Chemical Kinetics JEE Main PYQ: The Half-Life, Gas-Pressure and Arrhenius Traps That Decide This Chapter (2015-2026)
Chemical Kinetics JEE Main PYQ analysis (2015-2026). 189 questions mapped: half-life ratios, gas-pressure problems, Arrhenius shortcuts, 8 traps, and 12 exam-level NVQs and MCQs.
About one question every shift, and most of them end in a clean integer.
Chemical Kinetics is not the biggest chapter in JEE Main Chemistry. It is one of the most reliable. We counted 189 questions from 2015 to 2026. That works out to roughly one question in almost every shift. Each one rewards a fixed method, not a clever idea.
Three sub-topics do most of the work. Integrated rate laws, first-order applications, and the Arrhenius equation carry 145 of the 189 questions. That is 77 percent of the chapter. They also supply most of the NVQs, because their answers come out as neat numbers like 10, 16, or 400.
That neatness is the key to this chapter. Examiners build each NVQ backwards from an integer. They choose half-lives like 69.3 s and activation energies like 191.48 kJ mol⁻¹ on purpose. Once you see the trick, the arithmetic collapses in one line. Miss it, and you grind through logs with no calculator and no options to check.
We analysed every Chemical Kinetics question across JEE Main sessions from 2015 to 2026, including January and April 2026. This is part of Logic Bloom's JEE Main analysis series, after Thermodynamics, Solutions, and Electrochemistry. With this one, the Physical Chemistry core is complete.
| 🎯 We mapped every JEE Main Chemical Kinetics question of the decade. The app has them all, ready to play and practice. | |
|---|---|
| This chapter is won by drilling the same five setups until the shortcut is automatic. Logic Bloom's Playground turns it into practice: watch a first-order curve drop by equal half-lives, track partial pressures as a gas decomposes, and slide the temperature to see how k responds to Ea. Then drill every PYQ, including the compulsory NVQ type. When a missing 2.303 or a Celsius temperature catches you, TarQ teaches the fix, and your Mistake Book logs it. | Get the app → Free to start. |
The decade at a glance: where the marks sit, by format
Here is how the 189 questions split across the six sub-topics. The format column matters as much as the count. It tells you whether to expect four options or a blank NVQ box.
| Rank | Sub-topic | Questions | Share | Format lean |
|---|---|---|---|---|
| 1 | Integrated rate laws (zero and first order) | 56 | 29.6% | NVQ-heavy (36 NVQ, 20 MCQ) |
| 2 | First-order applications (gas pressure, decay) | 49 | 25.9% | NVQ-heavy (33 NVQ, 16 MCQ) |
| 3 | Arrhenius equation | 40 | 21.2% | NVQ-heavy (25 NVQ, 15 MCQ) |
| 4 | Rate law, order and molecularity | 24 | 12.7% | MCQ-heavy (7 NVQ, 17 MCQ) |
| 5 | Collision theory and catalysts | 11 | 5.8% | MCQ (2 NVQ, 9 MCQ) |
| 6 | Rate of reaction and stoichiometry | 9 | 4.8% | Balanced (4 NVQ, 5 MCQ) |
The volume has grown with the shift count. The offline years gave 4 or 5 questions a year. Since 2021, every year has given 16 to 27.
| Year | Questions | What repeated across shifts |
|---|---|---|
| 2015 to 2018 | 4 to 5 a year | Two-temperature Arrhenius, initial-rate tables, t75% vs t50% |
| 2019 | 15 | Pseudo-first-order hydrolysis, two-temperature Arrhenius |
| 2020 | 20 | k doubling or quadrupling between two temperatures; NVQs arrive in September |
| 2021 | 25 | t99.9% = 10 × t½; two species decaying to a set ratio |
| 2022 | 22 | The "t at x% = y × t½" NVQ stem, repeated across 4 shifts |
| 2023 | 27 | Gas-phase A → 2B + C pressure tables; parallel first-order decay |
| 2024 | 16 | Br-82 decay (t½ = 36 h); P at time t with P after a long time |
| 2025 | 23 | Zero-order time steps; catalyst Ea shifts; gas-phase pressure, again |
| 2026 | 24 | Rate coupling with BrO₃⁻ + 5Br⁻; k = A e^(−28000/T) giving Ea = 232 kJ mol⁻¹ |
Read the right-hand column as a warning. A template that shows up in one shift usually comes back. It returns with new numbers in two to four more shifts of that session.
What is still tested, and what is dead
| Still tested (revise this) | Deleted (do not study) |
|---|---|
| Rate and stoichiometry; rate law, order, molecularity, units of k; zero- and first-order integrated laws; half-life and x% times; gas-phase pressure problems; decay-style first-order problems; Arrhenius in every form; collision theory basics; catalyst effect on Ea | Adsorption isotherms, heterogeneous catalysis mechanisms, zeolite and enzyme catalysis (all left with Surface Chemistry in 2024); detailed collision-frequency derivations |
Integrated laws for second order and above are not on the syllabus as formulas. They appear only through graphs or through how half-life scales with concentration.
Sub-topic 1: Integrated rate laws, the biggest NVQ source
Zero and first order behave very differently, and JEE tests the difference. For first order, every half-life takes the same time. For zero order, each half-life is shorter than the one before. The picture below is worth memorising.
The reason is in the formulas. For zero order, t½ = [A]₀ / 2k, so half-life depends on how much is left. For first order, t½ = 0.693 / k, so it never changes. That single fact settles a whole family of questions.
The x%-completion ratios, the fastest marks in the chapter
For first order, the time to finish x% of the reaction is always a fixed multiple of t½. The concentration and even the value of k do not matter. Learn these ratios and many NVQs take ten seconds.
| 🧠 First-order completion times, as multiples of t½ | ||
|---|---|---|
| 50% done | ½ left | 1 × t½ |
| 67% done | ⅓ left | 1.6 × t½ (log 3 / log 2) |
| 75% done | ¼ left | 2 × t½ |
| 87.5% done | ⅛ left | 3 × t½ |
| 90% done | 1/10 left | 3.32 × t½ (3.33 if log 2 = 0.3) |
| 99% done | 1/100 left | 6.64 × t½, and t99% = 2 × t90% |
| 99.9% done | 1/1000 left | about 10 × t½, and t99.9% = 3 × t90% |
One warning on the 67% stem. Use what is left, not what reacted. If 67% has reacted, 33% remains, so the log term is log(100/33). Students who plug in 67 get 0.58 instead of 1.6, which turns 16 into 6 in the NVQ box.
The traps JEE Main repeats here
Trap 1: treating a zero-order half-life as constant. In 2025, a zero-order reaction had t½ = 1 h at 2.0 M. The question asked how long it takes to go from 0.50 M to 0.25 M. Students answered 1 h. But k = 2.0 / (2 × 1) = 1.0 M h⁻¹, so the time is (0.50 − 0.25) / 1.0 = 0.25 h. Zero order loses a fixed amount per hour, not a fixed fraction.
Trap 2: the missing or extra 2.303. Use 2.303 only when you work in log base 10. The natural-log form k = (1/t) ln([A]₀/[A]) needs no 2.303. Mixing the two forms is an instant wrong answer.
Trap 3: rounding the given logs. If a stem says "use log 2 = 0.3", use exactly 0.3. With 0.3, t90% / t50% is 3.33. With 0.301, it is 3.32. Near a rounding boundary, that difference changes the integer you enter.
Sub-topic 2: First-order applications, gas pressure and decay
This is the second biggest source, and the one that has grown fastest since 2023. Two setups dominate: a gas decomposing in a closed container, and a substance decaying over time.
Gas-phase decomposition: track A, not the total
The examiner gives you total pressure. The rate law needs the partial pressure of A only. Everything turns on converting one into the other. Say A starts at pressure Pᵢ and a pressure p of it has reacted. Count the gas molecules on each side, and the conversion falls out.
| Reaction | Total pressure at time t | A left, P_A | Pressure after a long time |
|---|---|---|---|
| A(g) → B(g) + C(g) | Pᵢ + p | 2Pᵢ − Pₜ | 2Pᵢ |
| A(g) → 2B(g) + C(g) | Pᵢ + 2p | (3Pᵢ − Pₜ) / 2 | 3Pᵢ |
| 2A(g) → 4B(g) + C(g) | Pᵢ + 1.5p | (5Pᵢ − 2Pₜ) / 3 | 2.5Pᵢ |
Then use the first-order law with P_A in place of concentration: k = (2.303 / t) log(Pᵢ / P_A). A newer twist since 2024 hides Pᵢ. The stem gives Pₜ and the pressure "after a very long time". Divide that final pressure by the gas multiplier to get Pᵢ back. For A → 2B + C, Pᵢ = P∞ / 3.
A worked JEE Main question
Q. (NVQ) The first-order reaction A(g) → 2B(g) + C(g) starts with pure A at 800 mm Hg in a rigid container. After 10 min the total pressure is 1600 mm Hg. Find the total pressure after 30 min, in mm Hg. Round off to the nearest integer.
Solution. Total pressure is 800 + 2p. At 10 min, 800 + 2p = 1600, so p = 400. That leaves P_A = 400, exactly half of 800. So t½ = 10 min. Then 30 min is three half-lives, and P_A = 800 / 8 = 100. The reacted pressure is 700, so the total is 800 + 2(700) = 2200 mm Hg.
Answer: 2200. Trap: using Pₜ = Pᵢ + p (the A → B + C formula) gives 1500. Doubling the total every 10 min gives 3200.
Decay problems that moved here from Physics
Radioactive decay is first order, so these are first-order problems in disguise. They have shown up often since 2024. Examples include Br-82 with t½ = 36 h, Sr-90 with t½ = 6.93 yr, and a 245-day half-life. Watch for half-lives like 6.93 or 69.3. They make k a round number, since 0.693 / 6.93 = 0.1.
Two more patterns come up. When two substances decay side by side, the time for their ratio to change is t = ln(final ratio / starting ratio) / (k_B − k_A). When one substance decays two ways at once, the rate constants add: k = k₁ + k₂.
Sub-topic 3: The Arrhenius equation and its built-in shortcut
The Arrhenius equation links the rate constant to temperature: k = A e^(−Ea/RT). JEE uses it in four forms. You must switch between them without stopping.
| Form | Equation | Use it when |
|---|---|---|
| Exponential | k = A e^(−Ea/RT) | Ea is hidden inside the exponent, like e^(−28000/T) |
| Natural log line | ln k = ln A − (Ea/R)(1/T) | A graph of ln k against 1/T is given |
| Base-10 line | log k = log A − Ea / (2.303 RT) | A graph of log k against 1/T is given |
| Two temperatures | log(k₂/k₁) = [Ea / 2.303R] × (T₂ − T₁) / (T₁T₂) | k is given at two temperatures |
The 19.15 shortcut. The product 2.303 × 8.314 is 19.15 J mol⁻¹ K⁻¹. When a stem gives Ea as 191.48 or 38.29 kJ mol⁻¹, that is no accident. Divide Ea by 19.15 × T and the exponent comes out as a whole number. In 2026, k = (5.5 × 10¹¹ s⁻¹) e^(−28000 K/T) gave Ea directly as 28000 × R, which is about 232 kJ mol⁻¹.
A worked JEE Main question
Q. (NVQ) A first-order isomerisation A → B has Ea = 191.48 kJ mol⁻¹ and A = 6.93 × 10²¹ s⁻¹. At 1000 K, how long does half of A take to convert, in picoseconds? Round off to the nearest integer. (Given 2.303R = 19.148 J K⁻¹ mol⁻¹)
Solution. Find the exponent: Ea / (2.303RT) = 191480 / (19.148 × 1000) = 10. So log k = log(6.93 × 10²¹) − 10, which gives k = 6.93 × 10¹¹ s⁻¹. Then t½ = 0.693 / (6.93 × 10¹¹) = 1.0 × 10⁻¹² s, which is 1 picosecond.
Answer: 1. Trap: reading picoseconds as nanoseconds gives 1000. Trying e^(−23.03) by hand usually means giving up.
The traps JEE Main repeats here
Trap 1: Celsius in the T₁T₂ term. At 27°C and 37°C, the gap T₂ − T₁ is 10 in both scales. But the product must be in kelvin: 300 × 310 = 93000, not 27 × 37 = 999. Using Celsius makes Ea about 93 times too small.
Trap 2: reading the graph's axis wrong. For ln k against 1/T, slope = −Ea/R. For log k against 1/T, slope = −Ea/(2.303R). With a slope of −5000 K on an ln k plot, Ea = 5000 × 8.314 = 41.6 kJ mol⁻¹. Adding 2.303 here gives a wrong 95.7 kJ mol⁻¹.
Trap 3: the sign of the slope. As 1/T rises, the temperature falls, so k falls. The ln k against 1/T line always slopes down for a positive Ea. An option with an upward line, or a negative Ea, is a trap.
Sub-topics 4 and 6: Rate law, order, units, and rate in terms of a species
These come mostly as MCQs. They test whether you get order from experiment rather than from the balanced equation.
Order is experimental, molecularity is not. Order comes from data and can be zero, a fraction, or even negative. Molecularity counts molecules in one elementary step, so it is 1, 2, or rarely 3. Sugar hydrolysis is bimolecular but behaves as first order, because water is in huge excess. That is a pseudo-first-order reaction.
Units of k fix the order. The unit is mol^(1−n) L^(n−1) s⁻¹. The power of L is always n − 1. So L² mol⁻² s⁻¹ means n − 1 = 2, and the reaction is third order. Students who read the powers upside down call it second order.
Rate of a species needs its coefficient. For BrO₃⁻ + 5Br⁻ + 6H⁺ → 3Br₂ + 3H₂O, the rate of reaction is (1/5) of the rate at which Br⁻ disappears. In 2026, Br⁻ disappeared at 2.0 × 10⁻³ M min⁻¹. That made the reaction rate 4.0 × 10⁻⁴ M min⁻¹. Matching it to a first-order reaction with [A] = 0.10 M gave k = 4.0 × 10⁻³ min⁻¹. Skipping the divide-by-5 step gives a wrong 2.0 × 10⁻² min⁻¹.
Volume change scales rate by mⁿ. Squeeze a gas mixture to 1/m of its volume and every concentration rises m times. The rate rises mⁿ times, where n is the true order. Use the order from the rate law, never the coefficients.
Sub-topic 5: Collision theory and catalysts
This is a small, MCQ-only area, usually written as statements to judge. The diagram below answers most of them.
Three facts carry this area. First, collision theory says k = P Z e^(−Ea/RT). The steric factor P counts the need for the right orientation. Second, if Ea = 0, then k = A, so k no longer depends on temperature. Third, A has the same units as k, because the exponential part has no units.
The most common wrong statement says a catalyst lowers the forward barrier and raises the backward one. It does not. Both drop together, so the reaction reaches the same equilibrium, just faster.
| ⚡ The 8 traps, one line each | |
|---|---|
| Order vs molecularity | Order comes from data, not from the balanced equation. |
| Units of k | Power of L is n − 1. L² means third order. |
| Zero-order half-life | It shrinks as concentration falls. Use Δ[A] / k. |
| ln vs log | 2.303 appears only with log base 10. |
| Celsius | T₁T₂ must be in kelvin, even when the gap is the same. |
| Slope sign | ln k against 1/T slopes down. Ea is positive. |
| Species coefficient | Divide a species' rate by its coefficient first. |
| Gas pressure | The rate law uses P_A, not total pressure. |
The formula sheet you need cold
No calculator, about two minutes a question. Have these at reflex speed.
| Quantity | Zero order | First order |
|---|---|---|
| Integrated law | [A] = [A]₀ − kt | k = (2.303/t) log([A]₀/[A]) |
| Half-life | [A]₀ / 2k, depends on [A]₀ | 0.693 / k, constant |
| Time to finish | [A]₀ / k = 2 × t½ | Never truly finishes |
| Units of k | mol L⁻¹ s⁻¹ | s⁻¹ |
| Straight-line plot | [A] against t, slope −k | log [A] against t, slope −k/2.303 |
| Quantity | Formula or value |
|---|---|
| Rate and coefficients | r = −(1/a) d[A]/dt = +(1/c) d[C]/dt |
| Units of k, order n | mol^(1−n) L^(n−1) s⁻¹ |
| Gas phase, A → 2B + C | k = (2.303/t) log[2Pᵢ / (3Pᵢ − Pₜ)] |
| Using P after a long time | k = (2.303/t) log[(P∞ − Pᵢ) / (P∞ − Pₜ)] |
| Two parallel paths | k = k₁ + k₂ |
| Arrhenius, two temperatures | log(k₂/k₁) = [Ea / 2.303R] × (T₂ − T₁)/(T₁T₂) |
| Catalyst, same A | log(k_cat / k) = ΔEa / (2.303RT) |
| Multi-step k = k₁k₂/k₃ | Ea = Ea₁ + Ea₂ − Ea₃ |
| Handy values | log 2 = 0.301, log 3 = 0.477, ln 2 = 0.693, 2.303R = 19.15 J mol⁻¹ K⁻¹ |
| 🎯 Drill the whole decade of Kinetics PYQs, sorted by sub-topic and format. | |
|---|---|
| Knowing the 19.15 shortcut is not the same as spotting it with no options and a −1 waiting. Logic Bloom's Playground has every JEE Main Chemical Kinetics PYQ from 2015 to 2026, tagged by sub-topic, format, and the exact trap it hides. Miss a coefficient or a kelvin conversion and TarQ walks you through the fix, then your Mistake Book brings it back until it sticks. Deleted Surface Chemistry topics are stripped out, so you never waste a rep. | Get the app → Free to start. |
12 must-attempt JEE Main questions, built on real PYQ patterns
These twelve follow the templates JEE Main repeats most. Seven are MCQs and five are NVQs, close to the real Section A and B mix. Each one needs at least two steps or turns on a trap. Try each before reading the answer.
1. (MCQ) For A + B → products, initial rates were measured at a fixed temperature:
Exp 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ mol L⁻¹ s⁻¹
Exp 2: [A] = 0.20 M, [B] = 0.10 M, rate = 4.0 × 10⁻³ mol L⁻¹ s⁻¹
Exp 3: [A] = 0.20 M, [B] = 0.20 M, rate = 1.6 × 10⁻² mol L⁻¹ s⁻¹
The overall order and the unit of k are:
(A) 3; L² mol⁻² s⁻¹ (B) 3; mol² L⁻² s⁻¹ (C) 2; L mol⁻¹ s⁻¹ (D) 3; mol L⁻¹ s⁻¹
Work. Exp 1 to 2: [A] doubles and the rate doubles, so the order in A is 1. Exp 2 to 3: [B] doubles and the rate rises 4 times, so the order in B is 2. Overall order is 3. The unit is mol^(1−3) L^(3−1) s⁻¹, which is L² mol⁻² s⁻¹.
Answer: (A). Trap: (B) flips the powers; (C) takes order from the coefficients 1 + 1; (D) uses the unit of rate itself.
2. (MCQ) For the gas reaction A + 2B → C, doubling [A] at fixed [B] makes the rate 4 times. Doubling [B] at fixed [A] leaves the rate unchanged. The mixture is squeezed to one-third of its volume at constant temperature. The rate becomes:
(A) 27 times (B) 9 times (C) 6 times (D) 3 times
Work. The data give rate = k[A]²[B]⁰, so the order is 2. One-third the volume makes every concentration 3 times larger. The rate rises 3² = 9 times.
Answer: (B). Trap: (A) uses the coefficient sum 1 + 2 = 3; (C) multiplies 3 × 2 instead of raising to the power.
3. (MCQ) A zero-order reaction has t½ = 40 min when [A]₀ = 0.80 M. How long does [A] take to fall from 0.40 M to 0.10 M?
(A) 80 min (B) 30 min (C) 15 min (D) 20 min
Work. For zero order, t½ = [A]₀ / 2k, so k = 0.80 / (2 × 40) = 0.010 M min⁻¹. The drop is 0.40 − 0.10 = 0.30 M. Time = 0.30 / 0.010 = 30 min.
Answer: (B). Trap: (A) counts two equal half-lives, a first-order habit; (C) uses k = [A]₀ / t½ without the 2; (D) stops after the first half-life at 0.40 M.
4. (NVQ) In a first-order reaction, [A] falls from 0.80 M to 0.10 M in 90 min. How many minutes does it take to fall from 0.10 M to 0.025 M? Round off to the nearest integer.
Work. 0.80 to 0.10 is three halvings, so 3 × t½ = 90 min and t½ = 30 min. 0.10 to 0.025 is two halvings, so it takes 2 × 30 = 60 min.
Answer: 60. Trap: assuming a straight-line drop of 0.70 M in 90 min gives about 10.
5. (MCQ) A first-order reaction is 90% complete in 46 min. How long does it take to be 99.9% complete?
(A) 138 min (B) 460 min (C) 51 min (D) 92 min
Work. t90% uses log(100/10) = 1. t99.9% uses log(100/0.1) = 3. So t99.9% = 3 × t90% = 3 × 46 = 138 min.
Answer: (A). Trap: (B) uses the t99.9% = 10 × t½ rule on t90%; (C) scales by 99.9/90; (D) uses t99% = 2 × t90%.
6. (NVQ) The first-order reaction A(g) → 2B(g) + C(g) runs at constant volume, starting with pure A. After 30 min the total pressure is 825 mm Hg. After a very long time it is 900 mm Hg. Find t½ in minutes. Round off to the nearest integer.
Work. At the end, all A has become 3 volumes of gas, so Pᵢ = 900 / 3 = 300. A left = (P∞ − Pₜ) / 2 = (900 − 825) / 2 = 37.5. That is 300 / 8, so three half-lives passed in 30 min. t½ = 10 min.
Answer: 10. Trap: treating it as A → B + C (Pᵢ = 450, A left = 75) gives about 12.
7. (NVQ) A radioactive substance decays by first-order kinetics with t½ = 6.93 years. What percentage of it remains after 23.03 years? Round off to the nearest integer. (Given ln 2 = 0.693, ln 10 = 2.303)
Work. k = 0.693 / 6.93 = 0.1 per year. Then kt = 0.1 × 23.03 = 2.303. So ln([A]₀/[A]) = 2.303 = ln 10, and one-tenth remains. That is 10%.
Answer: 10. Trap: reporting the amount decayed gives 90.
8. (NVQ) Substances A and B decay by first order, with half-lives of 30 min and 10 min. They start at equal concentration. After how many minutes is [A] four times [B]? Round off to the nearest integer.
Work. After time t, [A]/[B] = 2^(t/10) ÷ 2^(t/30) = 2^(t/15). Setting 2^(t/15) = 4 = 2² gives t/15 = 2. So t = 30 min.
Answer: 30. Trap: tracking only A, as if B stayed fixed, gives 60.
9. (MCQ) The rate constant of a reaction doubles when the temperature rises from 27°C to 37°C. The activation energy is closest to: (Given log 2 = 0.301, R = 8.314 J K⁻¹ mol⁻¹)
(A) 23.3 kJ mol⁻¹ (B) 0.58 kJ mol⁻¹ (C) 123.4 kJ mol⁻¹ (D) 53.6 kJ mol⁻¹
Work. Use kelvin: T₁ = 300 K, T₂ = 310 K. log 2 = [Ea / (2.303 × 8.314)] × 10 / (300 × 310). So Ea = 0.301 × 19.15 × 9300 ≈ 53600 J mol⁻¹ = 53.6 kJ mol⁻¹.
Answer: (D). Trap: (A) drops the 2.303; (B) uses 27 × 37 in place of 300 × 310; (C) puts ln 2 into the log form.
10. (MCQ) A plot of ln k against 1/T is a straight line with slope −6000 K. The activation energy is: (R = 8.314 J K⁻¹ mol⁻¹)
(A) 114.9 kJ mol⁻¹ (B) −49.9 kJ mol⁻¹ (C) 49.9 kJ mol⁻¹ (D) 6.0 kJ mol⁻¹
Work. For an ln k plot, slope = −Ea/R. So Ea = 6000 × 8.314 = 49884 J mol⁻¹, about 49.9 kJ mol⁻¹. Ea is positive, since the line slopes down.
Answer: (C). Trap: (A) adds 2.303, which belongs to a log k plot; (B) keeps the slope's minus sign; (D) forgets R.
11. (NVQ) At 500 K, a catalyst lowers the activation energy of a reaction by 38.29 kJ mol⁻¹. The pre-exponential factor stays the same. The rate constant rises by a factor of 10^x. Find x. Round off to the nearest integer. (Given 2.303R = 19.15 J K⁻¹ mol⁻¹)
Work. log(k_cat / k) = ΔEa / (2.303RT) = 38290 / (19.15 × 500) = 38290 / 9575 ≈ 4.0. So the rate constant rises 10⁴ times, and x = 4.
Answer: 4. Trap: dividing by R instead of 2.303R gives about 9.2; leaving Ea in kJ gives 0.004.
12. (MCQ) Consider two statements.
Statement I: A catalyst lowers the activation energy of the forward and backward reactions by the same amount.
Statement II: For an exothermic reaction, a catalyst makes ΔH more negative and raises the equilibrium constant.
Choose the correct answer.
(A) Both I and II are correct (B) Both I and II are incorrect (C) I is correct but II is incorrect (D) I is incorrect but II is correct
Work. A catalyst offers a lower path between the same start and end. Both barriers fall by the same amount, so Statement I is true. Since start and end do not move, ΔH and K stay the same. Statement II is false.
Answer: (C). Trap: (A) believes a catalyst shifts the equilibrium toward products.
What to expect in JEE Main 2027
The 2015 to 2026 pattern, plus compulsory NVQs, points to four moves.
1. Gas-phase problems that hide Pᵢ. Expect Pₜ at some time plus the pressure after a long time, often with a non-1:1 reaction like 2A → 4B + C. Your first step is Pᵢ = P∞ divided by the gas multiplier.
2. Arrhenius chained to a half-life. Single-step Ea questions are turning into two-step NVQs. Find k from the Arrhenius equation using the 19.15 trick, then turn it into t½. A multi-step mechanism with Ea = Ea₁ + Ea₂ − Ea₃ is also likely.
3. More decay problems. Decay-style first-order problems will stay a staple. Watch for two species decaying to a set ratio, and one species decaying two ways at once.
4. Graph and statement MCQs in Section A. When Kinetics lands in Section A, expect order read from graphs: [A] against t, t½ against [A]₀, or rate against [A]. These come mixed with collision-theory and catalyst statements.
How to split your time: 50% on first-order laws, x% ratios, gas pressure, and decay, all practised without a calculator. 30% on Arrhenius in every form, especially the kelvin and 19.15 habits. 20% on rate-law tables, zero order, coefficients, and catalyst statements. Spend none on adsorption or enzyme catalysis.
| 🚀 Turn this analysis into a real score. | |
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| You now know the five setups and the eight traps. Close the gap between knowing and scoring. Logic Bloom's Playground drills every JEE Main Chemical Kinetics PYQ of the decade, and Battleground lets you test the chapter in 1v1 duels against other aspirants. Understand through games. Score through practice. | Get the app → Free to start. |