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calculate the lattice energy of $KI,$ given that enthalphy of (i) sublimation of K is $90\;K\;J mol^{-1}$, (ii) dissociation of $I_2$ to $I^{-}$ is $213\;kJmol^{-1}$ (iii) ionisation of K to $K^{+}$ is $415\;kJ mol^{-1}$ (iv) electron affinity for I to $I^{-}$ is $-294.6 \;KJ\;mol^{-1}$ and $SH_f$ overall is $-327\;kJ\;mol^{-1}$

$(A)\;666\;KJ\;mol^{-1} \\ (B)\;644\;KJ\;mol^{-1} \\(C)\;633\;KJ\;mol^{-1} \\(D)\;650\;KJ\;mol^{-1} $
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$\Delta H_f=-327\;KJ\;mol^{-1}$
$\Delta H_S=90\;KJ\;mol^{-1}$
$IE =415\;KJ\;mol^{-1}$
$\Delta H_d=213\;KJ\;mol^{-1}$
$EA= -294.6 \;KJ\;mol^{-1}$
$\Delta H_f=\Delta H_S+IE +\large\frac{1}{2} $$\Delta H_d+EA+U$
$-327 =90 +415+\large\frac{1}{2} $$\times 213 -294.6 +U$
$U= -644\;KJ \;mol^{-1}$
Hence B is the correct answer.
answered Apr 4, 2014 by meena.p

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