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One mole of a...

Question

One mole of atoms of an element M absorbs 576 kJ energy for the formation of $M^{+}$ ions. The total amount of energy required for one mole of M atoms to form tripositive ions is 5140 kJ. 2nd and 3rd IP values of the metal are in 2 : 3 ratio. Calculate $I P_{2}$ and $I P_{3}$ .

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Solution

They gave IP $_{1}$ = 576 KJ/mol.......(1)
IP $_{1}$ +IP $_{2}$ +IP $_{3}$ = 5140 KJ/mol.......(2)
Because there M changes to M $^{3 +}$
and $\frac{I P_{2}}{I P_{3}}$ = $\frac{2}{3}$ IP $_{3}$ = $\frac{3}{2}$ IP $_{2}$ .........(3)
Using equation (2)
Ip $_{1}$ + IP $_{2}$ +IP $_{3}$ =5140 KJ/mol.
576 +IP $_{2}$ + $\frac{3}{2}$ IP $_{2}$ = 5140 [from (3)]
576 +IP $_{2}$ [ $\frac{5}{2}$ ] =5140.
$\frac{5}{2} \times$ IP $_{2}$ = 5140 -576.
$\frac{5}{2} \times$ IP $_{2}$ = 4564
IP $_{2}$ =1825.6 KJ/mol
IP $_{3}$ = $\frac{3}{2}$ IP $_{2}$ =2, 738.4 KJ/mol

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