol} \)Total ionization energy for one mole of Al to Al\(^{3+}\):\[ \mathrm{I_{total}} = 578 + 1817 + 2745 = 5140 \text{ kJ/mol} \]

Calculate Energy Required for 0.005 Moles

Now, calculate the energy required to convert 0.005 moles of Aluminium to Al\(^{3+}\):\[ \text{Energy for 0.005 moles} = 0.005 \text{ moles} \times 5140 \text{ kJ/mol} \]\[ = 25.7 \text{ kJ} \]

Select the Correct Option

The calculated energy needed is 25.7 kJ. Comparing this result with the given options, the correct option is (b) 25.7 kJ.

Key concepts

Aluminium

Aluminium is a chemical element with the symbol Al and atomic number 13. It's known for being a lightweight metal, making it highly useful in many industries like automotive and construction. Despite its abundance in the Earth's crust, it's never found free in nature. Instead, it's always combined in various minerals. Aluminium's chemically reactive nature means it's used in making a variety of alloys and compounds. When considering ionization energy, aluminium is an atom with electrons in three energy levels, and its electron configuration explains how it loses electrons to form positive ions.

Moles Calculation

Understanding the concept of moles is crucial in chemistry. The mole is a fundamental unit in the International System of Units (SI) and is used to measure the amount of substance. It represents Avogadro's number of entities, which is approximately \(6.022 \times 10^{23}\) atoms, molecules, or whatever particles you’re considering.

To calculate moles, you use the formula:

• Moles = \(\frac{\text{mass of substance (g)}}{\text{molar mass (g/mol)}}\)

In the given problem, you start by converting the mass of aluminium from milligrams to grams. Then you divide by the molar mass of aluminium, which is 27 g/mol, to find the number of moles. This key calculation helps in determining how much of the element you're dealing with on a molecular level.

Energy Conversion

Energy conversion in this context refers to computing the total ionization energy required to form a particular ion from its neutral atom. This is done using the ionization energies provided. Since ionization energy is the amount of energy needed to remove an electron from a gaseous atom or ion, it plays an essential role in understanding atomic interactions and electron removals.

• The energies for aluminium for each ionization phase (from Al to Al\(^{3+}\)) are 578 kJ/mol, 1817 kJ/mol, and 2745 kJ/mol for the first, second, and third ionization energy steps respectively.

The total ionization energy converts these into a single value for one mole of aluminium. By multiplying this total by the number of moles you've calculated, you find the energy required to ionize all the atoms you've got.

Ionization Steps

Ionization energy refers to the energy needed to remove electrons from an atom. For aluminium, which can lose up to three electrons, it goes through three distinct ionization steps.

Each step requires a higher amount of energy:

• \(I_1 = 578\) kJ/mol for removing the first electron.
• \(I_2 = 1817\) kJ/mol for the second electron.
• \(I_3 = 2745\) kJ/mol for the third electron to fully ionize to Al\(^{3+}\).

This increase happens because after each electron is removed, the remaining electrons are held more tightly by the nucleus. As such, more energy is needed to remove each subsequent electron. Understanding these steps helps clarify why the total energy calculated amounts to 5140 kJ for a full conversion from Al to Al\(^{3+}\).