Chemical Kinetics Class 12 Notes - CBSE Chemistry Chapter 4 (Free PDF)

Chapter summary

Chemical Kinetics studies how fast reactions go and what controls their speed, covering rate of reaction, rate law, order and molecularity, and the integrated rate equations for zero and first order reactions. It also explains half-life, the Arrhenius equation linking rate to temperature and activation energy, and how collision theory and catalysts work. It is a high-yield, formula and numerical heavy chapter that NEET tests almost every year.

Chapter notes
🃏 Flash Cards: Chemical Kinetics

Class 12 Chemistry · Chapter 4 – swipe through all 10 cards to understand the whole chapter.

⏱️Start here1/10

Rate of Reaction

The rate of a reaction is how fast a reactant disappears or a product forms per unit time.

Rate = −(1/a)·d[A]/dt = +(1/c)·d[C]/dt for aA → cC

Units: mol L⁻1 s⁻1. Reactants get a minus sign, products a plus sign.

  • Average rate = Δ[X]/Δt; instantaneous rate = slope of conc-time curve
  • Divide each species by its coefficient to get one common rate
  • For N2 + 3H2 → 2NH3, H2 vanishes 3× and NH3 forms 2× as fast as N2
📐Core law2/10

Rate Law

The rate law links rate to reactant concentrations and is found only by experiment.

Rate = k [A]ˣ [B]ʸ

k = rate constant (depends on temperature, not on concentration).

  • Exponents x, y are experimental, NOT the balanced-equation coefficients
  • For multistep reactions the slowest (rate-determining) step fixes the rate law
  • k changes with temperature; doubling [A] that doubles rate ⇒ order 1 in A
🔢Key idea3/10

Order vs Molecularity

Order is the experimental concentration dependence; molecularity counts colliding species in an elementary step.

Order = x + y ; Molecularity = 1, 2, 3 (whole number only)

Order can be 0, fractional or negative; molecularity is always a positive integer.

  • Order comes from experiment; molecularity comes from the mechanism
  • Molecularity is never zero or fractional
  • Higher order means more sensitive to concentration, not necessarily faster
🧮Units trick4/10

Units of the Rate Constant

The units of k depend on the overall order, so units alone reveal the order.

k = (mol L⁻1)1⁻ⁿ s⁻1 for overall order n

Because rate is always mol L⁻1 s⁻1, the units of k must balance it.

  • Zero order: k in mol L⁻1 s⁻1
  • First order: k in s⁻1
  • Second order: k in L mol⁻1 s⁻1
📉Integrated5/10

Zero-Order Integrated Rate

A zero-order reaction proceeds at a fixed rate independent of concentration.

[A] = [A]0 − k t

Plot of [A] vs t is a straight line with slope −k.

  • Rate is constant no matter how much reactant remains
  • Examples: decomposition of NH3 on hot Pt, many photochemical reactions
  • Slope of [A]-vs-t line gives −k directly
📊Integrated6/10

First-Order Integrated Rate

For first order the rate is proportional to the remaining reactant, so it slows as it proceeds.

k = (2.303/t)·log([A]0/[A]) ⇔ ln([A]0/[A]) = k t

Plot of log[A] vs t is linear with slope −k/2.303. The 2.303 converts ln to log10.

  • All radioactive decay is first order
  • k is independent of initial concentration and of concentration units
  • Use fraction REMAINING (not % reacted) in [A]0/[A]
Half-life7/10

Half-Life (t½) and Order

Half-life is the time for the reactant concentration to fall to half its value.

First order: t½ = 0.693/k Zero order: t½ = [A]0/(2k)

0.693 = ln 2. General rule: t½ ∝ [A]01⁻ⁿ.

  • First-order t½ is independent of initial concentration (its hallmark)
  • Zero-order t½ is directly proportional to [A]0
  • After n half-lives, fraction remaining = (1/2)ⁿ
🌡️Temperature8/10

Arrhenius Equation

Rate rises steeply with temperature because more molecules cross the activation barrier.

k = A e^(−Eₐ/RT) ⇔ log k = log A − Eₐ/(2.303 R T)

T in Kelvin; rule of thumb: rate roughly doubles per 10°C rise.

  • Plot of log k vs 1/T is a straight line of slope −Eₐ/(2.303 R)
  • Two-temperature form: log(k2/k1) = (Eₐ/2.303R)·[(T2−T1)/(T1T2)]
  • Lower Eₐ ⇒ faster reaction; heating shifts Maxwell-Boltzmann to higher energy
💥Theory9/10

Collision Theory

A collision leads to reaction only if it has enough energy and the correct orientation.

Rate = Z_AB · e^(−Eₐ/RT) · P

Z_AB = collision frequency; P = steric (orientation) factor, usually < 1.

  • Two conditions: energy ≥ Eₐ AND proper molecular orientation
  • e^(−Eₐ/RT) is the fraction of energetic (effective) collisions
  • P corrects collision theory, which otherwise overpredicts the rate
⚗️Advanced10/10

Catalysis

A catalyst speeds a reaction by offering an alternative path of lower activation energy.

Catalyst lowers Eₐ (forward & reverse equally) · ΔH and Kc unchanged

It is not consumed; it is regenerated at the end.

  • Does not change ΔH or the position of equilibrium
  • Lowers Eₐ for both directions, so equilibrium is reached faster
  • A small drop in Eₐ gives a huge rise in rate (Eₐ sits in an exponent)
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📝 Practice Chemical Kinetics — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
The slope of the Arrhenius plot (ln k vs 1/T) of a first order reaction is −5×10³ K. The value of Eₐ of the reaction is (R = 8.314 J K⁻¹ mol⁻¹):
Correct answer: A. For ln k vs 1/T, slope = −Eₐ/R. So Eₐ = −slope × R = −(−5×10³)(8.314) = 5×10³ × 8.314 = 41570 J ≈ 41.5 kJ mol⁻¹.
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Q2NEET 2020
The rate constant for a first order reaction is 4.606×10⁻³ s⁻¹. The time required to reduce 2.0 g of the reactant to 0.2 g is:
Correct answer: B. First order: t = (2.303/k)log(a/(a−x)) = (2.303/4.606×10⁻³)log(2.0/0.2) = (2.303/4.606×10⁻³)(1) = 500 s.
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Q3NEET 2020
In the collision theory of chemical reactions, Z_AB represents the:
Correct answer: B. In rate = P·Z_AB·e^(−Eₐ/RT), Z_AB is the collision frequency — the number of effective binary collisions between A and B per unit time per unit volume. P is the steric factor and e^(−Eₐ/RT) is the fraction of collisions with energy ≥ Eₐ.
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Q4NEET 2020
The half-life of a zero order reaction having 0.02 M initial concentration of reactant is 100 s. The rate constant (in mol L⁻¹ s⁻¹) for the reaction is:
Correct answer: A. For zero order, t₁/₂ = [A]₀/(2k), so k = [A]₀/(2·t₁/₂) = 0.02/(2×100) = 1.0×10⁻⁴ mol L⁻¹ s⁻¹.
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Q5NEET 2019
A first order reaction has a rate constant of 2.303×10⁻³ s⁻¹. The time required for 40 g of this reactant to reduce to 10 g will be (log10 2 = 0.3010):
Correct answer: D. t = (2.303/k)log(a/(a−x)) = (2.303/2.303×10⁻³)log(40/10) = 1000×log 4 = 1000×0.6020 = 602 s.
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Q6NEET 2019
If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by:
Correct answer: B. t = (2.303/k)log(a/(a−x)). For 99% completion, a = 100, a−x = 1, so a/(a−x) = 100. t = (2.303/k)log 100 = (2.303/k)(2) = 4.606/k.
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Q7NEET 2017
For the hypothetical reaction X₂ + Y₂ → 2XY proceeding by the mechanism (i) X₂ ⇌ X + X (fast); (ii) X + Y₂ → XY + Y (slow); (iii) X + Y → XY (fast), the overall order of the reaction is:
Correct answer: D. The slow (rate-determining) step gives rate = k₁[X][Y₂]. From the fast equilibrium X₂ ⇌ 2X, K_eq = [X]²/[X₂], so [X] = (K_eq[X₂])^(1/2). Substituting: rate = k₁(K_eq)^(1/2)[X₂]^(1/2)[Y₂]. Order = 1/2 + 1 = 3/2 = 1.5.
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Q8NEET 2016
The decomposition of phosphine (PH₃) on tungsten at low pressure is a first-order reaction. This is because the:
Correct answer: A. For surface-catalysed unimolecular decomposition, rate = kαp/(1+αp). At low pressure αp ≪ 1, so (1+αp) ≈ 1 and rate ≈ kαp, i.e. rate is proportional to pressure (surface coverage). This first-power dependence makes it first order.
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Q9NEET 2010
For the reaction N₂O₅(g) → 2NO₂(g) + ½O₂(g), the rate of disappearance of N₂O₅ is 6.25×10⁻³ mol L⁻¹ s⁻¹. The rates of formation of NO₂ and O₂ respectively are:
Correct answer: B. −d[N₂O₅]/dt = (1/2)d[NO₂]/dt = (2/1)d[O₂]/dt (from unified rate). d[NO₂]/dt = 2×6.25×10⁻³ = 1.25×10⁻². d[O₂]/dt = (1/2)×6.25×10⁻³ = 3.125×10⁻³ mol L⁻¹ s⁻¹.
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Q10NEET 2009
For the reaction BrO₃⁻(aq) + 5Br⁻(aq) + 6H⁺ → 3Br₂(l) + 3H₂O(l), the rate of appearance of bromine (Br₂) is related to the rate of disappearance of bromide ion as:
Correct answer: A. Unified rate = −(1/5)d[Br⁻]/dt = +(1/3)d[Br₂]/dt. Therefore d[Br₂]/dt = (3/5)(−d[Br⁻]/dt) = −(3/5)d[Br⁻]/dt (the minus keeps Br₂ formation positive since d[Br⁻]/dt is negative).
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Frequently Asked Questions

What is the rate of a reaction in Chemical Kinetics?

The rate of a reaction is the change in concentration of a reactant or product per unit time, with units of mol per litre per second. For a reaction aA giving cC it is written as minus one over a times d[A]/dt, equal to plus one over c times d[C]/dt, so a single common rate value is obtained.

What is the difference between order and molecularity?

Order is the sum of the powers of concentration in the experimentally found rate law and can be zero, fractional, or a whole number. Molecularity is the number of species colliding in a single elementary step and is always a positive integer, never zero or fractional.

What is the formula for half-life of a first order reaction?

For a first order reaction the half-life is t-half equal to 0.693 divided by k, where 0.693 is the natural log of 2. A key feature is that this half-life is independent of the initial concentration, which is a hallmark of first order reactions, and all radioactive decay is first order.

Is Chemical Kinetics important for NEET?

Yes, Chemical Kinetics is part of the NEET Class 12 Chemistry syllabus and is a reliable scoring chapter that usually contributes one to two questions every year. The questions are mostly numerical, based on rate law, order, integrated rate equations, half-life, and the Arrhenius equation.

How does temperature affect reaction rate and what is the Arrhenius equation?

Raising the temperature increases the rate sharply, and as a rule of thumb the rate roughly doubles for every 10 degree Celsius rise. This is captured by the Arrhenius equation, k equals A times e to the power minus Ea over RT, where A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.

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