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Class 12 Chemistry Chapter 3 Chemical Kinetics — Formulas & Key Points

Chemical Kinetics—Chapter 3 of NCERT Class 12 Chemistry—explores how fast reactions proceed and what factors control those speeds. From rate laws and reaction orders to the Arrhenius equation and collision theory, this chapter underpins industrial catalysis, drug design, and environmental chemistry. This formula sheet organizes every key equation, definition, and mnemonic into tables and worked examples so you can revise efficiently before board exams and competitive tests.

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Key takeaways

✓Rate of reaction = −(1/a)(d[A]/dt) = −(1/b)(d[B]/dt) = (1/c)(d[C]/dt) expresses stoichiometric coefficients in rate expressions.
✓Order of reaction is the sum of powers in the rate law (determined experimentally); molecularity is the number of reacting species in an elementary step (always a whole number).
✓Arrhenius equation k = A·e^(−Ea/RT) links rate constant to activation energy and temperature; log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)] is the two-temperature form.
✓Integrated rate laws differ by order: zero-order [A] = [A]₀ − kt (straight line [A] vs t); first-order ln[A] = ln[A]₀ − kt (straight line ln[A] vs t); second-order 1/[A] = 1/[A]₀ + kt (straight line 1/[A] vs t).
✓Half-life formulas: t₁/₂ = [A]₀/(2k) for zero-order (depends on [A]₀), t₁/₂ = 0.693/k for first-order (independent of [A]₀), t₁/₂ = 1/(k[A]₀) for second-order.
✓Collision theory states only collisions with energy ≥ Ea and proper orientation yield products; fraction of molecules with energy ≥ Ea is e^(−Ea/RT).
✓Common mistakes: mixing up order (experimental) with molecularity (theoretical), incorrect units for k (zero-order: mol L⁻¹ s⁻¹; first-order: s⁻¹; second-order: L mol⁻¹ s⁻¹), and sign errors in Δ[A]/Δt for reactants.

Core Formulas: Rate of Reaction and Rate Law

The instantaneous rate of reaction measures how quickly concentrations change. For a general reaction aA + bB → cC + dD, the rate is expressed using stoichiometric coefficients to ensure a single value. The rate law, Rate = k[A]^m[B]^n, captures the experimental dependence on concentrations. Here k is the rate constant (temperature-dependent), and the exponents m and n define the order with respect to each reactant. The overall order is m + n. Remember that rate laws must be determined from experiment—they cannot be deduced from the balanced equation unless the reaction is elementary. Units of k vary with order: for zero-order mol L⁻¹ s⁻¹, first-order s⁻¹, second-order L mol⁻¹ s⁻¹, and so on.

Rate = −(1/a)(d[A]/dt) = −(1/b)(d[B]/dt) = (1/c)(d[C]/dt) = (1/d)(d[D]/dt) — negative sign for reactants (decreasing).
Rate law: Rate = k[A]^m[B]^n, where m and n are orders determined experimentally.
Overall order = m + n + … (sum of all exponents).
Units of k depend on overall order n: (mol L⁻¹)^(1−n) s⁻¹.

Integrated Rate Laws for Zero, First, and Second Order

Integrating the differential rate law gives concentration as a function of time. These equations are essential for calculating how much reactant remains after a given interval and for determining order from experimental data. Zero-order reactions have Rate = k, so concentration decreases linearly: [A] = [A]₀ − kt. A plot of [A] versus t is a straight line with slope −k. First-order reactions follow Rate = k[A], yielding ln[A] = ln[A]₀ − kt; a plot of ln[A] versus t is linear with slope −k. Second-order reactions (Rate = k[A]²) integrate to 1/[A] = 1/[A]₀ + kt, so 1/[A] versus t is linear with slope k. Knowing which plot gives a straight line immediately reveals the order. These integrated forms are the workhorses of kinetics problems on CBSE board exams and JEE.

Zero-order: [A] = [A]₀ − kt; straight line [A] vs t, slope = −k, intercept = [A]₀.
First-order: ln[A] = ln[A]₀ − kt or [A] = [A]₀·e^(−kt); straight line ln[A] vs t, slope = −k.
Second-order: 1/[A] = 1/[A]₀ + kt; straight line 1/[A] vs t, slope = k.
To find order from data, test which plot (concentration, ln, or 1/concentration vs time) is linear.

Half-Life Formulas and Order Dependence

Half-life t₁/₂ is the time required for the concentration to drop to half its initial value. For zero-order reactions, t₁/₂ = [A]₀/(2k), meaning half-life depends on the starting concentration and decreases as the reaction proceeds—successive half-lives get shorter. First-order reactions have t₁/₂ = 0.693/k (or ln2/k), independent of [A]₀; each half-life is constant, a hallmark of radioactive decay and many drug eliminations. For second-order, t₁/₂ = 1/(k[A]₀), so half-life increases as concentration falls. Recognizing how t₁/₂ varies with [A]₀ is a quick diagnostic for order. CBSE board questions often ask you to derive or apply these expressions, so memorize the three forms and their dependencies clearly.

Zero-order: t₁/₂ = [A]₀/(2k) — proportional to [A]₀.
First-order: t₁/₂ = 0.693/k = ln2/k — independent of [A]₀.
Second-order: t₁/₂ = 1/(k[A]₀) — inversely proportional to [A]₀.
Mnemonic: 'Zero grows, First constant, Second shrinks' (half-life behavior as [A]₀ changes).

Arrhenius Equation and Temperature Dependence

The Arrhenius equation, k = A·e^(−Ea/RT), quantifies how the rate constant k increases with temperature T. Here A is the pre-exponential or frequency factor (units same as k), Ea is the activation energy (J mol⁻¹ or kJ mol⁻¹), R = 8.314 J mol⁻¹ K⁻¹ is the gas constant, and T is absolute temperature in Kelvin. Taking natural logarithms gives ln k = ln A − Ea/(RT), a linear form: plot ln k versus 1/T to get slope −Ea/R and intercept ln A. For two temperatures, the equation becomes log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)], invaluable for calculating how much faster a reaction runs at a higher temperature. Remember to convert Ea to joules if R is in J mol⁻¹ K⁻¹, and always use Kelvin for temperature. CBSE problems frequently test unit consistency and the two-temperature form.

k = A·e^(−Ea/RT) — exponential dependence on activation energy and temperature.
ln k = ln A − Ea/(RT) — linear plot ln k vs 1/T, slope = −Ea/R.
Two-temperature form: log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)].
Use R = 8.314 J mol⁻¹ K⁻¹ and Ea in J mol⁻¹ (or both in kJ); T in Kelvin.

Order vs Molecularity and Elementary Reactions

Order and molecularity are often confused. Order is the sum of exponents in the experimentally determined rate law; it can be zero, fractional, or an integer and applies to the overall reaction. Molecularity is the number of molecules (or atoms/ions) that collide simultaneously in a single elementary step; it is always a positive integer (1 = unimolecular, 2 = bimolecular, 3 = termolecular). For an elementary reaction, the stoichiometric coefficients do equal the orders, so order equals molecularity. For complex (multi-step) reactions, the observed order is governed by the rate-determining step and need not match the overall stoichiometry. CBSE exam questions test this distinction sharply—memorize that molecularity is a theoretical concept applicable only to elementary steps, while order is an experimental quantity for any reaction.

Order: sum of exponents in rate law; experimental; can be 0, fraction, or integer; applies to overall reaction.
Molecularity: number of reacting species in an elementary step; theoretical; always a whole number (1, 2, or 3).
For elementary reactions: order = molecularity.
For complex reactions: order ≠ molecularity; order determined by rate-determining step.

Collision Theory and Activation Energy

Collision theory explains reaction rates at the molecular level. For a reaction to occur, molecules must collide with kinetic energy equal to or greater than the activation energy Ea and with proper orientation. The fraction of molecules possessing energy ≥ Ea at temperature T is given by the Boltzmann factor e^(−Ea/RT). Not all collisions are effective; the steric or probability factor P (or A in the Arrhenius equation) accounts for orientation requirements. The rate is thus proportional to collision frequency Z, the energy factor e^(−Ea/RT), and the steric factor P: Rate ∝ Z·P·e^(−Ea/RT). Catalysts work by lowering Ea, increasing the fraction of successful collisions without being consumed. CBSE questions ask you to sketch energy profiles showing Ea, ΔH, and the effect of a catalyst, so practice drawing and labeling these diagrams clearly.

Effective collisions require (i) energy ≥ Ea and (ii) proper orientation.
Fraction of molecules with E ≥ Ea = e^(−Ea/RT).
Rate ∝ collision frequency × orientation factor × energy factor.
Catalyst lowers Ea → increases e^(−Ea/RT) → faster reaction without changing ΔH.

Pseudo First-Order Reactions and Practical Kinetics

In a pseudo first-order reaction, one reactant is present in such large excess that its concentration remains essentially constant during the reaction. For example, hydrolysis of an ester in dilute aqueous solution: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH. Water is in huge excess, so [H₂O] ≈ constant. The true rate law Rate = k[ester][H₂O] simplifies to Rate = k'[ester], where k' = k[H₂O] is the pseudo first-order rate constant. The reaction then follows first-order kinetics, making analysis straightforward. Pseudo-order conditions are common in biochemistry (enzyme-substrate reactions) and industry, allowing simpler data treatment. CBSE numerical problems on ester hydrolysis or inversion of sucrose routinely invoke pseudo first-order kinetics. Always identify when a reactant is in excess and define the effective rate constant clearly in your working.

Pseudo first-order: true rate law is second-order, but one reactant in huge excess → appears first-order.
Example: Rate = k[A][B] with [B] >> [A] → Rate = k'[A], where k' = k[B].
Common cases: acid/base-catalyzed hydrolyses, inversion of sucrose in water.
Treat as first-order: ln[A] = ln[A]₀ − k't, t₁/₂ = 0.693/k'.

Common Mistakes, Units, and Sign Conventions

Students often lose marks on Chemical Kinetics for unit errors and sign slips. Remember that for reactants d[A]/dt is negative (concentration decreases), so the rate of reaction is defined as −d[A]/dt to keep Rate positive. Always divide by the stoichiometric coefficient to ensure a unique rate value. When using Arrhenius equation, match units: if R = 8.314 J mol⁻¹ K⁻¹, Ea must be in J mol⁻¹ (multiply kJ by 1000). Temperature must be in Kelvin—never Celsius. The units of k change with order: zero-order mol L⁻¹ time⁻¹, first-order time⁻¹, second-order L mol⁻¹ time⁻¹. In logarithmic forms, use ln (natural log, base e) unless the question specifies log₁₀ (common log, base 10); the conversion factor is 2.303. Double-check which half-life formula applies to which order. These details matter enormously in CBSE board marking schemes, where method marks depend on correct units and clear working.

Rate is always positive: use −(1/a)(d[A]/dt) for reactants, +(1/c)(d[C]/dt) for products.
Temperature in Kelvin: T(K) = T(°C) + 273.15.
Match R and Ea units: R = 8.314 J mol⁻¹ K⁻¹ ⇒ Ea in J mol⁻¹.
Units of k: zero-order mol L⁻¹ s⁻¹, first s⁻¹, second L mol⁻¹ s⁻¹.
ln vs log₁₀: ln = 2.303 log₁₀; Arrhenius often uses ln, but CBSE sometimes asks for log₁₀ form.
Do not confuse order (experimental) with stoichiometric coefficients (except for elementary steps).

Memory Tricks and Mnemonics for Quick Revision

Mnemonics help lock formulas and concepts into long-term memory. For order dependence of half-life, remember 'Zero Depends, First Fixed, Second Shrinks'—zero-order t₁/₂ depends on [A]₀, first-order is fixed (constant), second-order shrinks (inversely proportional to [A]₀). For the shape of concentration-time plots, think 'Zero Linear, First Exponential, Second Hyperbolic.' To recall which integrated law is which, note the plot type: zero-order plots [A] vs t (straight), first ln[A] vs t (straight), second 1/[A] vs t (straight). The acronym 'EAR' for Arrhenius—Energy, Activation, Rate constant—reminds you k depends on Ea and T. For collision theory, 'OEET'—Orientation, Energy, Effective collisions, Temperature—covers the essentials. CBSETUTOR.ai's AI tutor reinforces these tricks with personalized quizzes; students upload their notebook pages and get instant feedback on which formulas they mix up, all for ₹999/month with a 3-day free trial covering Classes 6 to 12.

'Zero Depends, First Fixed, Second Shrinks' — half-life behavior with [A]₀.
'Zero Linear, First Exponential, Second Hyperbolic' — concentration vs time graph shapes.
'Plot Zero [A], First ln[A], Second 1/[A]' — which y-axis for straight-line test.
'EAR' — Energy-Activation-Rate for Arrhenius equation.
'OEET' — Orientation, Energy, Effective collisions, Temperature for collision theory.

Worked Example 1: Determining Order from Concentration–Time Data

A classic CBSE board question provides concentration measurements at different times and asks you to deduce the order. The strategy is to test each integrated rate law. Suppose [A] at t = 0, 10, 20, 30 min is 1.00, 0.50, 0.25, 0.125 M. Check zero-order: [A] = [A]₀ − kt → 0.50 = 1.00 − k×10 gives k = 0.05 M min⁻¹; at t = 20, [A] should be 1.00 − 0.05×20 = 0, not 0.25—ruled out. Check first-order: ln[A] at t = 0 is ln(1) = 0, at t = 10 is ln(0.5) = −0.693, at t = 20 is ln(0.25) = −1.386, at t = 30 is ln(0.125) = −2.079. These are evenly spaced by −0.693, confirming linearity in ln[A] vs t with slope −k = −0.693/10 = −0.0693 min⁻¹. Hence the reaction is first-order, k = 0.0693 min⁻¹, and t₁/₂ = 0.693/k = 10 min (consistent with [A] halving every 10 min). Always show the test for at least two orders to earn full method marks.

Step 1: List data and calculate candidate plots ([A], ln[A], 1/[A]).
Step 2: Check if differences or ratios are constant.
Step 3: Identify which plot is linear → determines order.
Step 4: Slope of that plot gives k (sign and units).

Worked Example 2: Arrhenius Equation Two-Temperature Problem

A reaction has k₁ = 1.0×10⁻³ s⁻¹ at T₁ = 300 K and k₂ = 4.0×10⁻³ s⁻¹ at T₂ = 320 K. Find Ea. Use log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)]. Compute k₂/k₁ = 4.0/1.0 = 4, so log₁₀(4) ≈ 0.602. Then 0.602 = [Ea/(2.303×8.314)]×[(320−300)/(300×320)] = [Ea/19.15]×[20/96000] = [Ea/19.15]×(2.083×10⁻⁴). Solving: Ea = 0.602×19.15/(2.083×10⁻⁴) ≈ 11.53/0.0002083 ≈ 55360 J mol⁻¹ = 55.4 kJ mol⁻¹. Always write units at each step and use consistent R. This type of problem appears frequently in CBSE board papers, often worth 3–5 marks, so practice unit conversions and calculator work to avoid rounding errors.

Given k₁, k₂, T₁, T₂ → find Ea using log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)].
Step 1: Calculate k₂/k₁ and take log₁₀.
Step 2: Compute (T₂−T₁)/(T₁T₂).
Step 3: Rearrange for Ea = log₁₀(k₂/k₁) × 2.303R × (T₁T₂)/(T₂−T₁).
Step 4: Convert J mol⁻¹ to kJ mol⁻¹ if required.

Last-Minute Revision Checklist: One-Glance Summary

Use this quick checklist 24 hours before your CBSE board exam. Ensure you can write down the integrated rate laws for zero, first, and second order without peeking. Recite the half-life formulas and note which depends on [A]₀. Write the Arrhenius equation in both exponential and logarithmic forms, and the two-temperature version. Define order (experimental, sum of exponents) versus molecularity (theoretical, whole number for elementary steps). Sketch an energy profile showing reactants, products, activation energy, and the effect of a catalyst (lower peak, same start and end). List units of k for each order. Remember that pseudo first-order means one reactant in huge excess, making a second-order reaction behave as first-order. Know the collision theory factors: energy ≥ Ea, proper orientation, and Boltzmann fraction e^(−Ea/RT). Finally, review sign conventions: rate is positive, so −d[A]/dt for reactants. Print or screenshot this checklist and tick off each item as you verify your command of it.

Zero-order: [A] = [A]₀ − kt, t₁/₂ = [A]₀/(2k), units of k: mol L⁻¹ s⁻¹.
First-order: ln[A] = ln[A]₀ − kt, t₁/₂ = 0.693/k, units of k: s⁻¹.
Second-order: 1/[A] = 1/[A]₀ + kt, t₁/₂ = 1/(k[A]₀), units of k: L mol⁻¹ s⁻¹.
Arrhenius: k = A·e^(−Ea/RT); ln k = ln A − Ea/(RT); log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)].
Order vs molecularity: order = experimental sum of exponents; molecularity = whole number, elementary step only.
Collision theory: fraction e^(−Ea/RT), orientation factor, catalyst lowers Ea.
Pseudo first-order: one reactant in excess, observed k' = k[excess reactant].
Energy profile: label Ea (activation energy), ΔH (enthalpy change), catalyst lowers Ea but not ΔH.
Sign: Rate = −(1/a)(d[A]/dt) ensures positive value.

Frequently asked questions

What is the difference between rate of reaction and rate constant in Chemical Kinetics?+

Rate of reaction is the change in concentration per unit time (mol L⁻¹ s⁻¹) and varies with reactant concentrations. Rate constant k is the proportionality factor in the rate law; it is independent of concentration but depends on temperature and the presence of a catalyst. Units of k vary with reaction order.

How do I determine the order of a reaction from experimental data?+

Plot [A] vs t (zero-order), ln[A] vs t (first-order), and 1/[A] vs t (second-order). Whichever plot gives a straight line reveals the order. Alternatively, use the half-life method: if t₁/₂ is constant, it is first-order; if t₁/₂ ∝ [A]₀, zero-order; if t₁/₂ ∝ 1/[A]₀, second-order.

Why is the half-life of a first-order reaction independent of initial concentration?+

For first-order kinetics, ln([A]₀/2) = ln[A]₀ − kt₁/₂ simplifies to ln2 = kt₁/₂, so t₁/₂ = 0.693/k. The [A]₀ cancels out, leaving t₁/₂ dependent only on k. This constancy is characteristic of radioactive decay and many biological processes.

What is the Arrhenius equation and how does temperature affect the rate constant?+

The Arrhenius equation k = A·e^(−Ea/RT) shows that k increases exponentially with temperature because the fraction of molecules with energy ≥ Ea grows. A 10 K rise typically doubles or triples k for reactions with moderate Ea, explaining why reactions speed up when heated.

How do I use the two-temperature form of the Arrhenius equation to find activation energy?+

Given k₁ at T₁ and k₂ at T₂, use log₁₀(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/(T₁T₂)]. Rearrange to Ea = log₁₀(k₂/k₁) × 2.303R × (T₁T₂)/(T₂−T₁). Ensure T is in Kelvin and R = 8.314 J mol⁻¹ K⁻¹, then convert Ea to kJ mol⁻¹ by dividing by 1000.

What is the difference between order and molecularity in Chemical Kinetics?+

Order is the sum of exponents in the experimentally determined rate law and can be zero, fractional, or an integer; it applies to the overall reaction. Molecularity is the number of reacting molecules in a single elementary step, always a whole number (1, 2, or 3), and is a theoretical concept applicable only to elementary reactions.

What is a pseudo first-order reaction and where is it used?+

A pseudo first-order reaction is actually second-order, but one reactant is in such large excess that its concentration stays nearly constant, making the observed rate proportional to the other reactant only. Common in ester hydrolysis and enzyme kinetics. Example: Rate = k[ester][H₂O] ≈ k'[ester] where k' = k[H₂O].

How does a catalyst affect the rate constant and activation energy?+

A catalyst lowers the activation energy Ea, which increases the fraction of molecules with sufficient energy (e^(−Ea/RT) becomes larger). This raises the rate constant k, speeding up the reaction. Importantly, a catalyst does not change the enthalpy change ΔH or the equilibrium position, only the rate.

Why do we use natural logarithm (ln) in first-order integrated rate law?+

Integrating d[A]/[A] = −k dt directly yields ln[A] − ln[A]₀ = −kt because the integral of 1/x dx is ln x. Natural logarithm (base e) is the mathematical result of this integration. To use common logarithm (log₁₀), multiply by 2.303: log₁₀[A] = log₁₀[A]₀ − (k/2.303)t.

What are the units of the rate constant for zero, first, and second-order reactions?+

For zero-order: mol L⁻¹ s⁻¹ (or mol L⁻¹ time⁻¹). For first-order: s⁻¹ (or time⁻¹). For second-order: L mol⁻¹ s⁻¹ (or L mol⁻¹ time⁻¹). The general formula is (concentration)^(1−n) × time⁻¹, where n is the overall order. Always check units match when solving numericals.

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