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Question #44667

Difficulty: 8/10 | 🟢 Free

The Question

Consider the acid-catalyzed hydrolysis of an ester \ce{A} in a highly concentrated aqueous solution at 298 K298\text{ K}: \ceA(aq)+H2O(l)−>[H+]B(aq)+C(aq)\ce{A(aq) + H2O(l) ->[H^+] B(aq) + C(aq)} The initial reaction mixture contains 1.0 mol1.0\text{ mol} of \ce{A} and 4.0 mol4.0\text{ mol} of \ce{H2O}. The partial molar volumes of the components, which are assumed to remain constant over the course of the reaction, are given as: Vˉ\ceA=128 mL/mol\bar{V}_{\ce{A}} = 128\text{ mL/mol}, Vˉ\ceH2O=18 mL/mol\bar{V}_{\ce{H2O}} = 18\text{ mL/mol}, Vˉ\ceB=26 mL/mol\bar{V}_{\ce{B}} = 26\text{ mL/mol}, and Vˉ\ceC=20 mL/mol\bar{V}_{\ce{C}} = 20\text{ mL/mol}.

The true rate law for the reaction is Rate=k[\ceA][\ceH2O]\text{Rate} = k[\ce{A}][\ce{H2O}]. In standard dilute solutions, this is approximated as a pseudo-first-order reaction with an apparent rate constant kobs=k[\ceH2O]k_{\text{obs}} = k[\ce{H2O}]. Due to the high concentration and significant volume contraction upon reaction, [\ceH2O][\ce{H2O}] is not constant. Calculate the exact ratio of the instantaneous pseudo-first-order rate constant at 50%50\% completion of \ce{A} to its initial value, i.e., kobs(50%)kobs(0%)\frac{k_{\text{obs}}(50\%)}{k_{\text{obs}}(0\%)}.

Options

  • A.

    7/6

  • B.

    7/8

  • C.

    1

  • D.

    6/7

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

Concentration Unit Interconversion|Pseudo-First Order Kinetics

Why This Question Is Tricky

Common Mistakes

This problem tests the 'Solvent Volume Contraction Effect' in kinetics.

Students are heavily conditioned to treat [\ceH2O][\ce{H2O}] as an absolute constant (yielding a ratio of 1) in pseudo-first-order hydrolysis reactions.

Even if they realize water is consumed, they often forget that the total volume of the solution is a dynamic variable in non-dilute solutions.

If they assume constant volume, they incorrectly calculate [\ceH2O]0.5=3.5/200[\ce{H2O}]_{0.5} = 3.5/200, yielding the distractor 7/8.

The problem forces students to bridge thermodynamics (partial molar volumes) with chemical kinetics to find the true instantaneous molarity.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterSome Basic Concepts of Chemistry - Class 11 Part 1
ConceptConcentration Unit Interconversion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree