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Question: When one mole of anhydrous \({\text{FeS}}{{\text{O}}_{\text{4}}}\) is dissolved in excess of water, ...

When one mole of anhydrous FeSO4{\text{FeS}}{{\text{O}}_{\text{4}}} is dissolved in excess of water, there is evolution of 58.2 kJ58.2{\text{ kJ}} of heat. But when one mole of FeSO4⋅5H2O{\text{FeS}}{{\text{O}}_{\text{4}}} \cdot 5{{\text{H}}_2}{\text{O}} is dissolved in water, the heat change is +8.6 kJ + 8.6{\text{ kJ}}. Calculate the enthalpy of hydration of anhydrous FeSO4{\text{FeS}}{{\text{O}}_{\text{4}}}.
A. −49.6 kJ - 49.6{\text{ kJ}}
B. −66.8 kJ - 66.8{\text{ kJ}}
C. +49.6 kJ + 49.6{\text{ kJ}}
D. +66.8 kJ + 66.8{\text{ kJ}}

Explanation

Solution

The enthalpy change that occurs when a sample is dissolved in solvent is known as the enthalpy of the solution. The molecule on dissolving in water gets hydrated and the heat released when an anhydrous molecule gets dehydrated is known as enthalpy of hydration. We can obtain the enthalpy of hydration by relating the given enthalpy of solution.

Complete step by stepanswer:
We will write the chemical equations for all the possible reactions.
The reaction when anhydrous FeSO4{\text{FeS}}{{\text{O}}_{\text{4}}} is dissolved in water is as follows:
FeSO4(s)+aq→FeSO4(aq){\text{FeS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right) + {\text{aq}} \to {\text{FeS}}{{\text{O}}_{\text{4}}}\left( {{\text{aq}}} \right)
The reaction when FeSO4⋅5H2O{\text{FeS}}{{\text{O}}_{\text{4}}} \cdot 5{{\text{H}}_2}{\text{O}} is dissolved in water is as follows:
FeSO4⋅5H2O(s)+aq→FeSO4+5H2O{\text{FeS}}{{\text{O}}_{\text{4}}} \cdot 5{{\text{H}}_2}{\text{O}}\left( {\text{s}} \right) + {\text{aq}} \to {\text{FeS}}{{\text{O}}_{\text{4}}} + 5{{\text{H}}_2}{\text{O}}
We have to calculate the enthalpy of hydration of anhydrous FeSO4{\text{FeS}}{{\text{O}}_{\text{4}}}. The reaction is as follows:
FeSO4+5H2O+aq→FeSO4⋅5H2O{\text{FeS}}{{\text{O}}_{\text{4}}} + 5{{\text{H}}_2}{\text{O}} + {\text{aq}} \to {\text{FeS}}{{\text{O}}_{\text{4}}} \cdot 5{{\text{H}}_2}{\text{O}}
Thus, we can calculate the enthalpy of hydration when we subtract the lattice energy from the heat of the solution. The formula for the enthalpy of hydration is as follows:
Enthalpy of hydration=ΔHsolution−ΔHlattice{\text{Enthalpy of hydration}} = \Delta {H_{{\text{solution}}}} - \Delta {H_{{\text{lattice}}}}
Where ΔHsolution\Delta {H_{{\text{solution}}}} is the heat of the solution,
ΔHlattice\Delta {H_{{\text{lattice}}}} is the lattice energy.
We are given that one mole of anhydrous FeSO4{\text{FeS}}{{\text{O}}_{\text{4}}} on dissolving in water evolves 58.2 kJ58.2{\text{ kJ}} of heat and when one mole of FeSO4⋅5H2O{\text{FeS}}{{\text{O}}_{\text{4}}} \cdot 5{{\text{H}}_2}{\text{O}} on dissolving in water evolves +8.6 kJ + 8.6{\text{ kJ}}. Thus,
⇒Enthalpy of hydration=(58.2 kJ)−(+8.6 kJ)\Rightarrow {\text{Enthalpy of hydration}} = \left( {58.2{\text{ kJ}}} \right) - \left( { + 8.6{\text{ kJ}}} \right)
⇒Enthalpy of hydration=+49.6 kJ\Rightarrow {\text{Enthalpy of hydration}} = + 49.6{\text{ kJ}}
Thus, the enthalpy of hydration of anhydrous FeSO4{\text{FeS}}{{\text{O}}_{\text{4}}} is +49.6 kJ + 49.6{\text{ kJ}}.

**Thus, the correct option is (C) +49.6 kJ + 49.6{\text{ kJ}}.

Note:**
The amount of energy which is released when one mole of any substance is dissolved in water to form an infinitely diluted solution. The extent of hydration depends on the charge density on the ion, size of the ion. Do not get confused between the relation of the reactions to obtain the required equation. Also, remember that the enthalpy of hydration and the enthalpy of solution are different.