The Born-Haber cycle for KCl is evaluated with the following data :
${\Delta _f}{H^\Theta }$ for KCl = $-$436.7 kJ mol$-$1 ;
${\Delta _{sub}}{H^\Theta }$ for K = 89.2 kJ mol$-$1 ;
${\Delta _{ionization}}{H^\Theta }$ for K = 419.0 kJ mol$-$1 ;
${\Delta _{electron\,gain}}{H^\Theta }$ for Cl(g) = $-$348.6 kJ mol$-$1 ;
${\Delta _{bond}}{H^\Theta }$ for Cl2 = 243.0 kJ mol$-$1
The magnitude of lattice enthalpy of KCl in kJ mol$-$1 is _____________ (Nearest integer)
Answer (integer)
718
Solution
$${\Delta _f}H_{KCl}^\Theta = {\Delta _{sub}}H_{(K)}^\Theta + {\Delta _{ioniztion}}H_{(K)}^\Theta + {1 \over 2}{\Delta _{bond}}H_{(C{l_2})}^\Theta + {\Delta _{electron\,gain}}H_{(Cl)}^\Theta + {\Delta _{lattice}}H_{(KCl)}^\Theta $$<br><br>$$ \Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta $$<br><br>$\Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$ kJ mol<sup>$-$1</sup><br><br>The magnitude of lattice enthalpy of KCl in kJ mol<sup>$-$1</sup> is 718 (Nearest integer).
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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