Question

The energy required to convert all the atoms of magnesium to magnesium ions present in 24 mg of magnesium vapours is (First and second ionization enthalpies of Mg are 737.76 and $1450.73 KJ \, mol^{-1}$ respectively).

• A. 2 KJ
• B. 102 K J
• C. 92 K J
• D. 92 KJ

Answer 1

Answer: (A)

According to the definition of successive ionization enthalpies.
$Mg_{\left(g\right)}+\Delta_{i} H_{1} \to Mg^{+}_{\left(g\right)}+e^{-} ; $
$\Delta_{i} H_{1}=735.76 KJ \, mol^{-1}$
$Mg_{\left(g\right)}^{+}+\Delta_{i} H_{2} \to Mg^{2+}_{\left(g\right)}+e^{-};$
$\Delta_{i}H_{2}=1450.73 KJ \, mol^{-1}$
$\therefore \, $ Total amount of energy needed to convert $Mg_{\left(g\right)}$ atom into $Mg^{2+}\left(g\right) ion=\Delta_{i}H_{1}+\Delta_{i}H_{2}$
$=737.76+1450.73 KJ mol^{-1}$
$24 mg \, of \, Mg=\frac{24}{1000} g=\frac{24}{1000\times24} mole=10^{-3} mole$
$\therefore \,$ Amount of energy needed to ionize $10^{-3}$ mole of Mg vapours = 2188.49 $\times$ $10^{-3}$ = 2.188 kJ ≈ 2 kJ