Question

$Cs{F}_{\left(s\right)}\to C{s}_{\left(aq\right)}^{+}+F_{\left(aq\right)}^{-}$ $\Delta H=-40$ kJ $mo{l}^{-1}$
If lattice energy of CsF is $750$ kJ $mo{l}^{-1}$. Then the summation of heat of hydration of $C{s}^{+}$ and $F^{+}$ ion is:

• A. $-750$ kJ $mo{l}^{-}$
• B. $-790$ kJ $mo{l}^{-}$
• C. $-710$ kJ $mo{l}^{-}$
• D. $-830$ kJ $mo{l}^{-}$

Answer 1

The correct option is B $-790$ kJ $mo{l}^{-}$

Solution:- (B) $-790kJ/mol$

${CsF}_{\left(s\right)}⟶{{Cs}^{+}}_{(aq.)}+{F^{-}}_{(aq.)}\Delta H=-40kJ/mol$

For the above reaction-

$\Delta H_R={\sum \Delta H^0}_{\left(product\right)}-{\sum \Delta H^0}_{\left(reactant\right)}$

$\Rightarrow \Delta H_R={\Delta H}_{CsF}-({\Delta H}_{{Cs}^{+}}+{\Delta H}_{F^{-}})$

$\Rightarrow ({\Delta H}_{{Cs}^{+}}+{\Delta H}_{F^{-}})={\Delta H}_{CsF}-\Delta H_R$

Given:-

$\Delta H_R=-40kJ/mol$

${\Delta H}_{CsF}=750kJ/mol$

$\therefore ({\Delta H}_{{Cs}^{+}}+{\Delta H}_{F^{-}})=(-40)-750=-790kJ/mol$