One mole of magnesium in the vapour state absorbed 1300 kJ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ mol − 1 respectively, the final composition of the mixture i

Question

One mole of magnesium in the vapour state absorbed 1300 kJ of energy. If the first and second ionization energies of Mg are 750 and 1450 kJ mol − 1 respectively, the final composition of the mixture is

TestHub · Accepted Answer

Mg ⟶ Mg⁺ + e⁻.

To ionize one mole of Mg completely, 750 kJ is spent, and the energy left is 550 kJ.

1450 kJ ionizes 1 mole of Mg⁺,

so 550 kJ ionizes 0.38 mole of Mg⁺.

Mg⁺ ⟶ Mg²⁺ + e⁻.

1

$1 - 0.38$ , 0.38

% of Mg⁺ = 62%, and that of Mg²⁺ = 38%.

Chemistry - Thermochemistry Question with Solution | TestHub

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