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The lattice enthalpy of solid `NaCI` is `772 kJ mol^(-1)` and enthalpy of solution is `2kJ mol^(-1)`. If the hydration enthalpy of `Na^(+) &CI^(-)` ions are in the ratio of 3: 25, what is the magnitude of enthalpy of hydration (in kJ) of chloride ion ? |
Answer» Correct Answer - [350] [Lattice enthalpy `= 772 Kj//Mol` `((DeltaH.E)_(Na^(+)))/((DeltaH.E)_(CI^(-))) = (30)/(25) = (6)/(5),underset((x))((DeltaHE)_(Na^(+))) =(6)/(5)underset((y))((DeltaHE)_(CI^(-)))` `x = (6)/(5)y` `(DeltaH)_(solution")=` Lattice enthalpy `+H.E.` of `na^(+) +H.E.` of `CI^(-)` `2 = 272 +(y) +(6)/(5)(y)` `- 770 = (11)/(5) y, y = (770 xx 5)/(11) = 350 KJ]` |
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