Question

Lattice enthalpy for NaCl is +788 kJ mol^-1 and ΔH°_hyd= -784 kJ mol^-1. Enthaply of solution of NaCl is

• A. +572 kJ mol^-1
• B. +4 kJ mol^-1
• C. -572 kJ mol^-1
• D. -4 kJ mol^-1

Answer 1

The Correct Option is B

Enthalpy of Solution for NaCl

Given: 

• Lattice enthalpy for NaCl, \( \Delta H_{\text{lattice}} = +788 \text{ kJ/mol} \)
• Enthalpy of solution for NaCl, \( \Delta H_{\text{solution}} = -784 \text{ kJ/mol} \)

 

Objective: Find the enthalpy of hydration (\( \Delta H_{\text{hydration}} \)).

The relationship between lattice enthalpy, enthalpy of hydration, and enthalpy of solution is given by: \[ \Delta H_{\text{solution}} = \Delta H_{\text{hydration}} - \Delta H_{\text{lattice}} \]

Rearranging to solve for \( \Delta H_{\text{hydration}} \): \[ \Delta H_{\text{hydration}} = \Delta H_{\text{solution}} + \Delta H_{\text{lattice}} \]

Substituting the given values: \[ \Delta H_{\text{hydration}} = -784 \text{ kJ/mol} + 788 \text{ kJ/mol} = 4 \text{ kJ/mol} \]

Conclusion: The enthalpy of hydration is \( \Delta H_{\text{hydration}} = +4 \text{ kJ/mol} \), which corresponds to option D. +4 kJ/mol.

Answer 2

The enthalpy of solution (\( \Delta H_{\text{sol}} \)) of NaCl can be calculated using the following formula: \[ \Delta H_{\text{sol}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hyd}} \] where:
\( \Delta H_{\text{lattice}} \) is the lattice enthalpy of NaCl, which is given as \( +788 \, \text{kJ mol}^{-1} \),
\( \Delta H_{\text{hyd}} \) is the enthalpy of hydration of NaCl, which is given as \( -784 \, \text{kJ mol}^{-1} \).

Substitute the given values into the formula: \[ \Delta H_{\text{sol}} = 788 \, \text{kJ mol}^{-1} + (-784 \, \text{kJ mol}^{-1}) \] \[ \Delta H_{\text{sol}} = 788 - 784 = +4 \, \text{kJ mol}^{-1} \] Thus, the enthalpy of solution of NaCl is \( +4 \, \text{kJ mol}^{-1} \).

The correct answer is: \[{\text{(B) } +4 \, \text{kJ mol}^{-1}} \]