First and second ionization energy of an element M are 600 kJ/mole and 1600 kJ/mole. If 1000 KJ of energy is supplied
Answers
Answer:
Amount of Mg atoms=
24
1
=4.167×10
−2
mol
Energy absorbed in ionizing Mg to Mg
+
=4.167×10
−2
×740
=30.84KJ
Energy absorbed in ionizing Mg
+
to Mg
2+
=(50−30.84)KJ
=19.16KJ
Amount of Mg
+
converted to Mg
2+
=
1450
19.16
=1.321×10
−2
mol
Amount of Mg
+
remaining as such=4.167×10
−2
−1.321×10
−2
=2.846×10
−2
mol
Composition of final mixture would be as follows:
% of Mg
+
=
4.167×10
−2
2.846×10
−2
×100=68.3%
% of Mg
2+
=100−68.3=31.7 %
The first and second ionization of energy will be 63.8% and 31.7%.
GIVEN: The energy of an element M is 600 kJ/mole and 1600 kJ/mole and 1000 KJ of energy is supplied.
TO FIND First and second ionization energy
SOLUTION:
As we know,
Amount of Mg atoms = 1/24 = 4.167×10⁻²mol
Energy absorbed in ionizing Mg to Mg⁺=4.167×10⁻²×740 =30.84KJ
Energy absorbed in ionizing Mg⁺ to Mg²⁺=(50−30.84)KJ = 19.16KJ
Amount of Mg⁺ converted to Mg²⁺
= 19.16/1450
=1.321×10⁻²mol
Amount of Mg⁺ remaining as such
= 4.167×10⁻² −1.321×10⁻²
=2.846×10−2mol
The composition of the final mixture would be as follows:
% of Mg⁺= 2.846×10⁻²/4.167 * 10⁻² * 100 = 68.3%
% of Mg²⁺ = 100−68.3=31.7 %
The first and second ionization of energy will be 63.8% and 31.7%.
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