Chemical Kinetics — how fast reactions happen and why
Class 12 · Chapter 3. From cooking rice to shelf-life of medicines to the catalytic converter under your car — the speed of a reaction is often more important than whether it happens at all.
What is chemical kinetics?
Nobel Prize in Chemistry, 1903
Framed the equation that governs how rate rises with temperature.
Thermodynamics tells you whether a reaction can happen. Kinetics tells you how fast. A diamond, at room temperature, is technically supposed to turn into graphite — thermodynamics says so. But the reaction is so slow it takes billions of years, which is why your grandmother's diamond ring still sparkles. That gap between what should happen and how quickly it does is the entire subject of this chapter.
The two questions of kinetics
- How fast? Measure concentration versus time, define rate, extract a rate law.
- Why that fast? Explain the number with a mechanism, a collision picture, and an activation energy.
Two questions, four core equations, one giant table of applications. That is this chapter.
Why kinetics is a career subject
Every drug company runs on kinetics — the expiry date on a paracetamol strip is a first-order half-life calculation. Every food manufacturer runs on kinetics — the 'best before' date on a milk carton is an Arrhenius extrapolation. Every industrial chemical plant runs on kinetics — the reactor size for the Haber ammonia process, the operating temperature for the catalytic converter in your car, the reason cement takes exactly twenty-eight days to reach full strength. If you can predict a rate, you can design a reactor. If you can design a reactor, you can build an industry.
The chapter roadmap
- Define rate — average, instantaneous, and how to measure it.
- Rate law and rate constant — the empirical equation of a reaction.
- Order and molecularity — the two numbers, and why they differ.
- Integrated rate equations for zero and first order.
- Half-life — and why first-order half-lives are constant.
- Temperature dependence — the Arrhenius equation.
- Collision theory — the atomic-level picture behind the rate.
- Catalysts — how a spectator changes everything.
2. Contents
Chapter video walkthrough
Source: authoritative-syllabus channel · verified 2026-07-28
This tab is an executive-summary teach of the full chapter β every subsection covered lucidly with a link to the deep-dive. A student reading only this tab comes away with the chapter arc.
What is chemical kinetics?
See the deep-dive interactive teaching in the linked tab below.
β Full deep-dive in the overview tabWhere you already know kinetics
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β Full deep-dive in the intuition tabWatch the rate change
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β Full deep-dive in the interactive tabEvery formula you need — and what it means
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β Full deep-dive in the hood tabApplications — where kinetics shows up in the world
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β Full deep-dive in the applications tabDive
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β Full deep-dive in the dive tabPractice — start here every day
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β Full deep-dive in the practice tabNinety-second recap
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β Full deep-dive in the reference tabWhere you already know kinetics
You have been running kinetics experiments in your kitchen and on your street since childhood. Every time you turn the flame up under the pressure cooker, every time you check whether the milk has spoiled, every time you notice that a hot cup of tea cools faster in December than in June — you are watching a rate change with a variable. Let us put words to those experiences before we build equations.
The pressure cooker — temperature accelerates everything
Rice takes forty minutes in an open pot but ten in a pressure cooker. The pressure cooker keeps water liquid at around a hundred and twenty degrees rather than a hundred, and just those extra twenty degrees roughly quadruples the rate of every hydrolysis reaction that softens the grain. This is the Arrhenius equation, in your kitchen, every single day. A rule of thumb from your grandmother: every ten degrees roughly doubles a reaction rate near room temperature. That rule is exactly what Arrhenius wrote down in 1889.
The refrigerator — temperature decelerates spoilage
Milk left on the counter turns sour in half a day. In the fridge, the same milk lasts a week. Nothing changed except the temperature dropped by about twenty-five degrees. The bacteria are still there; the sugars are still there. Every chemical step the bacteria run has slowed down by a factor of roughly ten. That is why cold storage was one of the most important inventions of the twentieth century — it did not stop chemistry, it just slowed it down enough for us to reach the market with fresh food.
Medicine dosage — half-life in your bloodstream
Every time a doctor writes 'take twice a day', they are doing a kinetics calculation in their head. A paracetamol tablet has a half-life of about four hours in the human body. That means after four hours, half the dose is gone; after eight hours, only a quarter remains. So you dose every six hours to keep the concentration in the therapeutic window — high enough to work, low enough to be safe. Every prescription you have ever taken was written with a rate constant in mind.
Fire — and why matches need striking
Petrol and air in your bike's fuel tank do not spontaneously catch fire, even though the reaction — petrol plus oxygen giving carbon dioxide and water — is enormously downhill. The reason is activation energy. The molecules need a certain minimum energy to start reacting, and at room temperature almost none of them have it. Strike a match, provide that starting energy in one small region, and the reaction ignites — and once ignited, it releases enough heat to keep going. This is why fires need a spark, why match heads have a friction primer, why the diesel engine compresses air until it is hot enough to ignite fuel on its own.
Rust — and why slow is still a rate
An iron gate takes years to rust visibly, yet it is definitely reacting. Rate does not mean fast; it means how much per second, however small. Chemists routinely measure rates over eleven orders of magnitude — from combustion, where a reaction finishes in microseconds, to radioactive decay of uranium, where half-lives run to four and a half billion years. Same physics, same equations, just different numbers.
Watch the rate change
Three interactive scenes. Slide the parameters, watch the animation, read the caption. Every scene is teaching you one specific idea — do not just look, watch what changes when you move the slider.
Scene 1 · Reaction rate simulator — concentration vs time
Red particles (reactant) convert to blue particles (product) over time. The graph on the side shows both concentrations. Change the rate constant with the slider and see how the whole curve reshapes.
Watch what stays the same when you change the rate. The shape of the curve — an exponential decay of reactant, an exponential build-up of product — does not change. Only the horizontal stretch changes. Doubling the rate constant halves the half-life. That is the signature of a first-order reaction, and once you see it here you will recognise it in every plot for the rest of the chapter.
Scene 2 · Arrhenius temperature slider — the exponential wall
Grey molecules bounce with a range of energies (colour-coded — blue slow, red fast). Only the red ones have enough energy to cross the activation-energy hill in the background. Watch how the red fraction grows explosively when you raise the temperature.
This is the whole secret of temperature dependence. Chemistry does not care about the average energy; it cares about the tail of the distribution — the small fraction of molecules moving fastest. Raise the temperature a little and that tail grows a lot, because it is exponentially sensitive to how much you multiply. This is why a ten-degree rise near room temperature can double the rate, and why a cold storage room can extend food shelf-life from days to weeks.
Scene 3 · Collision simulator — effective versus ineffective
Green ‘A’ molecules and orange ‘B’ molecules bouncing around a box. Every collision briefly flashes. Green flashes are effective collisions — enough energy and correct orientation — and produce a product molecule. Grey flashes are ineffective — a bump with no reaction.
The rate is not the collision frequency; it is the fraction of collisions that are effective. Most collisions are simply bounces — not enough energy, or the two molecules met at the wrong angle. Only the small fraction that meet with sufficient energy in the right orientation produce a product. This is why the pre-exponential factor $A$ in the Arrhenius equation is often smaller than the raw collision frequency — it accounts for the orientation penalty.
Every formula you need — and what it means
All the machinery of Class 12 kinetics. We take it one piece at a time. First the picture, then the equation, then the anchor, then the caveat.
1. Rate of a reaction — what we are actually measuring
The rate of a reaction is the change in concentration of a reactant or product per unit time. Because reactants disappear and products appear at different rates when stoichiometric coefficients differ, we define one canonical rate — divide each concentration change by its coefficient with a minus sign for reactants.
Units of rate are moles per litre per second in a solution, or moles per litre per minute in slower reactions. In the ozone-destruction reaction in the upper atmosphere, we might measure rates in molecules per cubic centimetre per second — but the idea is identical.
Where this idea breaks down: average rate over a big time interval hides everything interesting. Real kinetics uses the instantaneous rate — the slope of a tangent to the concentration-versus-time curve at one instant. A pen tangent on graph paper, or a calculator derivative if you have the equation.
2. Rate law — how the rate depends on concentration
The rate law is an empirical equation — you cannot guess it from the balanced equation, you have to measure it. It says: rate equals some rate constant multiplied by concentrations raised to certain powers. Those powers are called the orders with respect to each reactant.
Anchor: for the reaction of hydrogen with iodine, the balanced equation is $H_2 + I_2 \to 2HI$, and it happens to be first order in each — overall second order. But for hydrogen with bromine, the same-looking equation gives a rate law with a fractional order in bromine and an inhibition by product. Two reactions that look identical on paper have very different rate laws. This is why rate laws must be measured.
3. Order versus molecularity
Two different numbers, often confused, always tested. Order is empirical — the sum of exponents in the measured rate law. Molecularity is theoretical — the number of molecules that actually collide in a single elementary step. Order can be zero, fractional, negative, or greater than three. Molecularity is one, two, or, very rarely, three — because getting four molecules to meet simultaneously is a statistical impossibility.
An example that separates them: the decomposition of ozone runs through a two-step mechanism — first ozone breaks into oxygen molecule plus oxygen atom, then that atom attacks another ozone. Each step is molecularity two. But the overall rate law measured in the lab is second order in ozone, minus one order in oxygen — a strange net order that no single step could ever give. Order is what the world tells us; molecularity is what a single step involves.
4. Integrated rate equations — concentration versus time
The rate law tells you the derivative — how fast concentration is changing right now. To predict the concentration at a future time, we integrate. Class 12 focuses on two cases — zero order and first order.
Anchor: the decay of carbon-fourteen — the isotope used in carbon dating — is a perfect first-order reaction. Its half-life is five thousand seven hundred and thirty years. Measure the fraction of carbon-fourteen remaining in a bone or a piece of charcoal, and the first-order equation tells you exactly how long ago the tree was alive. This is how archaeologists dated the Mohenjodaro pottery and the Egyptian pyramids.
5. Half-life — the elegant signature of first order
Half-life is the time taken for the concentration of a reactant to drop to half its initial value. For first-order reactions, and only for them, the half-life is a constant — independent of the starting concentration. That is why we can quote a half-life for paracetamol without asking how much you took, and why radioactive isotopes have well-defined half-lives.
For zero-order the half-life shrinks as the concentration drops — $t_{1/2} = [A]_0 / (2k)$. This is why saturated systems like alcohol metabolism in the liver behave zero-order at high dose: your liver clears roughly a fixed amount per hour, so a bigger drink takes proportionally longer.
6. Arrhenius equation — the temperature dependence
Almost every rate constant depends on temperature in the same way. Svante Arrhenius, in 1889, wrote a single equation that fits combustion, enzyme catalysis, and drug degradation with the same form.
The everyday anchor: your grandmother's rule that every ten-degree rise doubles a rate. Plug numbers in: an activation energy of about fifty kilojoules per mole, a ten-degree rise near three hundred kelvin, and the exponential factor jumps by roughly a factor of two. Grandmother, without knowing it, was telling you the Arrhenius equation with a typical biochemical activation energy.
Where this idea breaks down: Arrhenius assumes $A$ and $E_a$ do not themselves depend on temperature. Over a small temperature range that is fine. Over hundreds of kelvin — combustion, plasma chemistry — more sophisticated forms take over. For your board exam, Arrhenius is exact.
7. Collision theory — where the rate constant comes from
The collision theory of bimolecular reactions says: for A and B to react, three things must happen. They must collide, they must have enough energy to break existing bonds, and they must meet in the correct orientation. The rate is the product of these three factors.
Anchor: consider two big organic molecules, each with a small reactive site. Even if they collide many times a second, most collisions bring the wrong ends together — steric factor is tiny, maybe one in a million. That is why enzymes matter: an enzyme grabs both molecules and holds them in exactly the right orientation, raising the steric factor to nearly one and giving a rate a million times higher than the uncatalysed reaction.
8. Catalysts — lowering the hill
A catalyst provides an alternative reaction pathway with a lower activation energy. It is not consumed — it appears on both sides of the mechanism — and it changes only the rate, never the equilibrium. Because the Arrhenius factor is exponential in activation energy, even a modest drop of thirty kilojoules per mole can accelerate a room-temperature reaction by a factor of ten thousand.
Anchor: the catalytic converter under every petrol car. Without it, the reaction of carbon monoxide with oxygen in the exhaust is millions of times too slow at exhaust temperature to clean the gas before it leaves the tailpipe. With a platinum-and-rhodium catalyst coated onto a ceramic honeycomb, the activation energy drops by more than half, and ninety-nine percent of the toxic gases are oxidised in the fraction of a second it takes the exhaust to pass through. This one device, mandatory since the 1970s, has saved millions of lives from urban air pollution.
Where this idea breaks down: a catalyst cannot make a thermodynamically forbidden reaction happen. It only speeds up reactions that were already going to happen — it just makes them happen faster. This is why you cannot catalyse water into gold, no matter how hard you try.
Applications — where kinetics shows up in the world
Every idea in this chapter is a working industry somewhere. Here are eight that matter most for your board, your JEE prep, and — if you go into science — your career.
1. Pharmaceutical shelf life — the expiry date is a half-life
Every drug slowly decomposes in its tablet. The regulator says a drug can be sold as long as at least ninety percent of the labelled dose is still active. So the shelf life is the time at which the drug drops to ninety percent — for a first-order reaction, that is roughly ten percent of the half-life. Pharma companies run accelerated stability tests at fifty degrees Celsius for a few weeks, use Arrhenius to extrapolate to room temperature, and print the expiry date from that calculation. That is why paracetamol lasts three years but eye drops last two months.
2. Food preservation — the cold chain
Milk spoils because of bacterial enzymes catalysing hydrolysis of lactose and proteins. Every enzyme obeys Arrhenius. Dropping the temperature from thirty to four degrees Celsius slows the enzymes by roughly a factor of ten — and therefore multiplies the shelf life by ten. That is why milk keeps for a week in the fridge but sours in a day on the counter. Freezing to minus twenty drops the rate another twenty-fold. Every packet of frozen food you have ever eaten is a demonstration of the Arrhenius equation.
3. Haber process — ammonia at just the right temperature
Nitrogen plus hydrogen giving ammonia is exothermic, so at high temperature the equilibrium moves backwards and you get less ammonia. But at low temperature the rate is too slow to be economic. So German engineers compromised at around four hundred and fifty degrees Celsius with an iron catalyst that drops the activation energy enough to make the rate practical. This trade-off — kinetics wants higher temperature, thermodynamics wants lower — feeds three billion people through synthetic fertiliser.
4. Contact process — sulphuric acid with vanadium pentoxide
Sulphur dioxide plus oxygen giving sulphur trioxide is another exothermic reaction where kinetics without a catalyst is hopelessly slow. Vanadium pentoxide as a heterogeneous catalyst lets the reaction run at around four hundred and fifty degrees at atmospheric pressure. Sulphuric acid is the most-produced industrial chemical in the world — used in fertilisers, batteries, dyes, detergents — and its whole production hinges on this one catalytic step.
5. Catalytic converter — kinetics on your commute
Under the floor of every petrol car sits a ceramic honeycomb coated with platinum, palladium, and rhodium. Exhaust passes through it in a fraction of a second. On the metal surface, carbon monoxide is oxidised to carbon dioxide, unburned hydrocarbons are burned off, and nitric oxide is reduced back to nitrogen. Without the catalyst the exhaust would leave the tailpipe still full of poisons. This device, invented in the 1970s, has saved millions of urban lives. It is the single most important application of chemical kinetics in your daily environment.
6. Enzyme kinetics — the biology behind digestion
Every enzyme in your body is a catalyst. Amylase in your saliva breaks starch into sugar — a reaction that would take days in a test tube happens in your mouth in seconds. Pepsin in your stomach denatures proteins; trypsin in your intestine cleaves them. Every one of these obeys the Michaelis-Menten form of enzyme kinetics, which is pseudo-first-order at low substrate and pseudo-zero-order at saturation. Your metabolism is kinetics from the moment you take a bite.
7. Explosives — chain reactions and runaway rates
An explosion is a chemical reaction so exothermic and so sensitive to temperature that once ignited it heats itself and the rate runs away in microseconds. TNT, gunpowder, dynamite — all rely on Arrhenius's exponential wall in reverse. Once the reaction starts, the released heat raises the local temperature by hundreds of kelvin, the rate jumps by orders of magnitude, and the whole mass converts to hot gas essentially instantaneously. The reason explosives can be stored safely is that at room temperature the rate is negligible — sitting quietly on the Arrhenius curve, just below the ignition threshold.
8. Cement setting — a slow twenty-eight-day reaction
Concrete gains strength over weeks because the hydration of calcium silicate is a diffusion-controlled reaction. The rate is highest in the first day, then falls as the reacted layer around each grain blocks further reaction. Civil engineers design bridges and dams to reach ninety percent strength at twenty-eight days — a standard age based directly on this rate. In cold climates, the whole schedule slips because the reaction slows with temperature — another everyday application of Arrhenius.
Board-pattern assessment
Everything below is either a real Board question from the last five years or a JEE-Main pattern. If you can do all of them cold, you are ready.
Section A · one-mark direct
- Define rate of a reaction.
- State the difference between molecularity and order.
- Write the units of the rate constant for a second-order reaction.
- What is a pseudo-first-order reaction?
- Name one homogeneous and one heterogeneous catalyst.
Section B · three-mark short
- Derive the integrated rate law for a first-order reaction and use it to define half-life.
- State and explain the Arrhenius equation. What is activation energy?
- Discuss the collision theory of bimolecular reactions.
Section C · five-mark long
- Explain how the rate of a reaction depends on concentration, temperature, and catalyst. Give one industrial example for each.
- Derive both zero-order and first-order integrated equations from the corresponding rate laws. Compare their half-life behaviour.
Common traps
- Order is not the sum of stoichiometric coefficients. It is measured, not guessed from the balanced equation.
- Molecularity is a property of an elementary step, never a fractional number, and applies only to a single step — not to an overall reaction.
- Half-life is constant only for first-order reactions. For zero and second order it depends on initial concentration.
- A catalyst does not change the equilibrium constant — it changes only the rate at which equilibrium is reached.
- The Arrhenius slope in a $\ln k$ versus $1/T$ plot is $-E_a/R$, not $-E_a$. Do not forget to multiply by $R$ at the end.
Past-year snapshot
Solution
Fraction remaining $= 0.7$. $k = (2.303/40) \log(1/0.7) = (2.303/40)(0.1549) = 8.92 \times 10^{-3}$ per minute. $t_{1/2} = 0.693/k = 77.7$ minutes.
Solution
$\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2) = (25000/8.314)(1/500 - 1/700) = 3008 \times 5.71 \times 10^{-4} = 1.72$. $k_2 = k_1 e^{1.72} = 1.6 \times 10^{-3} \times 5.58 = 8.9 \times 10^{-3}$ per second.
Solution
Order in $A$ = 1 (rate scales linearly). Order in $B$ = 0 (rate independent of B). Rate law: $\text{rate} = k[A]$.
Ninety-second recap
- Rate is change in concentration per unit time; rate law is an empirical equation relating rate to concentration.
- Order is measured, molecularity is theoretical; they can differ, especially in multi-step mechanisms.
- For first-order reactions, half-life is constant; for zero-order it drops with concentration.
- Arrhenius equation gives the temperature dependence of the rate constant; $\ln k$ versus $1/T$ is a straight line with slope $-E_a/R$.
- A catalyst lowers activation energy, changes only the rate — not the equilibrium constant.
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Study log — write your answers here
- Which single sentence in this chapter was the hardest to understand for you? Write it out. Re-read it after twenty-four hours.
- Which numerical example had the most steps? Redo it without looking. Compare.
- Explain in your own words to a friend who is not taking chemistry: what activation energy is, and why raising temperature accelerates almost every reaction.
Dive Deeper — JEE Main + Advanced Tier
This section takes you past the board syllabus into JEE Main and Advanced territory. Chemical kinetics is one of the most heavily-tested chapters in JEE — expect two to three questions every year. Master the ten problems and three worked examples below cold.
Beyond-Board Concepts for JEE
- Arrhenius equation and two-point $E_a$ calculation: $k = A \exp(-E_a/RT)$. Take log of two rate constants at two temperatures and subtract — $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$. This is the workhorse JEE numerical — given $k$ at $T_1$ and $T_2$, compute $E_a$; or given $E_a$ and one $k$, predict rate at any other temperature.
- Half-life for different orders: First-order $t_{1/2} = 0.693/k$ — independent of concentration (this is why radioactive decay is textbook first-order). Second-order $t_{1/2} = 1/(k[A]_0)$ — depends on initial concentration. Zero-order $t_{1/2} = [A]_0/(2k)$. JEE-Main routinely asks "which order has $t_{1/2}$ independent of $[A]_0$"; the answer is always first-order.
- Integrated rate laws in usable form: First-order — $\ln([A]_0/[A]) = kt$ — a plot of $\ln[A]$ vs $t$ gives a straight line of slope $-k$. Second-order — $1/[A] - 1/[A]_0 = kt$ — plot $1/[A]$ vs $t$. Zero-order — $[A]_0 - [A] = kt$ — plot $[A]$ vs $t$. Identifying the order from a table of $[A]$ vs $t$ is a favourite JEE-Advanced trick.
- Mechanism versus order: Order is what you measure from experiment; molecularity is what you count from the balanced mechanism. They agree only for elementary reactions. For a multistep mechanism, the order is set by the slowest (rate-determining) step, and can even be fractional or negative. JEE-Advanced 2022 asked: "Given a two-step mechanism, derive the rate law."
- Pseudo-first-order kinetics: If a reaction is truly second-order in $[A][B]$ but you swamp with $[B]$ so that $[B] \approx$ constant, the rate law reduces to $k'[A]$ where $k' = k[B]$. This is the "pseudo-first-order" trick — ester hydrolysis in excess water is the standard example. JEE-Main loves this pattern.
- Collision theory and steric factor: $k = P \cdot Z \cdot \exp(-E_a/RT)$, where $Z$ is the collision frequency and $P$ is the steric factor (usually $<< 1$ — molecules must collide with the right orientation, not just enough energy). This is why the Arrhenius pre-exponential $A$ is not simply the collision frequency.
- Catalyst effect on $E_a$: A catalyst provides an alternate pathway with lower $E_a$; it does not change $\Delta G$ or the equilibrium constant. If $E_a$ drops from 200 to 100 kJ mol$^{-1}$ at 300 K, the rate increases by $e^{100000/(8.314 \times 300)} \approx e^{40}$ — about $10^{17}$ times faster. JEE-Advanced 2023 had exactly this numerical.
JEE-tier Worked Examples (3)
Worked Example 1 · JEE Main 2023 pattern
Problem: For a first-order reaction, the rate constant is $2.3 \times 10^{-3}$ s$^{-1}$ at 300 K and $2.3 \times 10^{-2}$ s$^{-1}$ at 400 K. Find the activation energy $E_a$.
Approach: Two-point Arrhenius. $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$.
Solution: $\ln(10) = 2.303$. $(1/T_1 - 1/T_2) = 1/300 - 1/400 = (400-300)/(300 \times 400) = 100/120000 = 8.33 \times 10^{-4}$ K$^{-1}$. So $E_a = R \times 2.303 / (8.33 \times 10^{-4}) = 8.314 \times 2.303 / 8.33 \times 10^{-4} \approx 22{,}980$ J mol$^{-1}$ = 23 kJ mol$^{-1}$. Answer: about 23 kJ mol$^{-1}$.
Key trick: The Arrhenius equation is $\ln k = \ln A - E_a/RT$. The two-point version drops $\ln A$ — students who plug into the full form and forget to subtract get double-counted.
Worked Example 2 · JEE Main 2024 pattern
Problem: A first-order reaction has $t_{1/2} = 20$ minutes. Compute the time required for 90 percent completion.
Approach: First-order kinetics. Use $\ln([A]_0/[A]) = kt$ with $k = 0.693/t_{1/2}$.
Solution: $k = 0.693/20 = 0.0347$ min$^{-1}$. 90 percent completion means $[A] = 0.1 [A]_0$, so $[A]_0/[A] = 10$. Then $t = \ln(10)/k = 2.303/0.0347 = 66.4$ minutes. Answer: about 66 minutes, or equivalently $t = 2.303 \times t_{1/2} = 3.32 \times t_{1/2}$.
Key trick: Memorise "90 percent completion needs 3.32 half-lives" and "99 percent needs 6.64 half-lives" — JEE-Main asks these ratios year after year and shortcut-users save a full minute.
Worked Example 3 · JEE Advanced 2022 pattern
Problem: The rate of a reaction becomes eight times when the temperature is raised from 27°C to 87°C. Estimate the activation energy in kJ mol$^{-1}$. Use $R = 8.314$ J mol$^{-1}$ K$^{-1}$.
Approach: $k_2/k_1 = 8$, $T_1 = 300$ K, $T_2 = 360$ K. Apply two-point Arrhenius.
Solution: $\ln 8 = 3 \ln 2 = 2.079$. $(1/T_1 - 1/T_2) = 1/300 - 1/360 = 60/108000 = 5.556 \times 10^{-4}$ K$^{-1}$. $E_a = R \times 2.079 / (5.556 \times 10^{-4}) = 8.314 \times 2.079 / 5.556 \times 10^{-4} = 31{,}100$ J mol$^{-1}$ = 31.1 kJ mol$^{-1}$. Answer: about 31 kJ mol$^{-1}$.
Key trick: A common misfire is using log base 10 and forgetting the 2.303 factor — that gives $E_a$ ten times too small. Always check units mid-problem.
10 JEE-tier Practice Problems
JEE-Q1. [JEE Main 2022] For a first-order reaction, if 75 percent completion takes 40 minutes, then 50 percent completion takes:
Solution
75 percent completion = 2 half-lives. So $2 t_{1/2}$ = 40 min, giving $t_{1/2}$ = 20 min. First-order 50-percent time is exactly one half-life = 20 minutes.
JEE-Q2. [JEE Main 2023] The rate constant of a reaction is $5.4 \times 10^{-4}$ s$^{-1}$ at 27°C and increases four-fold at 47°C. The activation energy in kJ mol$^{-1}$ is closest to:
Solution
$\ln 4 = 1.386$. $(1/300 - 1/320) = 20/96000 = 2.083 \times 10^{-4}$. $E_a = 8.314 \times 1.386 / 2.083 \times 10^{-4} = 55{,}300$ J = 55 kJ.
JEE-Q3. [JEE Main pattern] For a zero-order reaction, $[A]$ vs $t$ is a straight line with negative slope. The half-life:
Solution
Zero-order: $t_{1/2} = [A]_0/(2k)$, directly proportional. Trap: many students mistake this for first-order behaviour.
JEE-Q4. [JEE Advanced 2023 pattern] For a reaction 2A + B → products, if the order with respect to A is 2 and with respect to B is 1, doubling both concentrations changes the rate by a factor of:
Solution
Rate = $k[A]^2[B]^1$. New rate = $k(2[A])^2(2[B]) = 8k[A]^2[B]$. Factor of 8.
JEE-Q5. [JEE Main 2024] Hydrolysis of an ester in dilute acid follows the rate law: rate = $k[\text{ester}][\text{H}^+]$. In excess acid, this becomes:
Solution
$[\text{H}^+]$ effectively constant → rate = $k'[\text{ester}]$ where $k' = k[\text{H}^+]$. First-order in ester — hence pseudo-first-order.
JEE-Q6. [JEE Main pattern] The temperature coefficient of a reaction is 2 (rate doubles per 10 K rise). What is the ratio of rates at 300 K and 340 K approximately?
Solution
40 K rise = 4 doublings = $2^4 = 16$. This is the "rule of thumb" version of Arrhenius; a rigorous computation uses the full equation.
JEE-Q7. [JEE Advanced pattern] A catalyst lowers $E_a$ from 75 to 50 kJ mol$^{-1}$ at 300 K. The rate increases by a factor of:
Solution
Rate ratio = $\exp((E_{a1}-E_{a2})/RT) = \exp(25000/(8.314 \times 300)) = \exp(10.02) \approx 2.24 \times 10^4$.
JEE-Q8. [JEE Main 2022] For the parallel first-order decomposition A → B ($k_1$) and A → C ($k_2$) with $k_1/k_2 = 2$, the fraction of A converted to B at long times is:
Solution
Fraction to B = $k_1/(k_1+k_2) = 2/(2+1) = 2/3$. This is the "branching ratio" — a JEE-Main staple.
JEE-Q9. [JEE Main 2024] For a first-order reaction, if the concentration falls from 800 to 50 in 20 min, the rate constant is:
Solution
$k = (1/t)\ln([A]_0/[A]) = (1/20)\ln(800/50) = (1/20)\ln(16) = (1/20)(2.773) = 0.1386$ min$^{-1}$.
JEE-Q10. [JEE Advanced pattern] Molecularity and order are equal for:
Solution
Molecularity = count of molecules in the elementary step. Order = experimental measurement. They coincide only for elementary reactions.
JEE Traps to Avoid
Practice — start here every day
Ten MCQs, three short-answer, two long-answer, three worked numericals. Every question tagged with the concept it tests. Do them cold — no textbook. Then check.
MCQ
Q1. For the reaction $2A + B \to$ products, the rate law is $\text{rate} = k[A][B]$. The overall order is:
Explanation
Overall order equals the sum of the exponents that actually appear in the empirical rate law: here $1 + 1 = 2$. The balanced-equation coefficients ($2 + 1 = 3$) are irrelevant to order.
Q2. The half-life of a first-order reaction is 6.93 minutes. The rate constant is:
Explanation
$k = 0.693 / t_{1/2} = 0.693 / 6.93 = 0.1$ per minute.
Q3. Which of the following is NOT true about a catalyst?
Explanation
A catalyst speeds up both forward and reverse reactions equally — equilibrium constant is unchanged.
Q4. The Arrhenius equation predicts that raising the temperature:
Explanation
$k = A e^{-E_a/RT}$ — the exponent grows less negative as $T$ rises, so $k$ grows exponentially.
Q5. The unit of the rate constant $k$ for a first-order reaction is:
Explanation
Rate has units of mol/L/s; $[A]^1$ has units of mol/L; so $k$ has units of per second.
Q6. Molecularity of a reaction is:
Explanation
Molecularity is a theoretical property of an elementary step. Order is measured. They can differ.
Q7. For a zero-order reaction, a plot of $[A]$ against time gives:
Explanation
Integrated zero-order: $[A]_t = [A]_0 - k t$. Straight line, slope $-k$.
Q8. If the rate constant doubles when temperature rises from 300 K to 310 K, the activation energy is approximately:
Explanation
$\ln 2 = (E_a/R)(1/300 - 1/310)$; solve $E_a \approx 53$ kJ/mol. The classic "ten-degree rule".
Q9. The pseudo-first-order rate of acid hydrolysis of methyl acetate in dilute acid depends on:
Explanation
Water is in vast excess; its concentration is effectively constant. So the observed rate depends only on ester concentration — pseudo-first order.
Q10. The Arrhenius plot of $\ln k$ versus $1/T$ has a slope of:
Explanation
Take log of Arrhenius: $\ln k = \ln A - (E_a/R)(1/T)$. Slope is $-E_a/R$; intercept is $\ln A$.
Short-answer questions
Q11
Distinguish between order and molecularity of a reaction with one example each.
Q12
Derive the integrated rate equation for a first-order reaction, and use it to show that the half-life is independent of the initial concentration.
Q13
Present the Arrhenius rate-constant relationship in full. Explain the physical significance of the pre-exponential factor $A$ and the barrier height $E_a$, and comment on why $A$ is often several orders larger than raw collision frequency.
Long-answer questions
Q14
Set out the collision-theory framework for two-molecule elementary steps. Introduce the activation-energy barrier and the orientation-dependent steric prefactor $P$, then discuss why the measured rate lies well below the raw geometric collision-frequency ceiling.
Q15
Describe how a catalyst speeds up a chemical transformation. Sketch a reaction-coordinate profile for the uncatalysed and catalysed routes, and argue why $K_{\text{eq}}$ is completely unaffected even though the forward and reverse rates both climb.
Worked numericals
Numerical 1
Problem. A first-order reaction has $k = 2.3 \times 10^{-3}$ per second. Compute the time required for 90 percent of the reactant to be consumed.
Solution
Fraction remaining = 0.1. Using $\ln([A]_0/[A]_t) = kt$: $\ln(1/0.1) = 2.303 = kt$; so $t = 2.303 / (2.3 \times 10^{-3}) = 1001$ seconds, about 16.7 minutes.
Numerical 2
Problem. The rate constant of a reaction is $1.5 \times 10^{-3}$ per second at 300 K and $6.0 \times 10^{-3}$ per second at 320 K. Compute the activation energy. ($R = 8.314$ J/K/mol.)
Solution
Using $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$: $\ln 4 = (E_a/8.314)(1/300 - 1/320) = (E_a/8.314)(2.083 \times 10^{-4})$. Solve: $E_a = 1.386 \times 8.314 / 2.083 \times 10^{-4} = 55{,}300$ J/mol, or about 55 kJ/mol.
Numerical 3
Problem. A radioactive isotope has a half-life of 20 years. What fraction of the original sample remains after 60 years?
Solution
60 years is three half-lives. After each half-life the fraction remaining halves: $1 \to 1/2 \to 1/4 \to 1/8$. Answer: one-eighth, or 12.5 percent.
10. Learn More
External resources: books, arXiv, journals, PhET simulations, MOOCs (NPTEL, Coursera, edX).
11. Reference & further reading
Curated pointers to primary sources, monographs, open courseware, and interactive simulations. Every entry has been checked against the topic scope of Class XII chemical kinetics.
Textbooks worth owning
- Atkins, Physical Chemistry — chapters on reaction kinetics and Arrhenius theory give the undergraduate derivation you can consult once your Class XII foundation is in place.
- Levine, Physical Chemistry — a heavier alternative that includes transition-state theory, useful if you plan on JEE-Advanced or IIT chemistry stream.
- Peter Atkins, Reactions — a lay reader's tour through chemical change; excellent bedside reading during exam prep to remember why the subject matters.
Free online courses
- NPTEL Physical Chemistry I (IIT Madras, Prof. B. L. Tembe) — the module on rate laws and Arrhenius plots exactly matches the JEE syllabus depth.
- MIT OpenCourseWare 5.60 Thermodynamics & Kinetics — Prof. Bawendi's lecture recordings; the second half of the course is dedicated to rate laws, integrated forms, and enzyme kinetics.
- Khan Academy: Kinetics — short animated explainers in Hindi and English; each concept is under six minutes so you can revise between tuition classes.
Interactive simulations
- PhET Reactions & Rates (University of Colorado Boulder) — drag temperature and concentration sliders, watch the Boltzmann distribution shift, count collisions with sufficient energy. This is the single best free tool for building an Arrhenius intuition.
- ChemCollective virtual labs — run a kinetics experiment with iodine clock, follow absorbance versus time, extract $k$ yourself. Uses only browser JavaScript.
- PraxiLabs kinetics module — a paid but school-friendly simulator that mirrors what you would do on NEET-style lab practicals.
Papers and primary literature
- Arrhenius, 1889 — the original paper that introduced the exponential temperature dependence. Only three pages long, and still readable in English translation.
- Michaelis & Menten, 1913 — the enzyme-kinetics paper whose form of rate law seeded modern biochemistry. Optional but a genuinely inspiring read.
- Eyring, 1935 — transition-state theory in one clean framework; skim this if you plan on chemistry as a degree.
Data tables for numerical work
- CRC Handbook of Chemistry and Physics — rate constants and activation energies for hundreds of standard reactions. Your library will have a print copy.
- NIST Chemical Kinetics Database — a free online database of gas-phase rate constants; useful for checking any numerical example you find in a coaching book.
Reviewer sign-off
Once you have finished this chapter, use the rehearsal box below to write a two-line note to yourself: what was easy, what was hard, what still confuses you. That single act cuts your revision time by half at exam week.