Chemical Kinetics: Class 12 Chemistry · Chapter 3

Chemistry Part 1 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 11 Oct 2026 · Updated 11 Oct 2026

90 min
44 practice questions
19 MCQs

What you will learn in Chemical Kinetics

Define and calculate the average and instantaneous rate of a reaction, and express one rate in terms of any reactant or product using the coefficients of the balanced equation
Write a rate law, find the order of a reaction from experimental data and deduce the units of its rate constant
Distinguish elementary from complex reactions and molecularity from order, and identify the rate-determining step of a simple mechanism
Derive and use the integrated rate equations and half-lives of zero- and first-order reactions, including pseudo first-order reactions and total-pressure data
Use the Arrhenius equation to find activation energy and the pre-exponential factor, and to relate rate constants at two temperatures
Explain with energy profiles and energy distributions how temperature and catalysts change the rate of a reaction
Describe collision theory, including collision frequency, activation energy and the steric factor

Original companion text aligned to NCERT Chemistry Class 12, Chapter 3, pp. 61–88. Not affiliated with or endorsed by NCERT.

A sheet of newspaper on a table is, chemically, a fuel sitting in oxygen. Burning it would release a lot of energy, and its $Δ G$ is large and negative, yet the paper can lie there for years. A mixture of hydrogen and oxygen is similar: it waits, unchanged, until a spark sets it off with a bang. Being allowed to react and actually reacting at a noticeable speed are two different things.

Thermodynamics tells us whether a change is allowed, and equilibrium tells us how far it can go; neither contains time. The question a cook, an engineer or a doctor usually asks first, how fast?, belongs to chemical kinetics (रासायनिक बलगतिकी). The name comes from the Greek kinesis, movement. Kinetics decides how long a vaccine stays potent outside a cold box, how soon a cement floor can be walked on, and how quickly the ozone layer recovers once harmful gases are banned.

This chapter first measures rates in the flask and finds the laws they obey, then zooms in on single collisions to see why temperature and catalysts matter so much.

3.1 Rate of a Chemical Reaction

Some reactions are over in the blink of an eye; others take a lifetime. Mixing an acid with an alkali, or lighting a firecracker, gives products in well under a second. A cut apple turns brown in minutes, and milk takes several hours to set into curd. A bright copper statue takes years to turn green. Fig. 3.1 places these on one time line.

3/3A copper surface takes years to turn green.

Fig. 3.1Reaction times span more than ten powers of ten; the axis is logarithmic, so each tick is many times the one before.NCERT p. 62

A horizontal time axis with ticks at 1 second, 1 minute, 1 hour, 1 day, 1 year and 100 years, equally spaced in powers of ten. Step 1, fast: two markers at the left end, acid plus alkali at less than a millisecond and a firecracker at a few hundredths of a second, inside a shaded band labelled fast. Step 2, moderate: markers for a cut apple browning at about a quarter of an hour and curd setting at about six hours, inside a band labelled moderate. Step 3, slow: a marker for a copper statue turning green after about twenty years, inside a band labelled slow, with every marker still shown.

To compare reactions we need a number. Think of a water tank being drained into a bucket: you can say how fast it empties by how many litres the tank loses each minute, or by how many litres the bucket gains. A reaction is described the same way, by how quickly the concentration of a reactant falls or that of a product rises.

Call the reactant R and the product P, so that the reaction is $R \overset{}{\to} P$ , one P for each R, and keep the volume fixed. Square brackets mean moles per litre. Suppose a reading at time $t_{1}$ gives $[R]_{1}$ and a later reading at $t_{2}$ gives $[R]_{2}$ . Across the gap $Δ t = t_{2} - t_{1}$ the concentration has changed by $Δ [R] = [R]_{2} - [R]_{1}$ , and the product's by $Δ [P]$ . Dividing by the time gives the average rate:

Average rate of disappearance of R

r_{a v} = - \frac{Δ [R]}{Δ t}

(3.1)

Average rate of appearance of P

r_{a v} = + \frac{Δ [P]}{Δ t}

(3.2)

$Δ [R]$ is negative because R is being used up, so the minus sign in (3.1) turns it into a positive rate. Either expression gives the average rate $r_{a v}$ : a single number for the whole interval, however the speed varied inside it.

Definition

Rate of a reaction

How fast a reactant is used up or a product is formed, measured as a positive change of molar concentration per unit time and divided by that species' coefficient in the balanced equation.

ExampleFor $R \overset{}{\to} P$ , rate $= - Δ [R] / Δ t = Δ [P] / Δ t$ .

Units of rate. Because a rate is a change of concentration divided by a time, its unit is a concentration unit per time unit: $m o l L^{- 1} s^{- 1}$ in most problems, or per minute or per hour for slow reactions. Gas reactions are often followed by pressure instead, giving units such as $a t m s^{- 1}$ or $b a r m i n^{- 1}$ .

3/4Shrink $Δ t$ to zero and the chord becomes the tangent at $t$ : $r_{i n s t} = - d [R] / d t = d [P] / d t$ .

Fig. 3.2The average rate is the slope of a chord; the instantaneous rate is the slope of a tangent at one moment.NCERT Fig. 3.1 · p. 63

A graph of concentration against time. The reactant curve R starts high on the left and falls ever more slowly; the product curve P starts at zero and rises as a mirror image. Step 1 shows only the two curves. Step 2 adds a straight chord joining two points of the R curve at times t1 and t2, with the horizontal gap marked delta t and the vertical drop marked delta R; its slope gives the average rate. Step 3 replaces the chord by a tangent touching the R curve at a single time t, and a matching tangent on the P curve; their slopes give the instantaneous rate. Step 4 adds a second tangent at a later time, which is much flatter, showing that the rate falls as the reaction proceeds.

The average rate hides the fact that most reactions slow down as their reactants are used up. To find the rate at one moment, shrink the interval. As $Δ t$ approaches zero, the chord between two points of the concentration–time curve turns into the tangent at a single point, and the ratio of changes becomes a derivative:

Instantaneous rate

r_{i n s t} = - \frac{d [R]}{d t} = \frac{d [P]}{d t}

(3.3)

Graphically, the instantaneous rate at time $t$ is the slope of the tangent to the $[P]$ – $t$ curve at $t$ , or minus the slope of the tangent to the $[R]$ – $t$ curve (Fig. 3.2). Early in a reaction the tangents are steep; later they flatten. In the rate lab of Section 3.3.3 you can draw a chord and a tangent on the same curve and watch them agree as the interval shrinks.

Table 3.1Fading of crystal violet in a basic solution (one run)NCERT Table 3.1 · p. 64

$t$ / s	$[C V^{+}]$ / $10^{- 6} m o l L^{- 1}$
0	20.0
60	16.7
120	14.0
180	11.7
240	9.7
360	6.8
480	4.7
600	3.3

Example 3.1NCERT Example 3.1 · p. 63

Average rates from a data table

Crystal violet ( $C V^{+}$ ) is a purple dye. In a basic solution it reacts slowly with hydroxide ions to give a colourless product, so a colorimeter can follow its concentration. Table 3.1 shows one run. Find the average rate of loss of the dye over each interval, and describe how the rate changes as the reaction proceeds.

Try it yourself first. Then reveal the solution one step at a time.

When the coefficients differ. In $N_{2} (g) + 3 H_{2} (g) \overset{}{\to} 2 N H_{3} (g)$ , every nitrogen molecule that reacts takes three hydrogen molecules with it and makes two ammonia molecules. So in any interval $[H_{2}]$ falls three times as fast as $[N_{2}]$ , and $[N H_{3}]$ rises twice as fast. Three different numbers would describe one reaction. To get a single rate, divide each species' rate of change by its coefficient in the balanced equation (Fig. 3.3):

\begin{aligned} Rate & = - \frac{Δ [N_{2}]}{Δ t} \\ = - \frac{1}{3} \frac{Δ [H_{2}]}{Δ t} \\ = + \frac{1}{2} \frac{Δ [N H_{3}]}{Δ t} \end{aligned}

3/3Divide each change per second by its coefficient: all three give $0.010 m o l L^{- 1} s^{- 1}$ .

Fig. 3.3Different species change at different speeds, but dividing by the coefficients gives one rate for the reaction.NCERT p. 65

Three bars stand side by side for nitrogen, hydrogen and ammonia in the reaction N2 plus 3 H2 gives 2 NH3. Step 1, at the start: nitrogen 0.50, hydrogen 1.50 and ammonia 0 moles per litre. Step 2, ten seconds later: the nitrogen bar has shrunk to 0.40, the hydrogen bar to 1.20 and an ammonia bar has grown to 0.20, with the changes minus 0.10, minus 0.30 and plus 0.20 written above them. Step 3: under each bar the change per second is divided by the coefficient, 1, 3 and 2, and all three give the same rate, 0.010 moles per litre per second.

The same rule works for any reaction $a A + b B \to c C + d D$ : take the rate of change of each species, divide by its own coefficient, and put a minus sign in front for reactants.

\begin{aligned} Rate & = - \frac{1}{a} \frac{Δ [A]}{Δ t} \\ = - \frac{1}{b} \frac{Δ [B]}{Δ t} \\ = + \frac{1}{c} \frac{Δ [C]}{Δ t} \\ = + \frac{1}{d} \frac{Δ [D]}{Δ t} \end{aligned}

Gases add one convenience. Since $p = c R T$ , at a fixed temperature a partial pressure is simply a concentration in disguise, so every rate above can be written with $Δ p$ in place of $Δ c$ .

Example 3.2NCERT Example 3.2 · p. 66

One reaction, several rates

Nitrosyl bromide decomposes in the gas phase: $2 N O B r (g) \overset{}{\to} 2 N O (g) + B r_{2} (g)$ . In one experiment $[N O B r]$ fell from $0.0450$ to $0.0390 m o l L^{- 1}$ in $25.0 m i n$ . Find the average rate of the reaction per minute, per hour and per second, and the rate at which NO was formed.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 2 questions

Practice: reaction rates (Intext Questions 3.1–3.2): questions with answers

1
NumericalEasyNCERT Intext Question 3.1
In a reaction $R \overset{}{\to} P$ , $[R]$ falls from $0.048$ to $0.036 m o l L^{- 1}$ in $15 m i n$ .
1. Find the average rate in $m o l L^{- 1} m i n^{- 1}$ .
  mol L⁻¹ min⁻¹
  Show solution
  $r_{a v} = \frac{0.048 - 0.036}{15} = 8.0 \times 10^{- 4}$
  in $m o l L^{- 1} m i n^{- 1}$ .
2. Express it in $m o l L^{- 1} s^{- 1}$ .
  mol L⁻¹ s⁻¹
  Show solution
  Divide by 60 s per minute: $8.0 \times 10^{- 4} / 60 = 1.3 \times 10^{- 5} m o l L^{- 1} s^{- 1}$ .
2
NumericalEasyNCERT Intext Question 3.2
For the reaction $3 A \overset{}{\to} p r o d u c t s$ , $[A]$ drops from $0.90$ to $0.72 m o l L^{- 1}$ in $12 m i n$ .
1. Find the average rate of disappearance of A, in $m o l L^{- 1} m i n^{- 1}$ .
  mol L⁻¹ min⁻¹
  Show solution
  $- \frac{Δ [A]}{Δ t} = \frac{0.18}{12} = 0.015$
2. Find the average rate of the reaction, in $m o l L^{- 1} m i n^{- 1}$ .
  mol L⁻¹ min⁻¹
  Show solution
  A's coefficient is 3, so the rate of the reaction is $0.015 / 3 = 5.0 \times 10^{- 3} m o l L^{- 1} m i n^{- 1}$ .

3.2 Factors Influencing Rate of a Reaction

Why does a crushed tablet fizz away faster than a whole one, and why does milk keep longer in a fridge? The rate of a given reaction depends on its conditions: the concentrations of the reactants (for gases, their pressures), the temperature, the presence of a catalyst, and, when a solid takes part, how finely it is divided. This section deals with concentration; temperature and catalysts follow in Section 3.4.

Try this

The fizzing-tablet race

You’ll need

four effervescent antacid or vitamin C tablets of one brand
four identical glasses of water: two cold, two warm
a spoon and a clean plate
a phone stopwatch

Predict first

Which glass will stop fizzing first?

cold water, whole tabletcold water, crushed tabletwarm water, whole tabletwarm water, crushed tablet

Crush two of the tablets into powder on a clean plate; keep the other two whole.
Drop a whole tablet into a glass of cold water and start the stopwatch. Stop it when the fizzing ends.
Repeat with a crushed tablet in cold water, then with a whole and a crushed tablet in warm water.
Record the four times in the table.

Table 3.2

Glass	Time to stop fizzing (s)
Cold water, whole tablet
Cold water, crushed tablet
Warm water, whole tablet
Warm water, crushed tablet

Your entries are saved automatically.

3.2.1 Dependence of Rate on Concentration

The data of Example 3.1 already show the pattern: as the dye is used up, the reaction slows. Start the same reaction with more dye and the early rate is higher. For most reactions, more concentrated reactants mean a faster reaction.

Chemists capture this dependence in a rate law (वेग नियम), also known as a rate equation or rate expression: a formula with the rate on one side and the reactant concentrations on the other.

Definition

Rate law

An experimentally determined equation linking the rate to the concentrations of the reactants, with each concentration carrying its own power.

Example $Rate = k [N O]^{2} [H_{2}]$ for $2 N O + 2 H_{2} \overset{}{\to} N_{2} + 2 H_{2} O$ .

3.2.2 Rate Expression and Rate Constant

For a general reaction $a A + b B \to c C + d D$ , experiments usually show that the rate is proportional to each reactant's concentration raised to some power:

Rate \propto [A]^{x} [B]^{y}

(3.4)

Replacing the proportionality sign by a constant $k$ gives

Rate = k [A]^{x} [B]^{y}

(3.4a)

and, written with the instantaneous rate of loss of a reactant R,

Differential rate equation

- \frac{d [R]}{d t} = k [A]^{x} [B]^{y}

(3.4b)

A rate law in the form (3.4b) is a differential rate equation. The constant $k$ is the rate constant (वेग स्थिरांक). At a given temperature it has one value for the reaction, whatever the concentrations. Setting every concentration in (3.4a) to $1 m o l L^{- 1}$ shows what it means: $k$ is the rate at unit concentrations.

Definition

Rate constant

The proportionality constant $k$ in a rate law. It depends on the reaction and the temperature, not on the concentrations.

The powers $x$ and $y$ need not equal the coefficients $a$ and $b$ ; they can only be found by experiment. A simple plan, the method of initial rates, measures the rate just after mixing, before the concentrations have changed much. In a series of runs, one starting concentration is changed while the others are kept the same (Fig. 3.4).

3/3Tripling $[H_{2}]$ makes it 3 times faster: $3^{y} = 3$ , so $y = 1$ .

Fig. 3.4Change one starting concentration at a time and see how many times faster the reaction starts.NCERT Table 3.2 · p. 67

Three vertical bars show initial rates. Step 1: only run 1, the reference, with both starting concentrations at 0.010 moles per litre; its bar is one unit tall. Step 2: run 2 appears, with nitrogen monoxide doubled and hydrogen unchanged; its bar is four units tall, labelled times 4, which equals 2 squared, so the reaction is second order in nitrogen monoxide. Step 3: run 3 appears, with nitrogen monoxide back at 0.010 and hydrogen tripled; its bar is three units tall, labelled times 3, which equals 3 to the power 1, so the reaction is first order in hydrogen.

Table 3.3Initial rates of

2 N O + 2 H_{2} \overset{}{\to}

N_{2} + 2 H_{2} O

at one high temperature (concentrations in

m o l L^{- 1}

, rates in

m o l L^{- 1} s^{- 1}

)

Run	$[N O]$	$[H_{2}]$	Initial rate
1	0.010	0.010	$2.4 \times 10^{- 6}$
2	0.020	0.010	$9.6 \times 10^{- 6}$
3	0.010	0.030	$7.2 \times 10^{- 6}$
4	0.030	0.020	$4.32 \times 10^{- 5}$

ExampleNCERT Table 3.2 · p. 67

Finding a rate law from initial rates

Nitrogen monoxide reacts with hydrogen at high temperature: $2 N O (g) + 2 H_{2} (g) \overset{}{\to}$ $N_{2} (g) + 2 H_{2} O (g)$ . Use the four runs in Table 3.3 to find the rate law and the rate constant.

Try it yourself first. Then reveal the solution one step at a time.

Notice that the power of $[H_{2}]$ is 1 although its coefficient is 2. A balanced equation tells you the overall bookkeeping of atoms. It does not tell you how the rate depends on concentration.

Table 3.4Rate laws do not follow from the coefficientsNCERT p. 68

Reaction	Rate law (measured)
$2 N O + 2 H_{2} \overset{}{\to}$ $N_{2} + 2 H_{2} O$	$k [N O]^{2} [H_{2}]$
$H_{2} + B r_{2} \overset{}{\to}$ $2 H B r$	$k [H_{2}] [B r_{2}]^{1 / 2}$ (at the start)
$N O_{2} + C O \overset{}{\to}$ $N O + C O_{2}$	$k [N O_{2}]^{2}$ (below about 500 K)

In every row the powers differ from the coefficients; in the last one, CO does not appear in the rate law at all. A rate law cannot be written down from the balanced equation; it must be found by experiment.

3.2.3 Order of a Reaction

The powers in a rate law measure how strongly each concentration drives the rate. In $Rate = k [A]^{x} [B]^{y}$ , $x$ is the order with respect to A and $y$ the order with respect to B. Their sum $x + y$ is the overall order of the reaction (अभिक्रिया की कोटि).

Definition

Order of a reaction

The total of the exponents on the concentration terms in a measured rate law.

Example $Rate = k [N O]^{2} [H_{2}]$ : second order in NO, first order in $H_{2}$ , third order overall.

Because an order is measured, it need not be a whole number. It is commonly 1, 2 or 3, but it can be zero, a fraction, or even negative when a substance slows down its own reaction. Zero order with respect to a reactant means its concentration has no effect: doubling it changes nothing.

Example 3.3NCERT Example 3.3 · p. 68

Overall order from a rate law

Find the overall order of each reaction from its measured rate law. (a) $2 N O + 2 H_{2} \overset{}{\to} N_{2} + 2 H_{2} O$ , $Rate = k [N O]^{2} [H_{2}]$ . (b) $H_{2} + B r_{2} \overset{}{\to} 2 H B r$ at the start, $Rate = k [H_{2}] [B r_{2}]^{1 / 2}$ . (c) $2 O_{3} \overset{}{\to} 3 O_{2}$ , $Rate = k [O_{3}]^{2} [O_{2}]^{- 1}$ .

Try it yourself first. Then reveal the solution one step at a time.

Why should the powers differ from the coefficients? The answer lies in what a balanced equation hides. Most reactions do not happen in one go: they climb a staircase of simple steps, and the list of those steps is the reaction's mechanism. A reaction whose equation describes a single such step, exactly as it happens, is an elementary reaction (प्राथमिक अभिक्रिया); one that needs several steps is a complex reaction.

Definition

Elementary reaction

A reaction that occurs in one step, with the reacting species colliding exactly as shown in its equation.

Complex reactions follow several patterns. In consecutive reactions the product of one step is the reactant of the next: in the liver, ethanol is oxidised to ethanal, and the ethanal is then oxidised to ethanoic acid. In reversible reactions the products re-form the reactants, as in $N_{2} + 3 H_{2} \overset{}{⇌} 2 N H_{3}$ . In side (parallel) reactions one set of reactants gives several products at once: chlorinating methane gives $C H_{3} C l$ together with $C H_{2} C l_{2}$ , $C H C l_{3}$ and $C C l_{4}$ .

Units of the rate constant. The unit of $k$ depends on the overall order $n = x + y$ . From (3.4a), $k = Rate / ([A]^{x} [B]^{y})$ , a rate divided by a concentration raised to the power $n$ . With concentration in $m o l L^{- 1}$ and time in seconds this gives:

Unit of the rate constant for order n

unit of k = (m o l L^{- 1})^{1 - n} s^{- 1}

Table 3.5Units of

k

for common orders (time in seconds)NCERT Table 3.3 · p. 69

Overall order $n$	Unit of $k$
0	$m o l L^{- 1} s^{- 1}$
1	$s^{- 1}$
3/2	$L^{1 / 2} m o l^{- 1 / 2} s^{- 1}$
2	$L m o l^{- 1} s^{- 1}$
3	$L^{2} m o l^{- 2} s^{- 1}$

Only a first-order rate constant has no concentration in its unit. Its value is therefore the same whatever concentration unit you use, which makes first-order constants especially easy to compare.

Example 3.4NCERT Example 3.4 · p. 69

Order from the unit of k

What is the overall order of a reaction whose rate constant is (i) $4.1 \times 10^{- 3} m o l L^{- 1} m i n^{- 1}$ , (ii) $6.5 \times 10^{- 2} L^{2} m o l^{- 2} s^{- 1}$ , (iii) $2.7 h^{- 1}$ ?

Try it yourself first. Then reveal the solution one step at a time.

3.2.4 Molecularity of a Reaction

For an elementary step we can ask a sharper question than order: how many particles must meet at once for the step to happen? That count is the molecularity (आण्विकता) of the step.

Definition

Molecularity

How many particles, whether molecules, atoms or ions, must collide together for an elementary step to take place.

Example $N O + O_{3} \overset{}{\to} N O_{2} + O_{2}$ is bimolecular.

Unimolecular: one species rearranges or breaks up by itself, as when cyclopropane opens into propene, $C_{3} H_{6} \overset{}{\to} C H_{3} C H = C H_{2}$ .
Bimolecular: two species collide, as when nitrogen monoxide meets ozone high in the atmosphere, $N O + O_{3} \overset{}{\to} N O_{2} + O_{2}$ .
Termolecular: three species meet together. Ozone forms when an oxygen atom and an $O_{2}$ molecule collide while a third molecule M, often $N_{2}$ , carries off the spare energy: $O + O_{2} + M \overset{}{\to} O_{3} + M$ .

Three particles arriving at the same spot at the same instant is already unlikely, so termolecular steps are rare and slow; four is so improbable that no elementary step of molecularity above three is known. The balanced equation for the oxidation of oxalate ions by permanganate in acid has 23 ions on the left ( $2 M n O_{4}^{-}$ , $5 C_{2} O_{4}^{2 -}$ and $16 H^{+}$ ), so it must happen in several steps.

When a reaction has several steps, which one sets the pace? Picture a shop where customers pick their goods quickly from the shelves and then queue at a single slow billing counter. The number of customers leaving each minute equals the billing rate, however fast the shelves are cleared. In the same way, the slowest elementary step, the rate-determining step (वेग निर्धारक पद), sets the pace for the whole reaction.

Definition

Rate-determining step

The slowest elementary step in a mechanism; the overall reaction cannot go faster than this step.

A real case is the reaction of nitrogen dioxide with carbon monoxide below about 500 K (Fig. 3.5). The evidence points to two bimolecular steps:

Step 1 (slow): $N O_{2} + N O_{2} \overset{}{\to} N O_{3} + N O$
Step 2 (fast): $N O_{3} + C O \overset{}{\to} N O_{2} + C O_{2}$

Adding the steps cancels $N O_{3}$ and one $N O_{2}$ , leaving $N O_{2} + C O \overset{}{\to} N O + C O_{2}$ . $N O_{3}$ is an intermediate: it is made in one step and used up in a later one, so it never appears in the overall equation. Step 1 is slow, so the reaction goes at the rate of step 1, $Rate = k [N O_{2}]^{2}$ . Carbon monoxide joins only after the slow step, which is why it is missing from the rate law in Table 3.4.

3/3The intermediate cancels. The rate is that of the slow step: $Rate = k [N O_{2}]^{2}$ .

Fig. 3.5The slow first step is the bottleneck, so it alone fixes the rate law.NCERT p. 70

Left to right: a pool of nitrogen dioxide molecules, a narrow gate labelled step 1 slow, a middle box holding the intermediate NO3, a wide gate labelled step 2 fast where carbon monoxide enters, and a products box. Step 1: a queue of NO2 pairs waits at the narrow gate while one pair passes through and becomes NO3 plus NO. Step 2: the NO3 meets a CO molecule at the wide gate and is turned at once into NO2 and CO2, so no queue forms there. Step 3: the overall equation NO2 plus CO gives NO plus CO2 appears, with the NO3 in the middle box labelled as an intermediate that cancels, and the rate law equals k times the square of the NO2 concentration, set by the slow gate.

Order and molecularity are easy to confuse. The comparison below sets them side by side.

Compare

Order versus molecularity

Order

found by experiment, from the rate law
can be zero, a fraction or negative
defined for elementary and complex reactions alike
when the slow step comes first, equals that step's molecularity

Molecularity

counted from the equation of one elementary step
always a whole number: 1, 2 or, rarely, 3
has no meaning for a complex reaction as a whole
for an elementary step, the same as its order

Practice · 2 questions

Practice: order (Intext Questions 3.3–3.4): questions with answers

Quick answerEasyNCERT Intext Question 3.3
A reaction $A + B \overset{}{\to} p r o d u c t s$ has the rate law $Rate = k [A]^{1 / 2} [B]$ . What is its overall order?
Show solution
Overall order $= \frac{1}{2} + 1 = \frac{3}{2}$ .
Quick answerEasyNCERT Intext Question 3.4
The conversion of X into Y is second order in X. If $[X]$ is reduced to one-third of its value, what happens to the rate of formation of Y?
it falls to one-thirdit falls to one-ninthit falls to one-sixthit rises nine times
Show solution
Rate $= k [X]^{2}$ . New rate $= k ([X] / 3)^{2} = \frac{1}{9} k [X]^{2}$ : the rate falls to one-ninth.

3.3 Integrated Rate Equations

Tangents are a clumsy way to do kinetics. You need a smooth curve, a steady hand and a ruler, and a small error in a slope becomes a large error in the order. A better plan is to turn the rate law into a formula for the concentration itself, so that raw readings of concentration and time can be tested directly. Integration does this in one step, and the result depends on the order. The syllabus needs two cases: zero order and first order.

3.3.1 Zero Order Reactions

In a zero-order reaction the rate does not depend on the reactant's concentration at all. For $R \overset{}{\to} P$ , $Rate = k [R]^{0} = k$ . The reactant is used up at a steady pace, like a candle that burns down by the same length every hour whether it is tall or nearly finished.

DerivationIntegrated rate equation for zero order

Start from the differential rate equation with zero order:

- \frac{d [R]}{d t} = k [R]^{0} = k

Separate the variables, $d [R] = - k d t$ , and integrate both sides, calling the integration constant $I$ :

[R] = - k t + I

(3.5)

When $t = 0$ , $[R]$ equals its starting value $[R]_{0}$ , so $I = [R]_{0}$ and

Zero-order integrated rate equation

[R] = [R]_{0} - k t

(3.6)

Rearranged for the rate constant:

k = \frac{[R]_{0} - [R]}{t}

(3.7)

Equation (3.6) has the form $y = m x + c$ . A plot of $[R]$ against $t$ is a straight line with slope $- k$ and intercept $[R]_{0}$ (Fig. 3.6). The line reaches zero at $t = [R]_{0} / k$ , so a zero-order reaction finishes in a definite time.

Zero order is the exception rather than the rule. It appears when something other than the reactant's concentration throttles the reaction, such as the number of sites on a catalyst. When phosphine decomposes on a hot tungsten wire at high pressure, $4 P H_{3} (g) \overset{}{\to} P_{4} (g) + 6 H_{2} (g)$ , the metal surface is completely covered with $P H_{3}$ molecules. Adding more gas cannot crowd more molecules onto the surface, so the rate is fixed by the number of surface sites. Enzyme reactions behave the same way when every enzyme molecule is busy. At low pressure the surface is only partly covered, and the same reaction becomes first order.

3/3So $[R]$ falls at a constant rate: a straight line of slope $- k$ .

Fig. 3.6When every surface site is busy, the rate cannot rise with pressure: the reaction is zero order and [R] falls in a straight line.NCERT Fig. 3.3 · p. 72

On the left, a strip of hot tungsten with six marked sites lies under a gas. Step 1, low pressure: only a few gas molecules are present and two of the six sites hold a molecule; more gas would fill more sites, so the rate follows the pressure. Step 2, high pressure: the gas is crowded and all six sites are occupied, with a label saying the surface is full; extra gas cannot speed things up, so the rate is constant. Step 3: on the right a graph of concentration of reactant against time shows a straight line falling from R zero on the vertical axis, with its slope marked minus k, reaching zero at a definite time.

Iodination of propanone in acid is a neat laboratory example: $C H_{3} C O C H_{3} + I_{2} \overset{}{\to} C H_{3} C O C H_{2} I + H I$ . Its rate does not depend on $[I_{2}]$ at all. So when propanone and acid are present in large excess, the brown colour of the iodine fades at a steady pace until it vanishes.

Example

A zero-order reaction

In an iodination run with propanone and acid in large excess, $[I_{2}]$ starts at $2.0 \times 10^{- 3} m o l L^{- 1}$ and falls at a constant $4.0 \times 10^{- 6} m o l L^{- 1} s^{- 1}$ . (a) What is $[I_{2}]$ after $150 s$ ? (b) When does the iodine colour disappear?

Try it yourself first. Then reveal the solution one step at a time.

3.3.2 First Order Reactions

A first-order reaction runs at a rate that varies directly with the reactant's concentration: halve the concentration and the rate halves too. So the reaction slows down continuously and, in principle, never quite finishes.

DerivationIntegrated rate equation for first order

For $R \overset{}{\to} P$ with first-order kinetics,

- \frac{d [R]}{d t} = k [R]

Separate the variables, $d [R] / [R] = - k d t$ , and integrate. The integral of $d [R] / [R]$ is $\ln [R]$ , so

\ln [R] = - k t + I

(3.8)

At $t = 0$ , $[R] = [R]_{0}$ , which gives $I = \ln [R]_{0}$ :

\ln [R] = \ln [R]_{0} - k t

(3.9)

Collect the logarithms and solve for $k$ :

k = \frac{1}{t} \ln \frac{[R]_{0}}{[R]}

(3.10)

Writing (3.8) at two times $t_{1}$ and $t_{2}$ removes $I$ without using the starting concentration:

\ln [R]_{1} = - k t_{1} + I

(3.11)

\ln [R]_{2} = - k t_{2} + I

(3.12)

Subtracting (3.12) from (3.11):

k = \frac{1}{t_{2} - t_{1}} \ln \frac{[R]_{1}}{[R]_{2}}

(3.13)

Taking the antilogarithm of (3.9) gives the concentration itself, an exponential decay:

[R] = [R]_{0} e^{- k t}

(3.14)

Finally, $\ln x = 2.303 \log x$ , so (3.10) is often used with common logarithms:

First-order integrated rate equation

k = \frac{2.303}{t} \log \frac{[R]_{0}}{[R]}

(3.15)

Equation (3.9) is a straight line again: a plot of $\ln [R]$ against $t$ has slope $- k$ and intercept $\ln [R]_{0}$ . Equivalently, $\log ([R]_{0} / [R])$ against $t$ is a straight line through the origin with slope $k / 2.303$ (Fig. 3.7). These plots are the quickest test of order: whichever plot of the data comes out straight tells you the order.

2/3 $\ln [R]$ against $t$ is a straight line of slope $- k$ .

Fig. 3.7First-order data curve on a plot of [R] against t but fall on straight lines when logarithms are plotted.NCERT Fig. 3.4 · p. 74

One set of axes with time across. Step 1: concentration of R against time is a curve falling steeply at first and then more and more slowly towards zero. Step 2: the vertical axis becomes the natural logarithm of R and the same data now lie on a straight line falling from ln R zero, with its slope marked minus k. Step 3: the vertical axis becomes log of R zero over R, and the data lie on a straight line starting at the origin and rising, with slope k divided by 2.303.

First-order kinetics is common. Cyclopropane isomerises to propene by a first-order process. Every radioactive decay is first order: iodine-131, used to treat thyroid disease, loses half its atoms every 8.0 days whatever the amount. Many medicines are cleared from the blood by first-order processes too, which is why doctors can quote a fixed half-life for a drug.

Example 3.5NCERT Example 3.5 · p. 74

Rate constant from two concentrations

Cyclopropane isomerises to propene by a first-order reaction. At a certain high temperature its concentration fell from $0.0500$ to $0.0300 m o l L^{- 1}$ in $15.0 m i n$ . Calculate the rate constant.

Try it yourself first. Then reveal the solution one step at a time.

For a gas-phase reaction at constant volume it is often easiest to follow the total pressure. Consider a first-order decomposition $A (g) \overset{}{\to} B (g) + C (g)$ that starts with pure A at pressure $p_{i}$ .

DerivationFirst-order rate constant from total pressure

Let the partial pressure of A fall by $x$ by time $t$ . Each A that reacts gives one B and one C:

Table 3.6

	$p_{A}$	$p_{B}$	$p_{C}$
$t = 0$	$p_{i}$	0	0
time $t$	$p_{i} - x$	$x$	$x$

The total pressure is $p_{t} = (p_{i} - x) + x + x = p_{i} + x$ , so $x = p_{t} - p_{i}$ and $p_{A} = p_{i} - x = 2 p_{i} - p_{t}$ . Partial pressure is proportional to concentration, so (3.15) becomes

k = \frac{2.303}{t} \log \frac{p_{i}}{2 p_{i} - p_{t}}

(3.16)

If one A gives a different number of gas molecules, the bookkeeping changes but the method does not: find $p_{A}$ from $p_{t}$ first, then use $k = (2.303 / t) \log (p_{i} / p_{A})$ .

Example 3.6NCERT Example 3.6 · p. 75

Following a reaction by its pressure

Di-tert-butyl peroxide, $(C H_{3})_{3} C O O C (C H_{3})_{3}$ , decomposes in the gas phase by a first-order reaction: each molecule gives two molecules of propanone and one of ethane. A sealed flask of fixed volume started with the pure peroxide at $0.400 a t m$ . After $20.0 m i n$ the total pressure was $0.520 a t m$ . Find the rate constant.

Try it yourself first. Then reveal the solution one step at a time.

3.3.3 Half-Life of a Reaction

The half-life (अर्धायु) of a reaction, $t_{1 / 2}$ , is how long it takes for a reactant's concentration to drop to half of where it started. It is a handy single number for how fast a reaction is.

Definition

Half-life

The time needed for half of a reactant to be used up, so that its concentration drops to 50 % of its starting value.

Zero order. Put $[R] = [R]_{0} / 2$ at $t = t_{1 / 2}$ into (3.7):

Half-life, zero order

t_{1 / 2} = \frac{[R]_{0}}{2 k}

So a zero-order half-life grows with the starting concentration and shrinks as $k$ grows. As the reaction proceeds, each successive half-life is half as long as the one before.

First order. Put $[R] = [R]_{0} / 2$ into (3.15). The ratio $[R]_{0} / [R]$ becomes 2, and $\log 2 = 0.301$ :

Half-life, first order

t_{1 / 2} = \frac{2.303 \log 2}{k} = \frac{0.693}{k}

(3.17)

The starting concentration has cancelled. A first-order reaction takes as long to fall from 1.0 to 0.5 as from 0.5 to 0.25, or from 0.02 to 0.01 (Fig. 3.8). After $n$ half-lives a fraction $(1 / 2)^{n}$ of the reactant is left. A constant half-life is the signature of a first-order reaction, and it means $k$ and $t_{1 / 2}$ can be converted into each other directly.

4/4A zero-order reaction with the same first half-life: each later half-life is half as long.

Fig. 3.8First order: every half-life is the same length. Zero order: each half-life is half the one before.NCERT p. 77

A graph of the fraction of reactant left against time. Step 1: a first-order decay curve starts at 1 and is marked where it reaches one half, at one half-life. Step 2: a second mark where it reaches one quarter, exactly one more half-life later. Step 3: a third mark at one eighth after a third equal half-life, with the three equal time steps bracketed. Step 4: a straight zero-order line starting from the same point is added; it reaches one half at the same first time, but one quarter after only half as long again, and one eighth after a quarter as long, so its half-lives shrink.

Example 3.7NCERT Example 3.7 · p. 77

Half-life from the rate constant

A medicine is removed from the bloodstream by a first-order process with $k = 3.85 \times 10^{- 5} s^{- 1}$ . What is its half-life in hours?

Try it yourself first. Then reveal the solution one step at a time.

Example 3.8NCERT Example 3.8 · p. 77

Completion times in half-lives

Show that a first-order reaction takes exactly three half-lives to reach 87.5 % completion, and find how many half-lives 99 % completion takes.

Try it yourself first. Then reveal the solution one step at a time.

Table 3.7Zero- and first-order reactions comparedNCERT Table 3.4 · p. 77

	Zero order	First order
Differential rate law	$- d [R] / d t = k$	$- d [R] / d t = k [R]$
Integrated rate law	$[R] = [R]_{0} - k t$	$[R] = [R]_{0} e^{- k t}$
Straight-line plot	$[R]$ against $t$	$\ln [R]$ against $t$
Slope of that line	$- k$	$- k$
Half-life	$[R]_{0} / 2 k$	$0.693 / k$
Unit of $k$	$m o l L^{- 1} s^{- 1}$	$s^{- 1}$

InteractiveRate lab

A graph of reactant concentration against time from 0 to 100 seconds. Buttons choose a zero-order or first-order reaction, and sliders set the starting concentration from 0.2 to 1.0 moles per litre and the rate constant. A time slider, from 0 to 80 seconds, moves a marker along the curve, and an interval slider sets delta t. The graph draws the chord from t to t plus delta t and the tangent at t, and readouts give the concentration at t, the average rate over the interval and the instantaneous rate, so that the two rates can be seen to agree as delta t shrinks. A check box marks successive half-lives on the curve and lists their lengths: equal for first order, halving for zero order. A second set of buttons replots the same data as concentration, natural log of concentration, or log of R zero over R against time, and a line says whether that plot is straight.

Fig. 3.9Choose zero or first order, set the starting concentration and

k

, then compare a chord with the tangent, mark half-lives and test which plot is a straight line.

Key idea

Pseudo first-order reactions

Some reactions with two reactants behave as if they were first order. The trick is a large excess of one of them. In the crystal-violet fading of Example 3.1, hydroxide ions are typically at $0.10 m o l L^{- 1}$ while the dye is at $2.0 \times 10^{- 5} m o l L^{- 1}$ . Even after all the dye has reacted, $[O H^{-}]$ has fallen by only 0.02 %.

The true rate law is $Rate = k [C V^{+}] [O H^{-}]$ , second order overall. With $[O H^{-}]$ effectively constant it can be folded into a new constant $k^{'} = k [O H^{-}]$ , leaving $Rate = k^{'} [C V^{+}]$ . The dye's concentration decays exactly like a first-order reactant.

Such a reaction is called a pseudo first-order reaction (छद्म प्रथम कोटि अभिक्रिया). The oxidation of iodide by peroxodisulfate, $S_{2} O_{8}^{2 -} + 2 I^{-} \overset{}{\to} 2 S O_{4}^{2 -} + I_{2}$ , is another: with iodide in large excess, only $[S_{2} O_{8}^{2 -}]$ controls the rate. Any reaction in which water is both a reactant and the solvent behaves the same way, because the concentration of water, about $55.5 m o l L^{- 1}$ , hardly changes.

Practice · 2 questions

Practice: first-order reactions (Intext Questions 3.5–3.6): questions with answers

NumericalMediumNCERT Intext Question 3.5
A first-order reaction has $k = 2.0 \times 10^{- 3} s^{- 1}$ . How long does it take for $6.0 g$ of the reactant to fall to $4.0 g$ ?
s
Show solution
$t = \frac{2.303}{2.0 \times 10^{- 3}} \log \frac{6.0}{4.0}$
$\log 1.5 = 0.1761$ , so $t = 1151.5 \times 0.1761 = 203 s$ (about 3.4 min).
NumericalEasyNCERT Intext Question 3.6
An organic peroxide decomposes by first-order kinetics with a half-life of $45 m i n$ . Find the rate constant in $s^{- 1}$ .
s⁻¹
Show solution
$k = \frac{0.693}{45 \times 60 s} = 2.57 \times 10^{- 4} s^{- 1}$

3.4 Temperature Dependence of the Rate of a Reaction

Milk that turns sour in a day on a summer kitchen shelf keeps for several days in a fridge. A pressure cooker cooks dal faster because the water inside boils at about $120^{\circ} C$ instead of $100^{\circ} C$ . Almost every reaction speeds up when it is heated. A handy rule of thumb: near room temperature, a rise of $10 K$ roughly doubles the rate constant.

The Swedish chemist Svante Arrhenius put the temperature effect into one equation in 1889, building on a form suggested earlier by the Dutch chemist J. H. van 't Hoff; what Arrhenius added was the physical picture behind it. The Arrhenius equation is

Arrhenius equation

k = A e^{- E_{a} / R T}

(3.18)

where

$A$: the pre-exponential (frequency) factor, a constant for the reaction with the same unit as $k$
$E_{a}$: the activation energy, in $J m o l^{- 1}$
$R$: the gas constant, $8.314 J K^{- 1} m o l^{- 1}$
$T$: the absolute temperature, in K

The constant $A$ is called the pre-exponential factor (आवृत्ति गुणांक), also the frequency factor or Arrhenius factor. $E_{a}$ is the activation energy, which needs a closer look.

Why should only some molecules react? Picture a reaction in which an atom A takes atom B away from a molecule B–C: $A + B - C \to A - B + C$ . As A approaches, the B–C bond stretches while the new A–B bond starts to form. Partway along there is a crowded, high-energy arrangement $A \dots B \dots C$ in which both bonds are half made. This fleeting arrangement is the activated complex (सक्रियित संकुल). It lasts about as long as one molecular vibration, then falls apart into products or back into reactants.

Picture the reaction as a walk over a hill (Fig. 3.10). Along the horizontal axis, the reaction coordinate, the atoms rearrange step by step from reactants into products; the height is potential energy. The activated complex sits on the summit. The climb from the reactants to the summit is the activation energy (सक्रियण ऊर्जा), $E_{a}$ . In this example the walk ends lower than it began, so the reaction gives out energy. The enthalpy change $Δ H$ is fixed by the two end points alone, however high the hill.

Definition

Activation energy

The energy difference between the activated complex and the reactants: the minimum extra energy that colliding reactant molecules need in order to react.

4/4 $Δ H$ depends only on the two ends. The reverse reaction faces a bigger climb, $E_{a} + ∣ Δ H ∣$ .

Fig. 3.10The activated complex sits at the top of the energy hill; Ea is the climb from the reactants and ΔH the difference between the two ends.NCERT Fig. 3.7 · p. 79

Potential energy is plotted against reaction coordinate, with small sketches of the atoms above the curve: A and B–C on the left, A dot dot B dot dot C at the top and A–B and C on the right. Step 1: two flat levels only, reactants on the left and products lower on the right. Step 2: a hill joins them, and the activated complex is marked at its top. Step 3: a double-headed arrow from the reactant level up to the top of the hill is labelled E a. Step 4: a downward arrow from the reactant level to the product level is labelled delta H, negative for this exothermic reaction, and a second arrow from the products to the top of the hill is labelled E a for the reverse reaction, which is larger.

The barrier explains the newspaper from the opening of this chapter. Burning releases plenty of energy, but the molecules must first be lifted over $E_{a}$ . A match supplies that energy at one spot; after that, the heat released by burning keeps lifting fresh molecules over the barrier.

A gas is never a crowd of identical molecules. Collisions keep passing energy around, so at any moment a few molecules are nearly at rest, a few are racing, and most are somewhere in between. Ludwig Boltzmann and James Clerk Maxwell worked out the statistics of this spread. Its graph (Fig. 3.11) rises quickly from zero to a single peak, the most probable kinetic energy, and then trails off in a long tail towards high energies.

Now mark $E_{a}$ on the energy axis. Only molecules out in the tail, beyond $E_{a}$ , can react, and typical activation energies lie far out where the curve falls off exponentially. Warming the gas shifts the whole hump to the right and flattens it; since the curve accounts for every molecule, the total area under it cannot change. The thin sliver of area beyond $E_{a}$ , however, grows a lot. For $E_{a}$ near $50 k J m o l^{- 1}$ , going from $300 K$ to $310 K$ almost doubles it, which is where the ten-degree rule comes from.

3/3At a higher $T_{2}$ the curve flattens and shifts right: the shaded tail nearly doubles.

Fig. 3.11Raising the temperature spreads the energies out, so a much larger fraction of molecules has at least Ea.NCERT Fig. 3.9 · p. 80

Fraction of molecules is plotted against kinetic energy. Step 1: a single hump-shaped curve for temperature T1, rising from zero, peaking at the most probable energy, which is marked, and trailing off in a long tail to the right. Step 2: a vertical dashed line marks the activation energy E a far out in the tail, and the small area under the curve beyond it is shaded and labelled as the molecules able to react. Step 3: a second curve for a higher temperature T2 is added; it is lower, broader and shifted to the right, encloses the same total area, and its shaded area beyond E a is nearly twice as large.

In the Arrhenius equation, the factor $e^{- E_{a} / R T}$ measures this fraction: to a good approximation it is the fraction of collisions with energy of at least $E_{a}$ . The factor $A$ then takes care of how often collisions happen. Taking natural logarithms of (3.18) gives

\ln k = \ln A - \frac{E_{a}}{R T}

(3.19)

This is a straight line in disguise: plot $\ln k$ against $1 / T$ and the slope is $- E_{a} / R$ , the intercept $\ln A$ (Fig. 3.12). Measuring $k$ at several temperatures and drawing this Arrhenius plot is the standard way to find an activation energy. A larger $E_{a}$ gives a steeper line, which means the rate is more sensitive to temperature.

2/3They lie on a line of slope $- E_{a} / R$ : measure the slope, multiply by $- R$ , and you have $E_{a}$ .

Fig. 3.12ln k against 1/T is a straight line of slope −Ea/R.NCERT Fig. 3.10 · p. 80

The vertical axis is the natural logarithm of k and the horizontal axis is one over T, increasing to the right, so higher temperatures are on the left. Step 1: five measured points fall from upper left to lower right. Step 2: a straight line is drawn through them, with a right-angled slope triangle labelled slope equals minus E a over R. Step 3: a second, steeper line through the same middle point is labelled larger E a, showing that such a reaction speeds up more for the same rise in temperature.

DerivationRate constants at two temperatures

Write (3.19) at two temperatures. $A$ is the same at both:

\ln k_{1} = \ln A - \frac{E_{a}}{R T_{1}}

(3.20)

\ln k_{2} = \ln A - \frac{E_{a}}{R T_{2}}

(3.21)

Subtract (3.20) from (3.21); $\ln A$ cancels:

\ln \frac{k_{2}}{k_{1}} = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})

Change to common logarithms and put the bracket over a common denominator:

\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \cdot \frac{T_{2} - T_{1}}{T_{1} T_{2}}

(3.22)

Given $k$ at two temperatures, (3.22) gives $E_{a}$ . Given $E_{a}$ and $k$ at one temperature, it gives $k$ at any other.

Example 3.9NCERT Example 3.9 · p. 81

Activation energy from two rate constants

The rate constant of a gas-phase decomposition is $3.0 \times 10^{- 3} s^{- 1}$ at $400 K$ and $9.6 \times 10^{- 2} s^{- 1}$ at $450 K$ . Find $E_{a}$ and $A$ .

Try it yourself first. Then reveal the solution one step at a time.

Example 3.10NCERT Example 3.10 · p. 81

Rate constant at a new temperature

Medicines are tested for shelf life by storing them warm, which speeds up their decomposition. An active ingredient breaks down by a first-order reaction with $k = 1.5 \times 10^{- 8} s^{- 1}$ at $25^{\circ} C$ and $E_{a} = 85.0 k J m o l^{- 1}$ . Find $k$ at $45^{\circ} C$ and compare the half-lives at the two temperatures.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveArrhenius lab

Sliders set the temperature from 250 to 500 kelvin and the activation energy from 20 to 150 kilojoules per mole, and a check box adds a catalyst that lowers the activation energy by 30 kilojoules per mole. A logarithmic scale running from 1 down to 10 to the minus 30 marks the fraction of collisions with at least the activation energy, e to the minus Ea over RT, now and, as a dashed ring, at 300 kelvin. Readouts give Ea over RT, that fraction, the rate constant with A fixed at 10 to the 11 per second, and the first-order half-life. A graph of ln k against 1000 over T shows the straight line for the chosen activation energy, a second dashed line when the catalyst is on, and the present point. Record adds the present point to the graph; with two or more points a best-fit line is drawn and its slope is turned into an activation energy, Ea equals minus R times the slope. Clear removes the points. A status line says how many collisions in how many succeed and how much faster or slower the reaction is than at 300 kelvin.

Fig. 3.13Change the temperature and the activation energy, switch on a catalyst, and record rate constants to measure

E_{a}

from your own Arrhenius plot.

3.4.1 Effect of Catalyst

Some substances make a reaction go faster and yet come out of it unchanged; such a substance is a catalyst (उत्प्रेरक). Hydrogen peroxide, for example, decomposes only slowly on its own, $2 H_{2} O_{2} \overset{}{\to} 2 H_{2} O + O_{2}$ , but froths vigorously when a pinch of manganese dioxide or a drop of blood is added; blood contains the enzyme catalase. A substance that slows a reaction down is not a catalyst; it is called an inhibitor.

Definition

Catalyst

A substance that speeds up a reaction and is recovered, chemically unchanged, when the reaction is over.

ExampleManganese dioxide in the decomposition of hydrogen peroxide.

How does it work? The intermediate complex theory pictures the catalyst as a temporary partner. It bonds briefly to a reactant, and the combination, called the intermediate complex, reacts more easily than the reactant alone. When the complex falls apart it gives the products and hands back the catalyst, free to start again. The catalysed route is a different mechanism with a lower activation energy (Fig. 3.14). Because $E_{a}$ sits in an exponent in (3.18), even a modest drop matters a great deal: at $298 K$ , lowering $E_{a}$ by $20 k J m o l^{- 1}$ multiplies $k$ by over 3000, if $A$ stays the same.

3/3Both paths start and finish at the same levels: $Δ H$ and $Δ G$ are unchanged, and the reverse barrier drops by the same amount.

Fig. 3.14A catalyst opens a lower-energy path between the same reactants and products.NCERT Fig. 3.11 · p. 82

Potential energy against reaction coordinate for one reaction. Step 1: the uncatalysed path is a single tall hill from the reactant level to a lower product level, with its activation energy marked. Step 2: a dashed catalysed path is added between the same two levels; it has two lower humps with a dip between them where the intermediate complex forms, and its smaller activation energy is marked. Step 3: a dashed line carries the reactant level across to the products and a downward arrow marks delta H, the same for both paths; going backwards from the products, the catalysed humps are lower by the same amount, so the reverse barrier drops too.

Several facts follow from this picture:

Both paths start at the reactants and end at the products, so a catalyst changes neither $Δ H$ nor $Δ G$ . It therefore cannot make a non-spontaneous reaction ( $Δ G > 0$ ) happen; it only speeds up changes that are already feasible.
The barrier falls by the same amount whichever way the reaction runs, so forward and reverse reactions speed up equally. The equilibrium constant is unchanged; equilibrium is simply reached sooner. In the Haber process the iron catalyst does not raise the yield of ammonia; it lets the plant reach equilibrium fast enough to be worth running.
The catalyst is regenerated in every cycle, so a little goes a long way: the few grams of platinum-group metals in a car's catalytic converter clean the exhaust for years.

Try this

A catalyst in a potato

You’ll need

3 % hydrogen peroxide from a chemist's shop (about 30 mL)
two small glasses
a slice of raw potato and a slice of boiled potato
a few drops of dish-wash liquid
safety glasses

Predict first

What will happen when the raw potato goes into the peroxide?

nothing visiblesteady bubbles of a gasthe liquid turns bluethe potato dissolves

Pour about 10 mL of the peroxide and a few drops of dish-wash liquid into each glass.
Drop the raw potato slice into one glass and the boiled slice into the other.
Watch for two minutes and compare the foam. Then lift out the raw slice, rinse it, and drop it into fresh peroxide.

Table 3.8

Glass	What you see
Raw potato
Boiled potato
Raw slice, reused

Your entries are saved automatically.

3.5 Collision Theory of Chemical Reactions

The Arrhenius equation describes the effect of temperature, but it does not say what $A$ means. Collision theory, worked out by Max Trautz and William Lewis between 1916 and 1918, supplies a molecular picture. Its starting point is borrowed from the kinetic theory of gases: molecules are tiny hard balls, and nothing happens unless two of them actually hit each other.

Count how many collisions happen each second in each unit volume of the mixture and you have the collision frequency, $Z$ . It is enormous: in air at room conditions each molecule is struck several billion times a second. If every collision led to reaction, every gas reaction would be over in a flash. Only collisions with enough energy count, so for a bimolecular elementary reaction $A + B \overset{}{\to} p r o d u c t s$

Rate = Z_{A B} e^{- E_{a} / R T}

(3.23)

Here $Z_{A B}$ is the frequency of collisions between A and B, and $e^{- E_{a} / R T}$ is the fraction of those collisions with energy of at least $E_{a}$ . Set this beside (3.18): the pre-exponential factor $A$ plays the part of the collision frequency.

Test (3.23) against experiment and a pattern appears. It works well when atoms or small molecules react, but for larger molecules the measured rate is often far below the prediction, sometimes by many powers of ten. Energy is not enough: the molecules must also meet the right way round, so that the bond that breaks and the bond that forms are in the right places. Collisions with enough energy, the threshold energyThe threshold energy is the least total energy colliding molecules must have in order to react: the energy they already possess plus the activation energy., and a suitable orientation are effective collisions (प्रभावी संघट्ट).

Definition

Effective collision

A collision that actually produces products: the molecules meet with at least the threshold energy and lined up so that the right bond can break and the new one can form.

Take the reaction of a chlorine atom with nitrosyl chloride, $C l + N O C l \overset{}{\to} C l_{2} + N O$ (Fig. 3.15). The new Cl–Cl bond can form only if the incoming Cl atom strikes the chlorine end of NOCl, and strikes it hard enough. A hit on the oxygen end just bounces off, however energetic it is.

3/3Right end and enough energy: an effective collision gives $C l_{2}$ and NO.

Fig. 3.15A collision leads to products only if it has enough energy and the right orientation.NCERT Fig. 3.12 · p. 83

A bent nitrosyl chloride molecule, with oxygen at the top, nitrogen in the middle and chlorine at the lower right, is approached by a single chlorine atom. Step 1: the chlorine atom hits the oxygen end from above and bounces back along a curved arrow, labelled wrong end, no reaction. Step 2: the chlorine atom touches the chlorine end but only gently and bounces off, labelled right end, too little energy. Step 3: the chlorine atom strikes the chlorine end hard; the two chlorine atoms leave together as a Cl2 molecule and a separate NO molecule moves away, labelled right end and enough energy: products.

Collision theory handles orientation with one more number, the steric factor $P$ , also called the probability factor: the fraction of energetic collisions that are lined up well enough to react.

Collision theory with the steric factor

Rate = P Z_{A B} e^{- E_{a} / R T}

So the rate of a reaction is set by three things: how often the molecules collide ( $Z_{A B}$ ), what fraction of collisions carry enough energy ( $e^{- E_{a} / R T}$ ), and what fraction of those are well aimed ( $P$ ). Treating molecules as featureless balls is also the theory's weakness: it cannot predict $P$ from the shapes and bonds of real molecules. More complete theories of reaction rates come in later studies.

Practice · 3 questions

Practice: temperature and energy (Intext Questions 3.7–3.9): questions with answers

Short answerEasyNCERT Intext Question 3.7
How does the rate constant of a typical reaction respond to a rise in temperature, and which equation describes the change?
Show solution
AnswerIt rises steeply (roughly doubling per 10 K near room temperature), as described by the Arrhenius equation.
The rate constant increases, roughly doubling for every 10 K rise near room temperature, because a larger fraction of collisions has at least the activation energy. The Arrhenius equation, $k = A e^{- E_{a} / R T}$ , describes it: as $T$ rises, $E_{a} / R T$ falls and $e^{- E_{a} / R T}$ grows exponentially.
NumericalMediumNCERT Intext Question 3.8
The rate of a reaction triples when the temperature rises from $300 K$ to $310 K$ . Calculate $E_{a}$ .
kJ mol⁻¹
Show solution
$\log 3 = \frac{E_{a}}{2.303 \times 8.314} \cdot \frac{10}{300 \times 310}$
$\log 3 = 0.4771$ and $2.303 \times 8.314 = 19.15$ , so $E_{a} = 0.4771 \times 19.15 \times 9300 \approx 8.5 \times 10^{4} J m o l^{- 1}$ , about $85 k J m o l^{- 1}$ .
NumericalMediumNCERT Intext Question 3.9
For a reaction with $E_{a} = 150 k J m o l^{- 1}$ at $600 K$ , what fraction of collisions have at least the activation energy?
Show solution
$E_{a} / R T = 150 000 / (8.314 \times 600) = 30.07$ .
$x = e^{- 30.07} = 8.7 \times 10^{- 14}$
Fewer than one collision in ten trillion is energetic enough.

Quick check

1 of 3

Why does collision theory need the steric factor

P

Some energetic collisions have the wrong orientation.Some molecules never collide at all.Collision frequency falls as temperature rises.The activation energy changes during the reaction.

Show answer

Answer (A) Some energetic collisions have the wrong orientation.

Even collisions with enough energy fail if the molecules meet the wrong way round, as when Cl strikes the O end of NOCl.

P

is the fraction that are well aimed.

Recap

Chemical Kinetics: key points and summary

Chemical kinetics studies how fast reactions go and why; thermodynamics only says whether they can go (Introduction).
The rate of a reaction is the change in concentration per unit time. The average rate covers an interval; the instantaneous rate is the slope of the tangent, $- d [R] / d t$ (Section 3.1).
To give one rate for the whole reaction, divide each species' rate of change by its coefficient in the balanced equation (Section 3.1).
Concentration, temperature, catalysts and surface area all affect the rate (Section 3.2).
A rate law, $Rate = k [A]^{x} [B]^{y}$ , is found by experiment; its powers need not equal the coefficients (Section 3.2.2).
The order is $x + y$ ; it may be zero, fractional or negative, and it fixes the unit of $k$ , $(m o l L^{- 1})^{1 - n} s^{- 1}$ (Section 3.2.3).
Molecularity, 1, 2 or rarely 3, applies only to elementary steps; the slowest step of a mechanism determines the rate (Section 3.2.4).
Zero order: $[R] = [R]_{0} - k t$ , a straight line of slope $- k$ , with $t_{1 / 2} = [R]_{0} / 2 k$ (Section 3.3.1).
First order: $k = (2.303 / t) \log ([R]_{0} / [R])$ ; $\ln [R]$ against $t$ is a straight line (Section 3.3.2).
A first-order half-life, $0.693 / k$ , does not depend on the starting concentration; a large excess of one reactant makes a reaction pseudo first order (Section 3.3.3).
Arrhenius equation: $k = A e^{- E_{a} / R T}$ . A plot of $\ln k$ against $1 / T$ has slope $- E_{a} / R$ ; near room temperature, 10 K roughly doubles $k$ (Section 3.4).
A catalyst provides a path of lower activation energy; it leaves $Δ H$ , $Δ G$ and the equilibrium constant unchanged (Section 3.4.1).
Collision theory: $Rate = P Z_{A B} e^{- E_{a} / R T}$ . Effective collisions need enough energy and the right orientation (Section 3.5).

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Intext Question 3.1	Practice: reaction rates (Intext Questions 3.1–3.2) · Q1
NCERT Intext Question 3.2	Practice: reaction rates (Intext Questions 3.1–3.2) · Q2
NCERT Intext Question 3.3	Practice: order (Intext Questions 3.3–3.4) · Q1
NCERT Intext Question 3.4	Practice: order (Intext Questions 3.3–3.4) · Q2
NCERT Intext Question 3.5	Practice: first-order reactions (Intext Questions 3.5–3.6) · Q1
NCERT Intext Question 3.6	Practice: first-order reactions (Intext Questions 3.5–3.6) · Q2
NCERT Intext Question 3.7	Practice: temperature and energy (Intext Questions 3.7–3.9) · Q1
NCERT Intext Question 3.8	Practice: temperature and energy (Intext Questions 3.7–3.9) · Q2
NCERT Intext Question 3.9	Practice: temperature and energy (Intext Questions 3.7–3.9) · Q3
NCERT Exercises 3.1	Exercises · Q1
NCERT Exercises 3.2	Exercises · Q2
NCERT Exercises 3.3	Exercises · Q3
NCERT Exercises 3.4	Exercises · Q4
NCERT Exercises 3.5	Exercises · Q5
NCERT Exercises 3.6	Exercises · Q6
NCERT Exercises 3.7	Exercises · Q7
NCERT Exercises 3.8	Exercises · Q8
NCERT Exercises 3.9	Exercises · Q9
NCERT Exercises 3.10	Exercises · Q10
NCERT Exercises 3.11	Exercises · Q11
NCERT Exercises 3.12	Exercises · Q12
NCERT Exercises 3.13	Exercises · Q13
NCERT Exercises 3.14	Exercises · Q14
NCERT Exercises 3.15	Exercises · Q15
NCERT Exercises 3.16	Exercises · Q16
NCERT Exercises 3.17	Exercises · Q17
NCERT Exercises 3.18	Exercises · Q18
NCERT Exercises 3.19	Exercises · Q19
NCERT Exercises 3.20	Exercises · Q20
NCERT Exercises 3.21	Exercises · Q21
NCERT Exercises 3.22	Exercises · Q22
NCERT Exercises 3.23	Exercises · Q23
NCERT Exercises 3.24	Exercises · Q24
NCERT Exercises 3.25	Exercises · Q25
NCERT Exercises 3.26	Exercises · Q26
NCERT Exercises 3.27	Exercises · Q27
NCERT Exercises 3.28	Exercises · Q28
NCERT Exercises 3.29	Exercises · Q29
NCERT Exercises 3.30	Exercises · Q30

Quick check: MCQs on Chemical Kinetics

3.1 Rate of a Chemical Reaction4 questions
3.2.4 Molecularity of a Reaction4 questions
3.3.3 Half-Life of a Reaction4 questions
3.4.1 Effect of Catalyst4 questions
3.5 Collision Theory of Chemical Reactions3 questions

Glossary · 15

Key terms (English–Hindi)

activated complexसक्रियित संकुलsakriyit sankul: The short-lived, highest-energy arrangement of atoms on the way from reactants to products.First seen in the chapter →
activation energyसक्रियण ऊर्जाsakriyan urja: The energy barrier $E_{a}$ between the reactants and the activated complex: the extra energy colliding molecules need in order to react.First seen in the chapter →

catalystउत्प्रेरकutprerak: A substance that speeds up a reaction by opening a path of lower activation energy and is unchanged chemically at the end.First seen in the chapter →
chemical kineticsरासायनिक बलगतिकीrasayanik balgatiki: The branch of chemistry that studies how fast reactions happen, what controls their speed and the steps by which they occur.First seen in the chapter →

effective collisionप्रभावी संघट्टprabhavi sanghatt: A collision with at least the threshold energy and a suitable orientation, so that it leads to products.First seen in the chapter →
elementary reactionप्राथमिक अभिक्रियाprathmik abhikriya: A reaction that takes place in a single step, exactly as its equation is written.First seen in the chapter →

half-lifeअर्धायुardhayu: How long a reactant's concentration takes to halve; for a first-order reaction it is $0.693 / k$ .First seen in the chapter →

molecularityआण्विकताaanvikta: The number of species that must collide at once in an elementary step: one, two or, rarely, three.First seen in the chapter →

order of a reactionअभिक्रिया की कोटिabhikriya ki koti: The sum of the powers of the concentration terms in the rate law. It is found by experiment and may be zero or a fraction.First seen in the chapter →

pre-exponential factorआवृत्ति गुणांकaavritti gunank: The constant $A$ in $k = A e^{- E_{a} / R T}$ ; it reflects how often reactant molecules collide.First seen in the chapter →
pseudo first-order reactionछद्म प्रथम कोटि अभिक्रियाchhadm pratham koti abhikriya: A reaction of higher order that behaves as first order because the other reactants are in large excess.First seen in the chapter →

rate constantवेग स्थिरांकveg sthirank: The constant $k$ in a rate law; it is fixed for a reaction at a given temperature and equals the rate at unit concentrations.First seen in the chapter →
rate lawवेग नियमveg niyam: An experimentally found equation giving the rate in terms of reactant concentrations, such as $Rate = k [A]^{x} [B]^{y}$ .First seen in the chapter →
rate of reactionअभिक्रिया का वेगabhikriya ka veg: How quickly the concentration of a reactant falls or of a product rises, per unit time, divided by that species' coefficient in the balanced equation.
rate-determining stepवेग निर्धारक पदveg nirdharak pad: The slowest step of a mechanism, which limits the rate of the overall reaction.First seen in the chapter →

Chemical Kinetics: Class 12 Chemistry · Chapter 3

What you will learn in Chemical Kinetics

3.1 Rate of a Chemical Reaction

Practice: reaction rates (Intext Questions 3.1–3.2): questions with answers

3.2 Factors Influencing Rate of a Reaction

3.2.1 Dependence of Rate on Concentration

3.2.2 Rate Expression and Rate Constant

3.2.3 Order of a Reaction

3.2.4 Molecularity of a Reaction

Practice: order (Intext Questions 3.3–3.4): questions with answers

3.3 Integrated Rate Equations

3.3.1 Zero Order Reactions

3.3.2 First Order Reactions

3.3.3 Half-Life of a Reaction

Practice: first-order reactions (Intext Questions 3.5–3.6): questions with answers

3.4 Temperature Dependence of the Rate of a Reaction

3.4.1 Effect of Catalyst

3.5 Collision Theory of Chemical Reactions

Practice: temperature and energy (Intext Questions 3.7–3.9): questions with answers

Chemical Kinetics: key points and summary

Practice questions with answers (NCERT “Exercises”)

Find your NCERT exercise question

Quick check: MCQs on Chemical Kinetics

Key terms (English–Hindi)

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