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Question #44062
The Question
Consider the reversible liquid-phase isomerization reaction: .
At , the liquid mixture of A and B exhibits non-ideal behavior. The system is simultaneously in vapor-liquid equilibrium and chemical equilibrium.
At this temperature, the mixture forms an azeotrope (where the vapor phase mole fractions equal the liquid phase mole fractions) at a liquid phase mole fraction of A, .
Given that the vapor pressures of pure liquid A and pure liquid B at are and , respectively, what is the standard Gibbs free energy change () for the isomerization reaction at ?
(Assume the vapor phase behaves ideally and the standard state for each component is the pure liquid at ).
Options
- A.
- B.
- C.
- D.
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Numerical_Precision
Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.
Concepts Behind This Question
Homogeneous vs. Heterogeneous Equilibrium|Concentration Units Dependency|Azeotrope Boiling Point Logic
Why This Question Is Tricky
Common Mistakes
This problem tests the critical distinction between formal concentration (mole fraction) and thermodynamic activity in a non-ideal solution.
Students typically fall into the trap of assuming ideal behavior () and calculate , leading to Option B.
By introducing an azeotrope, the problem provides a hidden pathway to determine the ratio of activity coefficients () without needing the total pressure.
It weaponizes the deviation from Raoult's law, forcing the student to derive the true thermodynamic equilibrium constant using pure vapor pressures.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Equilibrium - Class 11 Part 1 |
| Concept | Homogeneous vs. Heterogeneous Equilibrium |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |