The lattice energy of solid NaCl is 772 kJ mol − 1 and enthalpy of solution is 2 kJmol − 1 . If the hydration energy of Na + & Cl − ions are in the ratio of 3:2.5, what is the enthalpy of hydration of

Question

The lattice energy of solid NaCl is 772 kJ mol − 1 and enthalpy of solution is 2 kJmol − 1 . If the hydration energy of Na + & Cl − ions are in the ratio of 3:2.5, what is the enthalpy of hydration of chloride ion?

TestHub · Accepted Answer

Enthalpy of solution is given by:

ΔH(solution) = Lattice Energy + Hydration Energy

Given:

Lattice energy of NaCl = $772 kJ mo l^{- 1}$

Enthalpy of solution = $2 kJ mo l^{- 1}$

Therefore,

Hydration Energy = Enthalpy of solution - Lattice Energy

Hydration Energy = $2 - 772 = - 770 kJ mo l^{- 1}$

Let the hydration energy of $Na^{+}$ be $3 x$ and the hydration energy of $Cl^{-}$ be $2.5 x$ .

Total hydration energy = Hydration energy of $Na^{+}$ + Hydration energy of $Cl^{-}$

$- 770 = 3 x + 2.5 x$

$- 770 = 5.5 x$

$x = \frac{- 770}{5.5} = - 140 kJ mo l^{- 1}$

Hydration energy of $Cl^{-} = 2.5 x = 2.5 \times - 140 = - 350 kJ mo l^{- 1}$

Therefore, the enthalpy of hydration of chloride ion is $- 350 kJ mo l^{- 1}$ .

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