CLASS 12 CHEMISTRY · CHAPTER 3 · STUDY NOTES

Chemical Kinetics

An in-depth study of the rates of chemical reactions, the factors influencing these rates, and the mechanisms by which reactions proceed, covering average and instantaneous rates, rate laws, integrated rate equations, and collision theory.

EXPLORE

Introduction to Chemical Kinetics

Learn to measure a rate, infer a rate law and explain why temperature or a catalyst changes the time scale.

A downhill reaction can still have a barrier

This schematic energy profile has products below reactants but a barrier between them. Thermodynamic favourability does not tell you the speed. The horizontal reaction coordinate is progress along a pathway, not elapsed time. Barrier heights and vertical values here are qualitative.

Chemical kinetics: The study of reaction rates, their dependence on conditions and the mechanisms connecting reactants to products.

Thermodynamics concerns the driving force and equilibrium position; kinetics concerns how fast a system changes. A favourable conversion can remain slow because its pathway has an activation barrier. Diamond’s persistence despite graphite’s greater stability under ordinary conditions illustrates the distinction.

Precipitation on mixing can be rapid, acid-catalysed hydrolysis can take a measurable time, and corrosion often develops gradually. These are observations under particular conditions, not permanent speed labels. Concentration, pressure for gases, temperature, catalyst and sometimes surface area can change the rate.

Let’s look a little closer.

Go deeper: Connect observations with a molecular explanation

Macroscopic concentration measurements tell us what disappears and appears. A proposed molecular mechanism must account for the experimental rate law and other evidence; a balanced overall equation by itself does not supply a mechanism.

Changing a pathway or a barrier can change speed without changing the equilibrium constant at the same temperature. Conversely, an equilibrium calculation does not specify how long the approach to equilibrium takes.

Take a moment. What do you think?

PAUSE & TRY · think first, then check with a free account

Question

A conversion has a negative Gibbs energy but remains almost unobservable over a year. Is this contradictory? Explain using the visual energy profile.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 1, 2.

EXPLORE

Rate of a Chemical Reaction

Measure disappearance and appearance, then divide by the appropriate coefficient.

Measure what disappears and what appears

Time: 40 s

Original ideal R → P example at constant volume: [R]₀=0.80 mol/L, k=0.015 s⁻¹. At 40 s, [R]=0.4390 and [P]=0.3610 mol/L. Loss and gain have equal magnitude because both coefficients are one. The orange curve is P and blue curve is R.

Reaction rate: At constant volume, a stoichiometrically normalised concentration change per unit time for the written reaction.

Observed process	Typical observation	Qualification
AgCl precipitation	Rapid on mixing	Depends on mixing and conditions
Acid hydrolysis	Measurable time	Depends on acid, substrate and temperature
Iron corrosion	Gradual	Depends on wetness, oxygen and surface

For R → P with coefficients one, average rate over an interval is −Δ[R]/Δt = Δ[P]/Δt. Reactant concentration falls, so its change is negative; the minus sign makes the forward disappearance rate positive. Common concentration-rate units are mol L⁻¹ s⁻¹.

For aA+bB→cC+dD, the common instantaneous reaction rate is −(1/a)d[A]/dt = −(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt, at constant volume. Thus in 2HI→H₂+I₂, HI disappears twice as fast as either product appears, although the normalised reaction rate is the same.

Go deeper: Choose what the measurement actually reports

State whether a reported number is the disappearance rate of one species or the normalised reaction rate. Otherwise a coefficient of two can quietly create a factor-of-two error. Pressure changes may track gas concentration at fixed temperature and volume, but each gas’s partial pressure and stoichiometry still matter.

A positive forward rate convention describes net progress in the chosen direction. In a reversible system, reverse reactions can also occur. At equilibrium, opposing rates balance rather than all molecular reactions stopping.

PAUSE & TRY · think first, then check with a free account

Question

In 2A→B, [A] falls by 0.12 mol/L in 30 s at constant volume. Calculate A’s average disappearance rate and the common reaction rate.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 2, 3.

EXPLORE

Average vs Instantaneous Rate

A secant compares two times; a tangent describes the rate at one time.

A secant averages; a tangent measures now

Time: 40 sSampling interval: 40 s

Original first-order model [R]=0.80e^(−0.015t) mol/L. Blue is concentration. Orange secant between 20.0 and 60.0 s gives average disappearance rate 6.6850e-3 mol L⁻¹ s⁻¹. Violet tangent at 40 s gives instantaneous disappearance rate 6.5857e-3 mol L⁻¹ s⁻¹. A concentration slope is negative; disappearance rate includes the minus sign. Shrinking the interval approaches the tangent limit, not an exactly equal finite interval.

Instantaneous rate: The limiting concentration-change rate at a specified time as the sampling interval approaches zero.

The average disappearance rate −Δ[R]/Δt corresponds to minus the slope of a secant joining two points on a reactant concentration curve. It hides changes within the interval. The instantaneous disappearance rate −d[R]/dt is minus the tangent slope at the chosen time.

For the visual’s original first-order model [R]=0.80e^(−0.015t) mol/L, the instantaneous disappearance rate at 40 s is 0.015×0.80e^(−0.60) = 0.006586 mol L⁻¹ s⁻¹. Shrink the sampling interval to see the secant approach this derivative. The curve is concentration; its slope has concentration-per-time units.

Go deeper: Read a slope without confusing it with a height

The concentration value is the vertical coordinate; it is not the rate. Two curves can have the same concentration at an instant yet have different slopes. A tangent’s slope uses two points on the tangent line, which need not both lie on the curved concentration trace.

Rate need not always decrease: a zero-order regime can have a constant rate, and autocatalytic systems can behave differently. The falling rate in the visual follows the explicitly chosen first-order model. For other coefficients, divide the species derivative by its coefficient to obtain the common reaction rate.

PAUSE & TRY · think first, then check with a free account

Question

A reactant tangent falls from 0.50 to 0.38 mol/L over a horizontal span of 20 s. For R→P, find the instantaneous disappearance rate represented by that tangent.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 4, 5.

EXPLORE

Rate Law and Order of Reaction

Change one concentration at a time to discover the powers.

Let concentration experiments reveal the powers

A concentration multiplier: ×1B concentration multiplier: ×1

Original illustrative initial-rate law r=k[A]²[B], with k=1.00 L² mol⁻² s⁻¹. [A]=0.10 and [B]=0.20 mol/L gives r=0.0020 mol L⁻¹ s⁻¹, 1 times the baseline. Doubling A quadruples rate; doubling B doubles it. This empirical model is not inferred from a balanced equation.

Reaction order: An exponent in an experimentally established rate law; overall order is the sum of its concentration exponents.

For r=k[A]^x[B]^y, x and y are partial orders and x+y is overall order. Keep temperature and other relevant conditions fixed when comparing measurements. Orders can be zero, whole numbers or fractions; they cannot generally be read from coefficients in the balanced overall equation.

In the original model r=k[A]²[B], doubling A at fixed B multiplies rate by four, while doubling B at fixed A doubles it. The overall order is three. If rate has units mol L⁻¹ s⁻¹, k has units (mol L⁻¹)^(1−n)s⁻¹ for overall order n. First-order k has s⁻¹; second-order k has L mol⁻¹ s⁻¹.

Go deeper: Keep k separate from the changing rate

At a fixed temperature and within an applicable model, k is independent of the varied reactant concentrations, while rate usually changes with them. Temperature or a catalyst can change the measured k. Its numerical value also depends on the units used in the law.

For an elementary mass-action step under the model’s assumptions, the reacting-species powers follow that step’s stoichiometry. This does not licence assigning those powers to a complex overall equation. A zero-order law describes an applicable concentration range, not an assertion that a depleted reactant continues reacting.

PAUSE & TRY · think first, then check with a free account

Question

At fixed temperature, doubling A alone triples rate and doubling B alone doubles rate. Find the partial orders and state whether they must match the overall equation coefficients.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 6, 7, 8.

EXPLORE

Molecularity of a Reaction

Count one step, then distinguish that count from measured order.

Count the species in one elementary event

One elementary bimolecular event involves two reacting species. Molecularity is a positive whole-number count for an elementary step; it is not assigned to the complete multi-step equation. A curved motion arrow shows approach; displayed spheres and timing are schematic, not molecular dynamics.

Molecularity: The number of reacting species involved in one elementary reaction event.

Property	Order	Molecularity
Determination	Measured rate-law exponent	Count in a specified elementary event
Values	Zero, integer or fractional	Positive integer; two-species encounters common
Applicability	Measured elementary or overall laws	Elementary steps only

An elementary unimolecular step involves one reacting species; a bimolecular step involves two. Molecularity is a positive whole-number count assigned to a specified elementary step. Three-species encounters are uncommon; an overall multi-step reaction has no single molecularity.

The chapter’s simplified peroxide mechanism uses H₂O₂+I⁻→H₂O+IO⁻ followed by H₂O₂+IO⁻→H₂O+I⁻+O₂. Each shown step is bimolecular. I⁻ is consumed then regenerated as catalyst; IO⁻ is formed then consumed as intermediate. Adding the steps gives 2H₂O₂→2H₂O+O₂.

Go deeper: A mechanism is more than a count

A slow controlling step can help explain a rate law, but pre-equilibria and intermediate concentrations can matter. Do not automatically equate its molecularity with the measured overall order. For the chapter’s simplified controlling first step, a mass-action description can involve [H₂O₂][I⁻].

An intermediate occupies a chemical state between steps; a transition state lies at a barrier along a step and is not the same as a stable intermediate. The regenerated catalyst participates chemically even though it cancels from the net equation.

PAUSE & TRY · think first, then check with a free account

Question

In A+C→I followed by I+B→P+C, identify the catalyst, intermediate and molecularity of each elementary step.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 9, 10, 11.

EXPLORE

Integrated Rate Equations: Zero Order

Equal time intervals remove equal amounts, until the model stops applying.

Equal time removes equal amounts

Time: 40 sInitial concentration: 0.80 mol/L

For original ideal R → P with k=0.005 mol L⁻¹ s⁻¹, [R]=[R]₀−kt while this law applies and reactant remains. At 40 s, [R]=0.600 mol/L. Straight-line slope is −k; equal intervals consume equal amounts. Extending the line below zero is physically invalid. Saturation can fail near exhaustion, ending the zero-order regime.

Integrated zero-order equation: [R]=[R]₀−kt when −d[R]/dt=k and the zero-order regime remains applicable.

Integration of the constant disappearance rate gives [R]=[R]₀−kt. A concentration-versus-time plot is a straight line with intercept [R]₀ and slope −k. For the visual’s original k=0.005 mol L⁻¹ s⁻¹, every 20 s consumes 0.10 mol/L while reactant remains and the law holds.

Surface-catalysed ammonia decomposition and saturated enzyme systems can show a zero-order regime when occupied active sites limit throughput. Adding more reactant then need not create more available reacting sites. Saturation is a condition, not a universal property at every concentration.

Go deeper: Check the end of the straight line

The formal depletion time is [R]₀/k. A negative concentration from extrapolating beyond it is physically meaningless. A real surface can lose saturation before formal depletion and enter a different kinetic regime.

If [R]₀ is doubled while k is fixed, twice as much material takes twice as long to remove in the same zero-order regime. The initial half-life [R]₀/(2k) depends on the starting amount; subsequent halving intervals shorten as the remaining amount becomes smaller.

PAUSE & TRY · think first, then check with a free account

Question

A zero-order model begins at 0.60 mol/L with k=0.0040 mol L⁻¹ s⁻¹. Find [R] at 50 s and explain why extending the same formula to 200 s is invalid.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 11, 12.

EXPLORE

Integrated Rate Equations: First Order

Equal time intervals remove equal fractions of what remains.

Equal time removes equal fractions

Time: 40 s

Original R → P uses [R]=[R]₀e^(−kt), k=0.015 s⁻¹. At 40 s, [R]=0.4390 mol/L. ln([R]/[R]₀)=−kt gives a straight line with slope −0.015 s⁻¹; log₁₀([R]₀/[R]) has slope k/ln10. Equal times remove equal fractions, not equal absolute amounts. Positive concentration approaches zero asymptotically in this ideal continuous law.

Integrated first-order equation: [R]=[R]₀e^(−kt), equivalently ln([R]₀/[R])=kt, for −d[R]/dt=k[R].

The first-order concentration trace curves because its disappearance rate falls with [R]. Taking ln([R]/[R]₀) gives a straight line of slope −k; plotting log₁₀([R]₀/[R]) gives slope k/ln10. Use consistent time units so k has their inverse.

For A(g)→B(g)+C(g) initially containing only A at pressure pᵢ, fixed T and V give pA=2pᵢ−ptotal. Thus k=(1/t)ln[pᵢ/(2pᵢ−ptotal)]. The measured total pressure includes products and cannot simply replace remaining reactant pressure. Radioactive decay is another first-order loss model; growing populations are a different, positive-growth model.

Go deeper: Logarithms expose the model

A dimensionless concentration ratio avoids taking the logarithm of a unit-bearing quantity. If experimental logarithmic data do not lie close to a line within measurement uncertainty, reconsider the assumed first-order model or changing conditions.

An ideal continuous exponential approaches zero asymptotically. This mathematical curve is not a guarantee of detectable molecules at every finite time in a finite sample. For gas calculations, the pressure relation also depends on the written stoichiometry and initial composition.

PAUSE & TRY · think first, then check with a free account

Question

An original first-order sample falls from 0.90 to 0.30 mol/L in 60 s. Calculate k and the concentration after a further 60 s under the same conditions.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 12, 13, 14, 15.

EXPLORE

Half-Life of a Reaction

First-order half-lives repeat; zero-order half-lives depend on the starting amount.

The next half can take the same time

Initial concentration: 0.80 mol/L

First-order t½=ln2/0.015=46.21 s, independent of [R]₀. After n half-lives the fraction is (½)^n. For zero order with k₀=0.005 mol L⁻¹ s⁻¹ and [R]₀=0.80 mol/L, the initial half-life is 80.0 s; later halving intervals shorten. 99.9% conversion takes ln1000/ln2≈9.966 first-order half-lives, not exactly ten.

Half-life: The time for a specified starting reactant concentration to fall to half that value.

Order	Rate Constant Unit	Half-Life Formula
Zero	mol L⁻¹ s⁻¹	t₁/₂ = [R]₀ / 2k
First	s⁻¹	t₁/₂ = 0.693 / k

For first order, t½=ln2/k, independent of [R]₀. After n half-lives, the remaining fraction is 2^(−n). For zero order, t½=[R]₀/(2k), so a higher starting concentration gives a longer initial half-life at the same k.

A pseudo-first-order law can arise when another reactant is in large excess and changes very little. In r=k[A][B], approximately constant [B] gives r≈k′[A] with k′=k[B]. Acid-catalysed ester hydrolysis with water in large excess and sucrose inversion under fixed conditions illustrate apparent first-order descriptions.

Go deeper: Quantify both the fraction and the approximation

99.9% conversion leaves 0.001 of the original material. For first order, the required number of half-lives is ln1000/ln2≈9.966, rather than exactly ten. Ten halves leave 1/1024≈0.0009766.

Large excess is a testable approximation. For an A+B consumption model starting with 1 unit A and 100 units B, consuming all A changes B by only 1%; with 10 units B it changes by 10%. Rate-constant, catalyst, solvent and temperature conditions must also remain sufficiently steady.

PAUSE & TRY · think first, then check with a free account

Question

A first-order reaction has k=0.020 s⁻¹. Find t½ and the time to 87.5% conversion. Would doubling initial concentration change either time?

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 16, 17, 18.

EXPLORE

Temperature Dependence and Arrhenius Equation

Heat redistributes molecular energies; read the slope to estimate activation energy.

Temperature changes the distribution and the rate constant

Temperature: 300 KActivation model threshold: 12 kJ/mol

The plotted ideal 3D translational-energy density compares 300 K and 300 K, with energy u=E/(R×300 K). Each complete distribution integrates to one; its tail continues beyond the plot. The shaded tail beyond the selected 12 kJ/mol threshold illustrates a larger energetic population on heating. It is not identified numerically with the Arrhenius factor e^(−Ea/RT), which is a simplified activation model.

Arrhenius equation: k=Ae^(−Ea/RT), an activation model relating a rate constant to absolute temperature over an applicable range.

A is the pre-exponential factor, Ea the activation energy, R the gas constant and T the kelvin temperature. Use Ea in J/mol when R=8.314 J mol⁻¹ K⁻¹. At fixed A and positive Ea, increasing T increases k. A 10 K rise is not universally a doubling.

The logarithmic form ln(k/kref)=ln(A/kref)−Ea/(RT) has slope −Ea/R against 1/T. For two temperatures with approximately constant A and Ea, ln(k₂/k₁)=(Ea/R)(1/T₁−1/T₂). The displayed original model uses A=1000 s⁻¹ and a selectable Ea, rather than measured data for a named reaction.

Go deeper: Read energy diagrams and distributions carefully

A schematic reaction-coordinate profile shows a transition-state barrier; its horizontal coordinate is not elapsed time. An activated complex or transition state is not interchangeable with a chemically distinct intermediate between steps.

The heated Maxwell–Boltzmann energy distribution is broader and its high-energy tail grows beyond a fixed threshold. The plotted 3D translational-energy tail is not exactly the single exponential Arrhenius factor; the latter is a simplified activation description. The graph’s full distribution extends beyond the displayed range.

PAUSE & TRY · think first, then check with a free account

Question

A first-order k rises from 0.010 to 0.030 s⁻¹ between 300 and 330 K. Assuming the two-temperature Arrhenius model, estimate Ea.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 18, 19, 20, 21.

EXPLORE

Catalysis and Collision Theory

A useful encounter needs more than contact, and a catalyst changes the pathway.

Lower the barrier and meet in the right way

The catalysed pathway has a lower barrier but the same reactant and product endpoints. A catalyst changes the route and can take part in intermediate steps; it is regenerated overall. It does not change the reaction’s ΔG° or equilibrium constant. Both forward and reverse approaches to the same equilibrium can become faster.

Effective collision: An encounter that satisfies the energy and orientation requirements of a specified reaction model.

In a simplified collision description, only some encounters have both sufficient energy and a suitable orientation. A steric factor accounts for orientation in a model such as r≈PZABe^(−Ea/RT), with ZAB defined consistently as collision frequency. This simplified picture does not capture every solution or complex-reaction mechanism.

For OH⁻+CH₃Br→CH₃OH+Br⁻, the illustrated backside approach aligns the incoming oxygen with the C–Br axis as the old bond breaks. This is bromomethane, not bromoethane: the carbon skeleton must remain consistent. The slowed spheres illustrate geometry, not an actual molecular trajectory measurement.

Go deeper: A faster route has the same destination

A catalyst offers an alternative pathway, often with a lower activation barrier, and is regenerated overall. It may be consumed in one step and returned later. It changes kinetics, not the reaction’s standard Gibbs energy or equilibrium constant at fixed temperature.

Forward and reverse relaxation towards the same equilibrium can both become faster. A catalyst neither shifts equilibrium nor makes an uphill net conversion favourable. The visual’s two relaxation curves share an illustrative equilibrium product fraction of 0.70; their chosen time constants are teaching parameters.

PAUSE & TRY · think first, then check with a free account

Question

A catalyst makes a reversible reaction reach equilibrium in 2 minutes rather than 20. Explain what can change and what stays fixed at the same temperature.

Write your own answer and compare it with ours. It’s free.

NCERT reference: chapter PDF pages 22, 23, 24.

Chemical Kinetics Class 12 Notes

Chapter notes

Chemical Kinetics

Introduction to Chemical Kinetics

A downhill reaction can still have a barrier

Rate of a Chemical Reaction

Measure what disappears and what appears

Average vs Instantaneous Rate

A secant averages; a tangent measures now

Rate Law and Order of Reaction

Let concentration experiments reveal the powers

Molecularity of a Reaction

Count the species in one elementary event

Integrated Rate Equations: Zero Order

Equal time removes equal amounts

Integrated Rate Equations: First Order

Equal time removes equal fractions

Half-Life of a Reaction

The next half can take the same time

Temperature Dependence and Arrhenius Equation

Temperature changes the distribution and the rate constant

Catalysis and Collision Theory

Lower the barrier and meet in the right way

Now practise Chemical Kinetics, free