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  • One mole of anhydrous dissolves in water and liberates <img alt="20 \mathrm{~J} / \mathrm{mol}" src

One mole of anhydrous \mathrm{AB} dissolves in water and liberates 20 \mathrm{~J} / \mathrm{mol} of heat. The value of \Delta_{\mathrm{hyd}} \mathrm{H}= \text{of} \; \mathrm{AB}\; \text{is}\; -30 \mathrm{~J} / \mathrm{mol}. The heat of dissolution of hydrated (\text{in}\; \mathrm{KJ} / \mathrm{mol}) salt \mathrm{AB} .2 \mathrm{H}_2 \mathrm{O} is

Option: 1

-10


Option: 2

-50


Option: 3

10

 


Option: 4

-40


Answers (1)

best_answer

Given :

Heat of dissolution of anhydrous \mathrm{AB}=-20 \mathrm{~J} / \mathrm{mol}

Heat of hydration of \mathrm{AB}=\Delta_{\mathrm{hyd}} H=-30 \mathrm{~J} / \mathrm{mol}

To find : Heat of dissolution of hydrated salt \mathrm{AB} .2 \mathrm{H}_2 \mathrm{O}

We can break the process into two steps:

Dissolve anhydrous AB in water to form hydrated \mathrm{AB} . x \mathrm{H}_2 \mathrm{O} :

\begin{aligned} & \mathrm{A B(s) \rightarrow A B(a q)\ \ \ \Delta H=-20 \mathrm{~J} / \mathrm{mol}} \\ \\& \mathrm{AB}(\mathrm{aq})+\mathrm{xH} _2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{AB} \cdot \mathrm{xH}_2 \mathrm{O}(\mathrm{s})\ \ \ \Delta \mathrm{H}=? \\ \\& \text { Hydrate } \mathrm{AB} \cdot \mathrm{xH}_2 \mathrm{O} \text { to } \mathrm{AB} \cdot 2 \mathrm{H}_2 \mathrm{O}: \\ \\& \mathrm{AB} \cdot \mathrm{xH}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{AB} \cdot 2 \mathrm{H}_2 \mathrm{O}(\mathrm{s}) \ \ \Delta \mathrm{H}=? \end{aligned}

To find the value of \Delta \mathrm{H}_2 , we can use the relationship:

\Delta_{\text {hyd }} \mathrm{H}=\Delta \mathrm{H}_2-\Delta \mathrm{H}_1

Where \Delta \mathrm{H}_1 is the heat of dissolution of AB to form hydrated \mathrm{AB} \cdot x \mathrm{H}_2 \mathrm{O} and \Delta_{\text {hyd }} \mathrm{H}=\Delta \mathrm{H}_2-\Delta \mathrm{H}_1 is the heat of hydration of AB.

Rearranging the above equation, we get:

\Delta \mathrm{H}_2=\Delta \mathrm{H}_1-\Delta_{\mathrm{hyd}} \mathrm{H}

Substituting the given values, we get:

\Delta \mathrm{H}_2=(-20)-(-30)=+10 \mathrm{~J} / \mathrm{mol}

Therefore, the heat of the dissolution of hydrated salt \mathrm{A B .2 \mathrm{H}_2 \mathrm{O} \text { is }+10 \mathrm{~kJ} / \mathrm{mol} \text {. }}

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Rishi

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