Question

Magnesium has the first and second ionisation potential 7.646 and 15.035 eV respectively. What is the amount of energy required in kJ to convert all the magnesium atoms present in 24 mg to magnesium vapours ?​

Answer 1

The energy required to convert one Mg atom to Mg

2+

ion is the sum of first and second ionization potentials = 7.646+15.035=22.681eV

=96.5×22.681=2188.7kJ/mol

The atomic mass of Mg is 24 g/mol.

12 mg of Mg corresponds to 0.5 mmoles.

The amount of energy required to convert 0.5 mmol of Mg atoms to Mg

2+

ions = 0.5×10

−3

×2188.7=1.1kJ

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Answer 2

Given

• First ionization energy of Magnesium = 7.646 eV.

• Second ionization energy = 15.035 eV

To Find

Energy in KJ needed to convert 24 mg of magnesium into its vapour.

Solutions

The reaction that takes place are :-

$\green{Mg\: \longrightarrow\:Mg^+\:+\:e^-\:\:\:\:({I.E}_1\:=\:7.646\:eV)\\ \\ \\ Mg^+\:\longrightarrow\:Mg^{2+}\:+\:e^-\:\:\:({I.E.}_2\:=\:15.035\:eV)}\\ \\ \\ \textsf{Now net ionization of the reaction } \\ \\ \\ \green{Mg\:\longrightarrow\:Mg^{2+}\:+\:2e^-}\\\\= {I.E.}_1\:+\:{I.E.}_2\\ \\ \\=7.646\:+\:15.035\\ \\ \\=22.681\: eV \\ \\ \\ \orange{1 eV \:= \:96.49\: KJ \:mol^{-1}} \\ \\ \\ Thus,\\ \\ \\ 22.681\:eV\:=22.681\:\times\:96.49\:kJ\:mol^{-1}\:=2188.489\:kJ\:mol^{-1}\\ \\ \\ \24 \:mg \:of \:magnesium\: = \:10^{-3} \:moles\: magnesium } \\ \\ \\ \textsf{Thus, total energy required} \\ \\ \\ =\:2188.49\:\times\:10^{-3} \:= \:2.188 kJ.$

$\boxed{\green{\textsf{Net energy needed is 2.188 kJ}}}$