Before You Begin
Question #44290
The Question
Consider the following halogen exchange equilibria at 298 K: Reaction I: (Equilibrium constant ) Reaction II: (Equilibrium constant ) Assume entropic contributions () are negligible such that . Given the standard thermodynamic data at 298 K: Electron gain enthalpy (): , Hydration enthalpy (): , Which of the following correctly identifies the standard enthalpy changes () for Reactions I and II, and the relationship between their equilibrium constants?
Options
- A.
, ; ,
- B.
, ; ,
- C.
, ; ,
- D.
, ; ,
Check Your Answer
Correct Answer
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 will always displace to form .
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 results in a massive hydration enthalpy that heavily stabilizes the ion.
This solvation energy completely overrides the gas-phase EGE difference, making the displacement of by highly endothermic and thermodynamically unfavorable ().
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
| Subject | Chemistry |
| Chapter | Classification of Elements and Periodicity in Properties - Class 11 Part 1 |
| Concept | Atomic vs Ionic Radii |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |