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Chemistry · THERMODYNAMICS

हिन्दी

Given 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?

  1. A
    \( -2.82\text{ kJ mol}^{-1} \)
  2. B
    \( +2.82\text{ kJ mol}^{-1} \)
  3. C
    \( -0.76\text{ kJ mol}^{-1} \)
  4. D
    \( -2.06\text{ kJ mol}^{-1} \)
View Answer & Solution

Correct answer · Option A

\( -2.82\text{ kJ mol}^{-1} \)

Explanation

\( \Delta H_{\text{dil}} = \Delta H_3 - \Delta H_1 = -74.85 - (-72.03) = -2.82\text{ kJ mol}^{-1} \).
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