Q.No.22. A molecule in solution crystallizes into two different crystal forms with rate constants of 0.02 s-1 and 0.13 s-1. If the crystallization is assumed to be under kinetic control, then half-life (in seconds), rounded off to one decimal place) of the molecule _________.
Introduction to Crystal Polymorphism Kinetic Control Half-Life
Crystal polymorphism kinetic control half-life determines the dominant form in crystallization processes, crucial for CSIR NET Life Sciences. When a molecule crystallizes into polymorphs with rate constants like 0.02 s⁻¹ and 0.13 s⁻¹, kinetic control prioritizes the fastest rate for half-life calculation. This guide solves the exact PYQ, optimizes for competitive exams.
The half-life of the molecule under kinetic control of crystallization is 5.3 seconds
Crystallization kinetics follow first-order rate laws, where the half-life depends on the fastest rate constant.
Problem Breakdown
A molecule forms two polymorphs with rate constants k1 = 0.02 s⁻¹ and k2 = 0.13 s⁻¹. Under kinetic control, the faster process (higher k) dominates, as the product forms based on formation rates rather than stability. The half-life t1/2 for first-order crystallization is given by t1/2 = ln(2)/k, using k = 0.13 s⁻¹.
Step-by-Step Calculation
Use ln(2) ≈ 0.693. Then, t1/2 = 0.693 / 0.13 ≈ 5.33 seconds, rounded to 5.3 seconds. For the slower k = 0.02, t1/2 = 0.693 / 0.02 = 34.65 seconds (34.7 rounded), but irrelevant under kinetic control.
Kinetic vs Thermodynamic Control
Kinetic control favors the faster-forming polymorph (k2), observed on short timescales. Thermodynamic control would yield the more stable form after equilibration, independent of rates. In exams like CSIR NET or GATE, select the maximum rate constant for kinetic scenarios.
Core Concept: First-Order Crystallization Kinetics
Crystallization from solution often follows first-order kinetics, where rate = k [molecule]. Half-life t1/2 = 0.693 / k remains constant, unlike zero-order. In polymorphism, two paths compete; kinetic control uses max k.
- Faster polymorph: k = 0.13 s⁻¹, t1/2 = 5.3 s.
- Slower polymorph: k = 0.02 s⁻¹, t1/2 = 34.7 s (not selected).
Detailed Solution for CSIR NET/GATE PYQ
Question: Molecule crystallizes into two forms, k = 0.02 s⁻¹, 0.13 s⁻¹. Kinetic control half-life (1 decimal)?
- Identify kinetic control: Fastest k = 0.13.
- Apply formula: t1/2 = ln(2)/0.13 = 0.693147/0.13 = 5.331 s.
- Round: 5.3 seconds.
| Parameter | Value | Role in Kinetic Control |
|---|---|---|
| k1 | 0.02 s⁻¹ | Slower path, ignored |
| k2 | 0.13 s⁻¹ | Dominant path, used for t1/2 |
| t1/2 | 5.3 s | Final answer |
Why Not Other Options? Exam Insights
34.7 s: Uses slower k; thermodynamic assumption error. Average k: Incorrect; no parallel averaging in kinetic control. Sum k: Wrong; half-life not additive. Common CSIR NET trap. Always verify first-order assumption via units (s⁻¹).
Applications in Biotechnology & Pharma
Crystal polymorphism kinetic control half-life impacts drug stability, solubility. Faster polymorph may be metastable but preferred initially. For CSIR NET prep, practice derivations: From [A] = [A]0 e-kt, at t1/2, [A] = [A]0 / 2 yields formula.
Keywords: crystal polymorphism, kinetic control half-life, rate constants crystallization, CSIR NET kinetics, first-order half-life calculation.
