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Question #44659

Difficulty: 8/10 | 🟢 Free

The Question

An infinite pool of an equimolar mixture of \ce14NH3\ce{^{14}NH3} and \ce15NH3\ce{^{15}NH3} undergoes catalytic oxidation. The rate-determining step exhibits a primary kinetic isotope effect (KIE) such that the rate constants ratio is k14/k15=4.0k_{14}/k_{15} = 4.0. The reaction is quenched at <1%<1\% conversion. The resulting nitrogen oxides are quantitatively isolated and completely converted into pure \ceN2O4(g)\ce{N2O4(g)}.

A sample of this synthesized \ceN2O4(g)\ce{N2O4(g)} is placed in an evacuated, rigid container and heated to 350 K350\text{ K}. The system reaches thermodynamic equilibrium:

\ceN2O4(g)<=>2NO2(g)\ce{N2O4(g) <=> 2NO2(g)}

At equilibrium, the total pressure is 1.0 atm1.0\text{ atm} and the observed vapor density of the gaseous mixture is 30.830.8.

Calculate the equilibrium constant KpK_p for this dissociation at 350 K350\text{ K}.

(Given: Exact isotopic masses \ce14N=14.0 u\ce{^{14}N} = 14.0\text{ u}, \ce15N=15.0 u\ce{^{15}N} = 15.0\text{ u}, \ceO=16.0 u\ce{O} = 16.0\text{ u}. Assume ideal gas behavior and vapor density is relative to M=2.0 uM = 2.0\text{ u}.)

Options

  • A.

    1.33 atm1.33\text{ atm}

  • B.

    1.29 atm1.29\text{ atm}

  • C.

    1.40 atm1.40\text{ atm}

  • D.

    5.33 atm5.33\text{ atm}

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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

SubjectChemistry
ChapterSome Basic Concepts of Chemistry - Class 11 Part 1
ConceptIsotopic Average Atomic Mass
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree