InterviewSolution
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InterviewSolution
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Lattice enthalpies of `BeF_(2),MgF_(2),CaF_(2) and BaF_(2)` are `-2906,-2610,-2459 and -2367 kJ mol^(-1)` resoectively. Hydration enthalpies of `Be^(2+),Ca^(2+),Ba^(2+) and F^(Θ)` are `-2194,-1921,-1577,-130 and -457 kJ mol^(-1)` respectively. Which of the fluorides has the highest solublility in water? |
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Answer» For `BeF_(2)`, `Delta_(hyd)H^(Θ)(BeF_(2))=Delta_(hyd)H^(Θ)(Be^(2+))+2Delta_(hyd)G^(Θ)(F^(Θ))` `=[-2494+2(-457)]kJ mol^(-1)` `=-3408 kJ mol^(-1)` `Delta_(U)H^(Θ)(BeF_(2))=-2906 kJ mol^(-1)` `Delta_("Soln")H^(Θ)(BeF_(2))=Delta_(hyd)H^(Θ)(BeF_(2))-Delta_(U)H^(Θ)(BeF_(2))` `=[-3408-(-2906)]kJ mol^(-1)` `=-502 kJ mol^(-1)` For `MgF_(2)`, `Delta_(hyd)H^(Θ)(MgF_(2))=Delta_(hyd)H^(Θ)(Mg^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))` `=[-1921+2(-457)]kJ mol^(-1)` `=-2835 kJ mol^(-1)` `Delta_(U)H^(Θ)(MgF_(2))=-2610 kJ Mol^(-1)` `Delta_("soln")H^(Θ)(MgF_(2))=Delta_(hyd)H^(Θ)(MgF_(2))-Delta_(U)H^(Θ)(MgF_(2))` `=[-2835-(-2610)]kJ mol^(-1)=-225 kJ mol^(-1)` For `CaF_(2)`, `Delta_(hyd)H^(Θ)(CaF_(2))=Delta_(hyd)H^(Θ)(Ca^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))` `=[-1577+2(-457)] kJ mol^(-1)` `=-2491 kJ mol^(-1)` `Delta_(U)H^(Θ)(CaF_(2))=-2459 kJ mol^(-1)` `Delta_("soln")H^(Θ)(CaF_(2))=Delta_(hyd)H^(Θ) (CaF_(2))-Delta_(U)H^(Θ)(CaF_(2))` `=[-2491-(-2459)] kJ mol^(-1)` `=-32 kJ mol^(-1)` For `BaF_(2)`, `Delta_(hyd)H^(Θ)(BaF_(2)=Delta_(hyd)H^(Θ)(Ba^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))` `=-1305+2(-457)` `=-2219 kJ mol^(-1)` `Delta_(U)H^(Θ)(BaF_(2))=-2367 kJ mol^(-1)` `Delta_("soln")H^(Θ)(BaF_(2))=Delta_(hyd)H^(Θ)(BaF_(2))-Delta_(U)H^(Θ)(BaF_(2))` `=[-2219-(-2367)] kJ mol^(-1)` `=148 kJ mol^(-1)` Since enthalpy of solution `Delta_("soln")H^(Θ)` is maximum for `BeF_(2)`, it has highest solubility in water. |
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