Strike a matchstick and it is over in two seconds. Leave an iron gate out in the monsoon and it takes twenty years to rust through. Both are chemical reactions. The difference between them is not what happens but how fast it happens — and putting an honest number on that difference is the whole job of Chemical Kinetics. It carries 7 marks in the CBSE Class 12 board paper, and if the word “kinetics” has been sitting in your stomach like a stone, please put it down. This chapter contains far fewer ideas than it appears to.
Here is why it usually feels heavy. Most notes hand you six formulas that seem to have arrived from six different planets: an average rate here, a rate law there, two integrated equations, a half-life, and then suddenly a logarithm with a temperature in it. Nothing connects. You memorise all six, and in the exam hall you cannot remember which one the question wants.
So we are going to do it the other way round. One reaction. One set of readings. Six different questions asked of the same table of numbers. Everything in this chapter — average rate, instantaneous rate, order, rate constant, half-life, even activation energy — is just a different lens pointed at the same six measurements. By the end you will not be reciting formulas; you will be reading a graph and letting the graph tell you the order. That single habit answers most chemical kinetics Class 12 important questions, and it is what turns integrated rate equation numericals with solutions from guesswork into a two-step routine.
C12H22O11 + H2O → C6H12O6 (glucose) + C6H12O6 (fructose)
Every worked example below draws on the same six readings. The numbers are deliberately clean so the arithmetic never hides the idea — but every single step is exactly what you would do with messy real laboratory data.
| Time t / min | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| [sucrose] / mol L−1 | 0.800 | 0.400 | 0.200 | 0.100 | 0.050 | 0.025 |
Two things will make this chapter much easier, and both are one click away. First, you must be completely comfortable with molarity, because every “rate” in this chapter is a concentration divided by a time — if mol L−1 still feels shaky, spend twenty minutes on the Solutions chapter, where concentration terms are built from scratch. Second, “instantaneous rate” is nothing more than a derivative wearing a lab coat, so if the idea of a slope at a single point is new, Application of Derivatives explains rate of change and tangent slopes in exactly the language we need.
Meet Your Tutor
Chemical Kinetics is a chapter where the same few ideas keep returning in different clothes: rate, order, time and temperature. I will help you read the data before choosing a formula, keep units visible, and test every numerical with a quick reasonableness check so the equations feel like tools rather than a list to memorise.
What You’ll Learn
This is the complete CBSE 2026-27 Unit 3 scope, nothing added and nothing quietly dropped. Tap any line to jump straight to it.
- Rate of a Chemical Reaction: Average and Instantaneous Rate
- Factors Influencing the Rate of a Reaction
- Rate Law, Order and Molecularity (and the Difference Between Them)
- Integrated Rate Equations for Zero and First Order, and Half-Life
- Reading the Graph: Which Plot Is a Straight Line?
- Temperature Dependence: The Arrhenius Equation and Activation Energy
- Collision Theory of Chemical Reactions
- Chemical Kinetics Class 12 Formula List (One-Page Revision)
Your Game Plan
Do not read this chapter front to back in one sitting. Kinetics rewards short, repeated passes far more than one long heroic one. Here is the order that actually works:
- Get “rate” solid first. Understand that rate is a slope on a concentration-time graph, and that average rate and instantaneous rate are two different slopes on the same curve. Everything else stands on this.
- Learn to read the graph before you learn the formulas. Spend real time on the straight-line plots. Once you can look at a dataset and say “that one is first order”, the integrated equations become obvious rather than memorised.
- Separate order from molecularity properly. This is the single most common place students lose easy marks. Two minutes of care now saves you a whole question later.
- Drill the units of k. They are different for every order and examiners love them. Do not skip this; it is free marks.
- Do the Arrhenius numericals with a calculator in hand. Reading them is not the same as doing them. Get the 1/T subtraction right and the rest follows.
- Finish with collision theory. It is qualitative and it explains why everything above is true, so it lands best once the maths already makes sense.
- Then the worksheet at the bottom. Cover the answers. Attempt every question on paper before you open a single accordion.
Study Notes
Rate of a Chemical Reaction: Average and Instantaneous Rate
Think about a car journey from Amritsar to Delhi. If you cover 450 km in 9 hours, your average speed is 50 km/h. But at no point did the speedometer sit politely at 50 — you crawled through traffic, you opened up on the highway. The speedometer reading at any single moment is your instantaneous speed. Chemical reactions behave in exactly the same way, and kinetics uses exactly the same two ideas.
The rate of a reaction is the change in concentration of a reactant or product per unit time. Because a reactant is being used up, its concentration change is negative, so we stick a minus sign in front to keep the rate a positive number. For a general reaction A → B:
Instantaneous rate = −d[A]/dt — the rate at one instant. Geometrically it is the slope of the tangent drawn at that single point on the curve.
Units of rate: mol L−1 s−1 (or mol L−1 min−1). For gases measured by pressure, atm s−1 is also used.
That word “tangent” is the only place calculus sneaks into this chapter, and it is the same tangent you meet in maths. If you want that idea properly nailed down, the Application of Derivatives chapter builds rate of change and tangent slope step by step — a genuinely useful half hour, because chemistry and maths are describing the same picture here.
0 to 20 min: [A] falls from 0.800 to 0.400 mol L−1.
Average rate = (0.800 − 0.400) / (20 − 0) = 0.400 / 20 = 0.0200 mol L−1 min−1 = 2.00 × 10−2 mol L−1 min−1
60 to 80 min: [A] falls from 0.100 to 0.050 mol L−1.
Average rate = (0.100 − 0.050) / 20 = 0.050 / 20 = 0.00250 mol L−1 min−1 = 2.50 × 10−3 mol L−1 min−1
Why it works: the second answer is exactly one-eighth of the first, even though both intervals are 20 minutes long. The reaction has genuinely slowed down, because there is less sucrose left to react. This is the single most important habit to build: an “average rate” is meaningless unless you say over which interval.
Compare with the average rate over 0–40 min:
(0.800 − 0.200) / 40 = 0.600 / 40 = 1.50 × 10−2 mol L−1 min−1
The average over the interval is about 2.16 times larger than the instantaneous rate at the end of that interval. That is not an error — it is the whole point. The average is dragged upwards by the fast early minutes.
Why it works: the chord starts high on the curve and ends at t = 40. The tangent only cares about the steepness right at t = 40, where the curve has already flattened. Same curve, two different slopes, two different questions.
One more piece of bookkeeping. When the balanced equation has coefficients other than 1, different species appear and disappear at different speeds, so we divide by the coefficient to get a single unambiguous rate of reaction. For aA + bB → cC + dD:
Our spine reaction has a coefficient of 1 on sucrose, which is exactly why it was chosen — no dividing, no confusion.
Rate of reaction = (1/1) d[O2]/dt = 2.5 × 10−4 mol L−1 s−1
−d[N2O5]/dt = 2 × 2.5 × 10−4 = 5.0 × 10−4 mol L−1 s−1
d[NO2]/dt = 4 × 2.5 × 10−4 = 1.0 × 10−3 mol L−1 s−1
Why it works: for every one O2 molecule made, two N2O5 molecules had to die and four NO2 molecules appeared. The coefficients are a recipe, and the rates follow the recipe.
Factors Influencing the Rate of a Reaction
Before any formula, get the physical picture. A reaction happens when particles meet, meet hard enough, and meet in the right orientation. Every single factor below works by changing one of those three things. If you remember that sentence, you can reason out this whole section in an exam even if your memory has gone blank.
1. Concentration (and pressure, for gases)
More crowded means more collisions per second. Double the concentration of a reactant and, in the simplest case, you double the number of collisions it makes — so the rate rises. Our spine dataset shows the flip side of this beautifully: as sucrose is used up, the solution gets less crowded, and the reaction visibly slows. For gases, raising the pressure squeezes the same molecules into a smaller volume, which is just concentration by another name.
2. Temperature
Raise the temperature and two things happen at once: molecules move faster (so they collide more often) and, far more importantly, a much larger fraction of them carry enough energy to actually react. The rough laboratory rule is that a 10 K rise roughly doubles the rate for many reactions near room temperature. We will make this precise with the Arrhenius equation later; for now, notice that this is why milk keeps in a fridge and spoils on a counter.
330 − 300 = 30 K = three steps of 10 K.
Factor = 2 × 2 × 2 = 23 = 8 times faster.
Why it works: each 10 K step multiplies, it does not add. Three doublings is eight-fold, not six-fold.
The honest caveat: this is a rule of thumb, not a law. It happens to hold when the activation energy is around 50–60 kJ mol−1 near room temperature. For our spine reaction the true factor per 10 K turns out to be about 4, not 2 — we calculate that properly in the Arrhenius section. Quote the rule of thumb in a theory answer; never use it in a numerical.
3. Catalyst
A catalyst is a substance that speeds up a reaction without being consumed by it. It does this by offering the reaction an alternative path with a lower activation energy — a lower mountain pass through the same range. Two consequences follow immediately, and both are examinable: a catalyst speeds up the forward and backward reactions equally, so it does not shift the position of equilibrium; and it does not change ΔH, because it does not move the energy of the reactants or the products, only the height of the barrier between them.
You already met catalysis without the vocabulary. In Class 10 you watched manganese dioxide tear hydrogen peroxide apart and came out of it unchanged — the Chemical Reactions and Equations chapter sets up decomposition and the idea of reaction conditions, which is the foundation this section is standing on.
4. Surface area of a solid reactant
A reaction between a solid and a liquid or gas can only happen at the surface, so the more surface you expose, the faster it goes. A log burns slowly; the same wood as sawdust can explode. Powdered chalk fizzes in acid far faster than a lump of the same mass.
Original cube: surface area = 6 × (1 cm)2 = 6 cm2
After cutting: 1000 cubes means 10 × 10 × 10, so each small cube has side 1/10 = 0.1 cm.
Surface area of one small cube = 6 × (0.1)2 = 6 × 0.01 = 0.06 cm2
Total = 1000 × 0.06 = 60 cm2
Increase = 60 / 6 = 10 times
Why it works: the volume is unchanged (1000 × 0.001 cm3 = 1 cm3), but area scales with the square of length while volume scales with the cube. Cutting into n3 pieces multiplies the area by n. Here n = 10.
5. Nature of the reactants
Ionic reactions in solution are usually near-instantaneous, because no covalent bonds need to be broken — the ions are already free and simply need to find each other. Reactions that require strong covalent bonds to be broken are far slower. This is why silver nitrate and sodium chloride give an instant white precipitate, while rusting takes years.
Rate Law, Order and Molecularity (and the Difference Between Them)
Here is the idea that trips up more students than any other in this chapter, so let us go slowly. The balanced chemical equation tells you the stoichiometry — how much of each thing you need. It does not tell you how the rate depends on concentration. That has to be measured in a laboratory. There is no shortcut, no clever inspection, no way to read it off the equation.
Rate = k [A]x [B]y
where k is the rate constant, x is the order with respect to A, y is the order with respect to B, and the overall order is x + y. Crucially, x need not equal a, and y need not equal b. The exponents come from experiment, full stop.
Order can be zero, a whole number, a fraction, or even negative. Molecularity is a completely different animal: it is the number of species that must collide simultaneously in a single elementary step. It can only be 1, 2 or 3 (three is already rare, because getting three particles to collide at once with the right energy and orientation is genuinely unlikely), and it must be a whole number. Molecularity applies only to an elementary step; a multi-step reaction has no single molecularity at all.
| Point of comparison | Order of reaction | Molecularity |
|---|---|---|
| Where it comes from | Experiment only | The balanced elementary step (theory) |
| Possible values | 0, whole numbers, fractions, even negative | 1, 2 or 3 only — never zero, never fractional |
| Applies to | Overall reactions and elementary steps alike | Elementary steps only |
| Meaning | Sum of the powers in the rate law | Number of particles colliding at once |
| For a multi-step reaction | Set by the slowest (rate-determining) step | Undefined for the overall reaction |
Expt 1: [A] = 0.10, [B] = 0.10, rate = 2.0 × 10−4
Expt 2: [A] = 0.20, [B] = 0.10, rate = 8.0 × 10−4
Expt 3: [A] = 0.10, [B] = 0.20, rate = 4.0 × 10−4
(concentrations in mol L−1, rates in mol L−1 s−1)
Step 1 — order in A. Compare experiments 1 and 2: [B] is held fixed, [A] doubles, rate goes 2.0 → 8.0, a factor of 4. Since 2x = 4, x = 2.
Step 2 — order in B. Compare 1 and 3: [A] is held fixed, [B] doubles, rate goes 2.0 → 4.0, a factor of 2. Since 2y = 2, y = 1.
Step 3 — write the rate law. Rate = k[A]2[B], overall order = 3.
Step 4 — find k. Using experiment 1:
k = rate / ([A]2[B]) = (2.0 × 10−4) / ((0.10)2 × 0.10) = (2.0 × 10−4) / (1.0 × 10−3) = 0.20 L2 mol−2 s−1
Check with experiment 2: 0.20 × (0.20)2 × 0.10 = 0.20 × 0.04 × 0.10 = 8.0 × 10−4 ✓
Check with experiment 3: 0.20 × (0.10)2 × 0.20 = 0.20 × 0.01 × 0.20 = 4.0 × 10−4 ✓
Why it works: you change one concentration at a time and watch what the rate does. Always verify k against a row you did not use to find it — it costs ten seconds and catches arithmetic slips.
Now the units of k, which are worth easy marks and which almost everyone gets wrong at least once. Because rate always has units of mol L−1 s−1, and [A]n has units of (mol L−1)n, the units of k must make the equation balance:
Units of k = (mol L−1 s−1) / (mol L−1)n = mol(1−n) L(n−1) s−1
Substitute the orders one by one:
n = 0: mol1 L−1 s−1 = mol L−1 s−1 (same units as the rate itself)
n = 1: mol0 L0 s−1 = s−1 (no concentration units at all)
n = 2: mol−1 L1 s−1 = L mol−1 s−1
n = 3: mol−2 L2 s−1 = L2 mol−2 s−1
Why it works: you never need to memorise four separate results. Memorise the one formula mol(1−n) L(n−1) s−1 and substitute. And notice the check that catches every error: our Example 6 gave overall order 3, and its k came out in L2 mol−2 s−1 — exactly as this formula predicts. ✓
Finally, the case that makes order and molecularity look like they contradict each other — and which the board loves to ask about. Our own spine reaction is the classic example.
C12H22O11 + H2O → C6H12O6 + C6H12O6
Two species react, so the elementary step is bimolecular and the true rate law is Rate = k′[sucrose][H2O], which would be second order overall. Yet experiment finds it to be first order. Why?
Because water is the solvent and is present in enormous excess. Pure water is about 1000 g L−1 ÷ 18 g mol−1 = 55.5 mol L−1. Our sucrose starts at 0.800 mol L−1 — roughly 69 times less concentrated.
Even if every last sucrose molecule reacts, water falls from 55.5 to about 54.7 mol L−1, a change of about 1.4%. To any measuring instrument, [H2O] is constant. So we fold it into the constant:
Rate = k′[sucrose][H2O] = (k′ × 55.5)[sucrose] = k[sucrose]
The reaction is therefore second order in reality but first order in behaviour. We call this a pseudo first order reaction.
Why it works: order is about what the rate visibly depends on. A quantity that never changes cannot make the rate change, so it disappears into k. Molecularity (2) and observed order (1) disagree, and both are correct — they are answering different questions.
(a) A reaction of order 1.5. Possible. Fractional orders are common in complex multi-step reactions. Molecularity could never be 1.5.
(b) A reaction of order zero. Possible — for example, the decomposition of ammonia on a hot platinum surface. The metal surface is fully covered with ammonia, so adding more ammonia changes nothing. Molecularity can never be zero, because you cannot have a step in which no particles take part.
(c) An elementary step with molecularity 4. Not observed in practice. Requiring four particles to arrive at the same point, at the same instant, with enough energy and the right orientation is so improbable that such steps do not contribute measurably.
(d) A reaction whose order changes with temperature. Possible, if the change in temperature makes a different step become the slowest one.
Why it works: order is a description of measured behaviour and is free to be strange. Molecularity is a count of physical particles in one step and must therefore be a small positive whole number.
Integrated Rate Equations for Zero and First Order, and Half-Life
The rate law tells you the rate at one instant. That is useful, but it is not what a laboratory actually gives you. A laboratory gives you a table — concentration at 0 minutes, at 20 minutes, at 40 minutes. An integrated rate equation is what you get when you take the rate law and turn it into a direct relationship between concentration and time. It answers the question you actually have: “given where I started, where will I be after t minutes?”
Zero order reactions
A zero order reaction is one whose rate does not depend on the concentration of the reactant at all. Rate = k[A]0 = k, a constant. It sounds bizarre until you picture the usual cause: a reaction happening on a catalyst surface that is already completely covered. Adding more reactant does not help, because there is nowhere for it to sit. The decomposition of ammonia on a hot platinum surface behaves this way, as does the decomposition of HI on gold.
[A] = [A]0 − kt
That is the equation of a straight line (y = c + mx) if you plot [A] against t: intercept [A]0, slope −k.
Half-life: set [A] = [A]0/2 and solve for t:
t1/2 = [A]0 / 2k
Note carefully: for zero order the half-life depends on the starting concentration. Units of k: mol L−1 s−1.
t / min : 0 10 20 30 40 50
[A] / M : 0.500 0.400 0.300 0.200 0.100 0.000
Step 1 — is it zero order? Look at the drops: 0.100, 0.100, 0.100, 0.100, 0.100 mol L−1 in every 10-minute window. The concentration falls by the same amount in equal times, regardless of how much is left. That is the fingerprint of zero order.
Step 2 — find k. From [A] = [A]0 − kt, take t = 30:
0.200 = 0.500 − k(30) ⇒ k(30) = 0.300 ⇒ k = 0.0100 mol L−1 min−1
(Units confirm zero order ✓)
Step 3 — half-life. t1/2 = [A]0/2k = 0.500 / (2 × 0.0100) = 0.500 / 0.0200 = 25 min
Check against the data: at t = 25, [A] = 0.500 − 0.0100(25) = 0.250 mol L−1 — exactly half. ✓
Step 4 — the second half-life. Now start afresh from 0.250 mol L−1:
t1/2 = 0.250 / 0.0200 = 12.5 min
Successive half-lives are therefore 25, 12.5, 6.25 min — each one half the previous one.
Step 5 — time to disappear completely. Set [A] = 0: t = [A]0/k = 0.500 / 0.0100 = 50 min. A zero order reaction genuinely finishes; a first order one, mathematically, never quite does.
Why it works: the reaction chews through reactant at a fixed rate, like a machine eating at a fixed speed. Once you have less left, that fixed speed clears it faster in proportion — hence shrinking half-lives.
First order reactions
A first order reaction has Rate = k[A]. The rate is proportional to how much is left, which is the most natural behaviour in chemistry: twice as many molecules, twice as many reactions per second. Radioactive decay is first order. So is our sucrose inversion. So is the decomposition of N2O5.
ln([A]0/[A]) = kt or equivalently ln[A] = ln[A]0 − kt
In base-10 form, which is what most board questions use:
k = (2.303/t) log([A]0/[A])
Straight line: plot ln[A] against t — slope −k, intercept ln[A]0. (Or log[A] against t, slope −k/2.303.)
Half-life: put [A] = [A]0/2, so ln 2 = k t1/2:
t1/2 = 0.693 / k
Note carefully: for first order the half-life is independent of the starting concentration. Units of k: s−1 or min−1.
Using k = (2.303/t) log([A]0/[A]) with [A]0 = 0.800 mol L−1:
t = 20: k = (2.303/20) log(0.800/0.400) = (0.11515)(0.30103) = 0.03466
t = 40: k = (2.303/40) log(0.800/0.200) = (0.057575)(0.60206) = 0.03466
t = 60: k = (2.303/60) log(0.800/0.100) = (0.038383)(0.90309) = 0.03466
t = 80: k = (2.303/80) log(0.800/0.050) = (0.028788)(1.20412) = 0.03466
t = 100: k = (2.303/100) log(0.800/0.025) = (0.02303)(1.50515) = 0.03466
Five readings, one constant. k = 3.47 × 10−2 min−1 (3 s.f.), and the units — time−1 with no “mol” in sight — confirm first order. ✓
Half-life: t1/2 = 0.693/k = 0.693 / 0.034657 = 20.0 min
The visual shortcut you should have spotted ten minutes ago: look back at the raw table. 0.800 → 0.400 → 0.200 → 0.100 → 0.050 → 0.025. The concentration halves every single 20-minute window, no matter how much is left. A constant half-life is the signature of first order, and you can read it straight off the table without any calculation at all.
Why it works: if the rate is proportional to what remains, then the fractional loss per unit time is fixed. Losing half of a lot and losing half of a little take exactly the same time.
“75% complete” means 25% remains, so [A]0/[A] = 100/25 = 4.
t75% = (2.303/k) log 4 = (2.303/0.034657)(0.60206) = 66.45 × 0.60206 = 40.0 min
“90% complete” means 10% remains, so the ratio is 10.
t90% = (2.303/k) log 10 = (2.303/0.034657)(1) = 66.4 min
“99% complete” means 1% remains, so the ratio is 100.
t99% = (2.303/k) log 100 = 66.45 × 2 = 132.9 min
Two elegant relationships worth memorising:
t75% = 2 × t1/2 (here 2 × 20 = 40 min ✓)
t99% = 2 × t90% (here 2 × 66.4 = 132.9 min ✓)
Why it works: 75% complete is just two successive halvings (100 → 50 → 25), and each halving takes one half-life. Similarly, going from 100% to 10% and then from 10% to 1% are both a “divide by ten”, and for first order every equal ratio takes an equal time.
Step 1 — what remains? 30% has reacted, so 70% remains. [A]0/[A] = 100/70.
Step 2 — apply the integrated equation.
k = (2.303/25) log(100/70) = (0.09212)(0.154902) = 0.01427 min−1
(equivalently k = (1/25) ln(1/0.70) = (0.04)(0.356675) = 0.014267 min−1)
Step 3 — half-life.
t1/2 = 0.693 / 0.014267 = 48.6 min
Sanity check: the reaction is not yet halfway at 25 minutes (only 30% done), so the half-life must be longer than 25 minutes. 48.6 min ✓
Why it works: percentages are ratios, and first order kinetics only ever cares about ratios. You never need to know the actual concentration in mol L−1 — which is exactly why questions can give you a percentage and nothing else.
Reading the Graph: Which Plot Is a Straight Line?
This short section is the hinge of the whole chapter, and it is the one most notes skip. You now have two integrated equations. Rather than filing them away as two more formulas, notice what they have in common: both of them are the equation of a straight line. They just disagree about what you should put on the y-axis.
| Feature | Zero order | First order |
|---|---|---|
| Rate law | Rate = k (independent of [A]) | Rate = k[A] |
| Integrated form | [A] = [A]0 − kt | ln([A]0/[A]) = kt or k = (2.303/t) log([A]0/[A]) |
| Straight-line plot | [A] vs t | ln[A] vs t (or log[A] vs t) |
| Slope of that line | −k | −k (or −k/2.303 if using log10) |
| Intercept | [A]0 | ln[A]0 |
| Units of k | mol L−1 s−1 | s−1 |
| Half-life | t1/2 = [A]0/2k — depends on [A]0 | t1/2 = 0.693/k — independent of [A]0 |
| Successive half-lives | Shrink: 25, 12.5, 6.25 min … | Constant: 20, 20, 20 min … |
| Does it ever finish? | Yes, at t = [A]0/k | No — only ever gets closer to zero |
1. Do equal times give equal drops? (0.500 → 0.400 → 0.300 …) ⇒ ZERO order. Plot [A] vs t and it is straight.
2. Do equal times give equal ratios? (0.800 → 0.400 → 0.200 …) ⇒ FIRST order. Plot ln[A] vs t and it is straight.
3. Is the half-life constant? If yes, first order — every time, no exceptions.
Two words to carry into the exam hall: differences means zero, ratios means first.
Set P: t = 0, 5, 10, 15 min → [A] = 0.80, 0.60, 0.40, 0.20 M
Differences: 0.20, 0.20, 0.20 — equal drops. Zero order. Plot [A] vs t; slope = −k gives k = 0.20/5 = 0.040 mol L−1 min−1.
Set Q: t = 0, 5, 10, 15 min → [A] = 0.80, 0.40, 0.20, 0.10 M
Ratios: each is half the one before — equal ratios. First order. Plot ln[A] vs t; t1/2 = 5 min, so k = 0.693/5 = 0.139 min−1.
Set R: t = 0, 20, 40, 60, 80, 100 min → our spine data
Constant half-life of 20 min. First order, k = 0.693/20 = 0.0347 min−1 — matching the five-reading calculation in Example 11 exactly. ✓
Why it works: you did not integrate anything. You looked at the shape of the numbers. That is what the phrase “reading the graph tells you the order” actually means in practice.
Temperature Dependence: The Arrhenius Equation and Activation Energy
Everything so far has held the temperature fixed. Now we let it move, and we find something surprising: the effect is enormous and it is not linear. A 10 K rise — barely noticeable to your hand — can double or quadruple the rate. Something exponential must be going on.
Here is the picture. In any sample, molecules do not all have the same energy; they have a wide spread of energies. Only those carrying at least a certain minimum — the activation energy, Ea — can get over the barrier and become products. Raising the temperature does not lift every molecule a little; it dramatically swells the population in the high-energy tail of the distribution. A small temperature rise, a large change in how many molecules can react.
A = the pre-exponential or frequency factor (roughly, how often collisions happen with the right orientation)
Ea = activation energy in J mol−1
R = 8.314 J K−1 mol−1
T = absolute temperature in kelvin (never in °C)
Taking logs gives the straight-line form:
ln k = ln A − Ea/RT ⇒ plot ln k against 1/T, slope = −Ea/R, intercept = ln A
And for two temperatures, the workhorse of every board numerical:
log(k2/k1) = (Ea / 2.303R) × (1/T1 − 1/T2)
Step 1 — the ratio. k2/k1 = 0.139/0.0347 = 4.00 (the reaction is four times faster for a 10 K rise).
Step 2 — the temperature term. Do this carefully; it is where marks are lost.
1/T1 − 1/T2 = 1/313 − 1/323 = (323 − 313)/(313 × 323) = 10 / 101099 = 9.891 × 10−5 K−1
Step 3 — substitute.
log 4.00 = 0.6021
0.6021 = (Ea / (2.303 × 8.314)) × 9.891 × 10−5
0.6021 = Ea × (9.891 × 10−5 / 19.147)
0.6021 = Ea × 5.166 × 10−6
Ea = 0.6021 / (5.166 × 10−6) = 1.165 × 105 J mol−1
Ea ≈ 116.5 kJ mol−1
(Using natural logs directly: Ea = R ln4 / (9.891 × 10−5) = 8.314 × 1.3863 / 9.891 × 10−5 = 1.165 × 105 J mol−1 — the same answer. ✓)
Why it works: the ratio of rate constants depends only on Ea and the two temperatures — the frequency factor A cancels completely. That is why you never need to know A to find Ea.
Step 1 — temperature term.
1/313 − 1/333 = (333 − 313)/(313 × 333) = 20/104229 = 1.9188 × 10−4 K−1
Step 2 — substitute.
log(k2/k1) = (116523 / 19.147) × 1.9188 × 10−4 = 6086.0 × 1.9188 × 10−4 = 1.1678
k2/k1 = 101.1678 = 14.7
Step 3 — the answers.
k2 = 14.7 × 0.0347 = 0.510 min−1
t1/2 = 0.693 / 0.510 = 1.36 min
Why it works — and why it matters: a 20 K rise, from 40 °C to 60 °C, has taken the half-life from 20 minutes down to about 82 seconds. Nothing about the chemistry changed. Only the fraction of molecules able to clear a 116.5 kJ mol−1 barrier changed — and it changed by a factor of nearly fifteen.
From ln k = ln A − Ea/RT, the slope is −Ea/R.
−Ea/R = −1.40 × 104
Ea = 1.40 × 104 × 8.314 = 1.164 × 105 J mol−1 = 116.4 kJ mol−1
This agrees with the two-point answer of 116.5 kJ mol−1 to within experimental rounding. ✓
The trap to watch: if the student had plotted log10 k against 1/T instead, the slope would be −Ea/2.303R, and Ea = 2.303 × 8.314 × slope magnitude. Always check which logarithm the axis is using before you multiply.
Why it works: a graph uses every data point instead of just two, so it averages out experimental scatter. This is why real laboratories always take the graphical route.
Fraction = e−Ea/RT
At 313 K: Ea/RT = 116523 / (8.314 × 313) = 116523 / 2602.3 = 44.78
Fraction = e−44.78 = 3.6 × 10−20
At 323 K: Ea/RT = 116523 / (8.314 × 323) = 116523 / 2685.4 = 43.39
Fraction = e−43.39 = 1.4 × 10−19
Ratio = 4.0 — exactly the factor by which the rate constant increased in Example 15. ✓
Why it works, and why it is worth sitting with: at 313 K, fewer than four molecules in every hundred billion billion carry enough energy to react. Warm the flask by ten degrees — a change you could not detect by touch — and that number quadruples. The exponent moved by only 1.39, but because it sits in an exponential, the effect is dramatic. That is why temperature matters so much more than concentration in kinetics.
Collision Theory of Chemical Reactions
Collision theory is the story underneath all the algebra. It is mostly qualitative, it is genuinely satisfying once it clicks, and it is where the marks are for the “explain why” questions.
The starting assumption is simple: molecules react only when they collide. Treat them as hard spheres, work out how often they bump into each other, and you get the collision frequency Z. For a gas at ordinary pressure this number is staggering — of the order of 1030 collisions per litre per second. If every collision produced a reaction, every gas-phase reaction would be over before you finished setting up the apparatus. They are not. So most collisions must be doing nothing at all.
1. The energy condition. The colliding pair must together carry at least the activation energy Ea. The fraction that manages this is e−Ea/RT.
2. The orientation condition. They must be facing the right way when they meet. A molecule struck on its unreactive end simply bounces off. This is accounted for by the steric or probability factor P, a number between 0 and 1.
Putting both together:
Rate = P × Z × e−Ea/RT
Compare with k = A e−Ea/RT and you can see what the mysterious A in the Arrhenius equation actually is: A = P × Z. The two equations were describing the same thing all along.
The everyday version: imagine trying to unlock a door in a dark corridor full of people jostling past. Bumping into the door is not enough (that is a collision). You need to be moving hard enough to push it (that is the energy condition) and holding the key the right way round (that is the orientation condition). Most bumps achieve nothing.
If every collision reacted:
Rate = 1.0 × 1030 ÷ (6.022 × 1023) = 1.7 × 106 mol L−1 s−1 — physically absurd, the reaction would be instantaneous.
Applying the energy condition (fraction 3.6 × 10−20, from Example 18):
Rate = 1.7 × 106 × 3.6 × 10−20 = 5.9 × 10−14 mol L−1 s−1
Applying the orientation condition (P = 0.01):
Rate = 5.9 × 10−14 × 0.01 = 5.9 × 10−16 mol L−1 s−1
Why it works: collision frequency alone over-predicts the rate by about twenty-two orders of magnitude. The exponential energy term does almost all of the correcting; the steric factor trims the rest. Note honestly that Z and P here are illustrative order-of-magnitude values, not measured constants for a specific reaction — the purpose is to show you the scale of each correction.
kcat/kuncat = e−Ea,cat/RT ÷ e−Ea,uncat/RT = e(Ea,uncat − Ea,cat)/RT
ΔEa = 116500 − 70000 = 46500 J mol−1
Exponent = 46500 / (8.314 × 313) = 46500 / 2602.3 = 17.87
Ratio = e17.87 = 5.8 × 107
The catalysed reaction is roughly 58 million times faster. A half-life of 20 minutes becomes about 20 microseconds.
Why it works: Ea sits in an exponent. Shaving 40% off the barrier does not make the reaction 40% faster — it changes it by nearly eight orders of magnitude. This is the entire commercial logic of industrial catalysis.
Two honest limitations, worth a line in a long answer. First, collision theory treats molecules as featureless hard spheres, which is why it needs the fudge-like factor P to work for anything more complicated than two atoms. Second, it takes no account of internal vibrations or of the fact that energy has to end up in the right bond. Transition state theory, which you meet later, repairs both. For Class 12, collision theory is the right level of explanation — just do not claim it is exact.
Chemical Kinetics Class 12 Formula List (One-Page Revision)
Everything in one place for the night before the exam. If you can reconstruct this table from memory, and you can say in one sentence what each line is for, you are ready.
| Quantity | Formula | Use it when |
|---|---|---|
| Average rate | −Δ[A]/Δt | Two readings, a time interval |
| Instantaneous rate | −d[A]/dt (slope of tangent) | One instant, from a graph |
| Rate with coefficients | −(1/a)d[A]/dt = +(1/c)d[C]/dt | Coefficients are not all 1 |
| Rate law | Rate = k[A]x[B]y, order = x + y | Initial-rate data tables |
| Units of k | mol(1−n) L(n−1) s−1 | Any order n; also as a self-check |
| Zero order integrated | [A] = [A]0 − kt | Equal drops in equal times |
| Zero order half-life | t1/2 = [A]0/2k | Depends on [A]0 |
| First order integrated | k = (2.303/t) log([A]0/[A]) | Equal ratios in equal times |
| First order half-life | t1/2 = 0.693/k | Independent of [A]0 |
| Arrhenius equation | k = A e−Ea/RT | Any temperature question |
| Arrhenius, two temperatures | log(k2/k1) = (Ea/2.303R)(1/T1 − 1/T2) | Two k values, or find k2 |
| Arrhenius, graphical | ln k vs 1/T, slope = −Ea/R | Several k values |
| Collision theory | Rate = P × Z × e−Ea/RT, A = PZ | “Explain why” questions |
| Energy profile | Ea(forward) − Ea(reverse) = ΔH | Checking a diagram is consistent |
And always: T in kelvin, Ea in joules when you use R = 8.314.
Practice Worksheet
Ten questions, written in the style the board actually uses. Do them on paper first — genuinely, with a pen — and only then open the answer. Reading a solution feels like learning and is not. Take R = 8.314 J K−1 mol−1 throughout.
Q1. Using the spine dataset from the top of this page, calculate the average rate of consumption of sucrose between t = 20 min and t = 60 min.
Show Answer
Average rate = −Δ[A]/Δt = (0.400 − 0.100) / (60 − 20) = 0.300 / 40
= 7.50 × 10−3 mol L−1 min−1
Sensible-answer check: this sits between the 0–20 min average (2.00 × 10−2) and the 60–80 min average (2.50 × 10−3), exactly as it should for a reaction that is slowing down. ✓
Q2. Write the units of the rate constant k for a zero order, a first order and a second order reaction, and state the general formula you used.
Show Answer
Zero order (n = 0): mol L−1 s−1
First order (n = 1): s−1
Second order (n = 2): L mol−1 s−1 (equivalently mol−1 L s−1)
Notice the pattern: zero order k has the same units as the rate itself, and first order k has no concentration units at all.
Q3. Show that the spine dataset represents a first order reaction, calculate the rate constant, and hence find the half-life.
Show Answer
t = 20: k = (2.303/20) log(0.800/0.400) = 0.11515 × 0.3010 = 0.03466 min−1
t = 60: k = (2.303/60) log(0.800/0.100) = 0.038383 × 0.9031 = 0.03466 min−1
t = 100: k = (2.303/100) log(0.800/0.025) = 0.02303 × 1.5051 = 0.03466 min−1
k is constant, so the reaction is first order.
k = 3.47 × 10−2 min−1
Half-life: t1/2 = 0.693/k = 0.693/0.034657 = 20.0 min
Confirmed directly by the table — the concentration halves in every 20-minute window. ✓
Q4. A zero order reaction has [A]0 = 0.20 mol L−1 and k = 5.0 × 10−3 mol L−1 s−1. Find (a) the half-life and (b) the time for the reactant to disappear completely.
Show Answer
t1/2 = [A]0/2k = 0.20 / (2 × 5.0 × 10−3) = 0.20 / 0.010 = 20 s
(b) Complete disappearance. Set [A] = 0 in [A] = [A]0 − kt:
t = [A]0/k = 0.20 / (5.0 × 10−3) = 40 s
Note the neat relationship for zero order: the reaction takes exactly two half-lives to finish completely. A first order reaction, by contrast, never reaches zero.
Q5. A first order reaction is 99% complete in 230 minutes. Calculate the rate constant and the half-life.
Show Answer
k = (2.303/t) log([A]0/[A]) = (2.303/230) × log 100 = (2.303/230) × 2
k = 0.010013 × 2 = 2.00 × 10−2 min−1
t1/2 = 0.693/k = 0.693 / 0.02002 = 34.6 min
Check: t99% should be about 6.64 half-lives (since ln 100 / ln 2 = 6.64). Indeed 6.64 × 34.6 = 230 min. ✓
Q6. For A + B → products, the following initial rates were measured at constant temperature. Determine the rate law, the overall order and the value of k.
Expt 1: [A] = 0.10, [B] = 0.10, rate = 1.2 × 10−3
Expt 2: [A] = 0.20, [B] = 0.10, rate = 2.4 × 10−3
Expt 3: [A] = 0.10, [B] = 0.30, rate = 1.08 × 10−2
(concentrations mol L−1, rates mol L−1 s−1)
Show Answer
Order in B (compare 1 and 3, [A] fixed): [B] triples, rate goes 1.2 × 10−3 → 1.08 × 10−2, a factor of 9. So 3y = 9 and y = 2.
Rate law: Rate = k[A][B]2 — overall order = 3
Value of k (from experiment 1):
k = 1.2 × 10−3 / (0.10 × (0.10)2) = 1.2 × 10−3 / 1.0 × 10−3 = 1.2 L2 mol−2 s−1
Verify with experiment 3: 1.2 × 0.10 × (0.30)2 = 1.2 × 0.10 × 0.09 = 1.08 × 10−2 ✓
Units check: overall order 3 predicts L2 mol−2 s−1 ✓
Q7. The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Calculate the activation energy.
Show Answer
1/T1 − 1/T2 = 1/300 − 1/310 = (310 − 300)/(300 × 310) = 10/93000 = 1.0753 × 10−4 K−1
Step 2 — Arrhenius.
log(k2/k1) = log 2 = 0.3010
0.3010 = (Ea / (2.303 × 8.314)) × 1.0753 × 10−4
0.3010 = Ea × (1.0753 × 10−4 / 19.147) = Ea × 5.616 × 10−6
Ea = 0.3010 / 5.616 × 10−6 = 5.36 × 104 J mol−1
Ea ≈ 53.6 kJ mol−1
This is exactly the size of activation energy for which the familiar “rate doubles per 10 K” rule of thumb happens to hold near room temperature.
Q8. Give three clear differences between the order of a reaction and its molecularity, and explain why the inversion of cane sugar is called a pseudo first order reaction.
Show Answer
1. Order is determined experimentally from the rate law; molecularity is deduced theoretically from the balanced elementary step.
2. Order may be zero, fractional or even negative; molecularity can only be 1, 2 or 3 and is always a positive whole number.
3. Order applies to overall reactions as well as elementary steps; molecularity is meaningful only for a single elementary step and is undefined for a multi-step reaction.
Pseudo first order: the reaction is
C12H22O11 + H2O → C6H12O6 + C6H12O6
Two species react, so it is bimolecular and truly second order: Rate = k′[sucrose][H2O]. But water is the solvent at about 55.5 mol L−1, vastly in excess, so [H2O] is effectively constant. Absorbing it into the constant gives Rate = k[sucrose], which is first order in behaviour. Hence pseudo first order.
Q9. For a zero order and a first order reaction, state (a) which graph gives a straight line, (b) what the slope of that line represents, and (c) how the successive half-lives behave.
Show Answer
(a) Plot [A] against t — straight line, intercept [A]0.
(b) Slope = −k, so k is the magnitude of the slope (units mol L−1 s−1).
(c) t1/2 = [A]0/2k, which depends on the initial concentration. Successive half-lives halve each time: for [A]0 = 0.500 M and k = 0.0100 M min−1 they are 25, 12.5, 6.25 min.
First order
(a) Plot ln[A] against t — straight line, intercept ln[A]0. (If log10[A] is plotted, the slope is −k/2.303.)
(b) Slope = −k (units s−1 or min−1).
(c) t1/2 = 0.693/k, independent of the initial concentration. Successive half-lives are all equal — for our spine reaction, 20 min every time.
Q10. (a) A catalyst lowers the activation energy of a reaction by 20 kJ mol−1. By what factor does the rate constant increase at 300 K, assuming the frequency factor is unchanged? (b) Explain why the catalyst does not change ΔH for the reaction.
Show Answer
Exponent = 20000 / (8.314 × 300) = 20000 / 2494.2 = 8.019
Ratio = e8.019 = 3.0 × 103
The reaction becomes roughly 3000 times faster — from a 20 kJ mol−1 saving on a barrier, which is a striking demonstration that Ea sits inside an exponential.
(b) ΔH is fixed by the difference in energy between the reactants and the products. A catalyst provides an alternative pathway with a lower barrier, but it does not alter the energy of either the reactants or the products — it only lowers the peak between them. Since both end points are unmoved, ΔH is unchanged. A useful corollary: because the catalyst lowers Ea for the forward and reverse reactions by the same amount, it speeds both up equally and therefore does not shift the position of equilibrium; it only helps equilibrium arrive sooner.
One last thought before you go. Every reaction you have ever met sits somewhere on the speed scale we have just learned to measure — and the slowest one you know is probably the iron gate outside your house. Rusting is kinetics too, just with the dial turned right down, and if you want to see the chemistry behind it again, the Class 10 Metals and Non-metals chapter covers corrosion and its prevention. It is a good reminder that this chapter is not abstract: it is the reason food spoils, medicines expire, and engines need catalytic converters.
That is the whole idea of kaizen: not one heroic all-night session, but one small, honest improvement, repeated. Do the three readings tonight. Do the Arrhenius numerical tomorrow. Do the worksheet on Sunday. In a fortnight this chapter will feel like something you own rather than something you are bracing for — and you will barely notice it happening. Small steps, every day. That is how 7 marks quietly become 7 marks you can count on.