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Question

The IE value of, Al(g)Al+(g)+e is 577.5 kJ mol1 and ΔH for Al(g)Al3+(g)+3e is 5140 kJ mol1. If second and third IE values are in the ratio 2:3. IE2 + IE30.5 in kJ mol1 is:

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Solution

Given that

IE1 of Al=577.5kJmol1....(i)
[IE1+IE2+IE3] of Al=5140kJmol1....(ii)
Also given IE2=(2/3)IE3.....(iii)
solving eqs (i),(ii),(iii)
IE2=1825kJmol1
IE3=2737.5kJmol1

IE2+IE3=1825+2737.5

=4562.5 kJ/mol

IE1+IE20.5=4562.50.5=4562

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