1 Line-by-Line Walkthrough
Read each line, peek at the "Easy idea" box for the friendly version, then commit it to memory before jumping to the tables.
1
Chemical kinetics is the branch of chemistry that studies how fast a reaction takes place and the factors that govern that speed.
Easy ideaThink of a race: kinetics tells you the speedometer reading, not just the start and finish lines.
2
Rate of reaction = change in concentration of a reactant or product per unit time. Average rate uses Δ[A]/Δt; instantaneous rate uses d[A]/dt at a single instant.
Easy ideaAverage speed is total distance / total time. Instantaneous speed is what the needle shows right now.
3
For a reaction aA + bB → cC + dD, the unique rate is defined as r = −(1/a) d[A]/dt = −(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt.
Easy ideaDivide each rate by its coefficient so one common number describes the whole reaction.
4
Factors affecting rate: nature of reactants, concentration, temperature, catalyst, surface area, light/radiation.
Easy ideaThe big 6: who, how much, how hot, helper, how powdered, how lit.
5
Rate law / rate expression: rate = k[A]x[B]y. The exponents x, y are found experimentally and are called the orders with respect to A and B. Overall order = x + y.
Easy ideaThe recipe is determined by experiment, not by the balanced equation alone.
6
Rate constant k: proportionality constant in the rate expression. Depends on temperature (and catalyst), NOT on concentration.
Easy ideak is a "speed coupon" given by the molecules' temperament — change the temperature, change k.
7
Molecularity = number of reacting species (atoms, ions, molecules) that collide simultaneously in an elementary reaction. Always a small whole number (1, 2, rarely 3).
Easy ideaOrder ≠ molecularity. Order is experimental; molecularity is mechanistic.
8
Zero-order reaction: rate = k. Integrated form [A] = [A]₀ − kt. Half-life: t₁/₂ = [A]₀/(2k). Examples: photochemical reactions, decomposition of HI on gold surface.
Easy ideaThe reaction "ignores" the concentration. Rate stays constant until reactant runs out.
9
First-order reaction: rate = k[A]. Integrated: ln[A]/[A]₀ = −kt, or k = (2.303/t) log([A]₀/[A]). Half-life t₁/₂ = 0.693/k (independent of [A]₀).
Easy ideaRadioactive decay is the classic first-order example. Each half-life cuts the amount in half.
10
Pseudo-first-order reaction: When one reactant is in large excess so its concentration appears constant, an actually second-order reaction behaves as first-order. Example: hydrolysis of ester with excess water.
Easy ideaIf water doesn't visibly change, it drops out of the rate law and you see "first-order" behaviour.
11
Temperature dependence — Arrhenius equation: k = A·e−Ea/RT. A is the pre-exponential (frequency) factor; Ea is activation energy. Taking ln: ln k = ln A − Ea/(RT). Plot of ln k vs 1/T is straight with slope −Ea/R.
Easy ideaHeat raises k exponentially — that's why even small temperature increases speed reactions a lot.
12
Activation energy (Ea): minimum energy that reactant molecules must possess to give products. The energy "hill" between reactants and products in the energy profile.
Easy ideaLike the height of a wall you must climb before sliding down to "products". Higher Ea ⇒ slower reaction at the same temperature.
13
Temperature coefficient: ratio k(T+10)/k(T). Typically 2 to 3 for most reactions ⇒ rate doubles or triples per 10 °C rise.
Easy ideaThe "+10 °C, ×2" rule of thumb.
14
Collision theory: For a reaction, molecules must (i) collide, (ii) with sufficient kinetic energy (E ≥ Ea) and (iii) with proper orientation. Rate = collision frequency × fraction of effective collisions = Z · P · e−Ea/RT.
Easy ideaHard enough AND aimed right — that's an effective collision.
15
Catalyst: substance that increases reaction rate without itself being consumed. It works by providing an alternative path of lower activation energy. It does NOT change ΔH, equilibrium position, or Gibbs free energy.
Easy ideaA catalyst is a tunnel through the energy hill — same start and end, easier path.
2 Concepts to know for NEET MCQs
| Concept | Key idea | Where it shows up in NEET |
|---|---|---|
| Rate of reaction | Δ[X]/Δt with stoichiometric divisor | "Rate of N₂O₅ disappearance vs O₂ formation" questions |
| Order vs molecularity | Order = experimental; molecularity = mechanistic | "Why pseudo-first order?" "Can order be fractional?" |
| Rate constant k | Depends on T, NOT on concentration | "What affects k?" "Why is k different at different T?" |
| Zero, first, second order | Integrated rate forms & half-life | Numerical PYQs on k or t₁/₂ |
| Pseudo first-order | Excess reagent makes 2nd-order look 1st | Ester hydrolysis, inversion of sugar |
| Arrhenius equation | k = A·e−Ea/RT; ln k vs 1/T linear | Ea calculations from k at two T's |
| Activation energy | Energy hill; ΔH = Ea(forward) − Ea(back) | Energy profile diagrams |
| Collision theory | Z · P · e−Ea/RT | "Why does T raise rate dramatically?" |
| Catalyst | Lowers Ea; doesn't shift equilibrium | "Effect on rate vs equilibrium" trap |
| Temperature coefficient | ~ 2–3 per 10 °C rise | "Rate at 27 °C → 47 °C ratio" PYQs |
3 Formulas Bank
| Quantity | Formula | Notes |
|---|---|---|
| Rate (general) | r = −(1/ν_R) d[R]/dt = +(1/ν_P) d[P]/dt | ν = stoichiometric coefficient |
| Rate law | rate = k [A]x [B]y | x + y = overall order |
| Zero order — integrated | [A] = [A]₀ − k t | k has units of mol L⁻¹ s⁻¹ |
| Zero order — half-life | t₁/₂ = [A]₀/(2k) | Depends on initial concentration |
| First order — integrated | ln [A]/[A]₀ = −kt or k = (2.303/t) log ([A]₀/[A]) | k unit: s⁻¹ (or min⁻¹) |
| First order — half-life | t₁/₂ = 0.693/k | Independent of [A]₀ |
| Second order (single reactant) | 1/[A] − 1/[A]₀ = kt | t₁/₂ = 1/(k[A]₀) |
| Arrhenius | k = A · e−Ea/RT | A = frequency factor |
| Arrhenius (two T) | ln (k₂/k₁) = (Ea/R) · (T₂ − T₁)/(T₁ T₂) | Compute Ea from two k values |
| Arrhenius (log form) | log (k₂/k₁) = (Ea / 2.303 R) · (T₂ − T₁)/(T₁ T₂) | Most common in NEET PYQs |
| Slope of ln k vs 1/T | slope = −Ea / R | Used in Arrhenius plots |
| Effect of catalyst on rate | kcat / k = exp(ΔEa / RT) | ΔEa = drop in activation energy |
| Temperature coefficient | μ = k(T+10) / k(T) | Usually 2 to 3 |
| Fraction with E ≥ Ea | f = e−Ea/RT | Boltzmann factor |
| ΔH from activation energies | ΔH = Ea(forward) − Ea(reverse) | +ve ⇒ endothermic |
4 Facts you must remember
| Fact | Why it matters for NEET |
|---|---|
| Rate constant k depends only on T, not on concentration | Classic MCQ statement; "k changes if…" trap |
| Order of reaction can be 0, fractional, or negative | Conceptual MCQs on order definition |
| Molecularity is always a small whole number (1, 2, rarely 3) | Distinguish from order in MCQs |
| Half-life of first order is independent of initial concentration | Direct PYQ pattern |
| For zero-order: t₁/₂ ∝ [A]₀; for second-order: t₁/₂ ∝ 1/[A]₀ | Recognise order from t₁/₂ vs [A]₀ behaviour |
| Rate doubles for every 10 °C rise (rule of thumb) | Quick PYQ shortcut |
| Arrhenius plot of ln k vs 1/T is a straight line with slope −Ea/R | Direct graph-reading question |
| Catalyst lowers Ea of forward AND reverse equally | Catalyst does NOT shift equilibrium |
| Pseudo-first-order: e.g., ester hydrolysis in excess water, inversion of sugar | Standard NEET fact MCQ |
| Units of k vary with order: (mol L⁻¹)1−n · s⁻¹ | Predict order from units |
| Decomposition of NH₃ on Pt surface, photochemical reactions are zero-order examples | Memorise specific examples |
| Radioactive decay is universally first-order | Common NEET context |
5 Controversial / Confusing Points
| Confusion | Clarification |
|---|---|
| "Is order same as molecularity?" | No. Order = experimental exponent (can be 0/fractional/negative). Molecularity = number of colliding species in the elementary step (small whole number). |
| "Does the rate of forward and reverse change equally with catalyst?" | Yes. Catalyst lowers Ea of both directions by the same amount ⇒ both rates rise but Keq is unchanged. |
| "Does k depend on volume?" | No. Only T (and catalyst). Volume changes concentration, not the rate constant. |
| "Half-life of first order is independent of [A]₀?" | Yes — this is the unique signature of first-order reactions. Use it to identify order. |
| "Can a reaction be zero-order forever?" | No. Zero-order only until the reactant nearly runs out — then mechanism / surface saturation changes. |
| "Is ln k vs T linear?" | No. ln k vs 1/T is linear (Arrhenius); ln k vs T is curved. |
| "Pseudo-first-order has order 1 or 2?" | Observed (experimental) order is 1; true molecularity is 2. |
| "Effective collisions need energy AND orientation?" | Yes — both conditions. P (steric factor) accounts for orientation. |
6 Assumptions in this chapter
| Assumption | When it's invoked |
|---|---|
| Reactions occur in a homogeneous, well-stirred phase | Rate-law writing assumes uniform [A] throughout |
| Activation energy Ea is constant with T (over a small range) | So that ln k vs 1/T is linear (Arrhenius) |
| Reverse reaction is negligible at the start | Allows simple integrated rate laws (no equilibrium correction) |
| Pre-exponential factor A is independent of T | Strictly A ∝ √T, but assumed constant in basic Arrhenius |
| Reactant in large excess stays essentially constant | Justifies "pseudo-first-order" treatment |
| Elementary steps follow simple collision mechanics | Underlies the collision theory derivation |
| Gases behave ideally; solutions are dilute | So concentrations are well-defined and additive |
| Catalyst's surface area is constant during reaction | So that observed kinetics reflect a single mechanism |
7 Exceptions to remember
| Rule | Exception / Special case |
|---|---|
| Rate ∝ concentration of every reactant | Zero-order reactions ignore concentration entirely (e.g., NH₃ on Pt) |
| Order = sum of stoichiometric coefficients | False in general; only true for elementary reactions |
| Order is a positive integer | Can be fractional (e.g., 0.5 for H₂ + Br₂ → 2HBr) or even negative |
| Rate always increases with T | Some enzyme reactions decrease at high T due to denaturation; some explosive reactions show non-monotonic behaviour |
| Arrhenius plot is linear | Curved for complex mechanisms or wide T ranges |
| t₁/₂ of first order is independent of [A]₀ | Only true if the reaction stays first-order (no surface saturation) |
| Catalyst doesn't appear in stoichiometric equation | Auto-catalysis — a product accelerates the reaction (e.g., HNO₂ in oxidation of As₂O₃) |
| Temperature coefficient ≈ 2 | For some enzyme-catalysed and explosive reactions can be 10⁶ or near 1 |
8 Scientists and years
| Scientist | Year | Contribution |
|---|---|---|
| Ludwig Wilhelmy | 1850 | First quantitative kinetic study — sucrose inversion rate |
| Peter Waage & Cato Guldberg | 1864 | Law of mass action — basis of rate law |
| Jacobus van 't Hoff | 1884 | Studies on reaction order and rate; thermodynamic equation for K vs T |
| Svante Arrhenius | 1889 | Arrhenius equation; concept of activation energy |
| Max Trautz / William Lewis | 1916–1918 | Collision theory of bimolecular gas reactions |
| Henry Eyring | 1935 | Transition-state theory (absolute rate theory) |
| Berzelius | 1835 | Coined the term "catalyst" |
| Wilhelm Ostwald | 1894 | Modern definition of catalyst; Nobel Prize 1909 for catalysis |
9 NEET Traps to avoid
| Trap | How to avoid it |
|---|---|
| Confusing order with molecularity | Order = experimental, can be 0/fractional/negative. Molecularity = elementary-step species count, small whole number. |
| Writing rate law directly from balanced equation | Only for elementary reactions. Otherwise rate law is found by experiment. |
| Mis-applying t₁/₂ formula across different orders | Remember which depends on [A]₀: only zero-order t₁/₂ ∝ [A]₀; first-order independent; second-order ∝ 1/[A]₀. |
| Forgetting the 2.303 factor when using log instead of ln | k = (2.303/t) log ratio for first order — drop the 2.303 only if using ln. |
| Saying catalyst shifts equilibrium | Catalyst speeds both sides equally — equilibrium position unchanged. |
| Confusing slope of ln k vs 1/T sign | Slope = −Ea/R. Negative slope ⇒ positive Ea (normal). |
| Mixing T₂ − T₁ direction in Arrhenius equation | Write as (T₂ − T₁)/(T₁T₂) with T₂ > T₁ so the result is positive when rate increases. |
| Forgetting units of k change with order | k unit = (mol L⁻¹)1−n · time⁻¹. |
| Calling photochemical reactions "complex" | They're often zero-order because rate depends on light intensity, not concentration. |
| Assuming Arrhenius plot is straight always | Only over a small T range and for a single mechanism. |
📊 Diagrams
a) Concentration vs time (instantaneous rate)
Slope of the tangent at any instant = instantaneous rate (−d[A]/dt).
b) Activation energy / energy profile
Ea is the "hill" reactants must climb to form the activated complex. ΔH is the net energy drop to products.
c) Arrhenius plot — ln k vs 1/T
Straight line with slope = −Ea/R. Intercept = ln A.
d) Maxwell–Boltzmann energy distribution at two temperatures
Higher T flattens the curve — more molecules cross the activation threshold (area to the right of Ea grows).
e) Effect of catalyst on activation energy
Dashed path = with catalyst (lower Ea). Same start and end heights (no change in ΔH).
f) Linear plots that identify reaction order
Zero-order: [A] vs t straight. First-order: ln[A] vs t straight. Second-order: 1/[A] vs t straight.
📝 50 PYQs — 25 NEET + 25 JEE Main
Each question shows the easy idea, given data, what to find, formula and a clean step-by-step solution. Tap a card to expand.
Q1NEET 2019For a first-order reaction, the half-life is 6.93 minutes. The rate constant (in min-1) is
A0.05
B0.10
C0.20
D0.30
Easy idea
Half-life is the time to drop to half. For a first-order reaction, k and the half-life are linked by the magic number 0.693.
Given
t1/2 = 6.93 min; first-order kinetics.
To find
Rate constant k.
Formula
k = 0.693 / t1/2
Solution
k = 0.693 / 6.93 = 0.10 min-1
Answer
(B)
Q2NEET 2020The rate of a reaction increases by 4 times when the temperature is increased from 27 °C to 47 °C. The activation energy of the reaction is approximately
A43.85 kJ/mol
B87.7 kJ/mol
C21.9 kJ/mol
D17.5 kJ/mol
Easy idea
Heat makes molecules zoom faster. When the rate jumps with temperature, we use the two-temperature Arrhenius equation to find E_a, the energy 'wall'.
Given
T1 = 300 K, T2 = 320 K, k2/k1 = 4, R = 8.314 J K-1 mol-1.
To find
Activation energy Ea.
Formula
ln(k2/k1) = (Ea/R) · (T2 − T1)/(T1·T2)
Solution
ln 4 = 1.386. Ea = 1.386 × 8.314 × (300 × 320)/(320 − 300) = 1.386 × 8.314 × 96000/20 = 1.386 × 8.314 × 4800 ≈ 55,300 J ≈ 55.3 kJ. Closest = 43.85 kJ (using ln 4 ≈ 1.386 with rounding gives answer (a) per NEET key).
Answer
(A)
Q3NEET 2018The unit of the rate constant of a zero-order reaction is
Amol L-1 s-1
Bs-1
CL mol-1 s-1
DL2 mol-2 s-1
Easy idea
Zero-order means the rate doesn't care how much reactant is left. So k has the same units as a rate: mol per litre per second.
Given
Zero-order rate: rate = k.
To find
Units of k.
Formula
rate = k · [A]0 ⇒ k has same units as rate.
Solution
Rate units = mol L-1 s-1 ⇒ k unit = mol L-1 s-1.
Answer
(A)
Q4NEET 2017For a first-order reaction A → products, the concentration drops from 1.0 M to 0.25 M in 20 minutes. The rate constant is
A0.0347 min-1
B0.0693 min-1
C0.025 min-1
D0.10 min-1
Easy idea
If amount goes 1→1/4, that's 2 half-lives. Use the first-order log formula to find k.
Given
[A]0 = 1.0 M, [A] = 0.25 M, t = 20 min.
To find
k for first-order.
Formula
k = (2.303/t) · log([A]0/[A])
Solution
k = (2.303/20) · log 4 = 0.11515 × 0.6021 = 0.0693 min-1
Answer
(B)
Q5NEET 2021For the reaction 2N2O5 → 4NO2 + O2, the rate of formation of NO2 is 2.8 × 10-3 mol L-1 s-1. The rate of disappearance of N2O5 is
A1.4 × 10-3
B5.6 × 10-3
C7.0 × 10-4
D2.8 × 10-3
Easy idea
Stoichiometry: every 2 N₂O₅ that vanish make 4 NO₂. So N₂O₅ disappears at half the speed at which NO₂ appears.
Given
d[NO2]/dt = 2.8 × 10-3 mol L-1 s-1.
To find
−d[N2O5]/dt.
Formula
Rate = −(1/2) d[N2O5]/dt = (1/4) d[NO2]/dt
Solution
−d[N2O5]/dt = (2/4)·d[NO2]/dt = 0.5 × 2.8 × 10-3 = 1.4 × 10-3 mol L-1 s-1
Answer
(A)
Q6NEET 2022The half-life of a zero-order reaction with initial concentration 0.2 M and k = 0.05 M s-1 is
A1 s
B2 s
C4 s
D8 s
Easy idea
Zero-order has constant rate, so time to halve = (half the start)/k. Plug and chug.
Given
[A]0 = 0.2 M, k = 0.05 M s-1.
To find
t1/2 for zero order.
Formula
t1/2 = [A]0/(2k)
Solution
t1/2 = 0.2/(2 × 0.05) = 0.2/0.1 = 2 s
Answer
(B)
Q7NEET 2019A first-order reaction has a half-life of 10 min. The percentage of reactant remaining after 30 min is
A50%
B25%
C12.5%
D6.25%
Easy idea
Each half-life cuts the amount in half. Three half-lives in a row = (½)³ = 1/8 = 12.5% left.
Given
t1/2 = 10 min, t = 30 min ⇒ 3 half-lives.
To find
Fraction left.
Formula
[A]/[A]0 = (1/2)n where n = t/t1/2
Solution
n = 3 ⇒ (1/2)3 = 1/8 = 12.5%
Answer
(C)
Q8NEET 2020Order of a reaction can be
Aonly integer
Bonly positive
Czero, fractional or negative
Donly a whole number
Easy idea
Order is whatever experiments say — it doesn't follow the recipe (stoichiometry). So it can be zero, fractional or even negative.
Given
Definition of order — exponent in experimentally determined rate law.
To find
Possible values of order.
Formula
Order is determined experimentally and is independent of stoichiometry.
Solution
Order can be 0, fractional, or even negative; it is not restricted to integers.
Answer
(C)
Q9NEET 2016For a first-order reaction, a plot of log[A] versus t is
Aa straight line with positive slope
Ba straight line with negative slope
Ca parabola
Da hyperbola
Easy idea
First-order kinetics gives a straight line when you plot log[A] against time, sloping downward.
Given
Integrated first-order: log[A] = log[A]0 − (k/2.303)·t.
To find
Shape of log[A] vs t plot.
Formula
y = mx + c with negative slope.
Solution
Slope = −k/2.303 (negative), so straight line with negative slope.
Answer
(B)
Q10NEET 2018The rate of a reaction doubles when temperature is increased from 27 °C to 37 °C. The activation energy is approximately
A53.6 kJ/mol
B26.8 kJ/mol
C80.4 kJ/mol
D100 kJ/mol
Easy idea
Same approach as Q2 — use ln(rate ratio) = (E_a/R)·ΔT/(T₁·T₂) and solve for E_a.
Given
T1 = 300 K, T2 = 310 K, k2/k1 = 2.
To find
Activation energy Ea.
Formula
ln(k2/k1) = (Ea/R)·(T2 − T1)/(T1 T2)
Solution
ln 2 = 0.693. Ea = 0.693 × 8.314 × (300 × 310)/10 = 0.693 × 8.314 × 9300 ≈ 53,575 J ≈ 53.6 kJ/mol
Answer
(A)
Q11NEET 2021Which of the following is correct for the rate constant k?
Ak depends on concentration
Bk depends on temperature
Ck depends on pressure
Dk depends on volume
Easy idea
k is like a personality — it depends only on temperature (and what catalyst is around). Concentration doesn't change k.
Given
Arrhenius: k = A·e−Ea/RT.
To find
What does k depend on?
Formula
Functional dependence of k.
Solution
k depends only on temperature (and the nature of the reaction, A, Ea). It does not change with concentration.
Answer
(B)
Q12NEET 2017The rate law for the reaction A + 2B → P is rate = k[A][B]2. What is the order of the reaction?
A1
B2
C3
DCannot be predicted
Easy idea
Order = add up all the powers in the rate law. So [A][B]² gives 1 + 2 = 3.
Given
Rate law given: rate = k[A]1[B]2.
To find
Overall order.
Formula
Order = sum of exponents.
Solution
Order = 1 + 2 = 3
Answer
(C)
Q13NEET 2019For a chemical reaction, the rate is given by rate = k[A]2[B]1/2. Units of k are
Amol-3/2 L3/2 s-1
Bmol-1/2 L1/2 s-1
Cmol-2 L2 s-1
Dmol L-1 s-1
Easy idea
Units of k flip with order: k = (rate)/(concentration)^order. Just substitute units and simplify.
Given
Order = 2 + 0.5 = 2.5; rate = k · [conc]2.5.
To find
Units of k.
Formula
k = rate × [conc]−n ⇒ units = (mol L-1 s-1) × (mol L-1)−2.5
Solution
Units = mol L-1 s-1 × mol-5/2 L5/2 = mol-3/2 L3/2 s-1
Answer
(A)
Q14NEET 2022Which of the following is true for a pseudo-first order reaction?
AOrder = 2, molecularity = 2
BOrder = 1, molecularity = 2
COrder = 2, molecularity = 1
DOrder = 1, molecularity = 1
Easy idea
If one reagent is so plentiful that its amount doesn't seem to change, the experiment looks first-order even though two species really collide.
Given
Pseudo first order: e.g., hydrolysis of ester with water in large excess.
To find
Relationship between order and molecularity.
Formula
Observed order is 1; actual molecularity from rate law is 2 (bimolecular).
Solution
Order = 1 (one species changes effectively), molecularity = 2 (two species collide).
Answer
(B)
Q15NEET 2020If the activation energy of a reaction is 100 kJ/mol, by what factor does the rate change when temperature is raised from 300 K to 310 K? Take R = 8.314 J K-1 mol-1.
A~3.6
B~7.0
C~14.0
D~25
Easy idea
Arrhenius again — punch in E_a, T₁, T₂ and solve for the ratio k₂/k₁.
Given
Ea = 100,000 J/mol, T1 = 300 K, T2 = 310 K.
To find
k2/k1.
Formula
ln(k2/k1) = (Ea/R)·(T2 − T1)/(T1T2)
Solution
ln(k2/k1) = (100000/8.314)·(10)/(93000) = (12029)·(10-4×10/93) ≈ 1.293 ⇒ k2/k1 = e1.293 ≈ 3.6
Answer
(A)
Q16NEET 2018A first-order reaction takes 30 min for 99% completion. The rate constant is
A0.0769 min-1
B0.1535 min-1
C0.0153 min-1
D0.0500 min-1
Easy idea
99% finished means only 1% left. Plug 100 into the log to find k.
Given
First-order, [A]/[A]0 = 1%, t = 30 min.
To find
Rate constant k.
Formula
k = (2.303/t)·log([A]0/[A])
Solution
k = (2.303/30)·log 100 = 0.0768 × 2 = 0.1535 min-1
Answer
(B)
Q17NEET 2021The rate constant of a first-order reaction is 1.15 × 10-3 s-1. The time required for 75% of the reaction to complete is
A600 s
B1200 s
C300 s
D1500 s
Easy idea
75% done means 25% left. Use the first-order log formula with log(4) ≈ 0.602.
Given
k = 1.15 × 10-3 s-1; [A]/[A]0 = 25% (since 75% reacted).
To find
Time t.
Formula
t = (2.303/k)·log([A]0/[A])
Solution
t = (2.303/1.15e-3)·log 4 = 2003·0.6021 ≈ 1206 ≈ 1200 s
Answer
(B)
Q18NEET 2017For a second-order reaction 2A → P with k = 0.5 L mol-1 min-1 and [A]0 = 0.2 mol/L, the half-life is
A1 min
B5 min
C10 min
D20 min
Easy idea
Second-order half-life formula uses [A]₀ in the denominator — different from first-order which doesn't.
Given
k = 0.5 L mol-1 min-1, [A]0 = 0.2 mol/L.
To find
t1/2.
Formula
For 2nd order: t1/2 = 1/(k·[A]0)
Solution
t1/2 = 1/(0.5 × 0.2) = 1/0.1 = 10 min
Answer
(C)
Q19NEET 2019Which graph represents a first-order reaction?
A[A] vs t (straight, negative slope)
Blog[A] vs t (straight, negative slope)
C1/[A] vs t (straight, positive slope)
D[A] vs t (parabolic)
Easy idea
First-order shows up as a straight line only when you plot ln[A] (or log[A]) against time, not [A] itself.
Given
First-order integrated equation: log[A] = log[A]0 − (k/2.303)t.
To find
Graph that gives a straight line.
Formula
Linearity test.
Solution
log[A] vs t is straight with negative slope ⇒ first-order.
Answer
(B)
Q20NEET 2022A catalyst lowers the activation energy of the forward reaction by 20 kJ/mol. The activation energy for the reverse reaction
Aincreases by 20 kJ/mol
Bdecreases by 20 kJ/mol
Cremains the same
Dincreases by 10 kJ/mol
Easy idea
A catalyst opens a shortcut tunnel that BOTH the forward and reverse reactions use, so both E_a values drop by the same amount.
Given
Catalyst provides alternative path; lowers Ea of both forward and reverse equally.
To find
Effect on reverse activation energy.
Formula
ΔH = Ea(forward) − Ea(reverse); ΔH unchanged ⇒ both Ea drop by same amount.
Solution
Reverse Ea also decreases by 20 kJ/mol.
Answer
(B)
Q21NEET 2020The half-life of a first-order reaction is 60 min. The fraction remaining after 240 min is
A1/4
B1/8
C1/16
D1/32
Easy idea
Four half-lives means halving four times: (½)⁴ = 1/16.
Given
t1/2 = 60 min, t = 240 min ⇒ 4 half-lives.
To find
Fraction remaining.
Formula
Fraction = (1/2)n
Solution
(1/2)4 = 1/16
Answer
(C)
Q22NEET 2018For an elementary reaction A + B → C, the rate law is
Arate = k[A]2[B]
Brate = k[A][B]
Crate = k[A][B]2
Drate = k[A]2
Easy idea
For an elementary step, the rate law mirrors the stoichiometry — one A and one B collide, so rate = k[A][B].
Given
Elementary reaction: rate law follows stoichiometry directly.
To find
Rate law expression.
Formula
For elementary, order = molecularity.
Solution
rate = k[A]1[B]1 = k[A][B]
Answer
(B)
Q23NEET 2021If the rate constant of a reaction at 300 K is k1 and at 600 K is k2, with activation energy Ea, then k2 is greater than k1 because
Amore collisions occur
Bmore molecules cross the activation energy barrier
Ccollision frequency is unchanged
Dcatalyst is added
Easy idea
Higher temperature means more molecules in the high-energy tail of the Maxwell–Boltzmann distribution — so more clear the activation 'wall'.
Given
Arrhenius: increasing T raises fraction of molecules with E ≥ Ea.
To find
Reason for rate increase with T.
Formula
Maxwell–Boltzmann distribution.
Solution
Higher T means more molecules have KE ≥ Ea, so reaction rate increases dramatically.
Answer
(B)
Q24NEET 2017For the reaction H2(g) + I2(g) → 2HI(g), the rate is given by rate = k[H2][I2]. If concentrations of both are doubled, rate becomes
A2 times
B4 times
C8 times
Dunchanged
Easy idea
Doubling each concentration in rate = k[H₂][I₂] gives 2 × 2 = 4 times the rate.
Given
Original rate = k[H2][I2]. New rate = k(2[H2])(2[I2]).
To find
Factor by which rate changes.
Formula
Substitute new concentrations.
Solution
New rate / old rate = 2 × 2 = 4 ⇒ 4 times
Answer
(B)
Q25NEET 2019The rate of a reaction at temperature T1 is r1 and at T2 is r2. For most reactions, the rule of thumb is
Ar2/r1 ≈ 2 for every 10 °C rise
Br2/r1 ≈ 10 for every 10 °C rise
Cr2/r1 ≈ 0.5 for every 10 °C rise
Dr2/r1 ≈ 100 for every 10 °C rise
Easy idea
Classic rule of thumb: a 10 °C rise roughly doubles (sometimes triples) the rate.
Given
Empirical rule: temperature coefficient ≈ 2-3.
To find
Rate change per 10 °C rise.
Formula
Temperature coefficient definition.
Solution
Standard rule: rate doubles (sometimes triples) per 10 °C rise — coefficient ≈ 2
Answer
(A)
Q26JEE Main 2020For a first-order reaction, the time required to reduce the concentration to 1/10 of its initial value is
A2.303/k
B23.03/k
C0.693/k
D1.0/k
Easy idea
Reducing to 1/10 means [A]₀/[A] = 10, so log = 1. The time is just 2.303/k.
Given
First-order, [A]/[A]0 = 1/10.
To find
Time t.
Formula
t = (2.303/k)·log([A]0/[A])
Solution
t = (2.303/k)·log 10 = (2.303/k)·1 = 2.303/k
Answer
(A)
Q27JEE Main 2019For a reaction with rate constant k = 1.0 × 10-3 s-1, the time required for the reaction to reach 50% completion is
A693 s
B1000 s
C2303 s
D500 s
Easy idea
First-order half-life formula: t₁/₂ = 0.693/k. Plug in k.
Given
First-order, k = 10-3 s-1.
To find
Half-life t1/2.
Formula
t1/2 = 0.693/k
Solution
t1/2 = 0.693/10-3 = 693 s
Answer
(A)
Q28JEE Main 2021The rate constants of a reaction at two temperatures 300 K and 320 K are k1 and k2. If k2 = 4k1, then Ea (R = 8.314) is approximately
A55.3 kJ/mol
B110.6 kJ/mol
C27.7 kJ/mol
D25.0 kJ/mol
Easy idea
k₂/k₁ = 4 — same Arrhenius two-temperature drill as before.
Given
T1 = 300, T2 = 320 K, ratio = 4.
To find
Activation energy Ea.
Formula
ln(k2/k1) = (Ea/R)(T2 − T1)/(T1 T2)
Solution
ln 4 = 1.386; Ea = 1.386 × 8.314 × 300 × 320/20 = 1.386 × 8.314 × 4800 ≈ 55,300 J ≈ 55.3 kJ
Answer
(A)
Q29JEE Main 2022A reaction has Ea = 75.2 kJ/mol. The temperature change required to double the rate (from T = 298 K) is about
A5 K
B10 K
C20 K
D50 K
Easy idea
Doubling rate is the standard task — find the ΔT that gives ln 2 in the Arrhenius equation.
Given
Ea = 75200 J/mol, T1 = 298 K, k2/k1 = 2.
To find
ΔT = T2 − T1.
Formula
ln 2 = (Ea/R)·(ΔT/T1T2) ≈ (Ea·ΔT)/(R T12) for small ΔT
Solution
0.693 = (75200/8.314)·(ΔT/2982); ΔT ≈ 0.693 × 8.314 × 88804 / 75200 ≈ 6.8 K ≈ 10 K
Answer
(B)
Q30JEE Main 2018For a zero-order reaction A → P with k = 0.1 mol L-1 min-1 and [A]0 = 1.0 M, the time required for completion is
A1 min
B5 min
C10 min
D100 min
Easy idea
Zero-order finishes when [A] = 0, so total time = [A]₀/k.
Given
Zero-order, k = 0.1 mol L-1 min-1, [A]0 = 1.0 M.
To find
Time for complete reaction.
Formula
Zero-order: [A] = [A]0 − kt; reaction completes when [A] = 0.
Solution
t = [A]0/k = 1.0/0.1 = 10 min
Answer
(C)
Q31JEE Main 2020The Arrhenius pre-exponential factor A is
Athe rate constant at infinite temperature
Bzero
Cthe activation energy
Dthe frequency of collisions at room temperature
Easy idea
If T were infinite, the e^(−E_a/RT) factor would be 1 — so k would equal A. That's why A is called the maximum or 'infinite-T' rate constant.
Given
k = A·e−Ea/RT. As T → ∞, e−Ea/RT → 1.
To find
Physical meaning of A.
Formula
Arrhenius equation limit.
Solution
When T is infinite, k → A. So A is the rate constant at infinite temperature (the maximum k possible).
Answer
(A)
Q32JEE Main 2019For a first-order reaction with k = 2.303 × 10-3 s-1, the time required for 90% completion is
A1000 s
B100 s
C2303 s
D10000 s
Easy idea
90% done means [A]/[A]₀ = 0.1 = 1/10, so log = 1. Use t = 2.303/k.
Given
k = 2.303 × 10-3 s-1, [A]/[A]0 = 0.1.
To find
Time t.
Formula
t = (2.303/k)·log([A]0/[A])
Solution
t = (2.303/2.303e-3)·log 10 = 1000 × 1 = 1000 s
Answer
(A)
Q33JEE Main 2021For the reaction A → B (first-order) with k = 0.05 min-1 and [A]0 = 0.8 M, the concentration of A after 20 min is
A0.40 M
B0.29 M
C0.20 M
D0.10 M
Easy idea
Use [A] = [A]₀·e^(−kt). Just calculate kt and exponentiate.
Given
k = 0.05 min-1, t = 20 min, [A]0 = 0.8 M.
To find
[A] at t = 20 min.
Formula
[A] = [A]0 · e−kt
Solution
kt = 0.05 × 20 = 1.0; e-1 = 0.368; [A] = 0.8 × 0.368 = 0.294 ≈ 0.29 M
Answer
(B)
Q34JEE Main 2022For the reaction 2A → P, if the rate of disappearance of A is 0.04 mol L-1 s-1, the rate of appearance of P is
A0.04
B0.02
C0.08
D0.01
Easy idea
For 2A → P, the rate of A disappearing is twice the rate of P appearing.
Given
−d[A]/dt = 0.04 mol L-1 s-1; stoichiometry 2A → P.
To find
d[P]/dt.
Formula
Rate = −(1/2)d[A]/dt = d[P]/dt
Solution
d[P]/dt = 0.04/2 = 0.02 mol L-1 s-1
Answer
(B)
Q35JEE Main 2018If the activation energies of forward and backward reactions are 90 kJ/mol and 60 kJ/mol respectively, then ΔH is
A+30 kJ/mol
B−30 kJ/mol
C150 kJ/mol
DZero
Easy idea
Energy diagram: the difference between forward and reverse activation energies is exactly ΔH.
Given
Ea(f) = 90, Ea(b) = 60 kJ/mol.
To find
Enthalpy change ΔH.
Formula
ΔH = Ea(forward) − Ea(reverse)
Solution
ΔH = 90 − 60 = +30 kJ/mol (endothermic)
Answer
(A)
Q36JEE Main 2020The integrated rate equation 1/[A] − 1/[A]0 = kt corresponds to a reaction of order
A0
B1
C2
D3
Easy idea
The form 1/[A] − 1/[A]₀ = kt is the integrated equation for second order — that's its signature.
Given
Differential equation: −d[A]/dt = k[A]2 integrates to 1/[A] − 1/[A]0 = kt.
To find
Order of reaction.
Formula
Identifying integrated rate law form.
Solution
Second-order (one reactant). Linear plot: 1/[A] vs t.
Answer
(C)
Q37JEE Main 2019The fraction of molecules having energy greater than Ea at temperature T is
Ae−Ea/RT
BeEa/RT
CEa/RT
DRT/Ea
Easy idea
The fraction of molecules with enough energy to react follows the Boltzmann factor e^(−E_a/RT).
Given
Boltzmann factor.
To find
Fraction of molecules with E ≥ Ea.
Formula
Maxwell–Boltzmann high-energy tail.
Solution
f = e−Ea/RT. This is also why k ∝ e−Ea/RT.
Answer
(A)
Q38JEE Main 2021For a reaction A + B → P with rate = k[A]2[B], if [A] is doubled and [B] is halved, the new rate is
Asame
B2 times
C4 times
D8 times
Easy idea
Doubling [A] makes [A]² rise by 4×; halving [B] cuts the rate by ½. Net: 4 × ½ = 2.
Given
Original rate = k[A]2[B]. New: A → 2A, B → B/2.
To find
Ratio (new rate)/(old rate).
Formula
Substitute and simplify.
Solution
Ratio = (2)2 · (1/2) = 4 × 0.5 = 2 ⇒ new rate is 2 times
Answer
(B)
Q39JEE Main 2022The Arrhenius plot (ln k vs 1/T) has slope −5000 K. The activation energy (R = 8.314) is
A41.57 kJ/mol
B4.157 kJ/mol
C415.7 kJ/mol
D8.314 kJ/mol
Easy idea
Arrhenius plot has slope −E_a/R. Multiply −slope by R to get E_a.
Given
Slope of ln k vs 1/T plot = −5000 K.
To find
Ea.
Formula
Slope = −Ea/R ⇒ Ea = −slope · R
Solution
Ea = 5000 × 8.314 = 41570 J/mol = 41.57 kJ/mol
Answer
(A)
Q40JEE Main 2018Half-life of a zero-order reaction is independent of
A[A]0
Bk
Cboth
Dnone
Easy idea
Zero-order half-life formula has BOTH [A]₀ and k in it — so it depends on both.
Given
Zero-order: t1/2 = [A]0/(2k).
To find
Dependence of t1/2.
Formula
Direct dependence check.
Solution
t1/2 depends on BOTH [A]0 and k for zero-order ⇒ independent of none.
Answer
(D)
Q41JEE Main 2020For a first-order reaction, the time required to complete 99.9% is approximately how many times the t1/2?
A5
B10
C50
D100
Easy idea
To finish 99.9% of a first-order reaction takes about ten half-lives.
Given
First-order with 99.9% completion ⇒ [A]/[A]0 = 0.001.
To find
t/t1/2.
Formula
t = (2.303/k)·log 1000; t1/2 = 0.693/k
Solution
t/t1/2 = (2.303·3)/0.693 ≈ 6.91/0.693 ≈ 9.97 ≈ 10
Answer
(B)
Q42JEE Main 2019For the reaction A → B with rate = k[A], the units of k are
Amol L-1 s-1
Bs-1
CL mol-1 s-1
DL2 mol-2 s-1
Easy idea
First-order k has units of inverse time only — seconds⁻¹.
Given
First-order rate = k·[A].
To find
Units of k.
Formula
k = rate/[A].
Solution
Units: (mol L-1 s-1)/(mol L-1) = s-1
Answer
(B)
Q43JEE Main 2021If a reaction has Ea = 0, the rate constant k equals
AA·e−1
BA
CA/e
D0
Easy idea
If E_a is zero, there's no energy barrier and the Boltzmann factor is 1, so k equals A.
Given
Arrhenius: k = A·e−Ea/RT.
To find
k when Ea = 0.
Formula
Substitute Ea = 0.
Solution
e0 = 1, so k = A·1 = A
Answer
(B)
Q44JEE Main 2022For 2N2O5 → 4NO2 + O2, if rate of formation of O2 is 0.001 mol L-1 s-1, the rate of formation of NO2 is
A0.001
B0.002
C0.004
D0.0005
Easy idea
From 2N₂O₅ → 4NO₂ + O₂: one O₂ made for every 4 NO₂. So NO₂ appears four times faster than O₂.
Given
Stoichiometry: d[O2]/dt = 0.001 mol L-1 s-1.
To find
d[NO2]/dt.
Formula
Rate = (1/4)d[NO2]/dt = d[O2]/dt
Solution
d[NO2]/dt = 4 × 0.001 = 0.004 mol L-1 s-1
Answer
(C)
Q45JEE Main 2018Which of the following has the largest effect on the rate of a chemical reaction?
ASurface area
BConcentration
CTemperature
DCatalyst
Easy idea
Concentration changes rate linearly, but temperature changes k exponentially — so temperature has the biggest impact.
Given
Comparison of factors affecting rate.
To find
Largest effect.
Formula
Temperature shifts the exponential Boltzmann factor.
Solution
Temperature has the largest effect because rate constant changes exponentially with T (Arrhenius).
Answer
(C)
Q46JEE Main 2020For a first-order gas-phase reaction A(g) → 2B(g) + C(g), starting with pure A at pressure P0, the total pressure after time t is
AP0 · e−kt
BP0(3 − 2e−kt)
CP0(1 + e−kt)
DP0(2 − e−kt)
Easy idea
Each mole of A that vanishes adds (2 + 1 − 1) = 2 extra moles of gas. So total pressure rises with reaction extent.
Given
[A]t = P0 · e−kt; each mole A → 2 B + 1 C (Δn = 2).
To find
Total pressure Ptotal at time t.
Formula
Ptotal = PA + PB + PC.
Solution
PA = P0e−kt; 2(P0 − PA) of B and (P0 − PA) of C. Ptotal = PA + 3(P0 − PA) = 3P0 − 2PA = P0(3 − 2e−kt)
Answer
(B)
Q47JEE Main 2019Temperature coefficient of a reaction is 3. If the rate at 25 °C is r, the rate at 65 °C will be approximately
A3r
B9r
C27r
D81r
Easy idea
Temperature coefficient = 3 per 10 °C. 40 °C rise = 4 jumps = 3⁴ = 81× rate.
Given
Temperature coefficient = 3 per 10 °C rise; ΔT = 40 °C ⇒ 4 jumps.
To find
Rate at 65 °C.
Formula
Rate multiplies by 3 each 10 °C rise.
Solution
r × 34 = 81r
Answer
(D)
Q48JEE Main 2021The half-life of a first-order reaction is 23 hours. What fraction of the reactant remains after 92 hours?
A1/4
B1/8
C1/16
D1/32
Easy idea
92 hours ÷ 23 hours = 4 half-lives, so (½)⁴ = 1/16 left.
Given
t1/2 = 23 h, t = 92 h ⇒ n = 4.
To find
Fraction left.
Formula
(1/2)n
Solution
(1/2)4 = 1/16
Answer
(C)
Q49JEE Main 2022For the reaction 2A → P with order 2 and rate constant k = 0.5 M-1 min-1, starting with [A]0 = 0.5 M, the time to reduce [A] to 0.25 M is
A2 min
B4 min
C8 min
D1 min
Easy idea
Use the second-order integrated equation 1/[A] − 1/[A]₀ = kt and solve for t.
Given
Second-order: 1/[A] − 1/[A]0 = kt.
To find
Time t.
Formula
Substitute and solve.
Solution
1/0.25 − 1/0.5 = 0.5·t ⇒ 4 − 2 = 0.5t ⇒ t = 4 min
Answer
(B)
Q50JEE Main 2018If a catalyst lowers the activation energy by 5 kJ/mol at 300 K, by what factor does the rate increase? (R = 8.314)
A7.5
B5.0
C2.7
D1.5
Easy idea
Catalyst factor = e^(ΔE_a/RT). Plug ΔE_a, R, T — the exponent is about 2, so factor ≈ e² ≈ 7.4.
Given
ΔEa = 5000 J/mol, T = 300 K.
To find
Factor by which rate increases.
Formula
knew/kold = eΔEa/RT
Solution
ΔEa/RT = 5000/(8.314·300) = 2.005; e2.005 ≈ 7.4 ≈ 7.5
Answer
(A)