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Chemical Kinetics

NEET > Chemistry > Chemical Kinetics

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Chapter Snapshot - Chemical Kinetics

The chapter that quantifies how fast reactions occur and why. Rate of reaction, rate law and order, molecularity, integrated rate equations for zero and first order, half-life (t1/2 = 0.693/k for first order), Arrhenius equation (k = Ae^(-Ea/RT)), collision theory with energy and orientation barriers, effect of catalyst, and pseudo-unimolecular reactions are the tested pillars. NEET draws 2 to 3 questions from this chapter every year, making it one of the highest-yield chapters in physical chemistry.

āœ“ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
2-3
NEET consistently draws 2 to 3 questions: one on first-order rate constant or half-life calculation, one on Arrhenius equation (activation energy from two temperatures), and one conceptual on order vs molecularity or effect of catalyst.
Time Required (Practical)
ā±
12-14 hrs
Rate of reaction definitions 1 hr; molecularity vs order 1 hr; first order integrated rate equation and half-life 3 hrs; zero and second order 2 hrs; Arrhenius equation and collision theory 2 hrs; experimental methods for order 1 hr; photochemical reactions 1 hr; MCQ practice 2 hrs.
Difficulty Level
⚔
Moderate
First order kinetics is heavily formula-based but the formulas are consistent (k = 2.303/t log(a0/at), t1/2 = 0.693/k). The Arrhenius equation requires comfortable handling of logarithms and inverting temperatures. The conceptual distinction between order and molecularity is the most common source of errors.
Most Asked Style: Numerical MCQ: calculate the rate constant from concentration-time data for a first-order reaction; find t1/2 given k; calculate Ea from rate constants at two temperatures using the Arrhenius equation; identify the order of reaction from units of k.Biggest Trap: Confusing molecularity with order. Molecularity is always a positive integer derived from the elementary step mechanism. Order is experimentally determined and can be zero, fractional, or negative. For complex reactions the overall order has no simple relation to stoichiometric coefficients.Fast Win: Memorise three formulas: (1) k = 2.303/t log([A]0/[A]t) for first order, (2) t1/2 = 0.693/k for first order (independent of initial concentration), (3) log(k2/k1) = Ea/(2.303R) [1/T1 - 1/T2] for Arrhenius. These three cover 80% of all NEET kinetics numericals.Revision-Friendly: High. The three core formulas fit on one card. First order dominates NEET, so focus there. Quick relationships: t(75%) = 2 t(1/2); t(99.9%) = 10 t(1/2). Unit of k for nth order: (mol/L)^(1-n) time^(-1).

Subtopics - Chemical Kinetics (NEET)

Seven topic blocks: rate of reaction with factors affecting rate, molecularity vs order of reaction, first order kinetics with integrated rate equation and half-life, higher order reactions (zero, second, third), Arrhenius equation with collision theory and catalysis, experimental methods for order determination, and photochemical reactions.

Revision tip: For every kinetics numerical: (1) identify the order of reaction (from units of k or from the rate law), (2) select the correct integrated rate equation, (3) for first order, use k = 2.303/t log([A]0/[A]t) and t1/2 = 0.693/k. For Arrhenius: always convert temperatures to Kelvin before substituting.
NCERT LinesMCQsQuick Test

1) Rate of Reaction

Defines rate as change in concentration per unit time. For aA + bB to cC + dD: rate = -(1/a)(d[A]/dt) = (1/c)(d[C]/dt). Distinguishes average rate from instantaneous rate. Lists six factors affecting rate: nature of reactant, concentration, surface area, catalyst, temperature, and light.

Rate = -d[A]/dt for single reactantDivide by coefficient for multi-speciesAverage vs instantaneous rate6 factors affect rate
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Rate expression and stoichiometryFor the general reaction aA + bB to cC + dD, rate of reaction = -(1/a)(d[A]/dt) = -(1/b)(d[B]/dt) = (1/c)(d[C]/dt) = (1/d)(d[D]/dt). Negative sign for reactants (concentration decreasing). Dividing by the stoichiometric coefficient ensures the rate is the same regardless of which species is monitored. Units: mol L^-1 s^-1.
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Factors influencing rate of reactionSix factors: (1) Nature of reactant: reactions involving fewer or weaker bonds are faster. (2) Concentration: higher concentration leads to more frequent collisions. (3) Surface area: coal dust burns faster than a lump of coal. (4) Catalyst: lowers activation energy, allowing more molecules to react. (5) Temperature: a 10 degree C rise roughly doubles or triples the rate. (6) Light: certain reactions are photochemically initiated.

2) Molecularity and Order of Reaction

Molecularity is the number of reacting species in an elementary step (always a positive integer, rarely above 3). Order is the sum of the powers in the experimentally determined rate law (can be zero, fractional, or negative). For complex reactions, molecularity of the rate-determining step governs the kinetics. Includes pseudo-unimolecular reactions.

Molecularity: always positive integerOrder: experimental, can be 0 or fractionalRDS molecularity governs ratePseudo first order: excess solvent
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Molecularity vs orderMolecularity: minimum number of molecules, atoms, or ions that must collide for the elementary reaction to occur. Always a whole positive number (1, 2, or 3). Unimolecular (1), bimolecular (2), termolecular (3). Reactions with molecularity > 3 are rare because simultaneous collisions of 4+ species are improbable. Order: the sum of the concentration exponents in the experimentally determined rate law. It can be zero, fractional, integral, or even negative (for inhibitors). For elementary reactions, order equals molecularity. For complex reactions, order is determined by experiment and may differ from the stoichiometric coefficients.
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Pseudo-unimolecular reactionsWhen one reactant is in large excess, its concentration remains effectively constant throughout, and the reaction appears to follow first order kinetics with respect to the other reactant. Example: sucrose hydrolysis C12H22O11 + H2O to glucose + fructose is bimolecular but follows first order kinetics because [H2O] is nearly constant. The observed rate constant k' = k[H2O]. Acid-catalysed ester hydrolysis is another common example.

3) First Order Kinetics, Rate Equation, and Half-Life

The integrated first order rate equation: k = (2.303/t) log([A]0/[A]t). A plot of log[A]t vs t gives a straight line with slope -k/2.303. Half-life t1/2 = 0.693/k is independent of initial concentration. Time for 75% completion = 2 t1/2; for 99.9% = 10 t1/2. Gaseous first order reactions use partial pressure: k = (2.303/t) log(P1/P2). Multiple cases for total pressure calculations.

k = 2.303/t log([A]0/[A]t)t1/2 = 0.693/kt(75%) = 2 t1/2log[A]t vs t: slope = -k/2.303
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Integrated first order rate equationStarting from -d[A]/dt = k[A], integration from [A]0 to [A]t gives: ln([A]0/[A]t) = kt, or equivalently k = (2.303/t) log([A]0/[A]t). Rearranging: log[A]t = (-k/2.303)t + log[A]0, which is a straight line equation (y = mx + c). Plotting log[A]t vs t gives slope = -k/2.303 and intercept = log[A]0. Unit of k for first order: time^-1 (s^-1, min^-1).
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Half-life and completion timesHalf-life t1/2: time for [A] to drop to [A]0/2. For first order: t1/2 = 0.693/k. Independent of initial concentration. Key relationships: t(75% complete) = 2 t1/2 (two half-lives reduce to 25%). t(87.5%) = 3 t1/2. t(99.9%) = 10 t1/2. General: for nth order, t1/2 is proportional to [A]0^(1-n). For zero order: t1/2 = [A]0/(2k), proportional to [A]0. For second order: t1/2 = 1/(k[A]0), inversely proportional to [A]0.
›
Gaseous first order reactionsFor gaseous reactions, partial pressure replaces concentration (P proportional to [A] at constant T and V). Case I: given P(A) at t=0 and t=t, k = (1/t) ln(P1/P2). Case II: given total pressure at t=0 (P1) and t=t (P2), with A to B+C: PA at time t = 2P1 - P2, so k = (2.303/t) log(P1/(2P1-P2)). Case III: given total pressure at t=t and t=infinity, with P(infinity) = 2P1: k = (2.303/t) log(P3/(2(P3-P2))). These pressure formulas are heavily tested in NEET.

4) Zero, Second, and Third Order Reactions

Zero order: rate = k, [A]t = [A]0 - kt, t1/2 = [A]0/(2k). Second order: 1/[A]t = kt + 1/[A]0, t1/2 = 1/(k[A]0). Third order: more complex expression, t1/2 proportional to [A]0^(-2). General nth order: unit of k is (mol/L)^(1-n) time^(-1), and t1/2 proportional to [A]0^(1-n).

Zero: [A]t = [A]0 - ktSecond: 1/[A]t = kt + 1/[A]0t1/2 proportional to [A]0^(1-n)Unit: (mol/L)^(1-n) time^-1
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Zero order reactionsRate = k (constant, independent of concentration). Integrated form: [A]t = [A]0 - kt. Plot of [A]t vs t is a straight line with slope -k. Half-life: t1/2 = [A]0/(2k), directly proportional to initial concentration. Unit of k: mol L^-1 time^-1. Examples: decomposition of NH3 on hot platinum surface, enzyme-catalysed reactions at substrate saturation.
›
Second and third order reactionsSecond order (single reactant): 1/[A]t = kt + 1/[A]0. Plot of 1/[A] vs t is linear with slope k. Half-life: t1/2 = 1/(k[A]0), inversely proportional to [A]0. Unit: L mol^-1 time^-1. Examples: thermal decomposition of NO2, saponification of ester with NaOH. Third order: t1/2 = 3/(2k[A]0^2), proportional to [A]0^(-2). Unit: L^2 mol^-2 time^-1. Examples: 2NO + O2 to 2NO2, 2NO + Cl2 to 2NOCl.

5) Arrhenius Equation, Collision Theory, and Effect of Catalyst

The Arrhenius equation k = Ae^(-Ea/RT) quantifies the temperature dependence of rate constants. Log k vs 1/T gives a straight line with slope -Ea/(2.303R). Two-temperature form: log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2). Collision theory explains why only a fraction of collisions are effective (energy barrier + orientation barrier). A catalyst lowers Ea equally for forward and reverse reactions without shifting equilibrium.

k = Ae^(-Ea/RT)log(k2/k1) = Ea/2.303R (1/T1 - 1/T2)Catalyst lowers Ea both waysTemperature coefficient = 2-3
›
Arrhenius equationk = Ae^(-Ea/RT), where A = pre-exponential or frequency factor (effective collisions per unit time per unit volume), Ea = activation energy, R = gas constant, T = absolute temperature. Taking log: log k = log A - Ea/(2.303RT). Plot of log k vs 1/T: straight line with slope = -Ea/(2.303R) and intercept = log A. For two temperatures: log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2). Larger Ea means smaller k (slower reaction). The Boltzmann factor e^(-Ea/RT) represents the fraction of molecules with energy greater than or equal to Ea.
›
Collision theory and energy barrierFor a reaction to occur, molecules must collide. Not every collision is effective. Two barriers must be overcome: (1) Energy barrier: colliding molecules must have energy equal to or greater than the threshold energy (activation energy). (2) Orientation barrier: molecules must collide with correct spatial orientation so old bonds break and new bonds form. Temperature coefficient: rate approximately doubles or triples for every 10 degree C rise, because the fraction of molecules above Ea increases steeply.
›
Effect of catalystA catalyst increases reaction rate without being consumed. It provides an alternative reaction pathway with lower activation energy. The Ea for both forward and reverse reactions is lowered by the same amount. Therefore a catalyst does NOT change the position of equilibrium or the equilibrium constant K. It only speeds up the approach to equilibrium. If a catalyst doubles the forward rate, it also doubles the reverse rate.

6) Experimental Determination of Order

Four methods to determine order experimentally: integrated equation or hit-and-trial (substitute data into rate equations for different orders and check which gives constant k), fractional change method (compare half-lives at different initial concentrations), graphical method (plot rate vs concentration), and Ostwald isolation method (take all reactants except one in excess).

Hit and trial: constant k?t1/2 vs [A]0: slope gives 1-nlog(rate) vs log(conc): slope = nIsolation: one at a time
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Methods for order determinationFour standard methods: (1) Integrated equation method (hit and trial): substitute a, x, and t values into zero, first, and second order equations. The one giving constant k is the correct order. (2) Fractional change method: t(1/2) proportional to [A]0^(1-n). Running experiments at two initial concentrations and comparing half-lives: n = 1 + (log(t1/t2))/(log(a2/a1)). (3) Graphical method: plot rate (-dx/dt) vs (a-x), (a-x)^2, or (a-x)^3. The linear plot reveals the order. Or plot log(rate) vs log(a-x): slope = n. (4) Ostwald isolation method: take all reactants except one in large excess. The apparent order for the isolated reactant is found. Sum of individual orders gives overall order.

7) Photochemical Reactions

Reactions initiated by absorption of light. Einstein's law: each molecule absorbs one photon (E = hc/lambda per molecule, E = Nhc/lambda per mole). Rate depends on intensity of absorbed radiation, not temperature. Quantum yield phi = molecules reacted / photons absorbed. Photosensitisation: foreign substance absorbs light and transfers energy to reactant.

E = hc/lambda per photonRate depends on I(absorbed)Quantum yield phiTemperature has little effect
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Characteristics of photochemical reactionsKey features: (1) Each molecule absorbs exactly one photon (Einstein's law of photochemical equivalence). Energy per molecule = hc/lambda; per mole = Nhc/lambda. (2) Does not occur in the dark. (3) Each reaction requires a specific wavelength; UV light (short lambda, high energy) initiates reactions that visible light cannot. (4) Rate proportional to intensity of absorbed radiation: rate = k times I(abs). (5) delta-G may or may not be negative (light energy compensates). (6) Temperature has little effect on rate. (7) Quantum yield phi = number of molecules reacted / number of photons absorbed. HBr photosynthesis: phi = 0.01 (low). HCl photosynthesis: phi = 10^6 (chain reaction).

Chemical Kinetics Download Notes & Weightage Plan

For each topic in the Chemical Kinetics chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Rate of Reaction

Rate definition, stoichiometric rate expression, average vs instantaneous rate, six factors affecting rate.

Definitional topicQuick to reviseRate expression testedFactors are conceptual

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Rate = change in concentration per unit time. For aA + bB to cC + dD: -(1/a)(d[A]/dt) = (1/c)(d[C]/dt). Average rate: delta[A]/delta-t over a time interval. Instantaneous rate: d[A]/dt at a specific moment. Six factors: nature of reactant, concentration, surface area, catalyst, temperature, exposure to light.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write the general rate expression with stoichiometric coefficients. Know that rate for product appearance is positive, for reactant is negative. Memorise the six factors.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasional conceptual MCQ on writing rate expression with correct stoichiometric coefficients.
Time Required1 hr30 min on rate expression; 30 min on factors.
DifficultyEasyPure definitions and factual recall.
  • Scoring Focus: Writing the correct rate expression with stoichiometric coefficients. For 2A + B to 3C: rate = -(1/2)(d[A]/dt) = -d[B]/dt = (1/3)(d[C]/dt).
  • High-risk Area: Forgetting to divide by the stoichiometric coefficient. For 2N2O5 to 4NO2 + O2: the rate of disappearance of N2O5 is -(1/2)(d[N2O5]/dt), not -(d[N2O5]/dt). Missing the 1/2 introduces a factor-of-two error.
  • Best Practice Style: Always write the balanced equation first. Each d[species]/dt term is divided by its coefficient.
Priority rule: Low priority. Quick revision in 1 hour.

Molecularity and Order of Reaction

Distinguishing molecularity (theoretical, from mechanism) from order (experimental). Pseudo-unimolecular reactions.

Conceptual distinctionTested every yearOrder can be fractionalPseudo first order

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Molecularity: number of species in the elementary step. Always positive integer (1, 2, or 3). Cannot be zero or fractional. Order: sum of powers in the experimental rate law. Can be 0, fractional, integral, or negative. For elementary reactions: order = molecularity. For complex reactions: order is determined by experiment. Pseudo first order: one reactant in large excess (e.g., sucrose hydrolysis).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Make a comparison table: molecularity vs order. Key differences: molecularity is theoretical, order is experimental; molecularity must be integer, order need not be; molecularity applies to elementary steps only, order applies to overall reaction.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One conceptual MCQ distinguishing order from molecularity, or identifying a pseudo-first-order reaction.
Time Required1 hr30 min on definitions and comparison; 30 min MCQ practice.
DifficultyEasy-ModerateConceptual but requires precise understanding. Students confuse the two terms.
  • Scoring Focus: Molecularity can NEVER be zero or fractional. Order CAN be zero, fractional, or negative. Molecularity applies only to elementary steps. For complex reactions, only the rate-determining step's molecularity matters.
  • High-risk Area: Assuming order = molecularity for all reactions. This is true only for elementary reactions. For complex reactions, the overall order can differ significantly from the sum of stoichiometric coefficients. NEET tests this distinction directly.
  • Best Practice Style: If the question says 'elementary reaction', then order = molecularity. Otherwise, order must be determined experimentally.
Priority rule: Moderate priority. This conceptual distinction is tested almost every year.

First Order Kinetics, Rate Equation, and Half-Life

The most important topic for NEET. Integrated rate equation, half-life formula, pressure-based cases for gaseous reactions.

k = 2.303/t log([A]0/[A]t)t1/2 = 0.693/kGaseous: pressure replaces concentrationFour pressure cases

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)k = (2.303/t) log([A]0/[A]t). Plot of log[A]t vs t: straight line, slope = -k/2.303. t1/2 = 0.693/k, independent of [A]0. t(75%) = 2 t1/2; t(87.5%) = 3 t1/2; t(99.9%) = 10 t1/2. Gaseous cases: Case I: given P(A) at t=0 and t, k = (1/t) ln(P1/P2). Case II: given Ptotal at t=0 and t, A to B+C: PA(t) = 2P1 - Pt, k = (2.303/t) log(P1/(2P1-Pt)). Case III: given Ptotal at t and at infinity. Case IV: given pressure of products at t and infinity.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Practise three numericals: one concentration-based, one pressure-based (Case II), and one involving t(75%). Know that for first order, k has units of time^-1. The pressure cases are the most commonly tested variant.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2One or two numericals on first order kinetics: calculate k from concentration data, find t1/2, or use the pressure-based formula for a gaseous decomposition.
Time Required3 hrs1 hr on integrated rate equation derivation and practice; 1 hr on half-life and completion times; 1 hr on gaseous cases.
DifficultyModerateThe integrated equation involves logarithms. The gaseous pressure cases require setting up an ICE-like table for partial pressures. Case II and Case III are the trickiest.
  • Scoring Focus: Three high-yield results: (1) k = (2.303/t) log([A]0/[A]t), (2) t1/2 = 0.693/k (independent of [A]0 for first order), (3) for gaseous A to B+C: PA at time t = 2P(initial) - P(total at time t). Master these three and you can solve most first order problems.
  • High-risk Area: In gaseous Case II (A to B+C, given total pressure), students confuse P(total) with P(A). At time t, P(total) = P1 + x (where x = extent of reaction), not P1 - x. The pressure of A at time t is 2P1 - P(total), which is less than P1. Getting this algebra wrong inverts the log term.
  • Best Practice Style: For gaseous reactions: (1) write the ICE table in terms of partial pressures, (2) express P(A) at time t in terms of given total pressures, (3) substitute into k = (2.303/t) log(P(A,0)/P(A,t)).
Priority rule: Highest priority. First order kinetics is tested in every NEET paper. Spend 3 hours on practice.

Zero, Second, and Third Order Reactions

Zero order: rate is constant. Second order: 1/[A]t = kt + 1/[A]0. General nth order: unit of k and t1/2 proportional to [A]0^(1-n).

Zero: t1/2 proportional to [A]0Second: t1/2 proportional to 1/[A]0Unit of k: (mol/L)^(1-n) time^-1Rate becomes n^n times

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Zero order: rate = k. [A]t = [A]0 - kt. t1/2 = [A]0/(2k). Unit: mol L^-1 time^-1. Second order: 1/[A]t = kt + 1/[A]0. t1/2 = 1/(k[A]0). Unit: L mol^-1 time^-1. Two reactants with different concentrations: k = (2.303/t(a-b)) log(b(a-x)/(a(b-x))). Third order: t1/2 = 3/(2k[A]0^2). Unit: L^2 mol^-2 time^-1. General nth order: unit = (mol/L)^(1-n) time^-1. t1/2 proportional to [A]0^(1-n).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Make a comparison table: order 0, 1, 2, 3 with columns for integrated rate equation, half-life relation, units of k, and a linear plot. The half-life dependence on [A]0 is the fastest way to identify order.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasional MCQ asking for units of rate constant for a given order, or identifying order from half-life dependence on initial concentration.
Time Required2 hrs30 min on zero order; 1 hr on second order; 30 min on comparison table.
DifficultyModerateThe second order equation for two reactants with different concentrations is complex but rarely tested in NEET.
  • Scoring Focus: Two quick identification tools: (1) Unit of k: if k has units of time^-1, it is first order. If mol^-1 L time^-1, it is second order. (2) Half-life independence of [A]0 means first order. Half-life proportional to [A]0 means zero order. Half-life inversely proportional to [A]0 means second order.
  • High-risk Area: For zero order, [A]t = [A]0 - kt can give negative concentrations if t > [A]0/k. The reaction stops when [A] = 0, not when the formula gives a negative number. Students who extrapolate beyond completion time get physically meaningless results.
  • Best Practice Style: Use the half-life dependence on initial concentration as the fastest diagnostic: t1/2 proportional to [A]0^(1-n). n=0: proportional to [A]0. n=1: independent. n=2: inversely proportional.
Priority rule: Moderate priority. Second order is occasionally tested. Zero order more rarely. Focus on identification shortcuts.

Arrhenius Equation, Collision Theory, and Effect of Catalyst

k = Ae^(-Ea/RT). Two-temperature form for calculating Ea. Collision theory: energy barrier and orientation barrier. Catalyst lowers Ea without changing K(eq).

k = Ae^(-Ea/RT)Slope of log k vs 1/T = -Ea/2.303RCatalyst: lower Ea, same K(eq)Temperature coefficient: 2-3

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)k = Ae^(-Ea/RT). log k = log A - Ea/(2.303RT). Plot log k vs 1/T: slope = -Ea/(2.303R). Two temperatures: log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2). Temperature coefficient = k(T+10)/k(T) is usually 2-3. Collision theory: effective collision needs (1) energy >= Ea (threshold energy) and (2) correct orientation. Catalyst provides alternative path with lower Ea. Lowers Ea equally for forward and reverse reactions. Does not change equilibrium constant or position of equilibrium. If catalyst doubles forward rate, it doubles reverse rate too.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Practise the Arrhenius two-temperature calculation: given k1 at T1 and k2 at T2, find Ea. Or given Ea and k at one temperature, find k at another temperature. Always convert T to Kelvin.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question on Arrhenius equation: calculate Ea from rate constants at two temperatures, or a conceptual question on catalyst effect.
Time Required2 hrs45 min on Arrhenius equation and numericals; 30 min on collision theory; 30 min on catalyst; 15 min MCQ.
DifficultyModerateThe Arrhenius numerical requires careful handling of 1/T values (small numbers). The conceptual understanding of catalyst effect is straightforward.
  • Scoring Focus: The two-temperature Arrhenius formula is the most tested: log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2). Common NEET trick: providing temperatures in Celsius (must convert to Kelvin). Catalyst effect: lowers Ea, does not change delta-H or K(eq).
  • High-risk Area: Forgetting to convert Celsius to Kelvin in the Arrhenius equation. Using T = 25 instead of T = 298 K makes the 1/T term 12 times too large, giving an Ea that is absurdly small. NEET typically gives temperatures in Celsius to catch this error.
  • Best Practice Style: Step 1: convert all temperatures to Kelvin. Step 2: substitute into the two-temperature form. Step 3: use R = 8.314 J/mol K if Ea is in J/mol, or R = 2 cal/mol K if Ea is in cal/mol.
Priority rule: High priority. Arrhenius calculation appears frequently. Master the two-temperature form.

Experimental Determination of Order

Four methods: integrated equation (hit and trial), fractional change, graphical, and Ostwald isolation.

Hit and trial: constant k checkt1/2 vs [A]0: gives 1-nGraphical: log(rate) vs log(c)Isolation: one reactant isolated

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Four methods: (1) Integrated equation method: substitute experimental data into rate equations for different orders; the one giving constant k is correct. (2) Fractional change method: t1/2 proportional to [A]0^(1-n); run at two initial concentrations; n = 1 + log(t1'/t1'')/log(a2/a1). (3) Graphical: plot log(dx/dt) vs log(a-x), slope = n. Or plot [A] vs t (zero order), log[A] vs t (first order), 1/[A] vs t (second order) and check linearity. (4) Ostwald isolation: take all reactants except one in excess; determine the order for the isolated reactant; sum gives overall order.
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Know the graphical identifiers: linear [A] vs t = zero order; linear log[A] vs t = first order; linear 1/[A] vs t = second order. The Ostwald method is conceptual only at NEET level.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasional conceptual question on graphical identification of order or on the fractional change method.
Time Required1 hr30 min on graphical identification; 30 min on fractional change method.
DifficultyModerateThe graphical method is intuitive. The fractional change logarithmic formula requires careful algebra.
  • Scoring Focus: Fastest identification: if log[A] vs t is linear, the reaction is first order. If 1/[A] vs t is linear, it is second order. If [A] vs t is linear, it is zero order.
  • High-risk Area: Confusing the y-axis variables: [A] vs log[A] vs 1/[A]. Each corresponds to a different order. Plotting the wrong variable gives a curve instead of a straight line.
  • Best Practice Style: For graphical identification, try the first-order plot first (log[A] vs t) since first order is the most commonly tested. If linear, done.
Priority rule: Low-moderate priority. The graphical identification shortcut is useful.

Photochemical Reactions

Reactions initiated by light absorption. Einstein's law, quantum yield, characteristics distinguishing photochemical from thermal reactions.

E = Nhc/lambda per moleRate depends on light intensityQuantum yield phiT has little effect

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)Photochemical reactions: initiated by absorption of light. Einstein's law: each molecule absorbs one photon. E per mole = Nhc/lambda. Rate proportional to intensity of absorbed light: rate = k I(abs). Temperature has negligible effect. delta-G need not be negative (light provides energy). Quantum yield phi = molecules reacted / photons absorbed. HBr: phi = 0.01 (low efficiency). HCl: phi = 10^6 (chain reaction, high efficiency). Photosensitised reactions: photosensitiser absorbs light and transfers energy to reactant (e.g., chlorophyll in photosynthesis).
Download NotesPrintable PDF
ā˜…
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Know: photochemical rate depends on light intensity not temperature. E per mole of photons = Nhc/lambda. Quantum yield phi can be greater than 1 for chain reactions.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions0-1Occasional conceptual MCQ on characteristics of photochemical reactions or quantum yield.
Time Required1 hr30 min on characteristics; 30 min MCQ.
DifficultyEasyMostly factual recall. The concept of quantum yield is the only somewhat tricky idea.
  • Scoring Focus: Key distinction: photochemical rate depends on light intensity (not temperature). Quantum yield phi > 1 is possible for chain reactions (HCl synthesis: phi = 10^6).
  • High-risk Area: Assuming photochemical reactions always have phi = 1. Chain reactions like HCl synthesis have enormously high quantum yields because one absorbed photon triggers thousands of reaction cycles.
  • Best Practice Style: For photochemical MCQs: rate depends on I(abs), not T. delta-G can be positive. Each molecule absorbs exactly one photon.
Priority rule: Low priority. One question occasionally. Quick revision in 1 hour.

Chemical Kinetics Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Chemical Kinetics chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
First Order Rate Constant Calculation
NEETFirst orderRate constantHalf-life

Mistake Snapshot (What Students Do Wrong)

  • Using natural log formula but reading common log value: k = (1/t) ln([A]0/[A]t) uses natural log (ln), while k = (2.303/t) log([A]0/[A]t) uses common log (log10). Mixing the two gives k that is off by a factor of 2.303.
  • Thinking t(75%) = 1.5 times t(1/2): After one half-life, 50% remains. After two half-lives, 25% remains (75% complete). So t(75%) = 2 t(1/2), not 1.5 t(1/2). Students who confuse fraction remaining with fraction completed make this error.
2–3 Line Example (Typical Error)

A first order reaction has k = 0.01 min^-1. t1/2 = 0.693/0.01 = 69.3 min. t(75%) = 2 x 69.3 = 138.6 min. If one mistakenly uses t(75%) = 1.5 x 69.3 = 104 min, the answer is wrong because 75% completion means only 25% remains, which takes exactly two half-lives.

How NEET Frames The Trap

NEET gives k and asks for time to complete 75% or 87.5% of the reaction.

NEET-Style Trap Question Format

Q. The half-life of a first order reaction is 20 minutes. The time required for 75% completion is
A. 40 min   B. 30 min   C. 60 min   D. 20 min  
Trick: 75% completion means 25% remains, which is (1/2)^2 of the original. So it takes 2 half-lives = 40 min (Option A). Option B uses 1.5 x t1/2. Option C uses 3 x t1/2 (which would be 87.5%).

Quick rule: Fraction remaining after n half-lives = (1/2)^n. For 75% completion: 25% remains = (1/2)^2 = 2 half-lives. For 87.5%: 12.5% = (1/2)^3 = 3 half-lives. For 99.9%: (1/2)^10 approximately 0.1% remains = 10 half-lives.
Gaseous First Order Pressure Problems
NEETFirst orderPressureGaseous reaction

Mistake Snapshot (What Students Do Wrong)

  • Using total pressure instead of partial pressure of reactant: For A to B+C, the total pressure at time t is P(total) = P(A) + P(B) + P(C) = P1 + x. The partial pressure of A is P(A) = P1 - x = 2P1 - P(total). Substituting P(total) directly as P(A) gives wrong k.
  • Confusing P1 + x with P1 - x: Total pressure increases (P1 + x) because 1 mole of A produces 2 moles of products. But the partial pressure of A decreases (P1 - x). Students who assume total pressure decreases get the algebra inverted.
2–3 Line Example (Typical Error)

A(g) to B(g) + C(g). Initial pressure = 100 mmHg. Total pressure at time t = 130 mmHg. x = 130 - 100 = 30. P(A) = 100 - 30 = 70 mmHg. k = (2.303/t) log(100/70). If student uses P(total) as P(A): k = (2.303/t) log(100/130), which gives a negative log (impossible for k).

How NEET Frames The Trap

NEET gives initial pressure and total pressure at time t for a gaseous decomposition and asks for the rate constant.

NEET-Style Trap Question Format

Q. For the decomposition A(g) to 2B(g), initial pressure of A = 200 mmHg. Total pressure after 10 min = 300 mmHg. Rate constant is
A. 0.0693 min^-1   B. 0.0231 min^-1   C. 0.0462 min^-1   D. 0.1386 min^-1  
Trick: A to 2B: at time t, P(A) = P1 - x, P(B) = 2x. P(total) = P1 - x + 2x = P1 + x. So x = 300 - 200 = 100. P(A) = 200 - 100 = 100. k = (2.303/10) log(200/100) = 0.2303 x 0.301 = 0.0693 min^-1 (Option A). Option B uses P(total) as P(A).

Quick rule: For A to nB: total pressure = P1 + (n-1)x. So x = (P(total) - P1)/(n-1). P(A) = P1 - x. Then k = (2.303/t) log(P1/P(A)).
Arrhenius Equation Temperature Error
NEETArrheniusTemperatureActivation energy

Mistake Snapshot (What Students Do Wrong)

  • Using temperature in Celsius instead of Kelvin: The Arrhenius equation requires absolute temperature (Kelvin). Using Celsius (e.g., 25 instead of 298) makes 1/T about 12 times too large, giving Ea that is 12 times too small.
  • Swapping T1 and T2 in the two-temperature formula: The formula is log(k2/k1) = (Ea/2.303R)(1/T1 - 1/T2) where T2 > T1 and k2 > k1. Swapping gives a negative Ea, which is physically impossible for most reactions.
2–3 Line Example (Typical Error)

k = 2 x 10^-5 s^-1 at 300 K and 4 x 10^-5 s^-1 at 310 K. log(4/2) = (Ea/2.303 x 8.314)(1/300 - 1/310). 0.301 = (Ea/19.15)(10/93000) = Ea x 5.61 x 10^-6. Ea = 0.301/5.61 x 10^-6 = 53,655 J/mol = 53.7 kJ/mol. Using T = 27 and 37 instead of 300 and 310: 1/27 - 1/37 = 0.010, which gives Ea = 0.301/(0.010 x Ea/19.15), a completely wrong value.

How NEET Frames The Trap

NEET gives temperatures in Celsius and rate constants at those temperatures. Students must convert to Kelvin first.

NEET-Style Trap Question Format

Q. A reaction has Ea = 60 kJ/mol. If k at 27 degrees C is 1.5 x 10^-3 s^-1, what is k at 37 degrees C? (R = 8.314 J/mol K)
A. 3.0 x 10^-3   B. 1.5 x 10^-3   C. 6.0 x 10^-3   D. 0.75 x 10^-3  
Trick: Convert: T1 = 300 K, T2 = 310 K. log(k2/1.5 x 10^-3) = (60000/2.303 x 8.314)(1/300 - 1/310) = (3134)(1.075 x 10^-4) = 0.337. k2/k1 = 10^0.337 = 2.17. k2 = 2.17 x 1.5 x 10^-3 = 3.3 x 10^-3, closest to Option A (3.0 x 10^-3). Option B assumes no temperature effect. Option D swaps T1 and T2.

Quick rule: Always convert Celsius to Kelvin: K = C + 273. Then substitute. Check: higher T should give higher k.
Molecularity vs Order Confusion
NEETMolecularityOrderConceptual

Mistake Snapshot (What Students Do Wrong)

  • Stating molecularity can be zero or fractional: Molecularity is always a positive integer (1, 2, or 3). It is the number of species in the elementary step. Order CAN be zero, fractional, or negative.
  • Equating order with stoichiometric coefficients for complex reactions: For the overall reaction 2A + B to products, the order is NOT necessarily 3 (2+1). The order is determined experimentally and depends on the rate-determining step mechanism.
2–3 Line Example (Typical Error)

Overall reaction: 2N2O5 to 4NO2 + O2. This is a complex reaction with a multi-step mechanism. The experimental rate law is rate = k[N2O5], so the order is 1, not 2. Molecularity of the rate-determining step (unimolecular decomposition of N2O5) is 1.

How NEET Frames The Trap

NEET gives an overall reaction equation and asks for the order. The distractor is the sum of stoichiometric coefficients.

NEET-Style Trap Question Format

Q. Which of the following is true about molecularity?
A. It is always a whole number   B. It can be zero   C. It can be fractional   D. It can be negative  
Trick: Option A: Molecularity is always a positive whole number (1, 2, or 3). Options B, C, D describe properties of ORDER, not molecularity. Order can be zero (option B), fractional (option C), or negative (option D).

Quick rule: Molecularity: always positive integer, never zero or fractional. Order: anything goes (0, fractional, negative). For elementary reactions only, they are equal.
Catalyst and Equilibrium Misunderstanding
NEETCatalystEquilibriumActivation energy

Mistake Snapshot (What Students Do Wrong)

  • Thinking catalyst changes equilibrium position: A catalyst lowers Ea for both forward and reverse reactions equally. It speeds up both directions equally. The equilibrium constant K and equilibrium position remain unchanged.
  • Thinking catalyst changes delta-H of reaction: The catalyst changes the pathway but not the initial and final states. Since H is a state function, delta-H remains the same with or without catalyst.
2–3 Line Example (Typical Error)

For N2 + 3H2 to 2NH3, adding Fe catalyst: forward and reverse rates both increase equally. K(eq) remains the same. delta-H remains -92 kJ. Only Ea is lowered. Without catalyst, Ea might be 200 kJ; with Fe, it might be 120 kJ. But delta-H is always -92 kJ.

How NEET Frames The Trap

NEET asks what a catalyst changes: Ea, K, delta-H, or equilibrium position.

NEET-Style Trap Question Format

Q. Which of the following statements about a catalyst is correct?
A. It lowers the activation energy of the reaction   B. It changes the enthalpy of the reaction   C. It changes the equilibrium constant   D. It shifts the equilibrium position to the right  
Trick: Option A is correct. A catalyst provides an alternative pathway with lower Ea. It does NOT change delta-H (state function), K(eq) (depends only on delta-G), or equilibrium position. It only speeds up the approach to equilibrium.

Quick rule: Catalyst changes: Ea (lowers it), rate (increases it). Catalyst does NOT change: delta-H, K(eq), equilibrium position, delta-G.
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