Start Learning Free

Before You Begin

Question #44818

Difficulty: 8/10 | 🟢 Free

The Question

The electrophilic nitrosation of 4-pyridone (KK) at 298 K in an aqueous buffer proceeds exclusively via its minor enol tautomer, 4-hydroxypyridine (EE). The elementary steps are:

Step 1: K→k1EK \xrightarrow{k_1} E Step 2: E→k−1KE \xrightarrow{k_{-1}} K Step 3: E+NO+→k2P+H+E + NO^+ \xrightarrow{k_2} P + H^+

The rate constants are given as: k1=1.0×10−5 s−1k_1 = 1.0 \times 10^{-5} \text{ s}^{-1} k−1=1.0×102 s−1k_{-1} = 1.0 \times 10^2 \text{ s}^{-1} k2=5.0×104 M−1s−1k_2 = 5.0 \times 10^4 \text{ M}^{-1}\text{s}^{-1}

Assuming the Steady-State Approximation (SSA) is applicable to the enol intermediate, calculate the initial rate of formation of product PP when the initial concentrations are [K]0=0.1 M[K]_0 = 0.1 \text{ M} and [NO+]0=2.0×10−3 M[NO^+]_0 = 2.0 \times 10^{-3} \text{ M}. Additionally, determine the limiting kinetic order of the reaction with respect to [NO+][NO^+] at extremely high concentrations of NO+NO^+.

Options

  • A.

    5.0×10−7 M s−15.0 \times 10^{-7} \text{ M s}^{-1}; Zero order

  • B.

    1.0×10−6 M s−11.0 \times 10^{-6} \text{ M s}^{-1}; First order

  • C.

    5.0×10−7 M s−15.0 \times 10^{-7} \text{ M s}^{-1}; First order

  • D.

    1.0×10−6 M s−11.0 \times 10^{-6} \text{ M s}^{-1}; Zero order

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

Steady State Approximation|Nitrogenous Base Keto-Enol Tautomerism|Chemical Kinetics

Why This Question Is Tricky

Common Mistakes

This problem tests the application of the Steady-State Approximation (SSA) to a tautomeric pre-equilibrium that is not strictly 'fast'.

Students often fall into the trap of assuming tautomerism is an instantaneous dynamic equilibrium (Keq=[E]/[K]K_{eq} = [E]/[K]), which leads to the incorrect rate law v=Keqk2[K][NO+]v = K_{eq} k_2 [K][NO^+] (yielding 1.0×10−6 M s−11.0 \times 10^{-6} \text{ M s}^{-1}, Option B).

By forcing the use of SSA, the student must recognize that the rate of the second step (k2[NO+]k_2[NO^+]) is comparable to the reverse tautomerization (k−1k_{-1}), causing a deviation from pure pre-equilibrium kinetics.

Furthermore, it tests the conceptual understanding of kinetic shifts: at high electrophile concentrations, the rate-determining step shifts entirely to the enolization process, resulting in zero-order dependence on the electrophile.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Kinetics - Class 12
ConceptSteady State Approximation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree