Mg(g) +2F(9) - Mg2+ Mg2+6) + 2F 1.E, and I. E, of Mg() are 700 and 1400 k.J/mol. EA, for F() is – 350 kJ/mol The heat change for the reaction is 200 y then y is
Answers
Answered by
1
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 %
Similar questions