Question

A chemist has one mole of X-atom. He finds that on absorption of 410kJ, 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. Find the electron affinity of X.

Answer 1

$\frac{X}{2}\to \frac{X^{+}}{2}+e\frac{1}{2}.....\left(i\right)$
$e+\frac{X}{2}\to \frac{X^{-}}{2}\frac{1}{2}E.A(-ve).....\left(ii\right)$
(i)+(ii)
$x\to \frac{1}{2}{X}^{+}+\frac{1}{2}{X}^{-}\frac{1}{2}(I.E-E.A)=410kJ$
$(I.E-E.A)=820kJ$
Now $\frac{1}{2}{x}^{-}\to \frac{1}{2}{x}^{+}+2e^{-}....\left(iii\right)$
$\Delta H=735$
Now equation (iii) can be achieved by (i) + reverse (ii) and we will get
$\frac{1}{2}I.E+\frac{1}{2}E.A=735$
$I.E+E.A=\mathrm{1470....}\left(iv\right)$
2E.A=650
E.A-325 kJ/mol