The heat of solution of anhydrous $\mathrm{CuSO}_4$ and $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ are $-70 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $+12 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively.
The heat of hydration of $\mathrm{CuSO}_4$ to $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is $-x \mathrm{~kJ}$. The value of $x$ is ________. (nearest integer).
Answer (integer)
82
Solution
<p>(I) $\mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$
Solution $\Delta \mathrm{H}=-70 \mathrm{~kJ}$</p>
<p>(II) $$\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$$ Solution $\mathrm{\Delta H=12 \mathrm{~kJ}}$</p>
<p>(I) - (II)</p>
<p>$$\begin{aligned}
& \mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \\
& \Delta \mathrm{H}=-70-12=-82
\end{aligned}$$</p>
About this question
Subject: Chemistry · Chapter: Thermodynamics · Topic: Zeroth and First Law
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