Enthalpy of Solution ($$\text{CuSO}_4$$)

Formula / Core Reaction:

$$\text{CuSO}_4(s) + 5\text{H}_2\text{O}(l) \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \quad (\Delta H_\text{hydration} < 0)$$

$$\text{CuSO}_4(s) + aq \rightarrow \text{Cu}^{2+}(aq) + \text{SO}_4^{2-}(aq) \quad (\Delta H_\text{soln, anhydrous} < 0)$$

$$\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) + aq \rightarrow \text{Cu}^{2+}(aq) + \text{SO}_4^{2-}(aq) \quad (\Delta H_\text{soln, hydrated} > 0)$$

Concept: Dissolution of anhydrous $$\text{CuSO}_4$$ is exothermic because its lattice energy is smaller in magnitude than its high hydration energy. Dissolving hydrated $$\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$$ is endothermic because lattice breakdown dominates since hydration has already occurred.

Notation / Key Observation:

• $$\Delta H_\text{soln} (\text{Anhydrous CuSO}_4) < 0$$: Heat is released ($$\text{Exothermic}$$);
• $$\Delta H_\text{soln} (\text{CuSO}_4 \cdot 5\text{H}_2\text{O}) > 0$$: Heat is absorbed ($$\text{Endothermic}$$).