The enthalpy change of solution of is and the lattice dissociation enthalpy of is . If the enthalpy — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpy change of solution of $NaCl(s)$ is $-2 \text{ kJ mol}^{-1}$ and the lattice dissociation enthalpy of $NaCl$ is $+772 \text{ kJ mol}^{-1}$. If the enthalpy of hydration of $Na^+(g)$ is $-390 \text{ kJ mol}^{-1}$, what is the enthalpy of hydration of $Cl^-(g)$?
Answer: A
💡 Solution & Explanation
Enthalpy of solution is given by $\Delta H_{sol} = \Delta H_{lattice\_dissoc} + \Delta H_{hyd}(Na^+) + \Delta H_{hyd}(Cl^-)$. Substituting the given thermodynamic values: $-2 = 772 + (-390) + \Delta H_{hyd}(Cl^-)$. Solving this mathematically gives $\Delta H_{hyd}(Cl^-) = -2 - 382 = -384 \text{ kJ mol}^{-1}$.
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