Question

How much energy in kJ will be required to convert all the magnesium atoms into magnesium ions present in 12 mg of the metal vapours? $IE_{1}$, $IE_{2}$ for Mg being 7.646 and 15.035 eV respectively.

Answer 1

We know that,  

                        $1eV=1.602 \times10^{-22} kJ=96.49KJ/mol$

First and second ionization energies of Mg are $IE_{1} =7.646$ and $IE_{2}=15.035$ (given).

Therefore, the total energy required to convert magnesium atoms to magnesium ions is  

                                           $IE_{1}+IE_{2}=(7.646+15.035)eV$

                                           =22⋅681eV  

Total required energy in terms of kJ is   22⋅68×96⋅49=2188⋅39 kJ

Now, molar mass of magnesium is 24 gm/mol means 24 X 103 mg/mol

So, 12 mg of magnesium = $0.5\times10^{-3}mol$

Therefore, total energy required to convert all Mg atoms to $Mg^{2+}$ ions is $2188.39\times 0.5\times 10^{-3}=1.0942KJ$