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Standard XII

Chemistry

Electron Affinity

A chemist has...

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.

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Solution

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

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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.

X2→X+2+e12.....(i)e+X2→X−212E.A(−ve).....(ii)(i)+(ii)x→12X++12X−12(I.E−E.A)=410kJ(I.E−E.A)=820kJNow 12x−→12x++2e−....(iii) ΔH=735Now equation (iii) can be achieved by (i) + reverse (ii) and we will get 12I.E+12E.A=735I.E+E.A=1470....(iv)2E.A=650E.A-325 kJ/mol

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.