JEE MAIN - Chemistry (2021 - 26th August Morning Shift - No. 20)
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)
$${\Delta _f}{H^\Theta }$$ 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
718
Explanation
$${\Delta _f}H_{KCl}^\Theta = {\Delta _{sub}}H_{(K)}^\Theta + {\Delta _{ioniztion}}H_{(K)}^\Theta + {1 \over 2}{\Delta _{bond}}H_{(C{l_2})}^\Theta + {\Delta _{electron\,gain}}H_{(Cl)}^\Theta + {\Delta _{lattice}}H_{(KCl)}^\Theta $$
$$ \Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta $$
$$ \Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$$ kJ mol$$-$$1
The magnitude of lattice enthalpy of KCl in kJ mol$$-$$1 is 718 (Nearest integer).
$$ \Rightarrow - 436.7 = 89.2 + 419.0 + {1 \over 2}(243.0) + \{ - 348.6\} + {\Delta _{lattice}}H_{(KCl)}^\Theta $$
$$ \Rightarrow {\Delta _{lattice}}H_{(KCl)}^\Theta = - 717.8$$ kJ mol$$-$$1
The magnitude of lattice enthalpy of KCl in kJ mol$$-$$1 is 718 (Nearest integer).
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