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Question #44290

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following halogen exchange equilibria at 298 K: Reaction I: \ceCl(g)+F−(g)⇌Cl−(g)+F(g)\ce{Cl(g) + F^{-}(g) \rightleftharpoons Cl^{-}(g) + F(g)} (Equilibrium constant K1K_1) Reaction II: \ceCl(g)+F−(aq)⇌Cl−(aq)+F(g)\ce{Cl(g) + F^{-}(aq) \rightleftharpoons Cl^{-}(aq) + F(g)} (Equilibrium constant K2K_2) Assume entropic contributions (ΔS⊖\Delta S^\ominus) are negligible such that ΔG⊖≈ΔH⊖\Delta G^\ominus \approx \Delta H^\ominus. Given the standard thermodynamic data at 298 K: Electron gain enthalpy (ΔegH⊖\Delta_{eg}H^\ominus): \ceF=−333 kJ/mol\ce{F} = -333 \text{ kJ/mol}, \ceCl=−349 kJ/mol\ce{Cl} = -349 \text{ kJ/mol} Hydration enthalpy (ΔhydH⊖\Delta_{hyd}H^\ominus): \ceF−=−515 kJ/mol\ce{F^{-}} = -515 \text{ kJ/mol}, \ceCl−=−381 kJ/mol\ce{Cl^{-}} = -381 \text{ kJ/mol} Which of the following correctly identifies the standard enthalpy changes (ΔH⊖\Delta H^\ominus) for Reactions I and II, and the relationship between their equilibrium constants?

Options

  • A.

    ΔHI⊖=−16 kJ/mol\Delta H^\ominus_I = -16 \text{ kJ/mol}, ΔHII⊖=+118 kJ/mol\Delta H^\ominus_{II} = +118 \text{ kJ/mol}; K1>1K_1 > 1, K2<1K_2 < 1

  • B.

    ΔHI⊖=−16 kJ/mol\Delta H^\ominus_I = -16 \text{ kJ/mol}, ΔHII⊖=−150 kJ/mol\Delta H^\ominus_{II} = -150 \text{ kJ/mol}; K1>1K_1 > 1, K2>1K_2 > 1

  • C.

    ΔHI⊖=+16 kJ/mol\Delta H^\ominus_I = +16 \text{ kJ/mol}, ΔHII⊖=−118 kJ/mol\Delta H^\ominus_{II} = -118 \text{ kJ/mol}; K1<1K_1 < 1, K2>1K_2 > 1

  • D.

    ΔHI⊖=−16 kJ/mol\Delta H^\ominus_I = -16 \text{ kJ/mol}, ΔHII⊖=−16 kJ/mol\Delta H^\ominus_{II} = -16 \text{ kJ/mol}; K1>1K_1 > 1, K2>1K_2 > 1

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Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Atomic vs Ionic Radii|Electron Gain Enthalpy (EGE)|Solvent Effects on Reactivity

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Solvation Trap' in halogen thermodynamics.

Students often memorize that Chlorine has a more negative electron gain enthalpy than Fluorine due to interelectronic repulsions in Fluorine's compact 2p subshell.

Consequently, they assume \ceCl\ce{Cl} will always displace \ceF−\ce{F^{-}} to form \ceCl−\ce{Cl^{-}}.

However, this is only true in the gas phase (Reaction I).

In a polar protic solvent like water (Reaction II), the exceptionally small ionic radius of \ceF−\ce{F^{-}} results in a massive hydration enthalpy that heavily stabilizes the \ceF−(aq)\ce{F^{-}(aq)} ion.

This solvation energy completely overrides the gas-phase EGE difference, making the displacement of \ceF−(aq)\ce{F^{-}(aq)} by \ceCl(g)\ce{Cl(g)} highly endothermic and thermodynamically unfavorable (K2<1K_2 < 1).

Distractor B catches students who incorrectly apply Hess's law by subtracting the hydration enthalpies in the wrong direction.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterClassification of Elements and Periodicity in Properties - Class 11 Part 1
ConceptAtomic vs Ionic Radii
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree