Chemistry · THERMODYNAMICS
हिन्दीGiven the enthalpy changes for dissolving 1 mol of \( \text{HCl}(g) \): \( \text{HCl}(g) + 25\text{ aq.} \rightarrow \text{HCl}\cdot 25\text{ aq.}\quad \Delta H_1 = -72.03\text{ kJ mol}^{-1} \) \( \text{HCl}(g) + 40\text{ aq.} \rightarrow \text{HCl}\cdot 40\text{ aq.}\quad \Delta H_2 = -72.79\text{ kJ mol}^{-1} \) \( \text{HCl}(g) + \infty\text{ aq.} \rightarrow \text{HCl}\cdot \infty\text{ aq.}\quad \Delta H_3 = -74.85\text{ kJ mol}^{-1} \) What is the enthalpy of dilution when a solution containing 1 mol of \( \text{HCl} \) in \( 25\text{ moles} \) of water is diluted to infinite dilution?
Correct answer · Option 1
\( -2.82\text{ kJ mol}^{-1} \)
Explanation
\( \Delta H_{\text{dil}} = \Delta H_3 - \Delta H_1 = -74.85 - (-72.03) = -2.82\text{ kJ mol}^{-1} \).