Question

We consider formation of NaCl(s) using Born-Haber cycle which involves following steps:
(i) $Na\left(s\right)\frac{1}{2}{CI}_2⟶NaCI\left(s\right)\Delta H=Q$
(II) $Na\left(s\right)⟶Na\left(g\right)\Delta H=S$
(III) $Na\left(g\right)⟶{Na}^{+}\left(g\right)+e^{-}\Delta H=I$
(IV) $\frac{1}{2}{CI}_2\left(g\right)⟶CI\left(g\right)\Delta H=\frac{D}{2}$
(v) $CI\left(g\right)+e^{-}⟶C{I}^{-}\left(g\right)\Delta H=?$
(vI) ${Na}^{+}\left(g\right)+{CI}^{-}\left(g\right)⟶NaCI\left(s\right)\Delta H=-U$
Can you derive value of (EA) of CI(g) atom.

• A. $S+I+D/2$
• B. $Q+U-S-I-D/2$
• C. $I+Q-U$
• D. $S+Q$

Answer 1

Answer: (B)

$Q+U-S-I-D/2$

(V) involves addition of electron to $CI\left(g\right)$
$CI\left(g\right)+e^{-}⟶C{I}^{-}\left(g\right)$ EA=x
Energy released is (EA)
Adding (ii) to (vi) gives
$Na\left(s\right)\frac{1}{2}{CI}_2\left(g\right)⟶NaCI\left(s\right)$
Which is also the step (i), hence
$S+I+\frac{D}{2}+x\left(EA\right)-U=Q$
hence, x is calculated .