Question

1 mole of magnesium in the vapour state absorbed 1200 kJ $mo{l}^{-1}$ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ $mo{l}^{-1}$ respectively, the final composition of the mixture is

• A. 31 % $M{g}^{+}$ + 69 % $M{g}^{2+}$
• B. 69 % $M{g}^{+}$ + 31% $M{g}^{2+}$
• C. 86 % $M{g}^{+}$ + 14 % $M{g}^{2+}$
• D. 14 % $M{g}^{+}$ + 86 % $M{g}^{2+}$

Answer 1

The correct option is B 69 % $M{g}^{+}$ + 31% $M{g}^{2+}$

Energy absorbed in the ionization of 1 mole of Mg (g) to $M{g}^{+}$ (g) = 750 kJ.
Energy left unconsumed = 1200 - 750 = 450 kJ
This energy is used to convert $M{g}^{+}$ (g) to $M{g}^{2+}$(g), but we actually need to 1450 kJ to convert it entirely.

So, instead of all molecules absorbing a little bit of energy, a certain fraction of them only would absorb enough energy and get ionized while the rest remain in the same state. Remember, energy is quantized.

Thus,

% of$M{g}^{2+}$(g) = 450 x 100/ 1450 = 31 %

% of$M{g}^{+}$ (g) = 100 - 31 = 69%