Before You Begin
Question #44659
The Question
An infinite pool of an equimolar mixture of and undergoes catalytic oxidation. The rate-determining step exhibits a primary kinetic isotope effect (KIE) such that the rate constants ratio is . The reaction is quenched at conversion. The resulting nitrogen oxides are quantitatively isolated and completely converted into pure .
A sample of this synthesized is placed in an evacuated, rigid container and heated to . The system reaches thermodynamic equilibrium:
At equilibrium, the total pressure is and the observed vapor density of the gaseous mixture is .
Calculate the equilibrium constant for this dissociation at .
(Given: Exact isotopic masses , , . Assume ideal gas behavior and vapor density is relative to .)
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
Isotopic Average Atomic Mass|Relationship between Kp and Kc|Pseudo-First Order Kinetics
Why This Question Is Tricky
Common Mistakes
The problem tests the intersection of chemical kinetics (Kinetic Isotope Effect) and thermodynamics (Vapor Density & Equilibrium).
The 'Hidden Variable' is the effective molar mass of the reactant, which deviates from both the standard periodic table value and the equimolar precursor value due to the KIE in the rate-determining step of its synthesis.
Students often fall into the trap of either ignoring the isotopic mass shift entirely (yielding 1.29 atm), assuming the initial 1:1 stoichiometry is preserved (yielding 1.40 atm), or erroneously applying the kinetic rate ratio directly to the thermodynamic equilibrium constant (yielding 5.33 atm).
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Some Basic Concepts of Chemistry - Class 11 Part 1 |
| Concept | Isotopic Average Atomic Mass |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |