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 1kcal per mol. If the solvation energies of Na & Cl ions are in the ratio 6:5, the enthalpy of hydration of soduim ion is:

a)-85.6 Kcal/mol b)-97.5 Kcal/mol c) 82.6 kcal/mol d)100 Kcal/mol

Please post explanation also.

To determine the enthalpy of hydration of the sodium ion (Na⁺), we can use the information provided about the lattice energy of NaCl and the dissolution process. Let's break down the problem step by step.

Understanding the Concepts

Lattice energy is the energy released when gaseous ions combine to form a solid ionic compound. For NaCl, this value is given as 180 Kcal/mol. When NaCl dissolves in water, it dissociates into Na⁺ and Cl^- ions. The dissolution process is endothermic, meaning it absorbs energy from the surroundings, quantified here as 1 Kcal/mol.

Setting Up the Equation

The overall enthalpy change for the dissolution of NaCl can be expressed using Hess's law, which states that the total enthalpy change is the sum of the enthalpy changes for each step. The dissolution process can be represented as:

• NaCl(s) → Na⁺(aq) + Cl^-(aq)

The enthalpy change for this process (ΔH_dissolution) can be expressed as:

ΔH_dissolution = Lattice Energy + ΔH_solvation

Where:

• Lattice Energy = -180 Kcal/mol (since it is released)
• ΔH_solvation = Enthalpy of hydration of Na⁺ + Enthalpy of hydration of Cl^-

Calculating the Enthalpy of Hydration

Given that the dissolution is endothermic by 1 Kcal/mol, we can set up the equation:

1 Kcal/mol = -180 Kcal/mol + ΔH_solvation

Rearranging gives us:

ΔH_solvation = 1 Kcal/mol + 180 Kcal/mol = 181 Kcal/mol

Now, we know that the solvation energies of Na⁺ and Cl^- are in the ratio of 6:5. Let’s denote the enthalpy of hydration of Na⁺ as 6x and that of Cl^- as 5x. Therefore, we can express ΔH_solvation as:

ΔH_solvation = 6x + 5x = 11x

Finding the Value of x

Now we can set up the equation:

11x = 181 Kcal/mol

Solving for x gives:

x = 181 Kcal/mol / 11 = 16.4545 Kcal/mol

Now, substituting back to find the enthalpy of hydration of Na⁺:

Enthalpy of hydration of Na⁺ = 6x = 6 * 16.4545 Kcal/mol = 98.727 Kcal/mol

Rounding this value gives us approximately 97.5 Kcal/mol.

Final Answer

Thus, the enthalpy of hydration of the sodium ion is closest to option b) 97.5 Kcal/mol.