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Question

A chemist has one mole of X-atoms. He finds that on absorption of 410 kJ, half of X-atoms transfer one electron to the other half. If all the resulting X ions are subsequently converted to X+ ions, an addition of 735 kJ is required. The magnitude of electron affinity of X is.

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Solution

X2X+2+e 12I.E. ...(i)
e+X2X2 12E.A.(ve) ...(ii)
(i) + (ii)
X12X++12X 12(I.E.+E.A.)=410 kJ
I.E.+E.A.=820 kJ
Now 12X12X++2e ...(iii) H=735 kJ
Now equaiton (iii) can be achieved by (i) + reverse (ii) and we will get
12I.E.12E.A.=735
I.E.E.A.=1470 ...(iv)
2 E.A.=650
E.A. = -325 kJ/mol.

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