Start Learning Free

Before You Begin

Question #44231

Difficulty: 8/10 | 🟢 Free

The Question

Compound A, 2-(β\beta-D-glucopyranosyloxy)pyridine, undergoes enzymatic hydrolysis in an aqueous medium at 300 K300\text{ K} to yield D-glucose and 2-hydroxypyridine. The released 2-hydroxypyridine subsequently undergoes complete tautomerization to its more stable keto form, 2-pyridone.\n\nGiven the following thermodynamic data at 300 K300\text{ K}:\n\nStandard enthalpy of combustion (ΔHc∘\Delta H^\circ_c) in kJ/mol\text{kJ/mol}:\nCompound A (s) = −4500-4500\nD-Glucose (s) = −2800-2800\n2-Hydroxypyridine (s, enol form) = −1715-1715\n\nStandard enthalpy of solution (ΔHsol∘\Delta H^\circ_{sol}) in kJ/mol\text{kJ/mol}:\nCompound A (s) = +30+30\nD-Glucose (s) = +12+12\n2-Hydroxypyridine (s, enol form) = +8+8\n\nThermodynamic parameters for the tautomerization \ce2−Hydroxypyridine(aq)<=>2−Pyridone(aq)\ce{2-Hydroxypyridine (aq) <=> 2-Pyridone (aq)}:\nΔGtaut∘=−25.0 kJ/mol\Delta G^\circ_{taut} = -25.0\text{ kJ/mol}\nΔStaut∘=−30.0 J K−1mol−1\Delta S^\circ_{taut} = -30.0\text{ J K}^{-1}\text{mol}^{-1}\n\nCalculate the standard observed enthalpy of the overall reaction (ΔHobs∘\Delta H^\circ_{obs}) for the hydrolysis of Compound A (aq) yielding D-glucose (aq) and 2-pyridone (aq) at 300 K300\text{ K}.

Options

  • A.

    -29.0 kJ/mol

  • B.

    -20.0 kJ/mol

  • C.

    -19.0 kJ/mol

  • D.

    +5.0 kJ/mol

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

Hess's Law Application|Disaccharide Hydrolysis Products|Nitrogenous Base Keto-Enol Tautomerism

Why This Question Is Tricky

Common Mistakes

This problem tests the application of Hess's Law and the First Law of Thermodynamics to a multi-step biochemical reaction.

The 'hidden variable' is the thermodynamic stabilization provided by keto-enol tautomerization.

Students often calculate the enthalpy of hydrolysis up to the direct cleavage products (the enol form, +5.0 kJ/mol) but fail to account for the subsequent tautomerization to the keto form.

Additionally, it tests the correct integration of standard enthalpies of combustion and solution to transition from solid-state calorimetric data to aqueous reaction conditions, trapping students who confuse state functions or use Gibbs free energy directly instead of deriving the enthalpy of tautomerization.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptHess's Law Application
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree