NCERT Plus Two (Class 12) Chemistry • Chapter 3

Chemical Kinetics

Reaction Rates, Rate Law, Order vs Molecularity, Zero & First Order Integrated Equations, Half-Life, Arrhenius Equation & Collision Theory

📄 Official NCERT Textbook PDF

Topic 3.1: Rate of a Chemical Reaction: Average vs Instantaneous Rate & Stoichiometry

Did you know that thermodynamics says a sparkling diamond should spontaneously turn into pencil-lead graphite (because ΔG < 0)? Yet nobody worries about their diamond ring turning into pencil dust! Why? Because thermodynamics only predicts whether a reaction is feasible, whereas Chemical Kinetics (from Greek kinesis = movement) reveals how fast it happens—just like knowing a car can reach Kochi tells you nothing without checking its speedometer!

Chemical Kinetics studies reaction rates, the factors controlling them, and reaction mechanisms. Reactions span wildly different speeds: ionic precipitation (AgNO3 + NaCl → AgCl↓) is instantaneous, rusting of iron is very slow, while inversion of cane sugar and hydrolysis of starch proceed at moderate speeds.

1. Average Rate (rav) vs Instantaneous Rate (rinst)

For a reaction R → P at constant volume, rate equals the decrease in reactant concentration or increase in product concentration per unit time:

Average Rate: rav = − Δ[R] / Δt = + Δ[P] / Δt
Instantaneous Rate (Δt → 0): rinst = − d[R] / dt = + d[P] / dt = −(slope of tangent on [R] vs t curve)
Units: mol L-1 s-1 (solutions)  |  atm s-1 or bar s-1 (gases)

The minus sign before Δ[R] is multiplied because Δ[R] = [R]2 − [R]1 is negative, ensuring the reaction rate is always a positive quantity.

2. Rate Expression with Stoichiometric Coefficients

When coefficients are not 1, divide each species' rate of change by its stoichiometric coefficient so the overall rate of reaction has a single unique value:

For 2HI(g) → H2(g) + I2(g):   Rate = −(1/2) Δ[HI]/Δt = +Δ[H2]/Δt = +Δ[I2]/Δt
For 5Br- + BrO3- + 6H+ → 3Br2 + 3H2O:   Rate = −(1/5)Δ[Br-]/Δt = −Δ[BrO3-]/Δt = −(1/6)Δ[H+]/Δt = +(1/3)Δ[Br2]/Δt
NCERT Examples 3.1 & 3.2:
• Hydrolysis of butyl chloride (C4H9Cl): Average rate slows from 1.90 × 10-4 mol L-1 s-1 (0–50 s) to 0.4 × 10-4 mol L-1 s-1 (700–800 s). Tangent at t = 600 s gives rinst = 5.12 × 10-5 mol L-1 s-1.
• 2N2O5(g) → 4NO2(g) + O2(g): [N2O5] drops from 2.33 to 2.08 mol L-1 in 184 min.
  Average Rate = −(1/2)(2.08 − 2.33)/184 = 6.79 × 10-4 mol L-1 min-1 (1.13 × 10-5 mol L-1 s-1).
  Rate of NO2 production = 4 × Average Rate = 4 × 6.79 × 10-4 = 2.72 × 10-3 mol L-1 min-1!
Watch out when exams ask for "rate of production of NO2" vs "rate of reaction"! Rate of reaction divides by coefficient 4, whereas rate of production of NO2 (+Δ[NO2]/Δt) is 4 times the overall reaction rate!

Topic 3.2 – 3.2.2: Factors Influencing Rate, Rate Law & Rate Constant (k) Units

Imagine trying to bake bread faster: you can pack more dough ingredients (concentration), crank up the oven (temperature), or use active yeast (catalyst). But looking at a printed recipe alone never tells you which ingredient controls the oven bottleneck—you must test it in the kitchen! Similarly, a reaction's Rate Law can only be found experimentally, never by merely staring at the balanced equation.

Reaction rate depends on concentration of reactants (or pressure for gases), temperature, and catalyst. For a general reaction aA + bB → cC + dD:

Rate Law: Rate = −d[R]/dt = k [A]x [B]y   (Differential Rate Equation)
Overall Order of Reaction: n = x + y   (can be 0, 1, 2, 3, or even fractional!)
Rate Constant (k): Rate of reaction when concentration of every reactant is unity (1 mol L-1).

Experimental vs Theoretical Coefficients

  • 2NO(g) + O2(g) → 2NO2(g): From initial-rate data (Table 3.2: doubling [NO] quadruples rate; doubling [O2] doubles rate), Rate = k[NO]2[O2] (matches stoichiometry, order = 3).
  • CHCl3 + Cl2 → CCl4 + HCl: Rate = k[CHCl3][Cl2]1/2 (order = 1.5, does NOT match stoichiometry!).
  • CH3COOC2H5 + H2O → CH3COOH + C2H5OH: Rate = k[CH3COOC2H5]1[H2O]0 (order = 1).

Universal Units of Rate Constant (k)

k = Rate / [conc]n = (mol L-1 s-1) / (mol L-1)n = mol1−n Ln−1 s-1  (or bar1−n s-1 for gases)
Reaction Order (n)FormulaUnits of k (Solution)
Zero Order (n = 0)mol1 L-1 s-1mol L-1 s-1
First Order (n = 1)mol0 L0 s-1s-1 (or min-1)
Second Order (n = 2)mol-1 L1 s-1L mol-1 s-1
NCERT Examples 3.3 & 3.4:
• Rate = k[A]1/2[B]3/2 → Order = 1/2 + 3/2 = 2 (Second order).  |  Rate = k[A]3/2[B]-1 → Order = 3/2 − 1 = 1/2.
• Identify order from k: (i) k = 2.3 × 10-5 L mol-1 s-1 → 2nd order;  (ii) k = 3 × 10-4 s-1 → 1st order!
NEVER deduce reaction order from the stoichiometric coefficients of a balanced equation! Always remember the shortcut to spot order from the unit of k: look at the power of L in mol1−n Ln−1 s-1 — Order n = (power of L) + 1!

Topic 3.2.3 – 3.2.4: Elementary vs Complex Reactions, Molecularity & Rate-Determining Step

Picture a 4 × 100 m relay race. Even if three sprinters run like lightning, the team's finish time is dictated entirely by the slowest runner! Complex chemical reactions work the same way: instead of 10 molecules smashing together at one instant, the reaction marches through a sequence of simple steps—and the slowest step becomes the Rate-Determining Step!

Reactions taking place in a single step are elementary reactions. Reactions proceeding through a sequence of elementary steps (called a mechanism) are complex reactions.

1. Molecularity of a Reaction

The number of reacting species (atoms, ions, or molecules) taking part in an elementary reaction that must collide simultaneously to bring about a chemical reaction:

  • Unimolecular (1): NH4NO2 → N2 + 2H2O
  • Bimolecular (2): 2HI → H2 + I2
  • Trimolecular (3): 2NO + O2 → 2NO2 (rare and slow, because simultaneous collision of 3 species is unlikely; >3 is virtually impossible!).

2. Rate-Determining Step (RDS) in Mechanisms

For complex equations like KClO3 + 6FeSO4 + 3H2SO4 → KCl + 3Fe2(SO4)3 + 3H2O (apparently 10th order, actually 2nd order!), the slowest step controls the overall rate. Example: I--catalysed decomposition of H2O2 in alkaline medium:

Overall: 2H2O2 —(I-, OH-)→ 2H2O + O2   |   Rate = k[H2O2][I-] (Order = 2)
• Step 1 (Slow — RDS): H2O2 + I- → H2O + IO-  (IO- is an intermediate)
• Step 2 (Fast): H2O2 + IO- → H2O + I- + O2

3. Order vs Molecularity (Board Exam Classic)

PropertyOrder of a ReactionMolecularity of a Reaction
NaturePurely experimental quantityTheoretical concept from elementary step
Allowed ValuesCan be zero, integer, or fractionalOnly positive integers (1, 2, or 3); never 0 or fraction
ScopeApplies to both elementary & complex reactionsDefined only for elementary reactions (meaningless for complex)
Complex ReactionsDetermined by slowest step (RDS)Molecularity of slowest step = overall order
A favourite board trap: "Can molecularity be zero or 1.5?" NO! Zero molecules colliding means no reaction; half a molecule cannot collide! Only Order can be zero or fractional.

Topic 3.3 – 3.3.1: Integrated Rate Equations: Zero Order Reactions & Surface Catalysis

Picture a busy toll booth with only 6 lanes during a festival rush. Whether 100 cars or 1,000 cars queue behind the barrier, cars pass through at the exact same fixed rate—6 cars per minute—because every booth is completely saturated! That is how a Zero Order reaction works on a metal catalyst surface: once all surface sites are full, piling in more reactant gas changes nothing!

Measuring instantaneous tangent slopes at every time t is inconvenient, so we integrate the differential rate equation to link concentration directly with time.

Derivation for Zero Order Reaction (R → P)

Rate = −d[R]/dt = k [R]0 = k  ⇒  d[R] = −k dt
Integrating: [R] = −kt + I   |   At t = 0, [R] = [R]0 ⇒ I = [R]0
[R] = [R]0 − kt   ⇒   k = ([R]0 − [R]) / t

Key Graphical & Kinetic Features

  • Straight-Line Plot: Plotting [R] vs t gives a straight line with slope = −k and y-intercept = [R]0 (Fig. 3.3).
  • Zero-Order Half-Life (t1/2): At t = t1/2, [R] = [R]0/2 ⇒ t1/2 = [R]0 / 2k. Thus t1/2 ∝ [R]0!
  • 100% Completion Time: Zero-order reactions finish completely in finite time: t100% = [R]0 / k = 2 t1/2.
Classic NCERT Zero-Order Examples:
1. Decomposition of NH3 on hot Pt surface (1130 K, high pressure): 2NH3(g) —(Pt)→ N2(g) + 3H2(g); Rate = k[NH3]0 = k. At high pressure, the Pt surface saturates with NH3 molecules, making rate independent of gas pressure.
2. Thermal decomposition of HI on gold (Au) surface and certain enzyme-catalysed reactions at substrate saturation.
Notice carefully: for a zero-order reaction, half-life is directly proportional to [R]0 (t1/2 = [R]0/2k). If initial concentration doubles, the first half-life doubles—and each successive half-life is HALF the previous one!

Topic 3.3.2 – 3.3.3: First Order Integrated Equation, Gas-Phase Form & Constant Half-Life

Every radioactive isotope—like Carbon-14 ticking inside ancient wooden fossils—loses exactly half of its remaining atoms every 5,730 years, whether you start with a mountain of wood or a single toothpick! Why does its half-life refuse to depend on how much you started with? Because nuclear decay follows First Order Kinetics!

In a First Order reaction (R → P), rate is directly proportional to the first power of [R]: −d[R]/dt = k[R].

Exponential Form: [R] = [R]0 e−kt   |   ln[R] = −kt + ln[R]0
Logarithmic Form: k = (2.303 / t) log ([R]0 / [R])   or   k = [2.303 / (t2 − t1)] log ([R]1 / [R]2)
First-Order Half-Life: t1/2 = (2.303 log 2) / k = 0.693 / k  (Independent of [R]0!)

1. Straight-Line Graphs & Gas-Phase Pressure Form

  • ln[R] vs t: Straight line with slope = −k and intercept = ln[R]0 (Fig. 3.4).
  • log([R]0/[R]) vs t: Straight line passing through the origin with slope = k / 2.303 (Fig. 3.5).
  • Gas Phase (A(g) → B(g) + C(g)): If initial pressure is pi and total pressure at time t is pt = (pi−x) + x + x = pi + x, then pA = 2pi − pt ⇒ k = (2.303/t) log [pi / (2pi − pt)].
  • Examples: All natural/artificial radioactive decay (226Ra → 4He + 222Rn), decomposition of N2O5 & N2O, hydrogenation of ethene.
NCERT Worked Examples 3.5 – 3.8:
• Ex 3.5: N2O5 drops from 1.24 × 10-2 to 0.20 × 10-2 M in 60 min → k = (2.303/60) log(6.2) = 0.0304 min-1.
• Ex 3.6 (Gas Phase): 2N2O5 → 2N2O4 + O2; pi = 0.5 atm, at 100 s pt = 0.512 atm. Since pt = 0.5 + x, p(N2O5) = 1.5 − 2pt = 0.476 atm → k = (2.303/100) log(0.5/0.476) = 4.98 × 10-4 s-1.
• Ex 3.8 (99.9% vs t1/2): For 99.9% completion, [R] = 0.001[R]0 → t99.9% = (2.303/k) log(103) = 6.909/k = 10 × t1/2!
OrderDifferential LawIntegrated LawStraight PlotHalf-Life (t1/2)
0−d[R]/dt = kkt = [R]0 − [R][R] vs t (slope −k)[R]0 / 2k (∝ [R]0)
1−d[R]/dt = k[R]kt = 2.303 log([R]0/[R])ln[R] vs t (slope −k)0.693 / k (constant!)
In gas-phase problems, do NOT blindly memorize 2pi − pt! That holds ONLY for 1 mol gas → 2 mol gas (A → B + C). For 2N2O5 → 2N2O4 + O2 (2 mol → 3 mol), write the ICE pressure table: p(N2O5) = 3pi − 2pt!

Topic 3.3.3 (Pseudo First Order): Pseudo First Order Reactions: Ester Hydrolysis & Cane Sugar Inversion

Imagine tossing a single spoon of salt into the Arabian Sea and letting a chemical reaction consume the salt along with a matching molecule of water. Both salt and water react, yet did the ocean's water level change noticeably? Practically zero! Because water is in overwhelming excess, its concentration behaves like a constant—turning a true two-reactant reaction into a Pseudo First Order reaction!

The apparent order of a reaction can be altered by experimental conditions. A reaction that is truly higher order (e.g., second order, bimolecular) but behaves as a first-order reaction because one reactant is present in large excess is called a Pseudo First Order Reaction.

1. Acid-Catalysed Hydrolysis of Ethyl Acetate

CH3COOC2H5 + H2O —(H+)→ CH3COOH + C2H5OH
True Rate Law: Rate = k' [CH3COOC2H5] [H2O]  (Order = 2, molecularity = 2)
Since [H2O] is huge and stays nearly constant: Rate = k [CH3COOC2H5], where k = k'[H2O] (Apparent Order = 1!)
NCERT Quantitative Check (0.01 mol Ester + 10 mol Water):
• At t = 0: Ester = 0.01 mol,  H2O = 10.00 mol,  Acetic acid = 0,  Ethanol = 0.
• At completion (t): Ester = 0 mol,  H2O = 9.99 mol,  Acetic acid = 0.01 mol,  Ethanol = 0.01 mol.
Water concentration drops by just 0.1%—practically constant! Hence the reaction follows first-order kinetics with apparent k in s-1.

2. Inversion of Cane Sugar (Sucrose)

C12H22O11 + H2O —(H+)→ C6H12O6 (Glucose) + C6H12O6 (Fructose)
Rate = k [C12H22O11]   (Molecularity = 2, Apparent Order = 1)
Always check the excess condition! Hydrolysis of an ester is pseudo first order ONLY when water is in large excess (e.g., aqueous solvent). If ester and water are mixed in equal small moles inside an inert solvent, both concentrations change and it obeys second-order kinetics!

Topic 3.4 – 3.4.1: Temperature Dependence, Arrhenius Equation & Effect of Catalyst

Why does milk sour in hours on a hot Kerala afternoon yet stay fresh for days inside a refrigerator? Because for most reactions, every 10 °C rise in temperature nearly doubles the rate constant! Even more dramatically, a catalyst acts like a mountain tunnel—carving a much lower energy pass through the hill so hikers cross effortlessly, while starting village and ending valley stay at the exact same heights!

Heating accelerates reactions sharply (e.g., N2O5 half-life drops from 10 days at 0 °C to 5 hours at 25 °C to 12 minutes at 50 °C). Proposed by van't Hoff and justified by Arrhenius, temperature dependence obeys the Arrhenius Equation:

k = A e−Ea / RT   |   ln k = −(Ea / R)(1/T) + ln A   |   log(k2 / k1) = [Ea / (2.303 R)] [(T2 − T1) / (T1T2)]
  • A: Arrhenius / Frequency / Pre-exponential factor; Ea: Activation energy (J mol-1) needed to form the unstable activated complex (e.g., H2 + I2 → Activated Complex → 2HI); R = 8.314 J K-1 mol-1.
  • e−Ea/RT: Fraction of molecules having kinetic energy ≥ Ea (shaded tail in Maxwell–Boltzmann distribution, Fig. 3.8–3.9). Raising T by 10 K shifts and broadens the curve (total area stays constant = 1), nearly doubling this tail area!
  • Arrhenius Plot (Fig. 3.10): Plotting ln k vs 1/T gives a straight line with slope = −Ea / R and y-intercept = ln A.

Effect of Catalyst (3.4.1)

A catalyst increases reaction rate without undergoing permanent chemical change (e.g., 2KClO3 —(MnO2)→ 2KCl + 3O2). By Intermediate Complex Theory, it offers an alternate path with lower activation energy Ea (Fig. 3.11). An inhibitor reduces rate.

  • A catalyst does NOT alter Gibbs energy (ΔG), reaction enthalpy (ΔH), or Equilibrium Constant (K).
  • It lowers forward and backward Ea equally, helping attain equilibrium faster without shifting equilibrium position; it catalyses only spontaneous (ΔG < 0) reactions.
NCERT Example 3.9: At 500 K, k1 = 0.02 s-1; at 700 K, k2 = 0.07 s-1.
log(0.07/0.02) = [Ea / (2.303 × 8.314)] × [(700 − 500)/(500 × 700)] ⇒ 0.544 = (Ea / 19.15) × 5.714 × 10-4 ⇒ Ea = 18.23 kJ mol-1,  A = 1.61 s-1.
Board exam favourite trap: "How does a catalyst change ΔG and Equilibrium Constant K?" Answer: It does NOT change ΔG or K at all! Also watch units in Ea calculations: since R = 8.314 J K-1 mol-1, convert kJ to J (× 1000) before plugging into the two-temperature formula!

Topic 3.5: Collision Theory of Chemical Reactions: Threshold Energy & Steric Factor (P)

Picture trying to plug a USB-A drive into your laptop in the dark. Even if you shove hard enough (high energy), plugging it upside-down just bounces off the port! Only a push with enough force AND the right orientation clicks in! Molecules behave the exact same way—every reaction needs Threshold Energy plus Proper Orientation!

Developed by Max Trautz and William Lewis (1916–18) based on the kinetic theory of gases, Collision Theory treats reactant molecules as hard spheres that react when they collide.

1. Collision Frequency & Simple Model

Collision Frequency (Z) is the number of collisions per second per unit volume of the reaction mixture. For a bimolecular elementary reaction A + B → Products:

Simple Model: Rate = ZAB e−Ea/RT
Modified Collision Theory (with Orientation): Rate = P ZAB e−Ea/RT   (so Arrhenius A = P ZAB)

2. Two Criteria for an Effective Collision

Not every collision produces products. Only effective collisions succeed, requiring TWO simultaneous conditions:

  1. Energy Criterion (Threshold Energy): Colliding molecules must carry minimum total kinetic energy called Threshold Energy = Activation Energy (Ea) + Average energy possessed by reacting species.
  2. Orientation Criterion (Steric Factor P): Molecules must collide in proper geometric alignment so old bonds break and new bonds form! For example, in CH3Br + OH- → CH3OH + Br- (Fig. 3.12):
    • Proper Orientation: OH- attacks the δ+-Carbon from the backside opposite Br → bond forms!
    • Improper Orientation: OH- approaches the bulky δ--Br end → electrostatic/steric repulsion makes them simply bounce back unchanged!

Drawback of Collision Theory: It models atoms and complex molecules as rigid hard spheres and ignores their internal structural, vibrational, and rotational aspects.

Do NOT confuse Activation Energy (Ea) with Threshold Energy! Threshold energy is the total minimum energy colliding molecules must possess, whereas Ea = Threshold Energy − Average energy of reactants!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Most Asked) • 2 Marks): Write any four differences between Order and Molecularity of a reaction.

OrderMolecularity
1. Sum of powers of concentration terms in the rate law.1. Number of colliding species in an elementary reaction.
2. Experimental quantity.2. Theoretical concept.
3. Can be zero, integer, or fractional.3. Always a positive integer (1, 2, or 3); never zero or fraction.
4. Applicable to both elementary and complex reactions.4. Applicable only to elementary reactions.

Q2 (Kerala HSE March (Classic 3-Marker) • 3 Marks): Derive the integrated rate equation for a first-order reaction R → P, and deduce the expression for its half-life (t1/2).

For R → P, −d[R]/dt = k[R] ⇒ d[R]/[R] = −k dt.

Integrating both sides: ln[R] = −kt + I. At t = 0, [R] = [R]0 ⇒ I = ln[R]0.

Thus ln[R] = −kt + ln[R]0 ⇒ ln([R]0/[R]) = kt ⇒ k = (2.303 / t) log ([R]0 / [R]).

For half-life, at t = t1/2, [R] = [R]0/2 ⇒ t1/2 = (2.303 / k) log 2 = (2.303 × 0.3010) / k = 0.693 / k (independent of [R]0).

Q3 (Kerala HSE SAY / Model • 2 Marks): What is a pseudo first order reaction? Explain with two examples.

A reaction which is truly of higher order (e.g., second order) but obeys first-order kinetics when one of the reactants is taken in large excess is called a pseudo first order reaction.

Examples:
1. Acid hydrolysis of ethyl acetate: CH3COOC2H5 + H2O —(H+)→ CH3COOH + C2H5OH (Rate = k[CH3COOC2H5]).
2. Inversion of cane sugar: C12H22O11 + H2O —(H+)→ C6H12O6 + C6H12O6.

Q4 (Kerala HSE March • 1 Marks): Identify the reaction order from each rate constant: (i) k = 2.3 × 10-5 L mol-1 s-1  (ii) k = 3 × 10-4 s-1.

(i) Unit L mol-1 s-1 corresponds to a second-order reaction.
(ii) Unit s-1 corresponds to a first-order reaction.

Q5 (Kerala HSE SAY • 1 Marks): How does a catalyst affect (a) activation energy Ea, and (b) Gibbs energy ΔG and equilibrium constant K of a reaction?

(a) A catalyst lowers the activation energy (Ea) by providing an alternate reaction pathway.
(b) It causes no change in Gibbs energy (ΔG) or the equilibrium constant (K); it only helps attain equilibrium faster.

Q6 (Kerala HSE Model • 1 Marks): Name the straight-line plot obtained for a zero-order reaction and write the values of its slope and y-intercept.

Plot of [R] vs t gives a straight line with slope = −k and y-intercept = [R]0.

Q7 (Kerala HSE / NCERT Ex 3.8 • 3 Marks): Show that in a first-order reaction, the time required for 99.9% completion is 10 times its half-life (t1/2).

At 99.9% completion, [R] = [R]0 − 0.999[R]0 = 0.001[R]0.

t99.9% = (2.303 / k) log ([R]0 / 0.001[R]0) = (2.303 / k) log(103) = (2.303 × 3) / k = 6.909 / k.

Since t1/2 = 0.693 / k, ratio t99.9% / t1/2 = 6.909 / 0.693 ≈ 10. Hence t99.9% = 10 × t1/2.

Q8 (Kerala HSE March (Arrhenius Numerical) • 3 Marks): The rate of a chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate Ea. [R = 8.314 J K-1 mol-1]

Given T1 = 298 K, T2 = 308 K, k2/k1 = 2.

log(k2/k1) = [Ea / (2.303 R)] [(T2 − T1) / (T1T2)]

log 2 = [Ea / (2.303 × 8.314)] × [10 / (298 × 308)]

0.3010 = (Ea / 19.147) × 1.0895 × 10-4 ⇒ Ea = 52,897 J mol-1 ≈ 52.90 kJ mol-1.

Q9 (Kerala HSE SAY / NCERT • 2 Marks): For the reaction 5Br-(aq) + BrO3-(aq) + 6H+(aq) → 3Br2(aq) + 3H2O(l), write the rate expression in terms of each reactant and product.

Rate = −(1/5) Δ[Br-]/Δt = −Δ[BrO3-]/Δt = −(1/6) Δ[H+]/Δt = +(1/3) Δ[Br2]/Δt = +(1/3) Δ[H2O]/Δt.

Q10 (Kerala HSE March • 2 Marks): State the two essential conditions for an effective collision according to Collision Theory. What is the significance of steric factor P?

Two criteria for effective collision: (1) Colliding molecules must have energy ≥ Threshold Energy (Ea + average energy of reactants), and (2) they must collide in proper spatial orientation.

Steric factor (P) accounts for proper orientation in Rate = P ZAB e−Ea/RT so bond breaking and new bond formation occur smoothly.

Q11 (Kerala HSE Model / NCERT Ex 3.14 • 3 Marks): The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample.

Radioactive decay follows first-order kinetics: k = 0.693 / t1/2 = 0.693 / 5730 yr-1.

Here [R]0 = 100, [R] = 80. Using t = (2.303 / k) log ([R]0/[R]):

t = (2.303 × 5730 / 0.693) × log(100/80) = 19042.4 × 0.09691 ≈ 1845 years.

Q12 (Kerala HSE March / SAY • 4 Marks): (a) Derive the integrated rate equation and half-life for a zero-order reaction. Give one example. (b) A reaction is first order in A and second order in B: how is rate affected if [B] is tripled, and if both [A] and [B] are doubled?

(a) −d[R]/dt = k[R]0 = k ⇒ ∫d[R] = −k∫dt ⇒ [R] = −kt + [R]0 ⇒ k = ([R]0 − [R])/t. At t1/2, [R] = [R]0/2 ⇒ t1/2 = [R]0/(2k). Example: Decomposition of NH3 on hot Pt surface at high pressure.

(b) Rate = k[A][B]2.
(i) If [B] is tripled: Rate increases by 32 = 9 times.
(ii) If both [A] and [B] are doubled: Rate increases by 2 × 22 = 8 times.