JEE MAIN - Chemistry (2024 - 9th April Morning Shift - No. 29)
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).
Explanation
(I) $$\mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$$ Solution $$\Delta \mathrm{H}=-70 \mathrm{~kJ}$$
(II) $$\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4$$ Solution $$\mathrm{\Delta H=12 \mathrm{~kJ}}$$
(I) - (II)
$$\begin{aligned} & \mathrm{CuSO}_4+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} \\ & \Delta \mathrm{H}=-70-12=-82 \end{aligned}$$
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