The lattice enthalpy of solid `NaCl` is `772 kJ mol^(-1)` and enthalpy of solution is `2 kJmol^(-1)`. If the hydration enthaply of `Na^(+)` & `Cl^(-)`

Question

The lattice enthalpy of solid `NaCl` is `772 kJ mol^(-1)` and enthalpy of solution is `2 kJmol^(-1)`. If the hydration enthaply of `Na^(+)` & `Cl^(-)`

The lattice enthalpy of solid `NaCl` is `772 kJ mol^(-1)` and enthalpy of solution is `2 kJmol^(-1)`. If the hydration enthaply of `Na^(+)` & `Cl^(-)` ions are in the ratio of `3 : 2.5`, what is the enthalpy of hydration of chloride ion ?
A. `-140 kJmol^(-1)`
B. `-350 kJmol^(-1)`
C. `-351.81 mol^(-1)`
D. None

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B
`DeltaH_("solution")=DeltaH_("Hyd")(x)+DeltaH_("Lattice")`
`2=x+772`
`x=-770`
Enthalpy of hydration of `Cl^(-) =2.5/5.5 xx770=-350" kJ mol"^(-1)`

4 views Answered 2023-02-05 15:29