3.1 Rate of a Reaction and Rate Laws
How fast does a reaction go, and what does its speed depend on? Average and instantaneous rates, rate laws $\text{rate}=k[\mathrm{A}]^{x}[\mathrm{B}]^{y}$, order versus molecularity, and how to find a rate law from experiments.
Thermodynamics tells you whether a reaction can happen. It says nothing about how fast. Diamond turning into graphite is thermodynamically favoured, yet a diamond ring will outlast you by billions of years. A mixture of hydrogen and oxygen sits quietly in a balloon until a spark makes it explode in a fraction of a second. Iron rusts over months, milk turns sour in a day, and the silver bromide in a photographic film reacts in a thousandth of a second when light strikes it. Chemical kinetics is the study of reaction rates, of the factors that change them, and of the step-by-step pathways (mechanisms) by which reactions actually happen.
Rates matter everywhere. A pharmacist needs to know how fast a medicine decomposes to print an expiry date. An engineer designing an ammonia plant must make the reaction fast enough to be profitable. An atmospheric chemist must know how quickly nitrogen oxides destroy ozone. In this section you will learn to measure a rate from concentration data, to write the rate law of a reaction, to tell its order from its molecularity, and to find the rate law from a set of experiments by the method of initial rates.
Average and Instantaneous Rate
Consider a reaction $\mathrm{R}\rightarrow\mathrm{P}$. As time passes the concentration of the reactant $[\mathrm{R}]$ falls and that of the product $[\mathrm{P}]$ rises. The rate tells us how quickly these concentrations change. Because concentrations are usually in $\mathrm{mol\,L^{-1}}$ and time in seconds, rates are usually in $\mathrm{mol\,L^{-1}\,s^{-1}}$. For gases, partial pressures may be used instead of concentrations, giving rates in $\mathrm{atm\,s^{-1}}$ or $\mathrm{bar\,s^{-1}}$.
The rate of a reaction is the change in concentration of a reactant or product per unit time. For $\mathrm{R}\rightarrow\mathrm{P}$, if $[\mathrm{R}]$ changes by $\Delta[\mathrm{R}]$ and $[\mathrm{P}]$ by $\Delta[\mathrm{P}]$ in a time interval $\Delta t$, the average rate over that interval is
$$r_{av}=-\frac{\Delta[\mathrm{R}]}{\Delta t}=+\frac{\Delta[\mathrm{P}]}{\Delta t}.$$The minus sign makes the rate positive, since $\Delta[\mathrm{R}]$ is negative. The instantaneous rate is the rate at one particular moment $t$. It is the limit of the average rate as $\Delta t\rightarrow0$:
$$r_{inst}=-\frac{d[\mathrm{R}]}{dt}=+\frac{d[\mathrm{P}]}{dt}.$$On a graph of concentration against time, the average rate is the slope of the chord joining two points of the curve (with the sign changed for a reactant), and the instantaneous rate is the slope of the tangent at time $t$.
Most reactions slow down as they proceed, because the reactant is used up. So an average rate over a long interval hides how the rate really changes, and it depends on which interval you choose. The instantaneous rate is the meaningful quantity; average rates over short intervals are simply our best estimate of it from a table of data.
| $t$ / s | 0 | 300 | 600 | 900 | 1200 | 1800 | 2400 |
|---|---|---|---|---|---|---|---|
| $[\mathrm{N_{2}O_{5}}]$ / $\mathrm{mol\,L^{-1}}$ | 1.60 | 1.34 | 1.12 | 0.932 | 0.779 | 0.543 | 0.379 |
The table shows the decomposition of dinitrogen pentoxide in the gas phase at $318\ \mathrm{K}$, $\ \mathrm{2N_{2}O_{5}(g)\rightarrow4NO_{2}(g)+O_{2}(g)}$. We use it in Examples 1 and 3, in Interactive 1 and in the exercises.
Stoichiometry and rate. In $\mathrm{2N_{2}O_{5}(g)\rightarrow4NO_{2}(g)+O_{2}(g)}$, every time two molecules of $\mathrm{N_{2}O_{5}}$ disappear, four of $\mathrm{NO_{2}}$ and one of $\mathrm{O_{2}}$ appear. So $[\mathrm{NO_{2}}]$ rises twice as fast as $[\mathrm{N_{2}O_{5}}]$ falls, and $[\mathrm{O_{2}}]$ rises half as fast. To get one rate for the reaction, whichever species we watch, we divide each rate of change by its stoichiometric coefficient.
For a reaction $\ a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}$,
$$\text{rate}=-\frac{1}{a}\frac{d[\mathrm{A}]}{dt}=-\frac{1}{b}\frac{d[\mathrm{B}]}{dt}=+\frac{1}{c}\frac{d[\mathrm{C}]}{dt}=+\frac{1}{d}\frac{d[\mathrm{D}]}{dt}.$$The same relations hold for average rates with $\Delta$ in place of $d$. "Rate of disappearance of A" means $-d[\mathrm{A}]/dt$ and "rate of formation of C" means $+d[\mathrm{C}]/dt$; neither is divided by its coefficient. Only the "rate of reaction" is. This assumes no intermediate builds up in appreciable amounts, so the equation describes the change at every moment.
Factors Affecting Rate; the Rate Law
Four things decide how fast a given reaction goes:
- Concentration of the reactants (or pressure, for gases): more particles per litre means more encounters per second. This is the factor the rate law describes.
- Temperature: for many reactions the rate roughly doubles for every $10\ \mathrm{K}$ rise. Section 3.3 explains why.
- A catalyst: a substance that speeds up a reaction without being used up, by providing a different pathway (Section 3.3).
- Surface area (for reactions involving solids) and the nature of the reactants: powdered zinc dissolves in acid faster than a lump; ionic reactions in solution, such as $\mathrm{Ag^{+}(aq)+Cl^{-}(aq)\rightarrow AgCl(s)}$, are almost instantaneous because no covalent bonds need to break. Light starts some reactions, such as $\mathrm{H_{2}+Cl_{2}}$ and the reactions in photographic film.
How exactly does the rate depend on concentration? For $\mathrm{2NO(g)+O_{2}(g)\rightarrow2NO_{2}(g)}$, experiments show that doubling $[\mathrm{O_{2}}]$ doubles the rate, while doubling $[\mathrm{NO}]$ makes it four times faster. So the rate is proportional to $[\mathrm{NO}]^{2}[\mathrm{O_{2}}]$. A statement of this kind is a rate law.
For a reaction $\ a\mathrm{A}+b\mathrm{B}\rightarrow\text{products}$, the rate law (rate equation) is the experimentally found expression
$$\text{rate}=k\,[\mathrm{A}]^{x}\,[\mathrm{B}]^{y}.$$- $k$ is the rate constant: the rate when every concentration is $1\ \mathrm{mol\,L^{-1}}$. It depends on the reaction and the temperature, but not on the concentrations.
- $x$ and $y$ are the orders with respect to A and B; their sum $x+y$ is the overall order of the reaction.
- The orders are found by experiment. They need not equal the coefficients $a$ and $b$, and they may be zero, fractions or (rarely) negative.
Rate has units $\mathrm{mol\,L^{-1}\,s^{-1}}$, and $\text{rate}=k\,[\text{conc}]^{n}$ for overall order $n$. So
$$k=\frac{\text{rate}}{[\text{conc}]^{n}}\quad\Rightarrow\quad\text{units of }k=\mathrm{(mol\,L^{-1})^{1-n}\,s^{-1}}.$$| Order $n$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| Units of $k$ | $\mathrm{mol\,L^{-1}\,s^{-1}}$ | $\mathrm{s^{-1}}$ | $\mathrm{L\,mol^{-1}\,s^{-1}}$ | $\mathrm{L^{2}\,mol^{-2}\,s^{-1}}$ |
For gases with pressures in atm, replace $\mathrm{mol\,L^{-1}}$ by $\mathrm{atm}$: a second-order $k$ is then in $\mathrm{atm^{-1}\,s^{-1}}$. Only for a first-order reaction are the units the same in every concentration or pressure unit.
(a) $\mathrm{CHCl_{3}+Cl_{2}\rightarrow CCl_{4}+HCl}$, $\ \text{rate}=k[\mathrm{CHCl_{3}}][\mathrm{Cl_{2}}]^{1/2}$;
(b) $\mathrm{CH_{3}CHO\rightarrow CH_{4}+CO}$ (gas phase, about $800\ \mathrm{K}$), $\ \text{rate}=k[\mathrm{CH_{3}CHO}]^{3/2}$;
(c) $\mathrm{2NO(g)+O_{2}(g)\rightarrow2NO_{2}(g)}$, $\ \text{rate}=k[\mathrm{NO}]^{2}[\mathrm{O_{2}}]$.
(d) Three reactions have $k=2.3\times10^{-5}\ \mathrm{L\,mol^{-1}\,s^{-1}}$, $3.1\times10^{-4}\ \mathrm{s^{-1}}$ and $4.0\times10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. What are their orders?
1. Reading orders off the equation. For $\mathrm{CHCl_{3}+Cl_{2}}$ the coefficients are $1$ and $1$, but the order in $\mathrm{Cl_{2}}$ is $\tfrac12$. The rate law comes from experiment, never from the balanced equation (unless the reaction is known to be elementary, as below).
2. Confusing $k$ with the rate. The rate changes all the time as concentrations fall; $k$ stays constant at a given temperature.
3. Units of $k$. Work them out from the overall order each time, using Relations 3.2. "$\mathrm{mol\,L^{-1}\,s^{-1}}$" for every $k$ is a classic error.
4. Volume changes. Halving the volume of a gas mixture doubles every concentration, so a third-order reaction becomes $2^{3}=8$ times faster.
Molecularity, Elementary Reactions and the Rate-Determining Step
A balanced equation shows only the start and the end of a reaction. Most reactions actually happen in a series of simple steps, each involving one, two or (rarely) three particles. If a reaction happens in a single such step, we can say how many particles take part in it.
An elementary reaction is one that takes place in a single step, exactly as written. Its molecularity is the number of reacting species (atoms, ions or molecules) that must collide at the same time for the step to happen:
- unimolecular (one species): $\ \mathrm{NH_{4}NO_{2}\rightarrow N_{2}+2H_{2}O}$;
- bimolecular (two species): $\ \mathrm{2HI(g)\rightarrow H_{2}(g)+I_{2}(g)}$;
- termolecular (three species): $\ \mathrm{2NO(g)+O_{2}(g)\rightarrow2NO_{2}(g)}$.
Molecularity is always a small whole number ($1$, $2$ or $3$); it cannot be zero or a fraction. Steps with molecularity above three are practically unknown, because the chance that four particles meet at the same instant, with enough energy and the right orientation, is negligible. For an elementary step, and only for one, the rate law follows from the equation: $\ a\mathrm{A}+b\mathrm{B}\rightarrow\text{products}$ (elementary) has $\text{rate}=k[\mathrm{A}]^{a}[\mathrm{B}]^{b}$.
A complex reaction takes place in two or more elementary steps. The sequence of steps is its mechanism; the steps add up to the overall equation. A species formed in one step and used up in a later one is an intermediate; a species used up in one step and given back in a later one is a catalyst.
When one step is much slower than all the others, the overall reaction cannot go faster than that step, just as a queue moves no faster than its slowest counter. This slowest step is the rate-determining step, and (when it is the first step) the rate law of the whole reaction is the rate law of that step. Molecularity applies to each elementary step; for a complex reaction, only the order is meaningful.
| Order | Molecularity | |
|---|---|---|
| Defined as | sum of the powers of the concentrations in the rate law | number of species colliding in one elementary step |
| Found by | experiment | the proposed mechanism (theory) |
| Possible values | $0$, whole numbers, fractions (even negative) | $1$, $2$ or $3$ only |
| Applies to | elementary and complex reactions | elementary steps only |
Finding a Rate Law: the Method of Initial Rates
Since orders must be measured, we need a reliable experimental method. The cleanest is to run the reaction several times, changing the starting concentration of one reactant at a time, and to measure the rate at the very start, from the slope of the tangent at $t=0$. At that moment every concentration is exactly what we put in, and no product is yet present to interfere or react back.
- Write the general rate law $\text{rate}=k[\mathrm{A}]^{x}[\mathrm{B}]^{y}$ with unknown orders.
- Pick two experiments in which only $[\mathrm{A}]$ changes. Divide their rate laws: $k$ and $[\mathrm{B}]^{y}$ cancel, leaving $\dfrac{r_{2}}{r_{1}}=\left(\dfrac{[\mathrm{A}]_{2}}{[\mathrm{A}]_{1}}\right)^{x}$. Read off $x$ (doubling $[\mathrm{A}]$: rate $\times2\Rightarrow x=1$, $\times4\Rightarrow x=2$, unchanged $\Rightarrow x=0$). If the ratio is awkward, use $x=\dfrac{\log(r_{2}/r_{1})}{\log([\mathrm{A}]_{2}/[\mathrm{A}]_{1})}$.
- Repeat for B with two experiments at the same $[\mathrm{A}]$. If no such pair exists, divide anyway and substitute the order already found.
- Find $k$ from any one experiment, and check it with another. Give its units from the overall order.
| Experiment | $[\mathrm{NO}]_{0}$ / $\mathrm{mol\,L^{-1}}$ | $[\mathrm{Cl_{2}}]_{0}$ / $\mathrm{mol\,L^{-1}}$ | initial rate / $\mathrm{mol\,L^{-1}\,s^{-1}}$ |
|---|---|---|---|
| 1 | 0.010 | 0.010 | $1.2\times10^{-4}$ |
| 2 | 0.020 | 0.010 | $4.8\times10^{-4}$ |
| 3 | 0.010 | 0.020 | $2.4\times10^{-4}$ |
In JEE Main and NEET 2026 (elementary and complex reactions); not required by the CBSE syllabus, and JEE Advanced lists rate laws but not complex reactions.
If a fast, reversible step comes before the slow step, the rate law of the slow step contains an intermediate, which must be eliminated. For $\mathrm{2NO+O_{2}\rightarrow2NO_{2}}$ one proposed mechanism is: (1) $\mathrm{2NO\rightleftharpoons N_{2}O_{2}}$ (fast equilibrium, constant $K$); (2) $\mathrm{N_{2}O_{2}+O_{2}\rightarrow2NO_{2}}$ (slow, $k_{2}$). Step 2 gives $\text{rate}=k_{2}[\mathrm{N_{2}O_{2}}]$, and step 1 gives $[\mathrm{N_{2}O_{2}}]=K[\mathrm{NO}]^{2}$, so $\text{rate}=k_{2}K[\mathrm{NO}]^{2}[\mathrm{O_{2}}]$: the same third-order law as a single termolecular step. A rate law that fits the data supports a mechanism but never proves it.
Exercises
- 1 Write the rate of the reaction $\mathrm{2HI(g)\rightarrow H_{2}(g)+I_{2}(g)}$ in terms of the rate of change of the concentration of each species.
- 2 In a reaction $\mathrm{A}\rightarrow\text{products}$, $[\mathrm{A}]$ falls from $0.80$ to $0.62\ \mathrm{mol\,L^{-1}}$ in $30\ \mathrm{s}$. Find the average rate in $\mathrm{mol\,L^{-1}\,s^{-1}}$ and in $\mathrm{mol\,L^{-1}\,min^{-1}}$.
- 3 In $\mathrm{N_{2}(g)+3H_{2}(g)\rightarrow2NH_{3}(g)}$, ammonia forms at $2.0\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. Find the rates of disappearance of $\mathrm{N_{2}}$ and $\mathrm{H_{2}}$, and the rate of the reaction.
- 4 The rate of the reaction $\mathrm{2SO_{2}(g)+O_{2}(g)\rightarrow2SO_{3}(g)}$ is $2.5\times10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}}$. Find the rates of disappearance of $\mathrm{SO_{2}}$ and $\mathrm{O_{2}}$ and the rate of formation of $\mathrm{SO_{3}}$.
- 5 In a closed vessel, $\mathrm{2NO_{2}(g)\rightarrow2NO(g)+O_{2}(g)}$. The partial pressure of $\mathrm{O_{2}}$ rises by $0.012\ \mathrm{atm}$ in $20\ \mathrm{s}$. Find the average rate of disappearance of $\mathrm{NO_{2}}$ and the average rate of the reaction, in $\mathrm{atm\,s^{-1}}$.
- 6 Using the $\mathrm{N_{2}O_{5}}$ table, find the average rate of disappearance of $\mathrm{N_{2}O_{5}}$ and the average rate of formation of $\mathrm{O_{2}}$ between $900$ and $1800\ \mathrm{s}$.
- 7 The tangent to a $[\mathrm{R}]$–$t$ curve at a certain moment passes through the points $(0\ \mathrm{s},\ 0.50\ \mathrm{mol\,L^{-1}})$ and $(250\ \mathrm{s},\ 0.10\ \mathrm{mol\,L^{-1}})$. Find the instantaneous rate at that moment.
- 8 For a reaction that slows down as it proceeds, explain whether the instantaneous rate at $t=0$ is larger or smaller than the average rate over the first ten minutes. Use the shape of the curve.
- 1 For $\text{rate}=k[\mathrm{A}]^{1/2}[\mathrm{B}]^{3/2}$, give the order with respect to each reactant, the overall order and the units of $k$.
- 2 For $\text{rate}=k[\mathrm{A}][\mathrm{B}]^{2}$, by what factor does the rate change when (a) $[\mathrm{B}]$ is doubled; (b) both concentrations are doubled; (c) $[\mathrm{A}]$ is tripled and $[\mathrm{B}]$ is halved?
- 3 The reaction $\mathrm{2NO(g)+O_{2}(g)\rightarrow2NO_{2}(g)}$ has $\text{rate}=k[\mathrm{NO}]^{2}[\mathrm{O_{2}}]$. The volume of the vessel is suddenly reduced to one-third at constant temperature. How does the rate change?
- 4 Give the units of $k$ for (a) a zero-order reaction with time in minutes; (b) a second-order gas reaction with pressures in atm and time in seconds; (c) a third-order reaction with concentrations in $\mathrm{mol\,L^{-1}}$ and time in seconds.
- 5 Give the overall order of a reaction whose rate constant has units (a) $\mathrm{L\,mol^{-1}\,min^{-1}}$; (b) $\mathrm{mol^{-1/2}\,L^{1/2}\,s^{-1}}$; (c) $\mathrm{mol\,L^{-1}\,s^{-1}}$; (d) $\mathrm{h^{-1}}$.
- 6 For $\text{rate}=k[\mathrm{A}]^{n}$: (a) tripling $[\mathrm{A}]$ makes the rate $27$ times larger; (b) in another reaction, doubling $[\mathrm{A}]$ makes it $2.83$ times larger. Find $n$ in each case.
- 7 Explain, naming the factor involved: (a) zinc powder reacts with dilute $\mathrm{HCl}$ faster than zinc granules; (b) food keeps longer in a refrigerator; (c) a little $\mathrm{MnO_{2}}$ makes $\mathrm{KClO_{3}}$ give off oxygen much faster on heating, and is recovered unchanged.
- 8 For $\text{rate}=k[\mathrm{A}][\mathrm{B}]$ with $k=2.0\times10^{-2}\ \mathrm{L\,mol^{-1}\,s^{-1}}$, find the rate when $[\mathrm{A}]=0.25$ and $[\mathrm{B}]=0.40\ \mathrm{mol\,L^{-1}}$.
- 1 Give the molecularity of each elementary step: (a) $\mathrm{O_{3}\rightarrow O_{2}+O}$; (b) $\mathrm{H+Cl_{2}\rightarrow HCl+Cl}$; (c) $\mathrm{2NO+O_{2}\rightarrow2NO_{2}}$ (if elementary); (d) $\mathrm{Cl+Cl+M\rightarrow Cl_{2}+M}$, where M is any third molecule that carries away energy.
- 2 Explain why molecularity cannot be zero or fractional, and why elementary steps of molecularity four are practically never observed.
- 3 The mechanism $\mathrm{NO_{2}+F_{2}\rightarrow NO_{2}F+F}$ (slow), $\ \mathrm{F+NO_{2}\rightarrow NO_{2}F}$ (fast) has been proposed. Write the overall equation, name the intermediate, and predict the rate law and the overall order.
- 4 Below about $500\ \mathrm{K}$, the reaction $\mathrm{NO_{2}(g)+CO(g)\rightarrow NO(g)+CO_{2}(g)}$ is thought to go by the mechanism $\mathrm{NO_{2}+NO_{2}\rightarrow NO_{3}+NO}$ (slow), $\ \mathrm{NO_{3}+CO\rightarrow NO_{2}+CO_{2}}$ (fast). Show that the steps add up, and predict the rate law. What happens to the rate if $[\mathrm{CO}]$ is doubled?
- 5 True or false? Give a reason. (a) A reaction with $\text{rate}=k[\mathrm{A}][\mathrm{B}]$ must be an elementary bimolecular reaction. (b) For an elementary reaction, order and molecularity are equal. (c) A catalyst can appear in the rate law of a reaction.
- 6 The reaction $\mathrm{2A+B\rightarrow C}$ has $\text{rate}=k[\mathrm{A}][\mathrm{B}]$. Can it be elementary? Suggest a two-step mechanism that fits this rate law.
- 1 For $\mathrm{A}\rightarrow\text{products}$: $[\mathrm{A}]_{0}=0.10$ gives rate $3.0\times10^{-4}$ and $[\mathrm{A}]_{0}=0.20$ gives $1.2\times10^{-3}$. Find the order and $k$.
- 2 For $\mathrm{2A+B\rightarrow C}$: ($[\mathrm{A}]_{0}$, $[\mathrm{B}]_{0}$, rate) $=(0.20,\ 0.10,\ 4.0\times10^{-4})$, $(0.40,\ 0.10,\ 8.0\times10^{-4})$, $(0.40,\ 0.20,\ 3.2\times10^{-3})$. Find the rate law and $k$, and the rate when $[\mathrm{A}]_{0}=0.10$ and $[\mathrm{B}]_{0}=0.30$.
- 3 The acid-catalysed iodination of propanone (acetone), $\mathrm{CH_{3}COCH_{3}+I_{2}\xrightarrow{H^{+}}CH_{3}COCH_{2}I+HI}$, gives ($[\text{propanone}]$, $[\mathrm{I_{2}}]$, $[\mathrm{H^{+}}]$, rate): $(0.20,\ 0.010,\ 0.10,\ 5.6\times10^{-7})$, $(0.40,\ 0.010,\ 0.10,\ 1.12\times10^{-6})$, $(0.20,\ 0.020,\ 0.10,\ 5.6\times10^{-7})$, $(0.20,\ 0.010,\ 0.20,\ 1.12\times10^{-6})$. Find the order in each species, the rate law and $k$. What does the order in $\mathrm{I_{2}}$ tell you about the slow step?
- 4 Keeping $[\mathrm{B}]$ fixed, raising $[\mathrm{A}]$ from $0.050$ to $0.20$ raises the rate from $1.5\times10^{-3}$ to $3.0\times10^{-3}$. Find the order with respect to A.
- 5 Atmospheric chemistry. For $\mathrm{NO(g)+O_{3}(g)\rightarrow NO_{2}(g)+O_{2}(g)}$ at $298\ \mathrm{K}$: ($[\mathrm{NO}]$, $[\mathrm{O_{3}}]$, rate) $=(1.0\times10^{-6},\ 3.0\times10^{-6},\ 3.3\times10^{-5})$, $(2.0\times10^{-6},\ 3.0\times10^{-6},\ 6.6\times10^{-5})$, $(2.0\times10^{-6},\ 9.0\times10^{-6},\ 1.98\times10^{-4})$. Find the rate law and $k$.
- 6 Give two reasons why rates are measured at the start of a reaction, rather than part-way through, when finding a rate law.
E1. Board · 3 marks (a) Give two differences between the order and the molecularity of a reaction. (b) The rate law of $\mathrm{2NO_{2}(g)+F_{2}(g)\rightarrow2NO_{2}F(g)}$ is $\text{rate}=k[\mathrm{NO_{2}}][\mathrm{F_{2}}]$. Is the reaction elementary? Give a reason.
E2. JEE Main Numerical value. For the gas-phase reaction $\mathrm{3A\rightarrow2B}$, the rate of disappearance of A at a certain instant is $0.045\ \mathrm{mol\,L^{-1}\,s^{-1}}$. The rate of formation of B at that instant, in units of $10^{-3}\ \mathrm{mol\,L^{-1}\,s^{-1}}$, is ______.
E3. NEET The rate law of a reaction is $\text{rate}=k[\mathrm{A}]^{3/2}[\mathrm{B}]^{-1/2}$. The units of $k$ are (A) $\mathrm{s^{-1}}$ (B) $\mathrm{L\,mol^{-1}\,s^{-1}}$ (C) $\mathrm{mol\,L^{-1}\,s^{-1}}$ (D) $\mathrm{L^{1/2}\,mol^{-1/2}\,s^{-1}}$
E4. JEE Main NEET For $\mathrm{A+B\rightarrow P}$: ($[\mathrm{A}]_{0}$, $[\mathrm{B}]_{0}$, initial rate in $\mathrm{mol\,L^{-1}\,s^{-1}}$) $=(0.10,\ 0.10,\ 2.0\times10^{-3})$, $(0.30,\ 0.10,\ 6.0\times10^{-3})$, $(0.30,\ 0.30,\ 5.4\times10^{-2})$. The rate law is (A) $k[\mathrm{A}][\mathrm{B}]$ (B) $k[\mathrm{A}]^{2}[\mathrm{B}]$ (C) $k[\mathrm{A}][\mathrm{B}]^{2}$ (D) $k[\mathrm{A}]^{2}[\mathrm{B}]^{2}$
E5. JEE Advanced One or more correct. The decomposition $\mathrm{2NO_{2}Cl(g)\rightarrow2NO_{2}(g)+Cl_{2}(g)}$ is proposed to occur by $\mathrm{NO_{2}Cl\rightarrow NO_{2}+Cl}$ (slow) followed by $\mathrm{NO_{2}Cl+Cl\rightarrow NO_{2}+Cl_{2}}$ (fast). Which statements are correct? (A) $\text{rate}=k[\mathrm{NO_{2}Cl}]$ (B) Cl is a catalyst (C) the reaction is first order overall (D) the overall reaction is bimolecular
E6. Board NEET Assertion–reason. Assertion (A): The order of a reaction can be zero or fractional. Reason (R): The order of a reaction is the number of molecules that collide in its rate-determining step. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
Research question. Can a "clock reaction" — a reaction that suddenly turns blue — be used to measure a rate law with school equipment?
Why it matters. Real kinetics laboratories, from drug-stability testing in pharmaceutical companies to studies of atmospheric reactions, rarely watch concentrations directly. They time an event (a colour change, a pressure reached, a signal in a detector) and turn the time into a rate. In the persulphate–iodide clock, $\mathrm{S_{2}O_{8}^{2-}(aq)+2I^{-}(aq)\rightarrow2SO_{4}^{2-}(aq)+I_{2}(aq)}$, a small fixed amount of thiosulphate removes the iodine as fast as it forms; when the thiosulphate runs out, iodine meets starch and the solution turns blue. The same small amount of iodine has formed every time, so the initial rate is proportional to $1/t$.
What to do.
- (Illustrative data, constructed for this exercise.) A class kept $[\mathrm{S_{2}O_{8}^{2-}}]_{0}=0.040\ \mathrm{mol\,L^{-1}}$ and varied $[\mathrm{I^{-}}]_{0}$: $0.040$, $0.080$, $0.120$ and $0.160\ \mathrm{mol\,L^{-1}}$ turned blue after $64$, $32$, $21$ and $16\ \mathrm{s}$. Compute $1/t$ for each run.
- Plot $\log(1/t)$ against $\log[\mathrm{I^{-}}]_{0}$. Explain why the slope of this line is the order with respect to $\mathrm{I^{-}}$, and find it.
- With $[\mathrm{I^{-}}]_{0}=0.080\ \mathrm{mol\,L^{-1}}$ fixed, doubling $[\mathrm{S_{2}O_{8}^{2-}}]_{0}$ from $0.040$ to $0.080\ \mathrm{mol\,L^{-1}}$ cut the time from $32$ to $16\ \mathrm{s}$ (illustrative data, constructed for this exercise). Find the order with respect to persulphate and write the rate law.
- If your school has the chemicals (with your teacher's supervision), repeat one set of runs yourself and compare.
Tools: Calculator or spreadsheet; Interactive 2 and the check-lab under Exercises IV.
Hand in: Your table and log–log graph, the rate law, and two sources of error in timing a colour change by eye.
Going further. Repeat the timing at several temperatures; you will be able to find the activation energy of this reaction after Section 3.3.
🧪 Self-check quiz
Q1. For $\mathrm{2A\rightarrow3B}$, A disappears at $0.30\ \mathrm{mol\,L^{-1}\,s^{-1}}$. B is formed at
Q2. The rate constant of a first-order reaction has units
Q3. Which of these can have a fractional value?
Q4. For $\text{rate}=k[\mathrm{A}]^{2}[\mathrm{B}]$, $[\mathrm{A}]$ is doubled and $[\mathrm{B}]$ is halved. The rate
Q5. In a reaction mechanism, the rate-determining step is