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.
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 : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are

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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,