NEET Chemistry Chemical Kinetics 2027 is a high-scoring Class 12 chapter from Physical Chemistry that contributes 1–2 questions (4–8 marks) every year. This pillar guide compiles complete chapter notes on rate of reaction, rate law, order vs molecularity, integrated rate equations for zero, first and second-order reactions, half-life expressions, the Arrhenius equation, temperature coefficient, the role of catalysts, and the collision theory — followed by 10 NCERT-pattern MCQs with detailed solutions and two fully worked numerical problems on first-order kinetics and the Arrhenius equation. Last Updated: April 2026.
Chemical Kinetics sits at the boundary of physical chemistry and applied chemistry. Concepts here also feed directly into Electrochemistry (Tafel kinetics), Surface Chemistry (heterogeneous catalysis) and Biochemistry (enzyme kinetics, Michaelis–Menten approximation). Mastering kinetics gives compounding returns across the NEET Chemistry section.
NEET Weightage — Chemical Kinetics
| Year | Questions | Marks | Difficulty |
|---|---|---|---|
| NEET 2023 | 2 | 8 | Moderate |
| NEET 2022 | 1 | 4 | Easy |
| NEET 2021 | 2 | 8 | Moderate |
| NEET 2020 | 1 | 4 | Easy–Moderate |
| NEET 2019 | 2 | 8 | Moderate |
NCERT Class 12 Chemistry, Part I, Chapter 4, Pages 93–125. Master the integrated first-order rate equation and the Arrhenius equation — almost every year a numerical comes from one of these two.
1. Rate of a Chemical Reaction
The rate of a chemical reaction is the change in concentration of a reactant or product per unit time. For a reaction:
aA + bB → cC + dD
The rate is defined as:
Rate = -(1/a) d[A]/dt = -(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt
- Average rate = Δ[concentration] / Δt over a measurable interval.
- Instantaneous rate = d[concentration]/dt at a specific instant (slope of concentration-time curve).
- Units: mol L-1 s-1 (or mol L-1 min-1).
2. Factors Affecting Rate of Reaction
- Concentration of reactants — higher concentration → more collisions → faster rate.
- Temperature — rate increases with temperature; for most reactions a 10°C rise roughly doubles or triples the rate.
- Catalyst — provides an alternative pathway with lower activation energy.
- Surface area — for heterogeneous reactions, finer particles react faster.
- Nature of reactants — ionic reactions are generally faster than covalent.
- Pressure — for gaseous reactions, increased pressure increases concentration and rate.
3. Rate Law and Rate Constant
The rate law expresses the rate as a function of concentration of reactants:
Rate = k[A]x [B]y
Where k is the rate constant (specific reaction rate at unit concentration). Exponents x and y are determined experimentally and may differ from stoichiometric coefficients.
Order of a Reaction
Order = sum of exponents of concentration terms in the experimental rate law (x + y). It can be:
- Zero (rate independent of concentration)
- Fractional (e.g., 1/2, 3/2)
- Whole number (1, 2, 3)
Molecularity
Molecularity = number of reacting species (atoms, ions, molecules) participating in an elementary reaction. It is always a small whole number (1, 2 or rarely 3) and applies only to elementary reactions; for complex reactions, molecularity has meaning only for individual elementary steps.
Order vs Molecularity — key differences
| Order | Molecularity |
|---|---|
| Experimental quantity | Theoretical quantity |
| Can be zero, fractional or integral | Always a positive whole number |
| Applies to overall reaction | Applies only to elementary reactions |
| Determined from rate law | Determined from balanced elementary equation |
4. Integrated Rate Equations
Zero-order reaction
Rate = k[A]0 = k (independent of concentration). Integrated form:
[A] = [A0] – kt
Plot of [A] vs t is a straight line with slope -k. Half-life t1/2 = [A0]/2k. Examples: photochemical reaction of H2 + Cl2; decomposition of HI on gold surface; decomposition of NH3 on hot Pt surface.
First-order reaction
Rate = k[A]. Integrated form:
k = (2.303/t) log ([A0]/[A])
Plot of log[A] vs t is a straight line with slope -k/2.303. Half-life:
t1/2 = 0.693/k
Independent of initial concentration — a defining feature. Examples: radioactive decay; decomposition of N2O5; hydrolysis of methyl acetate in acidic medium (pseudo-first-order).
Pseudo-first-order reactions
Higher-order reactions that follow first-order kinetics due to large excess of one reactant. Example: hydrolysis of ester — rate = k[ester][H2O], but [H2O] is essentially constant, so observed rate = k'[ester].
5. Temperature Dependence — Arrhenius Equation
The rate constant varies with temperature according to the Arrhenius equation:
k = A · e-Ea/RT
Taking logarithm:
log k = log A – Ea/(2.303 RT)
- A = Arrhenius/frequency factor (related to collision frequency and orientation).
- Ea = activation energy — the minimum energy reactants must possess to undergo successful collision.
- R = gas constant (8.314 J mol-1 K-1); T in Kelvin.
Plot of log k vs 1/T is a straight line with slope -Ea/(2.303 R) — used to determine Ea graphically.
Temperature coefficient
Ratio of rate constants at two temperatures differing by 10°C:
μ = k(T+10) / kT ≈ 2 to 3
6. Collision Theory of Reaction Rates
Reactions occur through collisions of reactant molecules. For a successful (effective) collision, two conditions must be met:
- Energy criterion — colliding molecules must have energy ≥ Ea.
- Orientation criterion — molecules must collide in proper orientation; quantified by steric factor (P).
Rate = P · Z · e-Ea/RT, where Z is the collision frequency.
7. Catalyst — Role and Mechanism
A catalyst increases the rate of a reaction by providing an alternate pathway with lower activation energy. Key features:
- Does not appear in the overall stoichiometric equation; recovered chemically unchanged.
- Increases rate of forward and reverse reactions equally — does not change the equilibrium constant or the position of equilibrium.
- Does not alter ΔH of the reaction.
- Cannot initiate a reaction that is thermodynamically non-spontaneous.
Types of catalysis
- Homogeneous catalysis — catalyst and reactants in the same phase. Example: acid-catalysed hydrolysis of esters; oxidation of SO2 to SO3 by NO in the lead-chamber process.
- Heterogeneous catalysis — catalyst and reactants in different phases. Example: Haber’s process (Fe + Mo for N2 + 3H2 → 2NH3); Ostwald’s process (Pt for NH3 oxidation); contact process (V2O5 for SO2 + O2 → 2SO3); hydrogenation of vegetable oils with Ni.
- Enzymes — biological catalysts (proteins) showing extreme specificity, high efficiency (106–1020 times faster), optimum pH and temperature.
8. Worked Numerical — First-order Kinetics
Q. A first-order reaction has a rate constant k = 2.303 × 10-3 s-1. How long will it take for the initial concentration to reduce to one-fourth of its initial value?
Solution. For first order: t = (2.303/k) log ([A0]/[A]). With [A] = [A0]/4 → ratio = 4 → log 4 = 0.602.
t = (2.303 / 2.303 × 10-3) × 0.602 = 1000 × 0.602 = 602 seconds.
Alternative shortcut: One-fourth = (1/2)2 means two half-lives. t1/2 = 0.693/k = 0.693/(2.303 × 10-3) = 300.9 s; therefore 2 × t1/2 ≈ 602 s. Both methods agree.
9. Worked Numerical — Arrhenius Equation
Q. The rate of a reaction doubles when temperature changes from 300 K to 310 K. Calculate the activation energy.
Solution. Apply log (k2/k1) = (Ea/2.303 R) (T2 – T1)/(T1T2).
log 2 = 0.301; T1T2 = 300 × 310 = 93000; T2 – T1 = 10. R = 8.314 J mol-1 K-1.
Ea = (0.301 × 2.303 × 8.314 × 93000) / 10 ≈ 53.6 kJ mol-1.
10. High-Yield Memory Hooks for NEET
- Half-life formulae: zero-order t1/2 = [A0]/2k; first-order t1/2 = 0.693/k; second-order t1/2 = 1/(k[A0]).
- Units of k for n-th order: (mol L-1)1-n s-1.
- Plot for first order: log [A] vs t = straight line, slope = -k/2.303.
- Plot for Arrhenius: log k vs 1/T = straight line, slope = -Ea/(2.303 R).
- Average rate has positive sign for products, negative for reactants — coefficient division ensures unique value.
- Catalyst lowers Ea but does NOT change ΔH or Kc.
Practice MCQs — 10 NCERT-style Numericals
Test your understanding with the interactive quiz below.
[cg_quiz data=”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”]
FAQ
Q1. How is rate of reaction different from rate constant?
Rate depends on concentration and changes during the reaction. Rate constant (k) is the rate at unit concentration of all reactants and depends only on temperature and presence of a catalyst.
Q2. What is the unit of rate constant for a second-order reaction?
L mol-1 s-1 (or M-1 s-1). General formula: k has units of (mol L-1)1-n s-1, where n is the order.
Q3. Why is t1/2 of a first-order reaction independent of initial concentration?
Because t1/2 = 0.693/k — it depends only on the rate constant. This is why radioactive decay (first order) has a fixed half-life regardless of how much sample you start with.
Q4. Can a catalyst shift the equilibrium of a reaction?
No. A catalyst lowers Ea for both forward and reverse reactions equally and only helps the system reach equilibrium faster — it does not change Kc or Kp.
Q5. What is meant by activation energy?
The minimum energy that reacting molecules must possess (above the average energy) to undergo a chemical change on collision. Lower Ea → faster reaction at a given temperature.
Related NEET Chemistry Posts
- NEET Chemistry Electrochemistry 2027 — Formulas & MCQs
- NEET Chemistry Equilibrium 2027 — Le Chatelier & MCQs
- NEET Chemistry Coordination Compounds 2027 — MCQs
- NEET 2027 Syllabus — Complete Chapter-wise Guide
- NEET Chapter-wise Weightage 2027 — Data-Backed
Last Updated: April 2026 — NEET Gurukul Editorial Team.