Question

The heat of solution of anhydrous CuSO$_4$ and CuSO$_4\cdot$5H$_2$O are $-70$ kJ mol$^{-1}$ and $+12$ kJ mol$^{-1}$ respectively. The heat of hydration of CuSO$_4$ to CuSO$_4\cdot$5H$_2$O is $-x$ kJ. The value of $x$ is _____.

Answer 1

Correct Answer: 82

To find the heat of hydration \( -x \) for the reaction where anhydrous CuSO₄ becomes CuSO₄·5H₂O, we need to consider the heats of solution provided for each compound.

The problem provides:

Anhydrous CuSO₄: Heat of solution = \(-70 \text{ kJ/mol}\)

CuSO₄·5H₂O: Heat of solution = \(+12 \text{ kJ/mol}\)

The heat of hydration is the energy change when CuSO₄ is hydrated to form CuSO₄·5H₂O.

Considering the reactions:

(1) \(\text{CuSO}_4 \;(\text{anhydrous}) + \text{solvent} \rightarrow \text{solution}\) \quad \(\Delta H_1 = -70 \text{ kJ/mol}\)

(2) \(\text{CuSO}_4 \cdot 5\text{H}_2\text{O} + \text{solvent} \rightarrow \text{solution}\) \quad \(\Delta H_2 = +12 \text{ kJ/mol}\)

The heat of hydration involves the transition from CuSO₄ to CuSO₄·5H₂O:

\(\text{CuSO}_4 + 5\text{H}_2\text{O} \rightarrow \text{CuSO}_4 \cdot 5\text{H}_2\text{O}\)

This can be understood using Hess's law, which states the total enthalpy change for a reaction is the sum of the steps:

\(\Delta H_{\text{hydration}} = \Delta H_2 - \Delta H_1\)

Substitute the given enthalpy values:

\(-x = 12 \text{ kJ/mol} - (-70 \text{ kJ/mol})\)

\(-x = 12 \text{ kJ/mol} + 70 \text{ kJ/mol}\)

\(-x = 82 \text{ kJ/mol}\)

Thus, \(x = 82 \text{ kJ/mol}\).

This value fits exactly within the given range of 82, 82.

Therefore, the heat of hydration is \(x = 82\) kJ mol\(^{-1}\).

Answer 2

The heat of hydration of CuSO₄ to CuSO₄ · 5H₂O is given by the difference between the heat of solution of anhydrous CuSO₄ and CuSO₄ · 5H₂O.

Heat of hydration = Heat of solution of anhydrous CuSO₄ − Heat of solution of CuSO₄ · 5H₂O.

Substitute the values:

\( x = \left| -70 \, \text{kJ mol}^{-1} - \left( +12 \, \text{kJ mol}^{-1} \right) \right| \).

\( x = \left| -70 - 12 \right| = \left| -82 \right| = 82 \).

Final Answer: 82 kJ.