Chemical Kinetics — NEET Chemistry MCQs with answers

Chemical Kinetics is a Class 12 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 877 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Rate, rate law and order · Integrated rate equations and half-life · Arrhenius equation and activation energy · Collision theory and mechanisms · Catalysis and kinetics · First order half-life · First order graphical analysis · Rate law from stoichiometry · Half-life dependence on order · Factors affecting rate of reaction · Catalyst and activation energy · Second order integrated rate law · Rate as concentration approaches zero · Arrhenius plot · Half-life formula application · First order rate data · Catalyzed vs uncatalyzed activation energy · Temperature effect: rate constant vs equilibrium constant · Zero order reaction limit · Average vs instantaneous rate · Arrhenius equation setup · Concentration-time curves across orders · Temperature Dependence · Order of reaction · First order kinetics · Zero order kinetics · Molecularity vs order · Rate constant units · Effect of catalyst on activation energy · Pseudo first order reaction · Graph interpretation - first order reaction · Rate law and initial rate method · Instantaneous rate · Numerical - time for completion percentage (first order) · Rate law for elementary reactions · Assertion-reason on rate law · Numerical - relating rate of formation and consumption · Statement-based on reaction kinetics · Numerical - zero order rate constant from graph data · Radioactive decay as first order

8 free Chemical Kinetics practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Collision theory and mechanisms

   In a multi-step reaction mechanism, the overall rate of reaction is governed by which step?

   • A.The first step of the mechanism
   • B.The slowest step of the mechanism
   • C.The step with the highest molecularity
   • D.The average of all steps' rates
   Show answer and explanation

   Answer: B. The slowest step of the mechanism

   A reaction mechanism is a sequence of elementary steps; the slowest step acts as a bottleneck and controls the overall rate, so it is called the rate-determining step (RDS).

   Just as traffic through a series of gates moves only as fast as the slowest gate, a multi-step mechanism's overall rate cannot exceed the rate of its slowest elementary step. Faster steps before or after the RDS do not limit the overall rate; only the slowest step's rate law (in terms of species present up to that step) determines the observed rate law.

   Common mistake: Assuming the first step of the mechanism is always rate-determining.

2. Question 2 · easy · Collision theory and mechanisms

   For an elementary step A + 2B → products, what is its molecularity?

   • A.1
   • B.2
   • C.3
   • D.Cannot be determined without experimental data
   Show answer and explanation

   Answer: C. 3

   Molecularity is the number of reactant molecules/atoms/ions that collide simultaneously in an elementary step; here 1 molecule of A and 2 molecules of B collide together, giving molecularity 3.

   Molecularity is a theoretical, always-integer quantity read directly from the stoichiometry of a single elementary step. In A + 2B → products, one A molecule and two B molecules must collide simultaneously, so molecularity = 1 + 2 = 3 (a termolecular step). Unlike order, molecularity is never determined experimentally and is never zero or fractional for a genuine elementary step.

   Common mistake: Counting distinct species instead of total colliding molecules.

3. Question 3 · easy · Collision theory and mechanisms

   A reaction occurs by the two-step mechanism: Step 1 (slow): X → Y; Step 2 (fast): Y + X → Z. What is the predicted rate law for formation of Z?

   • A.Rate = k[X]
   • B.Rate = k[X]^2
   • C.Rate = k[Y]
   • D.Rate = k[X][Y]
   Show answer and explanation

   Answer: A. Rate = k[X]

   The rate law is fixed by the slow (rate-determining) step only; Step 1, X → Y, is unimolecular, so Rate = k[X].

   In a multi-step mechanism, only the slow step controls the overall rate. Step 1 (slow) is X → Y, an elementary unimolecular step, so its rate law is Rate = k[X]. Step 2 is fast and does not limit the rate, so it is irrelevant to the rate law even though it consumes an additional X.

   Common mistake: Using the overall stoichiometric equation instead of the slow step to write the rate law.

4. Question 4 · easy · Collision theory and mechanisms

   For the overall reaction 2NO(g) + O2(g) → 2NO2(g), a student writes the rate law directly as Rate = k[NO]^2[O2] using the balanced equation's coefficients as the powers, without knowing the mechanism. Is this approach generally valid?

   • A.Yes, because the order of any reaction always equals the stoichiometric coefficients of the balanced equation
   • B.Yes, because molecularity and order are always numerically identical for any reaction
   • C.No, in general the rate law must be determined experimentally or from the mechanism's rate-determining step, since most reactions are multi-step
   • D.No, because gas-phase reactions never follow simple power-law rate expressions
   Show answer and explanation

   Answer: C. No, in general the rate law must be determined experimentally or from the mechanism's rate-determining step, since most reactions are multi-step

   Reaction order can only be read from stoichiometric coefficients for elementary (single-step) reactions; most reactions, including 2NO + O2 → 2NO2, proceed via a mechanism, so the rate law must be found experimentally or derived from the rate-determining step.

   It is a common trap to assume that the coefficients in a balanced overall equation give the reaction order. This is true only when the reaction is elementary (occurs in one single step exactly as written). For multi-step mechanisms, the order reflects the rate-determining step, which may involve different species or powers than the overall equation. The reaction 2NO + O2 → 2NO2 is actually found experimentally to be third order overall (second order in NO, first order in O2), consistent with the coefficients here, but this must be confirmed experimentally/mechanistically, not assumed — many reactions (e.g., 2NO2 + F2 → 2NO2F) show orders that do NOT match their overall coefficients.

   Common mistake: Reading reaction order directly off the coefficients of the overall balanced equation.

5. Question 5 · medium · Collision theory and mechanisms

   For a bimolecular gas reaction, Ea/RT = 20 at 1000 K (given e^-20 = 2.06 × 10^-9). What fraction of colliding molecules possesses energy equal to or greater than the threshold (activation) energy?

   • A.2.06 × 10^-9
   • B.20
   • C.4.9 × 10^8
   • D.0.206
   Show answer and explanation

   Answer: A. 2.06 × 10^-9

   By collision theory, the fraction of molecules with energy ≥ Ea is given by the Boltzmann factor f = e^(-Ea/RT); substituting the given exponent directly gives f = e^-20 = 2.06 × 10^-9.

   Collision theory states that only molecules colliding with kinetic energy ≥ Ea (threshold energy) can react. The fraction of such molecules is f = e^(-Ea/RT). Here Ea/RT = 20 is given directly, so f = e^-20 = 2.06 × 10^-9 (using the supplied value). This tiny fraction explains why only a small proportion of collisions are effective even though molecules collide extremely frequently.

   Common mistake: Reporting the exponent value instead of evaluating the exponential itself.

6. Question 6 · medium · Collision theory and mechanisms

   For a reaction, the experimentally determined Arrhenius pre-exponential factor A(exp) is 1.0 × 10^11 L mol^-1 s^-1, while the theoretical collision frequency factor calculated from collision theory is 1.0 × 10^14 L mol^-1 s^-1. What is the steric (probability) factor P for this reaction?

   • A.1.0 × 10^3
   • B.1.0
   • C.1.0 × 10^-3
   • D.1.0 × 10^25
   Show answer and explanation

   Answer: C. 1.0 × 10^-3

   The steric factor is defined as P = A(experimental)/A(theoretical, collision); substituting gives P = 10^11/10^14 = 10^-3.

   Collision theory predicts a theoretical pre-exponential (collision frequency) factor from molecular size and speed, but many reactions react far slower than this predicts because molecules must also collide with correct orientation. This is captured by the steric factor P = A(exp)/A(theoretical) = (1.0 × 10^11)/(1.0 × 10^14) = 1.0 × 10^-3. A value of P much less than 1 indicates strict orientation requirements for effective collision.

   Common mistake: Dividing in the wrong order or assuming P should always be 1.

7. Question 7 · medium · Collision theory and mechanisms

   For a bimolecular reaction, the collision frequency Z is 5.0 × 10^10 L mol^-1 s^-1, the fraction of collisions with sufficient energy (E ≥ Ea) is 1.0 × 10^-8, and the steric factor P is 1.0. What is the rate constant k according to collision theory?

   • A.5.0 × 10^18 L mol^-1 s^-1
   • B.5.0 × 10^2 L mol^-1 s^-1
   • C.5.0 × 10^-2 L mol^-1 s^-1
   • D.5.0 × 10^10 L mol^-1 s^-1
   Show answer and explanation

   Answer: B. 5.0 × 10^2 L mol^-1 s^-1

   Collision theory gives k = P × Z × e^(-Ea/RT); substituting P = 1.0, Z = 5.0 × 10^10, and the energy fraction 1.0 × 10^-8 gives k = 5.0 × 10^2 L mol^-1 s^-1.

   The collision theory rate constant is k = P·Z·f, where f = e^(-Ea/RT) is the fraction of collisions with sufficient energy. Here P = 1.0 (no orientation restriction), Z = 5.0 × 10^10 L mol^-1 s^-1, and f = 1.0 × 10^-8. So k = 1.0 × (5.0 × 10^10) × (1.0 × 10^-8) = 5.0 × 10^2 L mol^-1 s^-1.

   Common mistake: Forgetting to multiply by the small energy fraction, or multiplying by its reciprocal.

8. Question 8 · medium · Collision theory and mechanisms

   Two gas-phase reactions occur at the same temperature with identical collision frequency Z and identical activation energy Ea. Reaction 1 has steric factor P1 = 1.0 and rate constant k1 = 2.0 × 10^-2 L mol^-1 s^-1. Reaction 2 has steric factor P2 = 4.0 × 10^-4. What is the rate constant k2?

   • A.8.0 × 10^-6 L mol^-1 s^-1
   • B.5.0 × 10^1 L mol^-1 s^-1
   • C.2.0 × 10^-2 L mol^-1 s^-1
   • D.8.0 × 10^2 L mol^-1 s^-1
   Show answer and explanation

   Answer: A. 8.0 × 10^-6 L mol^-1 s^-1

   Since k = P·Z·e^(-Ea/RT) and Z, Ea, T are identical for both reactions, k is directly proportional to P; so k2 = k1 × (P2/P1) = (2.0 × 10^-2) × (4.0 × 10^-4) = 8.0 × 10^-6 L mol^-1 s^-1.

   With Z and e^(-Ea/RT) identical for both reactions, k ∝ P. Thus k2/k1 = P2/P1, giving k2 = k1 × (P2/P1) = (2.0 × 10^-2 L mol^-1 s^-1) × (4.0 × 10^-4 / 1.0) = 8.0 × 10^-6 L mol^-1 s^-1. The much smaller steric factor for reaction 2 (stricter orientation requirement, e.g. a bulky or asymmetric molecule) sharply lowers its rate constant relative to reaction 1 despite identical collision frequency and activation energy.

   Common mistake: Assuming steric factor has no quantitative effect on the numeric value of k.

Questions about Chemical Kinetics for NEET

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