Class 12 Chemistry · Physical

CHEMICAL KINETICS

Complete Formula, Shortcut & Question Compendium

NEET · JEE Main · Board

Compiled from class notes — Day 1 to Day 6
Every formula · every trick · every solved question
Plus additional high-yield PYQ-style practice

1	Rate of Reaction — Basics & Stoichiometric Relations
2	Rate Law, Order & Molecularity
3	Zero Order Reactions
4	First Order Reactions — The Core of the Chapter
5	Half-Life & Fractional-Life Master Tricks
6	Second & nth Order — Comparison Table
7	Methods to Determine Order (4 methods + Trick Box)
8	Pressure-Based Numericals (Gaseous First Order)
9	Pseudo First Order Reactions
10	Collision / Transition State Theory, Activation Energy
11	Arrhenius Equation & Temperature Coefficient
12	Effect of Catalyst
13	Complex Reactions & Rate Determining Step
14	Graph Bank — Identify the Order at a Glance
15	MASTER TRICK SHEET (one-page revision)
16	Solved Questions from the Notes (Q1–Q30)
17	Additional High-Yield NEET/JEE Questions (Q31–Q60)
18	Assertion–Reason & Match the Following
19	Corrections to the Original Notes

1. Rate of Reaction — Basics & Stoichiometric Relations

Chemical kinetics deals with the speed (velocity) of a reaction, the factors that affect it, and the mechanism by which it proceeds.

Rate = Change in concentration / Time taken = ΔC / Δt

1.1 Types of reactions by speed

Type	Δt	Rate	Example / Remark
Fast (instantaneous)	Very small	Very high	Ionic reactions (AgNO₃ + NaCl). Rate cannot be measured accurately.
Slow	Very large	Very small	Rusting of iron. Rate cannot be measured accurately.
Moderate	Medium	Medium	2H₂O₂ → 2H₂O + O₂. Only these can be studied kinetically.

1.2 Average vs Instantaneous rate

Average rate (large Δt)	Instantaneous rate (dt → 0)
For reactant: r_av = −Δ[A]/Δt = −(C_f − C_i)/(t_f − t_i)	For reactant: r_inst = −d[A]/dt
For product: r_av = +Δ[P]/Δt	For product: r_inst = +d[P]/dt; obtained by drawing a tangent to the conc–time curve.

Why the minus sign? Reactant concentration falls, so Δ[A] is negative. Rate can never be negative — the minus sign in front of the formula flips it positive. E.g. 100 → 80 M in 5 s: ΔC/Δt = (80−100)/5 = −4, so rate = −(−4) = +4 M s⁻¹.

1.3 Relating rates of different species

For a balanced equation xA → yB:

Rate of reaction = −(1/x) d[A]/dt = +(1/y) d[B]/dt

For N₂ + 3H₂ → 2NH₃:

−d[N₂]/dt ÷ 1 = −d[H₂]/dt ÷ 3 = +d[NH₃]/dt ÷ 2 = Rate of reaction

Shortcut — "divide by coefficient" If rate of disappearance of N₂ = x, then rate of disappearance of H₂ = 3x and rate of appearance of NH₃ = 2x. The species with the biggest coefficient moves fastest.
Read the question carefully: "rate of reaction" needs the 1/coefficient; "rate of disappearance of X" does not.

Trap Rate data given in g L⁻¹ s⁻¹ must be converted to mol L⁻¹ s⁻¹ first using n = w/M. This is the single most common silent mistake in this topic.

2. Rate Law, Order & Molecularity

For aA + bB → products: Rate = k[A]^x[B]^y Overall order n = x + y

x, y are found experimentally — they equal a, b only for elementary (single-step) reactions.

2.1 Order vs Molecularity

Order	Molecularity
Experimental quantity	Theoretical quantity
Sum of powers in the rate law	Number of molecules colliding in an elementary step
Can be 0, fractional, or negative	Can never be zero, fractional, negative or infinite — only positive integers
Defined for overall reaction	Not defined for a complex overall reaction — defined only per elementary step
Can change with conditions (excess reagent, temperature)	Fixed for a given step; normally ≤ 3 (3 is rare, >3 practically impossible)

Key identity If order = molecularity, the reaction is elementary (single step). If order ≠ molecularity, it is a complex (multi-step) reaction and molecularity depends on the slowest step (RDS).

Molecularity examples

Reaction	Molecularity	Reaction	Molecularity
N₂O₄(g) → 2NO₂(g)	1 (unimolecular)	2HI → H₂ + I₂	2 (bimolecular)
2SO₂ + O₂ → 2SO₃	3 (termolecular)	A + 11B → C	12 — impossible as one step ⇒ must be complex

2.2 Units of rate constant — the universal formula

Units of k = mol⁽¹⁻ⁿ⁾ L⁽ⁿ⁻¹⁾ s⁻¹ (n = order)

Order n	0	1	2	3	1/2	3/2
Units of k	mol L⁻¹ s⁻¹	s⁻¹	L mol⁻¹ s⁻¹	L² mol⁻² s⁻¹	mol^1/2 L^−1/2 s⁻¹	mol^−1/2 L^1/2 s⁻¹

Reverse trick Given the units of k, you can instantly read off the order — a very common 4-mark-in-30-seconds MCQ. Units contain L mol⁻¹ raised to (n−1); count the power of L, add 1.

3. Zero Order Reactions

Rate = k[A]⁰ = k → −d[A]/dt = k → C₀ − C_t = kt or x = kt

t_1/2 = C₀ / 2k (t_1/2 ∝ C₀) t_completion = C₀/k = 2 t_1/2

Zero order fractional-life trick t_x% = (x/50) × t_1/2
⇒ t_50% = 1 t_1/2 · t_75% = 1.5 t_1/2 · t_100% = 2 t_1/2

Equal-time trick (zero order) In a zero order reaction the same amount is consumed in each equal time interval (not the same percentage).
100 → 60 in 10 s (40 units gone) → 20 in the next 10 s (another 40 units gone).
Compare with first order: 100 → 60 → 36 (same percentage each time).

Real examples: photochemical reaction H₂ + Cl₂ (sunlight); decomposition of NH₃ on hot Pt surface; decomposition of HI on gold surface; enzyme reactions at saturating substrate.

4. First Order Reactions — The Core of the Chapter

4.1 Full derivation (write this in board exams)

For A → B, with a = initial conc, (a − x) = conc at time t:

r = −d[A]/dt = k[A]¹ ⇒ −d[A]/[A] = k dt
Integrating from C₀ → C_t and 0 → t: −[ln A]_C₀^C_t = k[t]₀^t
ln C₀ − ln C_t = kt ⇒ ln(C₀/C_t) = kt

k = (2.303 / t) log (C₀ / C_t) = (2.303 / t) log [ a / (a − x) ]

C_t = C₀ e^−kt (exponential decay form)

Log helpers used: ∫dx/x = ln x · ln a − ln b = ln(a/b) · ln = 2.303 log · log a^b = b log a · log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021.

4.2 Limiting cases

At t =	C_t = C₀e^−kt gives	Conclusion
0	C_t = C₀e⁰ = C₀	Nothing has reacted yet.
∞	C_t = C₀e^−∞ = 0	A first order reactant is never completely consumed in finite time. The curve approaches zero only asymptotically.

Most asked one-liner "The reactant of a first order reaction can never be reduced to zero / can never be completed" — a direct NEET statement question. Contrast: a zero order reaction does finish, at t = C₀/k.

4.3 Half-life

Put C_t = C₀/2 (take a = 100, a−x = 50):

k = (2.303 / t_1/2) log 2 = (2.303 × 0.3010)/t_1/2 ⇒ k = 0.693 / t_1/2 = ln2 / t_1/2 ⇒ t_1/2 = 0.693 / k

Signature property For first order, t_1/2 is independent of initial concentration.
1 M → 0.5 M takes 10 s ⇒ 0.5 M → 0.25 M also takes 10 s.
(Zero order: 1 M → 0.5 M in 10 s ⇒ 0.5 M → 0.25 M takes only 5 s, because t_1/2 ∝ C₀.)

4.4 The n-half-lives shortcut (single most useful trick)

C_t = C₀ (1/2)ⁿ where n = total time / t_1/2

n (half-lives)	1	2	3	4	5	6	10
% left	50	25	12.5	6.25	3.125	1.5625	≈0.1
% consumed	50	75	87.5	93.75	96.875	98.44	≈99.9

Trap "% consumed" vs "% left" — the exam swaps these constantly. C_t from the formula is what is left; consumed = 100 − C_t.

4.5 Half-life when stoichiometric coefficient ≠ 1

When the rate law is written as rate of reaction (i.e. r = −(1/x)d[A]/dt = k[A]):

For xA → products: t_1/2 = 0.693 / (x k)
e.g. 2A → B ⇒ t_1/2 = 0.693/2k · 3A → B ⇒ t_1/2 = 0.693/3k

Read the question If instead the problem defines −d[A]/dt = k[A] directly (rate of disappearance of A), then plain t_1/2 = 0.693/k applies. Check which definition is given before dividing by x.

5. Half-Life & Fractional-Life Master Tricks

5.1 First order fractional life (memorise this table)

Quantity	In terms of k	In terms of t_1/2
t_50% (= t_1/2)	0.693 / k	1 t_1/2
t_75%	1.386 / k	2 t_1/2
t_87.5%	2.079 / k	3 t_1/2
t_93.75%	2.772 / k	4 t_1/2
t_96.875%	3.465 / k	5 t_1/2
t_90%	2.303 / k	3.32 t_1/2
t_99%	4.606 / k	6.64 t_1/2
t_99.9%	6.909 / k	≈ 10 t_1/2

Golden ratios (first order) t_99% = 2 × t_90% · t_99.9% = 3 × t_90% · t_1/2 : t_3/4 : t_7/8 = 1 : 2 : 3
Derivation of the first: t₉₉ = (2.303/k) log(100/1) = (2.303/k)×2 ; t₉₀ = (2.303/k) log(100/10) = (2.303/k)×1. Ratio = 2.

Always set a = 100 In every fractional-life problem take the initial amount as 100. Then (a − x) is simply "100 − percentage reacted": t_75% uses 100/25, t_90% uses 100/10, t_99% uses 100/1. No algebra needed.

5.2 Order from half-life dependence

t_1/2 ∝ 1 / a⁽ⁿ⁻¹⁾ (t_1/2)₁ / (t_1/2)₂ = (a₂ / a₁)ⁿ⁻¹

Order	0	1	2	3
t_1/2 ∝	a¹ (directly)	a⁰ (independent)	1/a	1/a²

6. Second & nth Order — Comparison Table

Feature	Zero order	First order	Second order	nth order
Rate law	r = k	r = k[A]	r = k[A]²	r = k[A]ⁿ
Integrated form	C₀ − C_t = kt	k = (2.303/t) log(C₀/C_t)	1/C_t − 1/C₀ = kt	—
Linear plot	[A] vs t	log[A] or ln[A] vs t	1/[A] vs t	1/[A]ⁿ⁻¹ vs t
Slope	−k	−k/2.303 (log) or −k (ln)	+k	—
t_1/2	C₀/2k	0.693/k	1/(k C₀)	∝ 1/C₀ⁿ⁻¹
Units of k	mol L⁻¹s⁻¹	s⁻¹	L mol⁻¹s⁻¹	mol¹⁻ⁿLⁿ⁻¹s⁻¹
Completion?	Yes, at t = C₀/k	Never (infinite time)	Never	—

7. Methods to Determine Order

Method 1 — Half-life method

Used when initial concentrations and their half-lives are given.

(t_1/2)₁ / (t_1/2)₂ = (a₂/a₁)ⁿ⁻¹

Worked: 0.1 M → t_1/2 = 10 s; 0.2 M → t_1/2 = 10 s.
10/10 = (0.2/0.1)ⁿ⁻¹ ⇒ 1 = 2ⁿ⁻¹ ⇒ n − 1 = 0 ⇒ n = 1 (first order)

Method 2 — Initial rate method

Used when initial concentrations and initial rates are tabulated.

Exp	[A]	[B]	Rate (mol L⁻¹s⁻¹)
1	0.1 M	0.1 M	10⁻³
2	0.2 M	0.1 M	2 × 10⁻³
3	0.1 M	0.2 M	4 × 10⁻³

Exp3/Exp1: 4 = 2^y ⇒ y = 2 (order w.r.t. B).
Exp2/Exp1: 2 = 2^x ⇒ x = 1 (order w.r.t. A).
Overall order = 3. And k = 10⁻³ / [(0.1)¹(0.1)²] = 10⁻³/10⁻³ = k = 1 L² mol⁻² s⁻¹

Shortcut Always divide the two experiments in which one reactant's concentration is unchanged — that species cancels and you isolate a single exponent in one line.

Method 3 — Integrated rate law / graphical / hit & trial

Used when concentration vs time data is given. Assume an order, compute k for each interval; the order for which k comes out constant is the answer.

Worked: [A] = 0.1 M (0 s), 0.06 M (10 s), 0.036 M (20 s).

Assume zero order (C₀ − C_t = kt)	Assume first order (k = 2.303/t · log C₀/C_t)
k′ = (0.1 − 0.06)/10 = 4 × 10⁻³ k″ = (0.06 − 0.036)/10 = 2.4 × 10⁻³ k′ ≠ k″ ⇒ not zero order	k′ = (2.303/10) log(0.1/0.06) = (2.303/10) log(10/6) k″ = (2.303/10) log(0.06/0.036) = (2.303/10) log(10/6) k′ = k″ ⇒ first order

Method 4 — Ostwald's isolation method

Rule Any reactant taken in large excess has its order treated as ZERO — its concentration stays effectively constant and merges into k.
Practical exam cue: if no data is given for one reactant in the table, that reactant is in excess and its order is zero.

For A + B + C → D with Rate = k[A]^x[B]^y[C]^z: if [C] is in excess, z = 0 and Rate = k′[A]^x[B]^y.

TRICK BOX — which method to use

What the question gives you	Method to apply
Initial concentration & half-lives	Half-life method
Initial concentration & rates	Initial rate method
Concentration & time	Hit & trial (integrated rate law)
One reactant clearly in excess / missing from table	Ostwald isolation → its order = 0

8. Pressure-Based Numericals (Gaseous First Order)

For A(g) → nB(g) where only A is taken initially, with P₀ = initial pressure, P_t = total pressure at time t, P_∞ = pressure after very long time:

Time	A	B	Total pressure
t = 0	P₀	0	P₀
t = t	P₀ − x	nx	P_t = P₀ + (n−1)x
t = ∞	0	nP₀	P_∞ = nP₀

k = (2.303/t) log [ (P_∞ − P₀) / (P_∞ − P_t) ]

k = (2.303/t) log [ (n−1) P₀ / (nP₀ − P_t) ] (use when P_∞ is not given)

Finding n n = (moles of gaseous product) / (moles of gaseous reactant)
A → 4B ⇒ n = 4/1 = 4 · 2A → 3B + C ⇒ n = 4/2 = 2 · A → 2B + C ⇒ n = 3/1 = 3

Two-line method (safer than the formula) 1. Get P₀ from P_∞ = nP₀.
2. Write P_t = P₀ + (n−1)x and solve for x.
3. Feed a = P₀, (a−x) = P₀ − x into the ordinary first-order formula.
This "method 2" also works when the question gives only partial pressures (e.g. P_A + P_B), where the P_∞ formula fails.

9. Pseudo First Order Reactions

Definition: Reactions which are not truly first order, but appear to be first order because one reactant is present in such large excess that its concentration is effectively constant.

Reaction	True rate law	Observed rate law
Hydrolysis of alkyl halide CH₃I + H₂O → CH₃OH + HI	r = k[CH₃I][H₂O]	r = k′[CH₃I] — H₂O in excess
Inversion of cane sugar C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆	r = k[C₁₂H₂₂O₁₁][H₂O]	r = k′[C₁₂H₂₂O₁₁]
Acidic hydrolysis of ester CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH	r = k[ester][H₂O]	r = k′[ester]

Exam points True molecularity = 2 (bimolecular) but observed order = 1. The products of cane sugar inversion are glucose + fructose. Ester hydrolysis is followed by titrating the acid formed; sugar inversion by a polarimeter (dextro → laevo).

10. Collision / Transition State Theory & Activation Energy

10.1 Four postulates

A reaction occurs due to collisions between reactant molecules.
As temperature increases, the number of collisions increases, so k and rate increase: T↑ ⇒ k↑ ⇒ R↑.
Not all collisions are effective. Only those with proper orientation lead to product.
A → A–B (A approaching the A end): effective | A → B–A (A approaching the B end): not effective.
Only molecules possessing energy greater than the threshold energy can react.

Threshold energy = Activation energy (E_a) + Average energy of reactants

10.2 Key definitions

Term	Definition
Threshold energy	The minimum energy a molecule must possess for product formation to start.
Activation energy E_a	The minimum extra energy that must be supplied to the reactant molecules so that the reaction occurs. Always positive.
Transition state / activated complex	The state in which old bonds are not completely broken and new bonds are not completely formed (A···A···B). Highly unstable, cannot be isolated, sits at the peak of the energy profile.

10.3 Energy profile & the ΔH relation

ΔH = E_a(forward) − E_a(backward)

Endothermic	Exothermic
H_P > H_R, ΔH = +ve, energy absorbed E_a(f) > E_a(b)	H_R > H_P, ΔH = −ve, energy released E_a(b) > E_a(f)
Written as: A + xJ → B \| A → B, ΔH = +xJ \| A → B − xJ	Written as: A → B + xJ \| A → B, ΔH = −xJ \| A − xJ → B

10.4 Maxwell–Boltzmann distribution (the two-curve graph)

Plot: fraction of molecules (y) vs kinetic energy (x). The peak corresponds to the most probable energy.
At higher temperature T₂ > T₁: peak becomes lower and shifts right; the curve broadens.
The shaded area beyond the threshold energy = fraction of molecules able to react. A rise of only 10°C roughly doubles this area — which is why rate doubles.

Fraction of molecules with energy > threshold energy = e^−E_a/RT

11. Arrhenius Equation & Temperature Coefficient

k = A e^−E_a/RT A = Arrhenius factor / pre-exponential factor / frequency factor = total number of collisions per second per unit volume.

Taking natural log on both sides:

ln k = ln A − E_a/RT → log k = log A − E_a/(2.303 RT)
Compare with y = mx + c: plot log k (y) vs 1/T (x)

Slope of ln k vs 1/T = −E_a/R · Slope of log k vs 1/T = −E_a/2.303R · Intercept = log A

11.1 Two-temperature form (most used numerically)

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

Remember Since rate ∝ k at fixed concentration, you may replace k₂/k₁ by r₂/r₁. Always convert °C to K (+273). R = 8.314 J K⁻¹mol⁻¹ = 2 cal K⁻¹mol⁻¹ — use whichever the question specifies, and the answer will be in J or cal accordingly.

11.2 Effects at a glance

Change	k	Rate	Reason
T increases	↑	↑	e^−Ea/RT increases; more molecules above threshold
E_a increases	↓	↓	Higher energy barrier
E_a decreases (catalyst)	↑	↑	Favourable condition
T → ∞	k → A	max	e⁰ = 1
E_a = 0	k = A	max	Every collision effective

11.3 Temperature coefficient (T.C. or μ)

T.C. = k_(T+10) / k_T usually measured from 25°C to 35°C; for most reactions μ lies between 2 and 3.

k_(t+Δt) / k_t = R_(t+Δt) / R_t = (T.C.)^Δt/10

T.C.	25°C	35°C	45°C	55°C
2	R	2R	4R	8R
3	R	3R	9R	27R

Shortcut Rate multiplier = μ raised to the power (temperature rise ÷ 10). A 40°C rise with μ = 2 gives 2⁴ = 16×; with μ = 3 gives 3⁴ = 81×.

12. Effect of Catalyst

A catalyst does not get consumed in the overall reaction but alters its rate by providing an alternative path of lower activation energy.

Positive catalyst	Negative catalyst (inhibitor)
Increases the rate of reaction	Decreases the rate of reaction
Lowers the energy barrier / E_a ⇒ k ↑	Raises the effective E_a ⇒ k ↓

What a catalyst does NOT change

K_c = k_f / k_b → catalyst makes k_f → n k_f and k_b → n k_b → K_c = n k_f / n k_b = unchanged

Both E_a(f) and E_a(b) are lowered by the same amount, so ΔH is unchanged.
Equilibrium constant is unchanged — only equilibrium is reached faster.
A catalyst cannot start a thermodynamically non-spontaneous reaction.

13. Complex Reactions & Rate Determining Step

Simple (elementary) reaction	Complex reaction
Occurs in a single step	Occurs in more than one step
Order = molecularity; rate law can be written directly from stoichiometry A + 2B → C ⇒ r = k[A]¹[B]², order = 3	Rate law must come from the mechanism, never from stoichiometry. Molecularity is undefined for the overall reaction (defined per step only).

Rate Determining Step (RDS) = the SLOWEST step of the mechanism. The overall rate law is written from the RDS.

13.1 Two standard cases

Case A — RDS given, no intermediate on its reactant side

A + 2B → C, with (i) A + B → D (slow, RDS), (ii) D + B → C (fast).
Rate is written straight from step (i): r = k[A][B], overall order = 2

Case B — RDS contains an intermediate ⇒ eliminate it using the fast equilibrium

A + 2B → C, with (i) A ⇌ D (fast, k₁/k₂), (ii) D + B → C (slow, k₃).

r = k₃[D][B]; K_c = [D]/[A] ⇒ [D] = K_c[A]
⇒ r = k₃K_c[A][B] = k′[A][B]

Classic example — decomposition of ozone

2O₃ → 3O₂: (i) O₃ ⇌ O₂ + O (fast), (ii) O₃ + O → 2O₂ (slow).

r = k₃[O₃][O]; K_c = [O₂][O]/[O₃] ⇒ [O] = K_c[O₃]/[O₂]
⇒ r = k′[O₃]²[O₂]⁻¹

Order w.r.t. O₃ = 2; order w.r.t. O₂ = −1; overall order = 1.

Why this is asked every year It is the standard illustration of a negative order — proof that order need not be a positive integer and that a product can retard a reaction.

14. Graph Bank — Identify the Order at a Glance

Plot	Zero order	First order	Second order
[A] vs t	Straight line, slope −k, meets x-axis (conc = 0)	Exponential decay, never touches x-axis	Steeper decay curve
log[A] or ln[A] vs t	Curve	Straight line; slope −k/2.303 (log), −k (ln); intercept log C₀	Curve
1/[A] vs t	Curve	Curve	Straight line, slope +k
Rate vs [A]	Horizontal line (rate independent of conc)	Straight line through origin, slope = k	Parabola
t_1/2 vs C₀	Straight line, positive slope	Horizontal line	Rectangular hyperbola
log k vs 1/T	Straight line for all orders; slope = −E_a/2.303R, intercept = log A

15. MASTER TRICK SHEET

Situation	Instant move
Rate given in g L⁻¹s⁻¹	Convert with n = w/M before anything else.
"Rate of reaction" asked	Divide by the stoichiometric coefficient.
Any % / fractional-life question	Set a = 100. Then a − x = 100 − %reacted.
First order + time is a multiple of t_1/2	Use C_t = C₀(1/2)ⁿ, n = t/t_1/2. Never use logs.
t_99% / t_90%	= 2. And t_99.9% = 3 t_90% ≈ 10 t_1/2.
Equal % consumed in equal time intervals	First order. (Equal amount consumed ⇒ zero order.)
t_1/2 unchanged when C₀ doubles	First order (n − 1 = 0).
t_1/2 halves when C₀ doubles	Second order. t_1/2 doubles ⇒ zero order.
One reactant missing from the data table	It is in excess ⇒ Ostwald ⇒ its order = 0.
Units of k given	Read off order directly: s⁻¹ = 1st, mol L⁻¹s⁻¹ = 0, L mol⁻¹s⁻¹ = 2nd.
Total pressure data for a gas-phase reaction	n = gaseous products/gaseous reactants; P_∞ = nP₀; then P_t = P₀ + (n−1)x.
Rate rises 4× for a 20°C rise	T.C. = 2 (since 2^20/10 = 4).
Fraction of molecules above threshold	e^−Ea/RT. If it works out to e^−0.693, answer is exactly 0.5 = 50%.
ΔH and one E_a given	ΔH = E_a(f) − E_a(b); watch the sign of ΔH.
"Effect of catalyst on K_c / ΔH"	No change. Only E_a falls and equilibrium arrives sooner.
Mechanism with a fast pre-equilibrium	Write r from the slow step, then substitute the intermediate using K_c of the fast step.
Order asked and it comes out fractional/negative	Perfectly valid — it just proves the reaction is complex.

16. Solved Questions from the Notes

Q1.

For N₂ + 3H₂ → 2NH₃, the rate of disappearance of H₂ is 6 g L⁻¹s⁻¹. Find the rate of formation of NH₃, the rate of disappearance of N₂, and the rate of reaction.

Convert: n = w/M = 6/2 = 3 mol L⁻¹s⁻¹ for H₂.
Rate of reaction = (1/3)(3) = 1 mol L⁻¹s⁻¹.
d[NH₃]/dt = 2 × 1 = 2 mol L⁻¹s⁻¹; −d[N₂]/dt = 1 mol L⁻¹s⁻¹; rate of reaction = 1 mol L⁻¹s⁻¹.

Q2.

For A → B with [A]₀ = 100 and [A] = 80 at t = 5 s, find the average rate.

r_av = −(80 − 100)/(5 − 0) = +4 units s⁻¹. For the product, r = (20 − 0)/5 = +4 — the two agree, as they must for 1:1 stoichiometry.

Q3.

Derive the integrated rate equation for a first order reaction and hence the expression for t_1/2.

See Section 4.1 and 4.3. Final: k = (2.303/t) log[a/(a−x)] and t_1/2 = 0.693/k.

Q4.

Show that a first order reactant is never completely consumed.

C_t = C₀e^−kt. At t = ∞, e^−∞ = 0 so C_t → 0, but only asymptotically — the concentration is zero only after infinite time. Hence completion never occurs in finite time.

Q5.

Find t_99%/t_90% for a first order reaction.

t₉₉ = (2.303/k) log(100/1) = (2.303/k)(2); t₉₀ = (2.303/k) log(100/10) = (2.303/k)(1).
t_99% = 2 × t_90%

Q6.

Express t_50%, t_75%, t_87.5%, t_93.75%, t_96.875% and t_99.9% in terms of t_1/2.

Successive halvings: 50 → 75 → 87.5 → 93.75 → 96.875 → 98.44 → 99.22 → 99.61 → 99.80 → 99.90.
1 t_1/2, 2 t_1/2, 3 t_1/2, 4 t_1/2, 5 t_1/2 and ≈ 10 t_1/2 respectively.

Q7.

A first order reaction goes 100 → 60 in 10 s. What remains after another 10 s?

Trick method: the same 40% is consumed in each equal interval. 60 × 40/100 = 24 consumed; 60 − 24 = 36.
Formula method: (2.303/10) log(100/60) = (2.303/10) log(60/x) ⇒ x = 60 × 60/100 = 36.

Q8.

A zero order reaction goes 100 → 60 in 10 s. What remains after another 10 s?

k = (100 − 60)/10 = 4 units s⁻¹. Next interval: 60 − C_t = 4 × 10 ⇒ C_t = 20.
Contrast with Q7 — the same wording gives 36 for first order, 20 for zero order.

Q9.

Zero order: 100 → 50 takes 10 s. How long for 50 → 25?

t_1/2 = C₀/2k ∝ C₀. Halving the starting concentration halves the half-life: 5 s.

Q10.

First order: 100 → 50 takes 10 s. How long for 50 → 25?

t_1/2 is independent of concentration: 10 s.

Q11.

A → 3B, k = 0.0693 s⁻¹. Calculate the % of reactant consumed after 40 s.

t_1/2 = 0.693/0.0693 = 10 s; n = 40/10 = 4.
C_t = 100(1/2)⁴ = 100/16 = 6.25 ⇒ consumed x = 100 − 6.25 = 93.75%.
Ladder check: 100 → 50 → 25 → 12.5 → 6.25.

Q12.

A → 3B, k = 0.0693 s⁻¹, [A]₀ = 0.1 M. Calculate the concentration of B after 30 s.

t_1/2 = 10 s; n = 30/10 = 3. C_t = 0.1(1/2)³ = 0.0125 M.
x = 0.1 − 0.0125 = 0.0875 M of A reacted. Since 1 A gives 3 B:
[B] = 3x = 3 × 0.0875 = 0.2625 M.

Q13.

Write t_1/2 for (a) 2A → B and (b) 3A → B, given rate = k[A].

(a) t_1/2 = 0.693/2k (b) t_1/2 = 0.693/3k — the stoichiometric coefficient multiplies k.

Q14.

A → B has t_1/2 = 10 s at [A]₀ = 0.1 M and t_1/2 = 10 s at [A]₀ = 0.2 M. Find the order.

10/10 = (0.2/0.1)ⁿ⁻¹ ⇒ 1 = 2ⁿ⁻¹ ⇒ 2⁰ = 1 ⇒ n − 1 = 0 ⇒ n = 1.

Q15.

A + B → C. From the table (0.1, 0.1, 10⁻³), (0.2, 0.1, 2×10⁻³), (0.1, 0.2, 4×10⁻³), find x, y and k.

Exp3/Exp1 ⇒ 4 = 2^y ⇒ y = 2. Exp2/Exp1 ⇒ 2 = 2^x ⇒ x = 1.
k = 10⁻³/[(0.1)(0.01)] = k = 1 L²mol⁻²s⁻¹; overall order 3.

Q16.

A → B with [A] = 0.1 M, 0.06 M, 0.036 M at t = 0, 10, 20 s. Find the order.

Zero order test: k₁ = 4×10⁻³, k₂ = 2.4×10⁻³ ⇒ unequal, rejected.
First order test: k₁ = (2.303/10)log(0.1/0.06); k₂ = (2.303/10)log(0.06/0.036) — both equal (2.303/10)log(10/6).
First order. Notice the concentration falls by a constant ratio (0.6×) each interval — the signature of first order.

Q17.

A + B + C → D, Rate = k[A]^x[B]^y. Given [A]₀ = 0.001 M and [B] = 0.1 M, with [A] = 0.001, 8×10⁻⁴, 6.4×10⁻⁴ at t = 0, 10, 20 s. Find the order.

No data for B and its concentration is much larger ⇒ B is in excess ⇒ order w.r.t. B = 0 (Ostwald).
Zero order test on A: k₁ = 2×10⁻⁵, k₂ = 1.6×10⁻⁵ ⇒ unequal, rejected.
First order test: k₁ = (2.303/10)log(10⁻³/8×10⁻⁴) = (2.303/10)log(5/4); k₂ = (2.303/10)log(8×10⁻⁴/6.4×10⁻⁴) = (2.303/10)log(5/4). Equal.
First order overall (order 1 in A, 0 in B and C).

Q18.

Experiment: [A] = 0.1 M (fixed, in excess); [B] = 0.001 → 0.002 M gives R = 10⁻⁵ → 4×10⁻⁵. Find the order.

[A] in excess ⇒ order w.r.t. A = 0. Then 4×10⁻⁵/10⁻⁵ = (0.002/0.001)^y ⇒ 4 = 2^y ⇒ y = 2, second order overall.

Q19.

Find n for (a) A(g) → 4B(g) and (b) 2A(g) → 3B(g) + C(g).

(a) n = 4/1 = 4 (b) n = (3+1)/2 = 2

Q20.

A(g) → 2B(g) + C(g), first order, only A taken initially. Total pressure after 20 s is 250 torr and after very long time is 300 torr. Calculate the half-life.

n = 3, so P_∞ = 3P₀ ⇒ 300 = 3P₀ ⇒ P₀ = 100 torr.
Method 1: k = (2.303/20) log[(300−100)/(300−250)] = (2.303/20) log 4 = (2.303/20)(0.602) = 0.0693 s⁻¹.
Method 2: P_t = P₀ + 2x ⇒ 250 = 100 + 2x ⇒ x = 75. k = (2.303/20) log(100/25) = (2.303/20)(2 log 2) = 0.0693 s⁻¹.
t_1/2 = 0.693/0.0693 = 10 s

Q21.

A(g) → 2B(g) + C(g), first order. Initially only A at 100 torr; after 20 s total pressure is 250 torr. Find the half-life.

n = 3. k = (2.303/20) log[(3×100 − 100)/(3×100 − 250)] = (2.303/20) log 4 = (2.303/20)×2×0.301 = 0.0693.
t_1/2 = 10 s

Q22.

A(g) → 2B(g) + C(g), first order. P_A + P_B = 175 torr at t = 20 s and 200 torr at t = ∞. Find t_1/2.

Here only partial pressures are given, so the P_∞ formula cannot be used — method 2 is compulsory.
At t: P_A + P_B = (P₀ − x) + 2x = P₀ + x = 175.
At ∞: P_A + P_B = 0 + 2P₀ = 200 ⇒ P₀ = 100 ⇒ x = 75.
k = (2.303/20) log(100/25) = 0.0693 ⇒ t_1/2 = 10 s

Q23.

A → B, ΔH = −30 kJ/mol, E_a(forward) = 10 kJ/mol. Calculate E_a for the backward reaction.

ΔH = E_a(f) − E_a(b) ⇒ −30 = 10 − E_a(b) ⇒ E_a(b) = 40 kJ/mol. (Exothermic, so E_a(b) > E_a(f) — consistent.)

Q24.

Calculate the fraction of molecules having energy more than the threshold energy if E_a = 0.693 kcal/mol at 227°C (R = 2 cal mol⁻¹K⁻¹).

T = 227 + 273 = 500 K; E_a = 693 cal/mol.
Fraction = e^−Ea/RT = e^{−693/(2×500)} = e^−0.693.
Since ln 2 = 0.693, e^−0.693 = 1/2 = 0.5, i.e. 50% of molecules.

Q25.

Temperature is raised from 227°C to 727°C and the rate becomes 10 times. Calculate E_a (R = 2 cal mol⁻¹K⁻¹).

T₁ = 500 K, T₂ = 1000 K, k₂/k₁ = 10.
log 10 = [E_a/(2.303×2)] × [(1000−500)/(1000×500)] ⇒ 1 = E_a/(4.606) × (1/1000)
E_a = 4.606 × 1000 = 4606 cal/mol = 4.606 kcal/mol

Q26.

The temperature coefficient of a reaction is 2. By what factor does the rate increase when the temperature is raised from 30°C to 50°C?

(T.C.)^Δt/10 = 2^(50−30)/10 = 2² = 4 times

Q27.

Two reactions have temperature coefficients 2 and 3 and equal rates at 25°C. Find the ratio of their rates at 65°C.

Δt = 40°C ⇒ exponent = 4.
Reaction 1: 2⁴ = 16. Reaction 2: 3⁴ = 81.
Ratio = 16 : 81

Q28.

A + 2B → C occurs as (i) A + B → D (slow), (ii) D + B → C (fast). Write the rate law and overall order.

Rate is set by the RDS: r = k[A][B]; order w.r.t. A = 1, w.r.t. B = 1, overall = 2. Note it is not k[A][B]² — stoichiometry must not be used for a complex reaction.

Q29.

A + 2B → C occurs as (i) A ⇌ D (fast), (ii) D + B → C (slow). Write the rate law.

r = k₃[D][B]. D is an intermediate: K_c = [D]/[A] ⇒ [D] = K_c[A].
r = k′[A][B], overall order 2

Q30.

For 2O₃ → 3O₂ with (i) O₃ ⇌ O₂ + O (fast) and (ii) O₃ + O → 2O₂ (slow), find the rate law and overall order.

r = k₃[O₃][O]; [O] = K_c[O₃]/[O₂].
r = k′[O₃]²[O₂]⁻¹; order in O₃ = 2, in O₂ = −1, overall order = 1

17. Additional High-Yield NEET / JEE Main Questions

Q31.

The rate constant of a reaction is 3 × 10⁻⁴ L mol⁻¹ s⁻¹. What is the order?

Units mol¹⁻ⁿLⁿ⁻¹s⁻¹ with L¹ ⇒ n − 1 = 1 ⇒ second order.

Q32.

A first order reaction is 75% complete in 60 minutes. Find t_1/2 and k.

75% complete = 2 half-lives ⇒ t_1/2 = 30 min; k = 0.693/30 = 0.0231 min⁻¹.

Q33.

A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2. (log 7 = 0.845)

k = (2.303/40) log(100/70) = (2.303/40)(2 − 1.845) = (2.303/40)(0.155) = 8.92 × 10⁻³ min⁻¹.
t_1/2 = 0.693/8.92×10⁻³ = ≈ 77.7 min

Q34.

For a zero order reaction, k = 2 × 10⁻² mol L⁻¹s⁻¹ and [A]₀ = 0.1 M. How long until the reaction is complete?

t = C₀/k = 0.1/(2×10⁻²) = 5 s (and t_1/2 = 2.5 s, i.e. half of it).

Q35.

The half-life of a second order reaction is 50 s at [A]₀ = 0.1 M. What is t_1/2 at [A]₀ = 0.4 M?

Second order: t_1/2 ∝ 1/C₀. Concentration ×4 ⇒ half-life ÷4 = 12.5 s.

Q36.

For a reaction, doubling [A] quadruples the rate and doubling [B] leaves the rate unchanged. Write the rate law and the overall order.

4 = 2^x ⇒ x = 2; 1 = 2^y ⇒ y = 0.
r = k[A]²[B]⁰ = k[A]²; overall order = 2

Q37.

The rate of a reaction becomes 8 times when temperature rises from 20°C to 50°C. Find the temperature coefficient.

8 = μ^30/10 = μ³ ⇒ μ = 2

Q38.

For a reaction, E_a = 0 and A = 3.6 × 10⁶ s⁻¹. What is k at 300 K?

k = Ae^−0/RT = A e⁰ = 3.6 × 10⁶ s⁻¹ — with zero activation energy every collision is effective and k is temperature independent.

Q39.

A plot of log k vs 1/T for a reaction gives slope = −5 × 10³ K. Calculate E_a (R = 8.314 J K⁻¹mol⁻¹).

Slope = −E_a/2.303R ⇒ E_a = 5×10³ × 2.303 × 8.314 = ≈ 95.7 kJ mol⁻¹

Q40.

The rate constants at 27°C and 37°C are k and 2k. Calculate E_a. (R = 8.314, log 2 = 0.3010)

log 2 = [E_a/(2.303×8.314)] × [10/(300×310)]
0.3010 = E_a × 10 / (19.147 × 93000) ⇒ E_a = ≈ 53.6 kJ mol⁻¹

Q41.

Which of the following is a pseudo first order reaction: (a) 2HI → H₂ + I₂, (b) acid hydrolysis of ethyl acetate, (c) 2N₂O₅ → 4NO₂ + O₂, (d) N₂ + 3H₂ → 2NH₃?

(b) — water is in large excess, so a truly bimolecular reaction behaves as first order. (c) is a genuine first order reaction, not pseudo.

Q42.

For the reaction 2N₂O₅ → 4NO₂ + O₂, the rate of formation of NO₂ is 2.8 × 10⁻³ M s⁻¹. Find the rate of the reaction and the rate of disappearance of N₂O₅.

Rate of reaction = (1/4)(2.8×10⁻³) = 7 × 10⁻⁴ M s⁻¹.
−d[N₂O₅]/dt = 2 × 7×10⁻⁴ = 1.4 × 10⁻³ M s⁻¹

Q43.

What is the effect of a catalyst on (i) ΔH, (ii) K_c, (iii) E_a(f), (iv) E_a(b), (v) the position of equilibrium?

(i) unchanged, (ii) unchanged, (iii) decreases, (iv) decreases by the same amount, (v) unchanged — only the time to reach equilibrium is reduced.

Q44.

The half-life of a radioactive isotope is 20 min. What fraction remains after 1 hour?

n = 60/20 = 3 ⇒ (1/2)³ = 1/8 = 12.5% (all radioactive decay is first order).

Q45.

The rate law for a reaction is r = k[A]^1/2[B]^3/2. Give the order and the units of k.

Order = 1/2 + 3/2 = 2; units = mol⁻¹L s⁻¹, i.e. L mol⁻¹s⁻¹.

Q46.

A reaction is 50% complete in 20 min and 75% complete in 40 min. What is its order?

t_75% = 2 × t_50% is the signature of first order. (Zero order would give t_75% = 1.5 t_1/2 = 30 min.)

Q47.

For a first order reaction, the time required for 99.9% completion is how many times that for half completion?

t_99.9% = 6.909/k, t_1/2 = 0.693/k ⇒ ratio ≈ 10.

Q48.

Identify order from graph: a plot of [A] against time is a straight line with negative slope.

Zero order; the magnitude of the slope equals k.

Q49.

For A + B → products, the rate becomes 2.25 times when both [A] and [B] are increased 1.5 times. Find the overall order.

2.25 = (1.5)ⁿ ⇒ (1.5)² = 2.25 ⇒ n = 2.

Q50.

Why is molecularity greater than 3 practically not observed?

The probability of four or more molecules colliding simultaneously with the correct orientation and sufficient energy is negligibly small.

Q51.

For 2A + B → C, if −d[A]/dt = 0.06 M s⁻¹, what is −d[B]/dt?

Rate of reaction = 0.06/2 = 0.03 ⇒ −d[B]/dt = 1 × 0.03 = 0.03 M s⁻¹.

Q52.

The activation energies of the forward and backward steps of a reaction are 60 and 45 kJ/mol. Is it exo- or endothermic, and what is ΔH?

ΔH = 60 − 45 = +15 kJ/mol ⇒ endothermic.

Q53.

Which quantity remains unchanged during the course of a first order reaction: rate, rate constant, concentration, or half-life? (Choose all that apply.)

Rate constant and half-life. Rate and concentration both fall with time.

Q54.

For a first order reaction, a plot of ln[A] vs t gives slope −0.0347 s⁻¹. Find t_1/2.

Slope = −k ⇒ k = 0.0347 s⁻¹; t_1/2 = 0.693/0.0347 = ≈ 20 s.

Q55.

Explain why the rate of a reaction increases sharply with temperature even though the average kinetic energy increases only slightly.

Because the number of molecules in the high-energy tail beyond the threshold energy (the shaded area of the Maxwell–Boltzmann curve) roughly doubles for a 10°C rise, even though the peak shifts only a little.

Q56.

For the mechanism (i) 2NO ⇌ N₂O₂ (fast), (ii) N₂O₂ + O₂ → 2NO₂ (slow), derive the rate law for 2NO + O₂ → 2NO₂.

r = k₂[N₂O₂][O₂]; K_c = [N₂O₂]/[NO]² ⇒ [N₂O₂] = K_c[NO]².
r = k′[NO]²[O₂], overall order = 3

Q57.

80% of a first order reaction is completed in 60 min. What percentage remains after 120 min?

After 60 min, 20% remains (fraction 0.2). After a second identical interval the same fraction applies: 0.2 × 0.2 = 0.04 ⇒ 4% remains, 96% completed.
This "multiply the surviving fraction" trick avoids logarithms entirely.

Q58.

A gaseous reaction A(g) → 2B(g) has its total pressure doubled at completion. Is this consistent with n = 2?

P_∞ = nP₀ = 2P₀ ⇒ yes, consistent. For A → 2B + C, P_∞ would be 3P₀.

Q59.

Arrange in increasing order of half-life dependence on initial concentration: zero, first, second order.

Second (t_1/2 ∝ 1/C₀) < First (independent) < Zero (t_1/2 ∝ C₀) — i.e. only zero order increases with concentration; only second order decreases.

Q60.

The rate constant of a first order reaction is 1.15 × 10⁻³ s⁻¹. How long will 5 g of the reactant take to reduce to 3 g? (log 1.667 = 0.2218)

t = (2.303/k) log(5/3) = (2.303/1.15×10⁻³) × 0.2218 = ≈ 444 s.
Note masses can be used directly in place of concentrations — the ratio is what matters.

18. Assertion–Reason & Match the Following

Assertion–Reason key: (a) Both A and R true, R is the correct explanation of A. (b) Both true, R is not the correct explanation. (c) A true, R false. (d) A false, R true.

AR1.

Assertion: The half-life of a first order reaction is independent of the initial concentration.
Reason: For a first order reaction t_1/2 = 0.693/k, and k does not depend on concentration.

Answer: (a)

AR2.

Assertion: Order of a reaction can be fractional.
Reason: Order is determined experimentally and need not match the stoichiometric coefficients.

Answer: (a)

AR3.

Assertion: A catalyst increases the equilibrium yield of the product.
Reason: A catalyst lowers the activation energy of the forward reaction.

Answer: (d) — the assertion is false; a catalyst lowers E_a of both directions equally, so K_c and hence yield are unchanged.

AR4.

Assertion: Molecularity of a complex reaction is not defined.
Reason: A complex reaction proceeds through several elementary steps, each with its own molecularity.

Answer: (a)

AR5.

Assertion: The rate of a zero order reaction falls as the reaction proceeds.
Reason: The concentration of the reactant decreases with time.

Answer: (d) — the assertion is false; for a zero order reaction the rate stays constant at k even though the concentration falls.

AR6.

Assertion: Inversion of cane sugar is a pseudo first order reaction.
Reason: Water, one of the reactants, is present in large excess so its concentration is effectively constant.

Answer: (a)

Match the Following — Set 1

	Column I		Column II
A	t_1/2 = C₀/2k	p	First order
B	k = (2.303/t) log(a/(a−x))	q	Zero order
C	Units L mol⁻¹s⁻¹	r	Second order
D	Rate independent of concentration	s	t_1/2 ∝ 1/C₀

Answer: A–q, B–p, C–r & s, D–q

Match the Following — Set 2

	Column I (Plot)		Column II (Meaning of slope)
A	ln k vs 1/T	p	−k
B	ln[A] vs t	q	−E_a/R
C	1/[A] vs t	r	+k
D	log k vs 1/T	s	−E_a/2.303R

Answer: A–q, B–p, C–r, D–s

Match the Following — Set 3

	Column I (Reaction)		Column II (Feature)
A	CH₃COOC₂H₅ + H₂O (acid)	p	Molecularity 3
B	2SO₂ + O₂ → 2SO₃	q	Pseudo first order
C	N₂O₄ → 2NO₂	r	Overall order 1 with a negative order term
D	2O₃ → 3O₂	s	Unimolecular

Answer: A–q, B–p, C–s, D–r

19. Corrections to the Original Notes

Three arithmetic slips appear in the handwritten notes. Worth marking in the notebook so the wrong version is not memorised.

Where	As written	Correct
Pressure numerical, A → 2B + C, P_∞ = 300, P_t = 250 (Method 1)	log 4 was taken as 0.3, giving t_1/2 = 20 s	log 4 = 0.602, so k = 0.0693 and t_1/2 = 10 s — matching Method 2 on the same page.
A → 3B, [A]₀ = 0.1 M, [B] after 30 s	One page reads C_t = 0.025, x = 0.075, [B] = 0.225	n = 3 ⇒ C_t = 0.1(1/2)³ = 0.0125, x = 0.0875, [B] = 0.2625 M.
Successive-halving ladder to 99.9%	Some intermediate sums drift (99.609 → 99.804 → ...)	Clean ladder: 50, 75, 87.5, 93.75, 96.875, 98.4375, 99.219, 99.609, 99.805, 99.902 at the 10th half-life.

One more clarification The note "first order reactants can never be consumed" is written next to the C_t = 0 result at t = ∞. Both statements are correct together: C_t → 0 only as t → ∞, so the reactant is never fully consumed in any finite time.

Final revision checklist

Can you write the first order integrated equation and derive t_1/2 = 0.693/k in under two minutes?
Do you know the units of k for orders 0, 1, 2, 3 and the reverse trick?
Can you produce t_75%, t_90%, t_99%, t_99.9% in terms of t_1/2 without deriving them?
Given a data table, can you name which of the four order-determination methods to use in one glance?
Can you set up the gaseous pressure table (t = 0, t, ∞) unaided?
Can you state the four things a catalyst does not change?
Can you eliminate an intermediate from an RDS using the fast-equilibrium K_c?
Do you know the 2O₃ → 3O₂ negative-order result by heart?

CHEMICAL KINETICS

Complete Formula, Shortcut & Question Compendium

NEET · JEE Main · Board

Contents

1. Rate of Reaction — Basics & Stoichiometric Relations

1.1 Types of reactions by speed

1.2 Average vs Instantaneous rate

1.3 Relating rates of different species

2. Rate Law, Order & Molecularity

2.1 Order vs Molecularity

Molecularity examples

2.2 Units of rate constant — the universal formula

3. Zero Order Reactions

4. First Order Reactions — The Core of the Chapter

4.1 Full derivation (write this in board exams)

4.2 Limiting cases

4.3 Half-life

4.4 The n-half-lives shortcut (single most useful trick)

4.5 Half-life when stoichiometric coefficient ≠ 1

5. Half-Life & Fractional-Life Master Tricks

5.1 First order fractional life (memorise this table)

5.2 Order from half-life dependence

6. Second & nth Order — Comparison Table

7. Methods to Determine Order

Method 1 — Half-life method

Method 2 — Initial rate method

Method 3 — Integrated rate law / graphical / hit & trial

Method 4 — Ostwald's isolation method

8. Pressure-Based Numericals (Gaseous First Order)

9. Pseudo First Order Reactions

10. Collision / Transition State Theory & Activation Energy

10.1 Four postulates

10.2 Key definitions

10.3 Energy profile & the ΔH relation

10.4 Maxwell–Boltzmann distribution (the two-curve graph)

11. Arrhenius Equation & Temperature Coefficient

11.1 Two-temperature form (most used numerically)

11.2 Effects at a glance

11.3 Temperature coefficient (T.C. or μ)

12. Effect of Catalyst

What a catalyst does NOT change

13. Complex Reactions & Rate Determining Step

13.1 Two standard cases

Case A — RDS given, no intermediate on its reactant side

Case B — RDS contains an intermediate ⇒ eliminate it using the fast equilibrium

Classic example — decomposition of ozone

14. Graph Bank — Identify the Order at a Glance

15. MASTER TRICK SHEET

16. Solved Questions from the Notes

17. Additional High-Yield NEET / JEE Main Questions

18. Assertion–Reason & Match the Following

Match the Following — Set 1

Match the Following — Set 2

Match the Following — Set 3

19. Corrections to the Original Notes

Final revision checklist