One of the questions that has stayed with me throughout my chemistry journey is why diamond doesn't spontaneously transform into graphite, despite graphite being the thermodynamically more stable form of carbon. That question is basically the whole reason chemical kinetics exists as a subject. Thermodynamics can tell you a reaction is favourable, but it says absolutely nothing about how fast it happens — and that gap between "possible" and "practical" is where this entire chapter lives.
This is one of the highest-yield chapters for IIT-JAM, BITSAT, GATE, CSIR-NET, and for anyone preparing TGT/PGT chemistry papers, because it blends conceptual clarity with calculation-heavy problems — exactly the combination that examiners love. I've tried to write this the way I'd actually explain it to a student sitting across from me, working through the logic rather than just listing formulas.
Table of Contents
- What Exactly Is Chemical Kinetics?
- Rate of a Reaction — Average and Instantaneous
- Units of Rate and Solved Rate Problems
- Expressing Rate for Reactions with Unequal Stoichiometric Coefficients
- Factors Influencing Reaction Rate
- Rate Law and Rate Constant
- Order of a Reaction
- Elementary and Complex Reactions
- Molecularity vs Order — The Classic Exam Trap
- Integrated Rate Equations — Zero Order
- Integrated Rate Equations — First Order
- First Order Reactions in Gas Phase (Pressure Method)
- Half-Life of a Reaction
- Pseudo First Order Reactions
- Temperature Dependence — The Arrhenius Equation
- Effect of Catalyst
- Collision Theory of Reaction Rates
- Quick Revision Summary
- Practice MCQs (60+ with Answers)
1. What Exactly Is Chemical Kinetics?
Every chemical reaction raises three separate questions, and it's worth keeping them apart in your head because exam papers love to test whether you can distinguish them. First, is the reaction feasible at all — something thermodynamics answers through Gibbs energy, where a reaction proceeds spontaneously at constant temperature and pressure only when ΔG is negative. Second, how far will it go before reaching equilibrium — that's the domain of chemical equilibrium. Third, and this is where kinetics comes in, how quickly does it get there?
The word "kinetics" itself comes from the Greek kinesis, meaning movement. Chemical kinetics is the branch of chemistry dealing with reaction rates and the mechanisms by which reactions occur. Here's the classic illustration examiners love: thermodynamic data tell us diamond should convert to graphite, yet in practice this conversion is so slow that it's imperceptible on any human timescale — which is exactly why people say "diamonds are forever." A thermodynamically favourable process can still be kinetically frozen.
Reactions span an enormous range of speeds. Precipitation of silver chloride when you mix silver nitrate and sodium chloride solutions happens almost instantly — that's an ionic reaction. Rusting of iron, by contrast, creeps along over months or years. Somewhere in between sit moderate-speed reactions like the inversion of cane sugar or hydrolysis of starch. Kinetics gives you the tools to quantify all three categories and — just as importantly — to understand what factors you could tweak to speed up or slow down a given reaction.
2. Rate of a Reaction — Average and Instantaneous
Think about how you'd describe a car's speed. You could say "it covered 100 km in 2 hours," which gives you an average speed, or you could look at the speedometer at one specific moment for the instantaneous reading. Reaction rate works on exactly the same logic, just swapping distance for concentration.
For a simple hypothetical reaction R → P, where the volume stays constant, suppose [R]₁ and [P]₁ are concentrations at time t₁, and [R]₂, [P]₂ are the concentrations at a later time t₂. Then Δt = t₂ − t₁, Δ[R] = [R]₂ − [R]₁, and Δ[P] = [P]₂ − [P]₁.
Since the reactant is being consumed, Δ[R] comes out negative — but rate is supposed to be a positive quantity, so we multiply by −1:
Rate of appearance of P = +Δ[P]/Δt ...(3.2)
These two expressions define the average rate, rav. It depends entirely on which time interval you pick — that's precisely its limitation. Average rate can't tell you the rate at one specific instant, because it stays constant across whatever interval you calculated it over.
To get the rate at a single moment, you shrink that interval down until Δt approaches zero — that's the instantaneous rate:
Graphically, this is the slope of the tangent drawn to the concentration-versus-time curve at that particular point. If you plot [R] against t, the curve for a reactant slopes downward, so the instantaneous rate equals minus the slope of the tangent; for a product, whose curve rises, the rate equals the slope directly.
Here's something worth internalising rather than memorising: the instantaneous rate isn't just an abstract idea used to make the topic harder — it's what a rate law actually describes at any given concentration, whereas average rate is only a convenient approximation over a finite interval. Numerical problems typically hand you a data table and ask you to compute rav over each interval, then separately ask for rinst at a specific time using a tangent or a given formula.
3. Units of Rate and Solved Rate Problems
Since rate is concentration divided by time, its units are concentration time⁻¹. With concentration in mol L⁻¹ and time in seconds, you get mol L⁻¹ s⁻¹. For gas-phase reactions where concentration is tracked via partial pressure instead, the rate units become atm s⁻¹ (or bar s⁻¹, depending on the pressure unit used).
For C₄H₉Cl + H₂O → C₄H₉OH + HCl, concentration data at various times let you compute the average rate over each interval by taking −Δ[C₄H₉Cl]/Δt. The data shows something instructive: the average rate steadily falls, from about 1.90 × 10⁻⁴ mol L⁻¹s⁻¹ over the first 50 s interval down to roughly 0.4 × 10⁻⁴ mol L⁻¹s⁻¹ between 700–800 s. This decline as the reaction proceeds is a direct visual proof that rate depends on reactant concentration — as [C₄H₉Cl] drops, so does the rate.
2N₂O₅(g) → 4NO₂(g) + O₂(g). Initial [N₂O₅] = 2.33 mol L⁻¹, and after 184 minutes it drops to 2.08 mol L⁻¹. Since the stoichiometric coefficient of N₂O₅ is 2, the average rate is:
Average rate = −(1/2)(Δ[N₂O₅]/Δt) = −(1/2) × [(2.08 − 2.33)/184] mol L⁻¹ min⁻¹ = 6.79 × 10⁻⁴ mol L⁻¹ min⁻¹
Converting: this equals 4.07 × 10⁻² mol L⁻¹ h⁻¹, or 1.13 × 10⁻⁵ mol L⁻¹ s⁻¹.
For the rate of NO₂ production, since NO₂ has coefficient 4: Rate = (1/4)(Δ[NO₂]/Δt), so Δ[NO₂]/Δt = 6.79 × 10⁻⁴ × 4 = 2.72 × 10⁻³ mol L⁻¹ min⁻¹.
Notice the pattern in both examples — whenever a species has a stoichiometric coefficient other than 1, you divide its concentration-change rate by that coefficient before equating it to the "rate of reaction." That single habit resolves almost every rate-expression question you'll encounter.
4. Expressing Rate for Reactions with Unequal Stoichiometric Coefficients
For Hg(l) + Cl₂(g) → HgCl₂(s), all coefficients equal 1, so the rate of disappearance of Hg, the rate of disappearance of Cl₂, and the rate of appearance of HgCl₂ are all numerically identical:
But take 2HI(g) → H₂(g) + I₂(g). Two moles of HI vanish for every one mole each of H₂ and I₂ formed, so the rate of HI consumption runs twice as fast as either product's formation rate. To make all three expressions numerically equal, you divide the HI term by 2:
The same logic extends to more complicated stoichiometry. For 5Br⁻(aq) + BrO₃⁻(aq) + 6H⁺(aq) → 3Br₂(aq) + 3H₂O(l):
Why does this matter so much for exams? Because a very frequent MCQ format gives you the rate of disappearance of one species and asks for the rate of appearance of another, purely through this coefficient-ratio logic — no rate law needed at all.
5. Factors Influencing Reaction Rate
Four experimental conditions govern how fast a reaction proceeds: concentration of reactants (or partial pressure, for gases), temperature, and the presence of a catalyst. The next several sections unpack each of these in turn, starting with concentration dependence — which is where the rate law comes in.
6. Rate Law and Rate Constant
For a general reaction aA + bB → cC + dD, experiments show:
−d[R]/dt = k[A]x[B]y ...(3.4b, differential rate equation)
Here x and y may or may not equal the stoichiometric coefficients a and b — that's the single most important sentence in this section, and examiners test it relentlessly. The constant k is the rate constant, and equation 3.4 itself is called the rate law or rate expression: the expression in which reaction rate is given in terms of molar concentration of reactants, each raised to some power that may or may not match its stoichiometric coefficient in the balanced equation.
Consider 2NO(g) + O₂(g) → 2NO₂(g). Experimental data (varying [NO] and [O₂] independently) shows that doubling [NO] while holding [O₂] fixed quadruples the rate — a fourfold jump signals second-order dependence on NO. Doubling [O₂] alone doubles the rate, meaning first-order dependence on O₂. So:
In this particular case the exponents happen to match the stoichiometric coefficients. But that's a coincidence, not a rule. Look at these two counter-examples:
| Reaction | Experimental Rate Expression |
|---|---|
| CHCl₃ + Cl₂ → CCl₄ + HCl | Rate = k[CHCl₃][Cl₂]1/2 |
| CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH | Rate = k[CH₃COOC₂H₅]¹[H₂O]⁰ |
Neither case matches the stoichiometric coefficients. This is why the golden rule of this topic is: rate law can never be predicted merely by inspecting a balanced equation — it has to be determined experimentally. This single line is worth memorising verbatim; it shows up as a direct statement-true/false question surprisingly often.
7. Order of a Reaction
In Rate = k[A]x[B]y, x tells you how sensitive the rate is to changes in [A], and y does the same for [B]. Individually, x is the "order with respect to A" and y the "order with respect to B." Add them together and you get the overall order: x + y.
The sum of the powers of concentration terms in the rate law is called the order of that chemical reaction. Order can be zero, a positive integer, or even a fraction — 0, 1, 2, 3, or something like 1/2 or 3/2. A zero-order reaction simply means the rate doesn't depend on reactant concentration at all.
(a) Order = 1/2 + 3/2 = 2 → second order.
(b) Order = 3/2 + (−1) = 1/2 → half order.
Units of the Rate Constant
Because k = Rate / ([A]x[B]y), its units change depending on the overall order n = x + y. This table is genuinely worth memorising cold, because a huge number of MCQs simply give you the unit of k and ask you to identify the order — no calculation required if you know this table.
| Order | Units of Rate Constant |
|---|---|
| Zero order | mol L⁻¹ s⁻¹ |
| First order | s⁻¹ |
| Second order | L mol⁻¹ s⁻¹ (or mol⁻¹ L s⁻¹) |
(i) Units match second order → second order reaction.
(ii) Units match first order → first order reaction.
8. Elementary and Complex Reactions
A balanced chemical equation is, in a sense, a lie about mechanism — or at least an incomplete truth. Reactions rarely happen in exactly one step. Reactions that do occur in a single step are called elementary reactions. When a sequence of elementary steps (collectively called a mechanism) leads to the overall products, the process is called a complex reaction.
These sequences can take several forms: consecutive reactions (oxidation of ethane to CO₂ and H₂O passes through intermediate alcohol, aldehyde, and acid stages), reverse reactions, and side reactions (nitration of phenol yielding both ortho- and para-nitrophenol simultaneously).
9. Molecularity vs Order — The Classic Exam Trap
Molecularity is the number of reacting species — atoms, ions, or molecules — that must collide simultaneously for an elementary reaction to occur. Reactions get classified by how many species collide:
- Unimolecular — one species reacts alone, e.g. decomposition of ammonium nitrite: NH₄NO₂ → N₂ + 2H₂O
- Bimolecular — two species collide simultaneously, e.g. 2HI → H₂ + I₂
- Trimolecular (termolecular) — three species collide at once, e.g. 2NO + O₂ → 2NO₂
Why do trimolecular reactions stay rare? Simply because the probability of three separate species arriving at the same point in space at exactly the same instant, with the right orientation and energy, is vanishingly small compared to two-body collisions. Consequently, molecularity higher than three essentially doesn't occur — any reaction that looks like it needs more than three molecules colliding together must actually be proceeding through multiple elementary steps.
Take KClO₃ + 6FeSO₄ + 3H₂SO₄ → KCl + 3Fe₂(SO₄)₃ + 3H₂O. On paper this looks like a tenth-order, ten-molecule collision — but experimentally it behaves as a second-order reaction, proving it must occur through several elementary steps rather than one giant collision.
This raises a natural question: if a reaction happens in several steps, which step controls the overall rate? Think of a relay race — the team's overall time is set by its slowest runner, not the fastest. Similarly, the rate-determining step is the slowest step in the mechanism, and it controls the rate of the entire reaction.
2H₂O₂ →(I⁻, alkaline medium)→ 2H₂O + O₂, with Rate = k[H₂O₂][I⁻] — first order in both. Evidence points to a two-step mechanism:
Step 1: H₂O₂ + I⁻ → H₂O + IO⁻ (slow — this is the rate-determining step)
Step 2: H₂O₂ + IO⁻ → H₂O + I⁻ + O₂ (fast)
Both steps are bimolecular elementary reactions. IO⁻ is an intermediate — formed during the reaction but absent from the overall balanced equation. Because step 1 is slow, it alone dictates the rate, which is exactly why the experimental rate law matches step 1's molecularity rather than the overall stoichiometry.
Three conclusions fall directly out of this discussion, and each one is exam gold on its own:
- Order is an experimental quantity — it can be zero or fractional, but molecularity can never be zero or non-integral.
- Order applies to both elementary and complex reactions; molecularity applies only to elementary reactions and has no meaning for a complex, multi-step reaction as a whole.
- For a complex reaction, the order is determined by the slowest step, and the molecularity of that slowest step equals the order of the overall reaction.
10. Integrated Rate Equations — Zero Order
Measuring instantaneous rate directly means drawing tangents on a curve, which is fiddly and imprecise. It's far more practical to work with an integrated rate equation — a relation connecting concentration at a given time directly to the rate constant, built by integrating the differential rate law. NCERT derives these fully only for zero and first order, which conveniently happen to be the two orders tested most heavily across every competitive exam.
For a zero order reaction R → P, the rate is proportional to the zeroth power of concentration — meaning it doesn't change with concentration at all:
⇒ d[R] = −k dt
Integrating both sides gives [R] = −kt + I, where I is the integration constant. At t = 0, [R] = [R]₀, so substituting gives I = [R]₀. This yields the zero order integrated rate law:
k = ([R]₀ − [R]) / t ...(3.7)
Compare this with the equation of a straight line, y = mx + c. Plotting [R] against t gives a straight line with slope = −k and intercept [R]₀ — a graph you should be able to sketch from memory, because "identify the order from this graph" questions are extremely common.
Zero order kinetics is relatively uncommon in nature, but it shows up in specific, testable scenarios — enzyme-catalysed reactions and reactions occurring on saturated metal surfaces. The textbook example is decomposition of ammonia on a hot platinum surface at high pressure:
Why does this happen? At high pressure, the platinum surface becomes completely saturated with gas molecules. Beyond that saturation point, changing the ammonia concentration in the gas phase can't change how much ammonia is actually sitting on the catalyst surface — so the rate stops depending on concentration entirely. Thermal decomposition of HI on a gold surface follows the same logic.
11. Integrated Rate Equations — First Order
For a first order reaction R → P, rate is directly proportional to [R]:
Integrating gives ln[R] = −kt + I. Applying the initial condition [R] = [R]₀ at t = 0 gives I = ln[R]₀, so:
ln([R]/[R]₀) = −kt, or k = (1/t) ln([R]₀/[R]) ...(3.10)
Between two arbitrary times t₁ and t₂ with concentrations [R]₁ and [R]₂, subtracting the two integrated forms gives a very useful two-point version:
Taking the antilog of equation 3.9 gives the exponential decay form that you'll recognise instantly from radioactivity:
Plotting ln[R] against t gives a straight line of slope −k and intercept ln[R]₀. Converting to base-10 logarithms (which is usually more convenient for hand calculation) gives the form you'll actually use in most numericals:
A plot of log([R]₀/[R]) versus t gives a straight line through the origin with slope k/2.303 — another graph worth being able to sketch on sight. Hydrogenation of ethene, C₂H₄(g) + H₂(g) → C₂H₆(g) with Rate = k[C₂H₄], is a standard first order example, and — this is worth remembering as a standalone fact — all natural and artificial radioactive decay follows first order kinetics, including decompositions like N₂O₅ and N₂O.
k = (2.303/60) × log(1.24×10⁻²/0.20×10⁻²) = (2.303/60) × log(6.2) = 0.0304 min⁻¹
12. First Order Reactions in Gas Phase (Pressure Method)
Gas-phase first order reactions are frequently monitored through total pressure rather than concentration directly, since pressure is often easier to measure in a closed vessel. For A(g) → B(g) + C(g), let pi be the initial pressure of A and pt the total pressure at time t. If x is the pressure drop in A, and one mole each of B and C forms, both B and C gain x atm:
| A(g) | B(g) | C(g) | |
|---|---|---|---|
| t = 0 | pi | 0 | 0 |
| time t | (pi − x) | x | x |
Since pt = (pi − x) + x + x = pi + x, we get x = pt − pi, and hence pA = pi − x = 2pi − pt. Substituting into the first order rate equation:
k = (2.303/100) log(0.5/0.476) = (2.303/100) × 0.0216 = 4.98 × 10⁻⁴ s⁻¹
13. Half-Life of a Reaction
The half-life, t1/2, is the time needed for a reactant's concentration to fall to exactly half its initial value. This single concept generates a disproportionate number of exam questions, mostly because zero and first order behave so differently.
For zero order, starting from k = ([R]₀ − [R])/t and substituting [R] = [R]₀/2 at t = t1/2:
Notice that t1/2 for zero order is directly proportional to initial concentration — double the starting amount, and the half-life doubles too.
For first order, starting from k = (2.303/t)log([R]₀/[R]) and substituting [R] = [R]₀/2:
This is the single most memorised formula in the entire chapter, and for good reason — t1/2 for a first order reaction is completely independent of initial concentration. Whether you start with 10 mol or 0.001 mol, the time to reach half that amount is exactly the same, fixed only by k.
t1/2 = 0.693/(5.5×10⁻¹⁴) = 1.26 × 10¹³ s
When 99.9% is complete, [R]remaining = [R]₀ − 0.999[R]₀ = 0.001[R]₀. So k = (2.303/t) log([R]₀/0.001[R]₀) = (2.303/t) log(10³), giving t = 6.909/k.
Since t1/2 = 0.693/k, the ratio t/t1/2 = (6.909/k) × (k/0.693) = 10. Confirmed.
| Order | Differential Rate Law | Integrated Rate Law | Straight-Line Plot | Half-Life | Units of k |
|---|---|---|---|---|---|
| 0 | d[R]/dt = −k | kt = [R]₀ − [R] | [R] vs t | [R]₀/2k | mol L⁻¹s⁻¹ |
| 1 | d[R]/dt = −k[R] | kt = ln([R]₀/[R]) | ln[R] vs t | 0.693/k | s⁻¹ |
14. Pseudo First Order Reactions
Sometimes a reaction that's genuinely higher order in reality still obeys a first order rate law experimentally, because of the specific conditions used. The hydrolysis of ethyl acetate is the classic case:
This is genuinely a second order reaction — both ethyl acetate and water concentrations affect the rate. But water is used in such large excess for hydrolysis that its concentration barely shifts over the course of the reaction. In a concrete example: starting with 0.01 mol ethyl acetate and 10 mol water, by completion you're left with 0 mol ethyl acetate but still 9.99 mol water — a negligible fractional change. Since [H₂O] is effectively constant, the rate law simplifies down to depending only on [ethyl acetate], and the reaction behaves exactly like a first order process. These are called pseudo first order reactions.
Inversion of cane sugar is another standard pseudo first order example:
Rate = k[C₁₂H₂₂O₁₁]
15. Temperature Dependence — The Arrhenius Equation
Almost every reaction speeds up when you heat it. A useful rule of thumb worth memorising: for most reactions, a 10° rise in temperature roughly doubles the rate constant. The decomposition of N₂O₅ illustrates the scale of this effect vividly — the time for half the material to decompose is 12 minutes at 50°C, 5 hours at 25°C, and a full 10 days at 0°C.
The quantitative relationship is the Arrhenius equation, first proposed by van't Hoff and given physical interpretation by Arrhenius:
where A is the Arrhenius factor (also called the pre-exponential or frequency factor, a constant specific to each reaction), R is the gas constant, and Ea is the activation energy in J mol⁻¹.
Take H₂(g) + I₂(g) → 2HI(g) as the classic teaching example. According to Arrhenius, this reaction proceeds only when a hydrogen molecule and an iodine molecule collide to form an unstable intermediate — the activated complex
.
This complex exists briefly before breaking apart into two HI molecules. The energy required to reach this activated complex from the reactants is the activation energy, visualised as the peak of a potential-energy-versus-reaction-coordinate curve. Energy is released as the complex collapses into products, and the net enthalpy change depends on the relative energies of reactants and products.
Not every molecule in a sample carries the same kinetic energy — Maxwell and Boltzmann worked out the statistical distribution of molecular energies. Plotting the fraction of molecules NE/NT against kinetic energy E gives a characteristic curve peaking at the most probable kinetic energy. Raise the temperature, and this peak shifts to higher energy while the curve broadens — meaning a larger fraction of molecules now possess energy exceeding Ea. Since the total area under the curve must stay constant (total probability = 1), this broadening directly explains why raising temperature by 10° roughly doubles the fraction of molecules with enough energy to react, which in turn roughly doubles the rate.
Taking the natural log of the Arrhenius equation linearises it beautifully:
Plotting ln k against 1/T gives a straight line with slope −Ea/R and intercept ln A — this is exactly how Ea and A get determined experimentally from rate-constant data at different temperatures. Writing equation 3.19 at two temperatures T₁ and T₂ and subtracting gives the two-temperature form used in almost every numerical on this topic:
log(0.07/0.02) = [Ea/(2.303 × 8.314)] × [(700−500)/(700×500)]
0.544 = Ea × 5.714×10⁻⁴/19.15 ⇒ Ea = 18230.8 J
Then from k = Ae−Ea/RT: 0.02 = A × e−18230.8/(8.314×500), giving A = 0.02/0.012 = 1.61
log k₂ = log(1.60×10⁻⁵) + [209000/(2.303×8.314)] × [(1/600) − (1/700)]
log k₂ = −4.796 + 2.599 = −2.197 ⇒ k₂ = 6.36 × 10⁻³ s⁻¹
16. Effect of Catalyst
A catalyst speeds up a reaction without itself undergoing any permanent chemical change. MnO₂ catalysing the decomposition of KClO₃ is the standard textbook example: 2KClO₃ →(MnO₂)→ 2KCl + 3O₂. Worth noting as a language point: if an added substance instead slows a reaction down, it's called an inhibitor, not a catalyst — "catalyst" specifically means rate-accelerating.
The mechanism is explained by the intermediate complex theory: the catalyst forms temporary bonds with reactants, creating an intermediate complex that has only a transitory existence before decomposing to give products plus the regenerated catalyst. The net effect is that the catalyst opens up an alternate reaction pathway with a lower activation energy than the uncatalysed route — and since the Arrhenius equation shows k rises exponentially as Ea falls, even a modest reduction in activation energy translates into a substantial rate increase.
A few catalyst properties are worth locking in, because they're tested as standalone true/false statements: a small amount of catalyst can act on a large amount of reactant; a catalyst does not alter the Gibbs energy change (ΔG) of a reaction; it catalyses spontaneous reactions but cannot make a non-spontaneous reaction proceed; and — this is the one students most often get wrong — a catalyst does not change the equilibrium constant of a reaction. What it does instead is speed up the forward and backward reactions to the same extent, so equilibrium is reached faster without shifting where that equilibrium actually sits.
17. Collision Theory of Reaction Rates
The Arrhenius equation works well empirically, but collision theory, developed by Max Trautz and William Lewis between 1916 and 1918, digs into the mechanistic "why" behind it, building on the kinetic theory of gases. It treats reactant molecules as hard spheres and assumes reaction only occurs when they physically collide.
The collision frequency (Z) is the number of collisions occurring per second per unit volume of the reaction mixture. For a bimolecular elementary reaction A + B → Products:
Here ZAB is the collision frequency of A and B, and e−Ea/RT represents the fraction of collisions energetic enough to exceed Ea. Comparing this with the Arrhenius equation shows that A corresponds physically to the collision frequency.
This basic form predicts rate constants fairly accurately for reactions between atoms or simple molecules, but it breaks down noticeably for complex molecules. The missing piece: not every energetically-sufficient collision actually produces product. Molecules also need proper orientation at the moment of collision — the reacting parts of each molecule need to be pointed the right way so that old bonds can break and new bonds can form. Collisions meeting both the energy threshold (called threshold energy, equal to activation energy plus whatever energy the reacting species already possess) and the orientation requirement are called effective collisions.
The formation of methanol from bromoethane is the textbook illustration: CH₃Br + OH⁻ → CH₃OH + Br⁻. If the OH⁻ ion approaches from the correct side, it can displace Br⁻ and form the product; if it approaches from the wrong orientation, the molecules simply bounce apart with no reaction at all.
To capture this, collision theory introduces a probability or steric factor P, modifying the rate expression to:
So under collision theory, both activation energy and molecular orientation jointly determine whether a collision is effective, and therefore determine the overall rate. It's worth being upfront about the theory's limitation too, since this occasionally appears as a conceptual question: collision theory treats molecules as featureless hard spheres and ignores their actual structural complexity — a simplification that later, more advanced theories (beyond this syllabus) work to correct.
18. Quick Revision Summary
- Chemical kinetics studies reaction rates, the factors controlling them, and reaction mechanisms — distinct from thermodynamics (feasibility) and equilibrium (extent).
- Rate can be average (over an interval) or instantaneous (at a point, via tangent slope); units are concentration time⁻¹.
- Rate law/rate expression connects rate to concentration via exponents that must be found experimentally, never assumed from stoichiometry.
- Order = sum of concentration exponents in the rate law; can be zero, integer, or fractional.
- Molecularity = number of colliding species in an elementary step; limited to 1, 2, or (rarely) 3; undefined for overall complex reactions.
- Zero order: [R] = [R]₀ − kt, t1/2 = [R]₀/2k (depends on initial concentration).
- First order: [R] = [R]₀e⁻ᵏᵗ, t1/2 = 0.693/k (independent of initial concentration).
- Pseudo first order reactions are truly higher order but behave as first order when one reactant is in large excess.
- Arrhenius equation k = Ae⁻ᴱᵃ/ᴿᵀ links k to temperature and activation energy; a 10° rise roughly doubles k.
- Catalysts lower activation energy via an alternate pathway, without changing ΔG or the equilibrium constant.
- Collision theory: Rate = PZABe⁻ᴱᵃ/ᴿᵀ, where effective collisions require both sufficient energy and correct orientation.
19. Practice MCQs (65 Questions with Answers)
I've grouped these roughly by topic so you can target weak areas, but treat them as a mixed set for a timed self-test too. Answers are given right after each question — cover them with your hand for genuine practice.
Rate of Reaction & Basic Concepts
Q1. The rate of a reaction is expressed in units of:
(a) mol L⁻¹ (b) mol L⁻¹ s⁻¹ (c) s⁻¹ only (d) L mol⁻¹ s⁻¹
Answer: (b)
Q2. For the reaction R → P, the average rate is given by:
(a) +Δ[R]/Δt (b) −Δ[R]/Δt (c) Δt/Δ[R] (d) [R]×t
Answer: (b)
Q3. Instantaneous rate is obtained mathematically by:
(a) taking Δt → ∞ (b) taking Δt → 0 (c) averaging over the whole reaction (d) doubling the average rate
Answer: (b)
Q4. Which of these reactions is typically the fastest?
(a) Rusting of iron (b) Hydrolysis of starch (c) Precipitation of AgCl from AgNO₃ and NaCl (d) Inversion of cane sugar
Answer: (c)
Q5. For a gaseous reaction, if concentration is expressed as partial pressure, the rate has units of:
(a) mol L⁻¹s⁻¹ (b) atm s⁻¹ (c) atm² (d) mol atm⁻¹
Answer: (b)
Q6. For 2HI(g) → H₂(g) + I₂(g), if the rate of disappearance of HI is 4 × 10⁻⁴ mol L⁻¹s⁻¹, the rate of formation of H₂ is:
(a) 8 × 10⁻⁴ (b) 4 × 10⁻⁴ (c) 2 × 10⁻⁴ (d) 1 × 10⁻⁴ mol L⁻¹s⁻¹
Answer: (c), since rate of HI loss is divided by 2
Q7. For 5Br⁻ + BrO₃⁻ + 6H⁺ → 3Br₂ + 3H₂O, if −d[Br⁻]/dt = 5 × 10⁻³ mol L⁻¹s⁻¹, then d[Br₂]/dt equals:
(a) 5 × 10⁻³ (b) 3 × 10⁻³ (c) 1 × 10⁻³ (d) 15 × 10⁻³ mol L⁻¹s⁻¹
Answer: (b), since rate = (1/5)(−d[Br⁻]/dt) = (1/3)(d[Br₂]/dt) → d[Br₂]/dt = 3 × rate = 3 × 10⁻³
Q8. Average rate of a reaction cannot be used to find:
(a) rate at any given instant (b) overall trend of the reaction (c) rate over a fixed interval (d) rate constant units
Answer: (a)
Rate Law, Rate Constant & Order
Q9. Rate law of a reaction can be determined by:
(a) balanced chemical equation (b) stoichiometric coefficients (c) experiment only (d) thermodynamic data
Answer: (c)
Q10. If Rate = k[A]²[B], the order with respect to B is:
(a) 0 (b) 1 (c) 2 (d) 3
Answer: (b)
Q11. For Rate = k[A]1/2[B]1/2, the overall order is:
(a) 0 (b) 1 (c) 1/2 (d) 2
Answer: (b)
Q12. A reaction having units of k as mol⁻¹ L s⁻¹ is of order:
(a) zero (b) first (c) second (d) third
Answer: (c)
Q13. A reaction having units of k as s⁻¹ is of order:
(a) zero (b) first (c) second (d) third
Answer: (b)
Q14. If doubling [A] doubles the rate while [B] has no effect, the rate law is:
(a) Rate = k[A]²[B] (b) Rate = k[A][B] (c) Rate = k[A] (d) Rate = k[B]
Answer: (c)
Q15. For 2A + B → C, if doubling [A] quadruples the rate and doubling [B] leaves rate unchanged, the order in A is:
(a) 1 (b) 2 (c) 0 (d) 1/2
Answer: (b)
Q16. A zero order reaction means:
(a) rate is independent of time (b) rate is independent of concentration of reactants (c) rate is infinite (d) reaction never completes
Answer: (b)
Q17. For CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, the experimental rate law is Rate = k[CH₃COOC₂H₅][H₂O]⁰. The order with respect to water is:
(a) 1 (b) 0 (c) 2 (d) cannot be determined
Answer: (b)
Q18. Order of reaction can never be:
(a) zero (b) fractional (c) negative for a reactant only (d) undefined/non-numerical
Answer: (d)
Molecularity & Mechanism
Q19. Molecularity of an elementary reaction can be:
(a) zero or fractional (b) 1, 2 or 3 only (c) any positive integer (d) negative
Answer: (b)
Q20. NH₄NO₂ → N₂ + 2H₂O is an example of a:
(a) bimolecular reaction (b) unimolecular reaction (c) termolecular reaction (d) zero order reaction only
Answer: (b)
Q21. Molecularity of the reaction 2NO + O₂ → 2NO₂, if elementary, is:
(a) 1 (b) 2 (c) 3 (d) 4
Answer: (c)
Q22. Molecularity is not applicable to:
(a) elementary reactions (b) complex (multi-step) reactions as a whole (c) unimolecular reactions (d) bimolecular reactions
Answer: (b)
Q23. The overall rate of a multi-step reaction is governed by:
(a) the fastest step (b) the slowest step (c) the average of all steps (d) the last step always
Answer: (b)
Q24. In the iodide-catalysed decomposition of H₂O₂, the species IO⁻ is best described as:
(a) a catalyst (b) an intermediate (c) an inhibitor (d) a product
Answer: (b)
Q25. KClO₃ + 6FeSO₄ + 3H₂SO₄ → KCl + 3Fe₂(SO₄)₃ + 3H₂O appears to be tenth order from stoichiometry but is experimentally found to be:
(a) first order (b) second order (c) third order (d) zero order
Answer: (b)
Integrated Rate Equations & Half-Life
Q26. For a zero order reaction, a plot of [R] vs t gives a straight line with slope:
(a) +k (b) −k (c) k/2.303 (d) 0
Answer: (b)
Q27. For a first order reaction, a plot of ln[R] vs t gives a straight line with intercept:
(a) k (b) [R]₀ (c) ln[R]₀ (d) −k
Answer: (c)
Q28. Half-life of a zero order reaction is:
(a) 0.693/k (b) [R]₀/2k (c) independent of [R]₀ (d) k/2.303
Answer: (b)
Q29. Half-life of a first order reaction is:
(a) directly proportional to [R]₀ (b) inversely proportional to [R]₀ (c) independent of [R]₀ (d) equal to k
Answer: (c)
Q30. A first order reaction has k = 0.693 min⁻¹. Its half-life is:
(a) 0.5 min (b) 1 min (c) 2 min (d) 0.693 min
Answer: (b)
Q31. If a first order reaction is 50% complete in 20 minutes, it will be 75% complete in:
(a) 30 min (b) 40 min (c) 60 min (d) 80 min
Answer: (b), since 75% completion = 2 half-lives = 40 min
Q32. For a first order reaction, time required for 99.9% completion is:
(a) same as t1/2 (b) 5 × t1/2 (c) 10 × t1/2 (d) 100 × t1/2
Answer: (c)
Q33. A first order reaction has rate constant 1.15 × 10⁻³ s⁻¹. The time for 5 g of reactant to reduce to 3 g is closest to:
(a) 111 s (b) 344 s (c) 444 s (d) 555 s
Answer: (c), t = (2.303/1.15×10⁻³) log(5/3) ≈ 444 s
Q34. Decomposition of SO₂Cl₂ takes 60 min for half its initial amount to decompose (first order). The rate constant is approximately:
(a) 1.92 × 10⁻⁴ s⁻¹ (b) 1.15 × 10⁻² s⁻¹ (c) 0.0304 min⁻¹ (d) 0.693 s⁻¹
Answer: (a)
Q35. For radioactive decay of ¹⁴C (t1/2 = 5730 years), a sample retaining 80% of original ¹⁴C is approximately how old? (log 1.25 ≈ 0.097)
(a) About 1845 years (b) About 2231 years (c) About 3200 years (d) About 5730 years
Answer: (a), using k = 0.693/5730 and t = (2.303/k)log(100/80)
Q36. The rate constant of a first order reaction is 60 s⁻¹. Time required to reduce concentration to 1/16th of its initial value is:
(a) t1/2 (b) 2 × t1/2 (c) 4 × t1/2 (d) 16 × t1/2
Answer: (c), since (1/2)⁴ = 1/16, so 4 half-lives are needed
Q37. Sucrose hydrolysis follows first order kinetics with t1/2 = 3 hours. The fraction remaining after 9 hours is:
(a) 1/2 (b) 1/4 (c) 1/8 (d) 1/16
Answer: (c), 9 hours = 3 half-lives → (1/2)³ = 1/8
Gas-Phase Kinetics (Pressure Method)
Q38. For a first order gas-phase reaction A(g) → B(g) + C(g), if pi is the initial pressure and pt the total pressure at time t, the partial pressure of A remaining is:
(a) pt − pi (b) 2pi − pt (c) pi + pt (d) pt/2
Answer: (b)
Q39. In the pressure-based first order rate equation, if pt increases with time, this indicates:
(a) reactant moles exceed product moles (b) product moles exceed reactant moles (c) reaction has stopped (d) reaction is zero order
Answer: (b)
Temperature Dependence & Arrhenius Equation
Q40. The Arrhenius equation is:
(a) k = A + e⁻ᴱᵃ/ᴿᵀ (b) k = Ae⁻ᴱᵃ/ᴿᵀ (c) k = A/e⁻ᴱᵃ/ᴿᵀ (d) k = A − Ea/RT
Answer: (b)
Q41. In the Arrhenius equation, A represents:
(a) activation energy (b) frequency/pre-exponential factor (c) gas constant (d) rate of reaction
Answer: (b)
Q42. A plot of ln k vs 1/T gives a straight line with slope:
(a) −Ea/R (b) Ea/R (c) −Ea/2.303R (d) ln A
Answer: (a)
Q43. For most reactions, a rise in temperature by 10° roughly:
(a) halves the rate constant (b) doubles the rate constant (c) has no effect (d) triples the rate constant
Answer: (b)
Q44. If the rate of a reaction quadruples when temperature rises from 293 K to 313 K, which equation is used to find Ea?
(a) k = 0.693/t1/2 (b) log(k₂/k₁) = (Ea/2.303R)[(T₂−T₁)/T₁T₂] (c) Rate = k[A]ⁿ (d) t1/2 = [R]₀/2k
Answer: (b)
Q45. Increasing temperature increases the rate of reaction mainly because:
(a) activation energy decreases (b) fraction of molecules with energy ≥ Ea increases (c) collision frequency alone decreases (d) molecules become larger
Answer: (b)
Q46. The area under the Maxwell-Boltzmann energy distribution curve represents:
(a) activation energy (b) total probability (always constant = 1) (c) rate constant (d) collision frequency
Answer: (b)
Q47. As temperature increases, the Maxwell-Boltzmann curve's peak:
(a) shifts to lower energy and narrows (b) shifts to higher energy and broadens (c) stays fixed (d) disappears
Answer: (b)
Q48. A reaction's rate constant is 4.5 × 10³ s⁻¹ at 10°C with Ea = 60 kJ/mol. To find the temperature at which k = 1.5 × 10⁴ s⁻¹, which relation is directly applicable?
(a) t1/2 formula (b) Two-temperature Arrhenius equation (3.22) (c) Collision theory equation (d) Rate law expression
Answer: (b)
Q49. Given k = (4.5 × 10¹¹ s⁻¹) e⁻²⁸⁰⁰⁰ᴷ/ᵀ, the activation energy Ea equals (R = 8.314 J/mol K):
(a) 28000 J/mol (b) 28000 × 8.314 J/mol ≈ 232.8 kJ/mol (c) 28000/8.314 J/mol (d) 8.314 J/mol
Answer: (b), since Ea/R = 28000 K
Catalysts
Q50. A catalyst increases reaction rate by:
(a) increasing activation energy (b) decreasing activation energy via an alternate pathway (c) changing ΔG of reaction (d) shifting equilibrium position
Answer: (b)
Q51. A catalyst does NOT:
(a) provide an alternate mechanism (b) get consumed permanently in the reaction (c) lower activation energy (d) increase rate constant
Answer: (b)
Q52. The theory that explains catalytic action via temporary bonding with reactants is:
(a) collision theory (b) intermediate complex theory (c) Arrhenius theory (d) transition state theory only
Answer: (b)
Q53. A catalyst affects which of the following?
(a) Equilibrium constant (b) ΔG of reaction (c) Time to reach equilibrium (d) Enthalpy of products
Answer: (c)
Q54. A substance that decreases the rate of a reaction is called:
(a) a catalyst (b) an inhibitor (c) an intermediate (d) an activator
Answer: (b)
Collision Theory
Q55. Collision theory was developed by:
(a) Arrhenius and van't Hoff (b) Max Trautz and William Lewis (c) Maxwell and Boltzmann (d) Le Chatelier
Answer: (b)
Q56. Effective collisions require:
(a) sufficient energy only (b) proper orientation only (c) both sufficient energy and proper orientation (d) neither, collisions always react
Answer: (c)
Q57. In Rate = PZABe⁻ᴱᵃ/ᴿᵀ, P stands for:
(a) pressure (b) probability/steric factor (c) product concentration (d) proportionality constant unrelated to orientation
Answer: (b)
Q58. Threshold energy is defined as:
(a) activation energy alone (b) activation energy + energy already possessed by reacting species (c) energy released in the reaction (d) energy of activated complex minus product energy
Answer: (b)
Q59. A key limitation of collision theory is that it:
(a) ignores activation energy (b) treats molecules as structureless hard spheres (c) cannot be applied to gas reactions (d) contradicts the Arrhenius equation
Answer: (b)
Mixed / Numerical Recall
Q60. For 3NO(g) → N₂O(g), given Rate = k[NO]², the order and units of k are:
(a) order 2, units mol⁻¹ L s⁻¹ (b) order 1, units s⁻¹ (c) order 3, units mol⁻² L² s⁻¹ (d) order 0, units mol L⁻¹s⁻¹
Answer: (a)
Q61. For H₂O₂ + 3I⁻ + 2H⁺ → 2H₂O + I₃⁻, Rate = k[H₂O₂][I⁻]. The overall order is:
(a) 1 (b) 2 (c) 3 (d) 0
Answer: (b)
Q62. For CH₃CHO(g) → CH₄(g) + CO(g), Rate = k[CH₃CHO]3/2. Units of k (concentration in mol L⁻¹, time in s) are:
(a) s⁻¹ (b) mol⁻¹/² L¹/² s⁻¹ (c) mol L⁻¹ s⁻¹ (d) mol⁻² L² s⁻¹
Answer: (b)
Q63. A reaction is second order overall and first order in each of two reactants A and B. If [A] and [B] are both doubled, the rate increases by a factor of:
(a) 2 (b) 4 (c) 8 (d) 1
Answer: (b)
Q64. A reaction is first order in A and second order in B. Tripling [B] alone changes the rate by a factor of:
(a) 3 (b) 6 (c) 9 (d) 27
Answer: (c)
Q65. For a reaction second order overall in one reactant, doubling that reactant's concentration increases the rate by a factor of:
(a) 2 (b) 4 (c) 1/2 (d) 8
Answer: (b)
