Consider the given data : (a) $\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow…
(a) $\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.10 \mathrm{H}_2 \mathrm{O}$
$\Delta \mathrm{H}=-69.01 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(b) $\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.40 \mathrm{H}_2 \mathrm{O}$
$\Delta \mathrm{H}=-72.79 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Choose the correct statement :
- Dissolution of gas in water is an endothermic process
- The heat of solution depends on the amount of solvent.
- The heat of dilution for the $\mathrm{HCl}\left(\mathrm{HCl} \cdot 10 \mathrm{H}_2 \mathrm{O}\right.$ to $\left.\mathrm{HCl}.40 \mathrm{H}_2 \mathrm{O}\right)$ is $3.78 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
- The heat of formation of HCl solution is represented by both (a) and (b)
Solution
$\Delta \mathrm{H}$ is negative so it means dissolution of gas $\mathrm{HCl}(\mathrm{g})$ is exothermic.
$\mathrm{HCl}(\mathrm{g})+10 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.10 \mathrm{H}_2 \mathrm{O}$ ...(1)
$\Delta \mathrm{H}_1=-69.01 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$
$\mathrm{HCl}(\mathrm{g})+40 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}.40 \mathrm{H}_2 \mathrm{O}$ ...(2)
$\Delta \mathrm{H}_2=-72.79 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$
Hence heat of solution depends upon amount of solvent
By equation..(2) - equation
$\mathrm{HCl}.10 \mathrm{H}_2 \mathrm{O}+30 \mathrm{H}_2 \mathrm{O}(\ell) \rightarrow \mathrm{HCl} \cdot 40 \mathrm{H}_2 \mathrm{O}$
So Heat of dilution $=-72.79-(-69.01)$
$=-3.78 \frac{\mathrm{~kJ}}{\mathrm{~mol}}$
Hence option (3) is incorrect.
For heat of formation reactant should be in elemental form hence option (4) is incorrect
Asked in: JEE Main 2025 (04 Apr Shift 2)