The lattice energy of solid \(NaCl\) is \(180\, \ kcal/mol\). The dissolution of the solid in water, in the form of ions is endothermic to the extent of 1 kcal/mol. If the solvation energies of \(Na^+\) and \(Cl^–\) ions are in the ratio \(6:5\), what is the enthalpy of hydration of sodium ion?

–97.6 kcal/mol

The correct answer is option 2. –97.6 kcal/mol.

To find the enthalpy of hydration of the sodium ion, we need to use the given information and the relationships between the different energy changes involved in the dissolution process of \( NaCl \) in water.

Given Data:

Lattice energy of \( NaCl \): \( U = 180 \, \text{kcal/mol} \)

Dissolution process is endothermic: \( \Delta H_{\text{dissolution}} = 1 \, \text{kcal/mol} \)Ratio of solvation energies of \( Na^+ \) and \( Cl^- \) ions: \(6:5\)

Relevant Equations and Relationships:

When \( NaCl \) dissolves in water, the overall enthalpy change (\( \Delta H_{\text{dissolution}} \)) can be expressed as:

\(\Delta H_{\text{dissolution}} = \Delta H_{\text{lattice energy}} + \Delta H_{\text{hydration}}\)

where:

\( \Delta H_{\text{lattice energy}} \) is the energy required to break the lattice, which is equal to the lattice energy \( U \), but it is an endothermic process, so it is \( +U \).

\( \Delta H_{\text{hydration}} \) is the energy released during the hydration (solvation) of the ions.

Given that the solvation energies of \( Na^+ \) and \( Cl^- \) ions are in the ratio 6:5, let's denote the enthalpy of hydration of \( Na^+ \) as \( \Delta H_{\text{hydration}}(Na^+) = x \). Then, the enthalpy of hydration of \( Cl^- \) would be \( \Delta H_{\text{hydration}}(Cl^-) = \frac{5}{6}x \).

The total enthalpy of hydration \( \Delta H_{\text{hydration}} \) for \( NaCl \) can be written as:

\(\Delta H_{\text{hydration}} = x + \frac{5}{6}x = \frac{11}{6}x\)

Step-by-Step Calculation:

From the dissolution process:

\(\Delta H_{\text{dissolution}} = \Delta H_{\text{lattice energy}} + \Delta H_{\text{hydration}}\)

Substituting the given values

\(1 \, \text{kcal/mol} = 180 \, \text{kcal/mol} + \Delta H_{\text{hydration}}\)

Solve for \( \Delta H_{\text{hydration}} \):

\(\Delta H_{\text{hydration}} = 1 \, \text{kcal/mol} - 180 \, \text{kcal/mol} = -179 \, \text{kcal/mol}\)

Now, we know:

\(\Delta H_{\text{hydration}} = -179 \, \text{kcal/mol}\)  

And,

\(\Delta H_{\text{hydration}} = \frac{11}{6}x\)

Solve for \( x \) (enthalpy of hydration of \( Na^+ \)):

\(-179 \, \text{kcal/mol} = \frac{11}{6}x\)

\(x = -179 \, \text{kcal/mol} \times \frac{6}{11}\)

\(x = -97.636 \, \text{kcal/mol}\)

Thus, the enthalpy of hydration of the sodium ion (\( Na^+ \)) is approximately: \(-97.6 \, \text{kcal/mol}\)