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Question: First and second ionization energies of Mg are \(7.646\) and \(15.035\)eV respectively. The amount o...

First and second ionization energies of Mg are 7.6467.646 and 15.03515.035eV respectively. The amount of energy in kJ needed to convert all the atoms of magnesium into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} ions present in 1212 mg of magnesium vapour is:
(Given 11 eV = 96.596.5kJ/mol)
A. 1.51.5
B. 2.02.0
C. 1.11.1
D. 0.50.5

Explanation

Solution

We have to convert the in 1212 mg of magnesium into Mg2+Mg^{2+} ions so, first we will determine the energy required to convert one mole of Mg into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} ions. Then we will calculate the mole of Mg present in 1212 mg of magnesium. Then by comparing the calculated moles with energy of one mole we will determine the energy required to convert the 1212 mg of magnesium.

Complete step-by-step solution: It is given that first and second ionization energies of Mg are 7.6467.646and 15.03515.035eV respectively. So, after two ionization Mg converts into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} so, the total energy required to convert the Mg into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} will be the sum of first and second ionization energy.
So,
7.646 +15.035 = 22.6817.646\, + 15.035\, = \,22.681eV
Now, we will convert the above energy from eV into kJ as follows:
Given, 11 eV = 96.596.5kJ/mol
22.68122.681eV = 2188.72188.7kJ/mol
So, the energy required to convert one mole Mg atom into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} ions is 96.596.5kJ/mol.
Now, we will determine the mole of Mg present in 1212 mg of magnesium vapour as follows:
First we will convert the amount of magnesium vapour form mg to gram as follows:
11 mg = 10−3{10^{ - 3}} gram
1212 mg = 12×10−312 \times {10^{ - 3}} gram
The mole formula is as follows:
mole  =  massmolar mass{\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{{\text{molar}}\,{\text{mass}}}}
Molar mass of Mg is 2424 g/mol.
On substituting 12×10−312 \times {10^{ - 3}} gram for mass and 2424 g/mol for molar mass,
mole  =  12×10−324{\text{mole}}\,{\text{ = }}\,\dfrac{{12 \times {{10}^{ - 3}}}}{{24}}
mole  =  0.5×10−3{\text{mole}}\,{\text{ = }}\,0.5 \times {10^{ - 3}}
So, we know that the energy required to convert one mole Mg atom into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} ions is 96.596.5kJ/mol so, the energy will be required to convert the 0.5×10−30.5 \times {10^{ - 3}} mole is,
One mole Mg = 96.596.5kJ
0.5×10−30.5 \times {10^{ - 3}} mol = 1.11.1kJ
So, the amount of energy in kJ needed to convert all the atoms of magnesium into Mg2 + {\text{M}}{{\text{g}}^{{\text{2 + }}}} ions present in 1212 mg of magnesium vapour is 1.11.1kJ.

Therefore, option (C) 1.11.1kJ is correct.

Note: The energy required to remove an electron from an isolated gaseous atom is known as ionization energy. To remove the first electron the energy required is known as first ionization energy and energy required for the second electron is known as second ionization energy. Ionization energy is additive. If we do not convert the mg amount of magnesium vapour into gram then we will get our answer of moles in mmol.