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\(^{-1}\).
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.
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The reaction : \(A_2 \rightleftharpoons 2A\)

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Which of the following is not correct?
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)