The enthalpy change of solution of is -2 kJ/mol and the lattice dissociation enthalpy of is +772 kJ/ — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy change of solution of $NaCl(s)$ is -2 kJ/mol and the lattice dissociation enthalpy of $NaCl$ is +772 kJ/mol. If the enthalpy of hydration of $Na^+(g)$ is -390 kJ/mol, what is the enthalpy of hydration of $Cl^-(g)$?
Answer: A
💡 Solution & Explanation
Enthalpy of solution is the sum of lattice dissociation enthalpy and the hydration enthalpies of the separated ions. $\Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hyd}(Na^+) + \Delta H_{hyd}(Cl^-)$. $-2 = 772 + (-390) + \Delta H_{hyd}(Cl^-)$. Solving gives $-384$ kJ/mol.
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