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Properties of Enthalpy

One gram of M...

Question

One gram of $M g$ atoms in vapour phase absorbs $50 kJ$ of energy. Find the percentage composition of $M g^{2 +}$ formed as a result of absorption of energy. $I E_{1}$ and $I E_{2}$ for $M g$ are $720$ and $1000 kJ/mole$ respectively.
(Given, atomic mass of $M g = 24 u$ )

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Solution

Energy required to convert $\frac{1}{24} mol$ of $M g^{+} = \frac{1}{24} \times 720 = 30.00 kJ$
Remaining energy $= 50 - 30 = 20 kJ$
Number of moles of $M g^{+}$ converted to $M g^{2 +} = \frac{20}{1000} = 0.02 mol$
$%$ of $M g^{2 +} = \frac{0.02}{\frac{1}{24}} \times 100 = 48 %$

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(IE)1 and (IE) 2 of Mg are 740,1540kJ mol-' calculate the percentage of Mg⁺and Mg²⁺ if 1g of Mg absorbs 50.0KJ of energy.

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