Calculate the energy required to convert all atoms in \(4.8 \mathrm{~g}\) of \(\mathrm{Mg}\) to…
Calculate the energy required to convert all atoms in \(4.8 \mathrm{~g}\) of \(\mathrm{Mg}\) to \(\mathrm{Mg}^{2+}\) in the vapour state. \(\mathrm{IE}_1\) and \(\mathrm{IE}_2\) of \(\mathrm{Mg}\) are \(740 \mathrm{~kJ} / \mathrm{mol}\) and \(1450 \mathrm{~kJ} / \mathrm{mol}\) respectively.
\(+740 \mathrm{~kJ} / \mathrm{mol}\)
\(-740 \mathrm{~kJ} / \mathrm{mol}\)
\(-1450 \mathrm{~kJ} / \mathrm{mol}\)
\(+438 \mathrm{~kJ} / \mathrm{mol}\)
Solution
For 1 mole of \(\left(\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}\right)\) \(\mathrm{IE}=\mathbb{E E}_1+\mathrm{IE}_2=(740+1450)=2190 \mathrm{~kJ} / \mathrm{mol}\).
Number of mole in \(4.8 \mathrm{~g}\) of \(\mathrm{Mg}=4.8 / 24=0.2 \mathrm{~mol}\)
For 1 mole energy required \(=2190 \mathrm{~kJ} / \mathrm{mol}\)
For 0.2 mole energy required \(=2190 \times 0.2=438 \mathrm{~kJ}\)
Thus, for \(4.8 \mathrm{~g}\) of \(\mathrm{Mg}\) to \(\mathrm{Mg}^{2+}\) conversion, energy required is \(438 \mathrm{~kJ}\).
Hence, the correct option is (d).