Chemical Kinetics Formula Sheet — JEE Main Chemistry

Every key Chemical Kinetics formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 11 sub-topics

Rate of a chemical reaction
Factors affecting rate: concentration, temperature, pressure, catalyst
Elementary and complex reactions
Order and molecularity of reactions
Rate law and rate constant with units
Differential and integral forms of zero and first-order reactions
Characteristics and half-lives
Effect of temperature on rate
Arrhenius theory
Activation energy and calculation
Collision theory of bi-molecular gaseous reactions

Rate of Reaction & Rate Expression

Rate of a reaction: The change in concentration of a reactant or product per unit time.

Average rate

r_{a v} = - \frac{Δ [ R ]}{Δ t} = + \frac{Δ [ P ]}{Δ t}

over an interval;

instantaneous rate

is its limit as

Δ t \to 0

— the tangent slope of the concentration–time curve.

Concentration versus time curves: reactant falling and product rising, with a tangent line giving the instantaneous rate and a chord giving the average rate — Reactant falls, product rises; tangent = instantaneous, chord = average rate.

Rate from stoichiometry

- \frac{1}{a} \frac{d [ A ]}{d t} = - \frac{1}{b} \frac{d [ B ]}{d t} = + \frac{1}{c} \frac{d [ C ]}{d t} = + \frac{1}{d} \frac{d [ D ]}{d t}

For

a A + b B \to c C + d D

. Divide each term by its coefficient so the unique rate is obtained. E.g.

2 HI \to H_{2} + I_{2}

: rate

= - \frac{1}{2} \frac{d [ HI ]}{d t}

Essentials

▸Units of rate: $mol L^{- 1} s^{- 1}$ (or $atm s^{- 1}$ for gases via partial pressure).
▸ $Δ [R]$ is negative, so a $- 1$ makes the rate positive.
▸Rate is $not$ constant — it falls as reactants are used up.

🎯 Exam · Factors that change the rate

Concentration

(pressure for gases),

temperature

, and a

catalyst

are the main factors; also surface area of solids and, for some reactions, light. Higher concentration/temperature

\Rightarrow

faster reaction.

✎ Example · Average rate

For

2 N_{2} O_{5} \to 4 N O_{2} + O_{2}

[N_{2} O_{5}]

falls

2.33 \to 2.08

mol

L^{- 1}

in 184 min. Find the average rate.

Rate $= - \frac{1}{2} \frac{Δ [ N _{2} O _{5} ]}{Δ t} = - \frac{1}{2} \cdot \frac{( 2.08 - 2.33 )}{184}$ .
$= 6.79 \times 1 0^{- 4}$ mol $L^{- 1}$ mi $n^{- 1}$ .

∴

1.13 \times 1 0^{- 5} mol L^{- 1} s^{- 1}

🚫 Examiner Trap · Examiner traps

(1) Divide each species' rate by its

coefficient

for the unique reaction rate (

2

HI: rate

= - \frac{1}{2} d [HI] / d t

). (2) Reactant terms carry a

-

sign so the rate stays positive. (3) Rate is NOT constant — it falls as reactants deplete. (4) Average rate uses a chord (

Δ

), instantaneous uses a tangent (d).

Rate Law, Order & Molecularity

Rate law (rate expression): The experimentally found dependence of rate on concentration:

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

. The powers x,y

cannot

be read off the balanced equation — they must be determined by experiment.

Differential rate equation

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

k

= rate constant (specific reaction rate); it is independent of concentration but depends on temperature and catalyst.

Order	$k = Rate / []^{n}$	Units of $k$
0	$\frac{mol L ^{- 1} s ^{- 1}}{1}$	$mol L^{- 1} s^{- 1}$
1	$\frac{mol L ^{- 1} s ^{- 1}}{mol L ^{- 1}}$	$s^{- 1}$
2	$\frac{mol L ^{- 1} s ^{- 1}}{( mol L ^{- 1} ) ^{2}}$	$mo l^{- 1} L s^{- 1}$

Elementary vs complex

▸ $Elementary$ reaction: one step; molecularity is defined.
▸ $Complex$ reaction: a sequence of elementary steps (a mechanism).
▸The slowest step is the $rate-determining step$ ; it fixes the overall rate and order.
▸Molecularity has no meaning for an overall complex reaction.

💡 Tip · Order from a rate law

Rate

= k [A]^{1/2} [B]^{3/2} \Rightarrow

order

= \frac{1}{2} + \frac{3}{2} = 2

. The

unit of k

also fixes the order:

s^{- 1} \Rightarrow

first order,

mo l^{- 1} L s^{- 1} \Rightarrow

second order.

🚫 Examiner Trap · Examiner traps

(1) Orders x,y are found by

experiment

, NOT from the balanced equation. (2) Order can be 0, fractional or negative;

molecularity

is an integer

1

–

3

for

elementary

steps only. (3) The unit of k reveals the order (

s^{- 1} \Rightarrow

first). (4) The slowest (rate-determining) step fixes the overall order.

Zero-Order Reactions

Zero-order reaction: Rate is independent of reactant concentration:

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

. The rate stays constant until the reactant runs out.

Integrated rate law

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

Form

y = m x + c

: a plot of

[R]

vs t is a straight line of slope

- k

and intercept

[R]_{0}

. Units of k:

mol L^{- 1} s^{- 1}

Straight-line plot of concentration of R versus time, starting at R-zero and decreasing linearly with slope minus k, showing rate independent of concentration — $[R]$ vs t is linear with slope $- k$ .

Half-life (zero order)

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

Directly proportional to the initial concentration

[R]_{0}

and inversely proportional to k.

Where zero order occurs

▸ $Surface-saturated$ catalytic reactions — the surface holds a fixed amount of reactant.
▸ $2 N H_{3} Pt, high P N_{2} + 3 H_{2}$ (hot platinum surface).
▸Thermal decomposition of HI on a gold surface.
▸Many enzyme reactions at saturating substrate.

★ Remember · Tell-tale sign

If a graph of

[R]

against t is a

straight line

, the reaction is zero order — the concentration drops by equal amounts in equal times.

🚫 Examiner Trap · Examiner traps

(1) Zero order: rate is INDEPENDENT of

[R]

(

k =

rate). (2)

[R]

vs t is a

straight line

(slope

- k

); units of k are

mol L^{- 1} s^{- 1}

. (3)

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

proportional

to initial concentration. (4) Occurs on saturated surfaces/enzymes — not common otherwise.

First-Order Reactions

Integrated rate law (first order)

k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]} [R] = [R]_{0} e^{- k t}

Equivalently

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

. Units of k:

s^{- 1}

(concentration-independent).

Left: concentration decaying exponentially with time; right: natural log of concentration falling as a straight line of slope minus k — $[R]$ decays exponentially; $ln [R]$ vs t is linear, slope $- k$ .

✎ Example · First-order rate constant

N_{2} O_{5}

decomposition (1st order):

[N_{2} O_{5}]

falls

1.24 \times 1 0^{- 2} \to 0.20 \times 1 0^{- 2}

mol

L^{- 1}

in 60 min.

$k = \frac{2.303}{60} lo g \frac{1.24 \times 1 0 ^{- 2}}{0.20 \times 1 0 ^{- 2}} = \frac{2.303}{60} lo g 6.2$ .

∴

k = 0.0304 mi n^{- 1}

Half-life (first order)

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

Independent

of initial concentration — a hallmark of first-order kinetics (e.g. all radioactive decay).

First-order examples

▸All natural & artificial $radioactive decay$ ( $_{88}^{226} Ra \to_{2}^{4} He +_{86}^{222} Rn$ ).
▸Hydrogenation of ethene; decomposition of $N_{2} O_{5}$ , $N_{2} O$ .
▸Gas-phase: track total pressure, $k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}$ .

🎯 Exam · Pseudo-first-order reactions

A higher-order reaction looks first order when one reactant is in large excess. Hydrolysis of an ester: Rate

= k^{'} [ester] [H_{2} O]

, but

[H_{2} O]

is ~constant, so Rate

= k [ester]

with

k = k^{'} [H_{2} O]

. Inversion of cane sugar is another example.

🚫 Examiner Trap · Examiner traps

(1) First order:

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

;

ln [R]

vs t is linear (slope

- k

). (2)

t_{1/2} = 0.693/ k

independent

of initial concentration (all radioactive decay). (3) Units of k are

s^{- 1}

. (4)

Pseudo-first-order

: a higher-order reaction looks first order when one reactant is in large excess (ester hydrolysis).

Half-Life & Integrated-Law Summary

Half-life (

t_{1/2}

): The time for the reactant concentration to fall to half its initial value. After n half-lives the fraction left is

(1/2)^{n}

, so a first-order reactant is never fully consumed.

Order	Integrated law	Linear plot	$t_{1/2}$
0	$[R] = [R]_{0} - k t$	$[R]$ vs t	$\frac{[ R ] _{0}}{2 k}$
1	$ln [R] = ln [R]_{0} - k t$	$ln [R]$ vs t	$\frac{0.693}{k}$

Reading the order off a graph

▸ $[R]$ vs t linear $\Rightarrow$ $zero$ order (slope $= - k$ ).
▸ $ln [R]$ (or $lo g [R]$ ) vs t linear $\Rightarrow$ $first$ order.
▸Constant $t_{1/2}$ regardless of $[R]_{0}$ $\Rightarrow$ first order.
▸Units of k also reveal order: $mol L^{- 1} s^{- 1}$ (0), $s^{- 1}$ (1).

✎ Example · 99.9% completion (first order)

Show the time for 99.9% completion is

10 t_{1/2}

$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ] _{0} - 0.999 [ R ] _{0}} = \frac{2.303}{k} lo g 1 0^{3} = \frac{6.909}{k}$ .
$\frac{t}{t _{1/2}} = \frac{6.909/ k}{0.693/ k}$ .

∴

\frac{t}{t _{1/2}} = 10

🚫 Examiner Trap · Examiner traps

(1)

t_{1/2} \propto [R]_{0}

for zero order but is

constant

for first order — the quickest way to tell them apart. (2) Linear

[R]

t \Rightarrow

zero; linear

ln [R]

t \Rightarrow

first. (3) After n half-lives, fraction left

= (1/2)^{n}

— never fully consumed. (4) 99.9% completion takes

10 t_{1/2}

(first order).

Temperature Dependence: Arrhenius & Activation Energy

Arrhenius equation

k = A e^{- E_{a} / RT}

A = frequency (pre-exponential) factor,

E_{a}

= activation energy,

R = 8.314 J K^{- 1} mo l^{- 1}

. The factor

e^{- E_{a} / RT}

is the fraction of molecules with energy

\geq E_{a}

Straight-line plot of natural log of k versus one over T with slope minus Ea over R and intercept ln A, beside the Arrhenius equation and its two-temperature form — $ln k$ vs $1/ T$ is linear: slope $= - E_{a} / R$ , intercept $= ln A$ .

Two-temperature form

lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} [\frac{T _{2} - T _{1}}{T _{1} T _{2}}]

Obtained by writing

ln k = ln A - \frac{E _{a}}{RT}

at two temperatures and subtracting; used to find

E_{a}

from two rate constants (or k at a new T).

What it tells you

▸A rise of $10^{\circ}$ C roughly $doubles$ the rate.
▸Higher T or lower $E_{a}$ $\Rightarrow$ larger k $\Rightarrow$ faster reaction.
▸ $E_{a}$ = energy gap between the activated complex and the reactants.
▸On a Maxwell–Boltzmann curve, raising T broadens it so more molecules clear $E_{a}$ .

✎ Example ·

k

at a new temperature

C_{2} H_{5} I \to C_{2} H_{4} + HI

k_{1} = 1.60 \times 1 0^{- 5} s^{- 1}

at 600 K,

E_{a} = 209

kJ mo

l^{- 1}

. Find

k_{2}

at 700 K.

$lo g k_{2} = lo g k_{1} + \frac{E _{a}}{2.303 R} [\frac{1}{T _{1}} - \frac{1}{T _{2}}]$ .
$lo g k_{2} = - 4.796 + 2.599 = - 2.197$ .

∴

k_{2} = 6.36 \times 1 0^{- 3} s^{- 1}

★ Remember · Catalyst

A catalyst provides an alternate path with

lower E_{a}

, so it speeds the reaction. It does

not

change

Δ G

K_{c}

— it speeds forward and backward equally, reaching equilibrium faster.

🚫 Examiner Trap · Examiner traps

(1)

ln k

1/ T

slope

= - E_{a} / R

(NOT

- E_{a}

). (2) A

1 0^{\circ}

C rise roughly DOUBLES the rate. (3) A

catalyst

lowers

E_{a}

(alternate path) but does NOT change

Δ G

or K — it speeds forward & back equally. (4) Use

R = 8.314

J/mol K with

E_{a}

in J (convert from kJ).

Collision Theory & Energy Profile

Collision theory: Reactant molecules (treated as hard spheres) react when they collide. The

collision frequency

Z is the number of collisions per second per unit volume; only a fraction of collisions actually lead to product.

Rate from collision theory

Rate = Z_{AB} e^{- E_{a} / RT} orientation Rate = P Z_{AB} e^{- E_{a} / RT}

Z_{AB}

= collision frequency,

e^{- E_{a} / RT}

= fraction of collisions with enough energy, P = steric (probability) factor for correct orientation.

Potential energy versus reaction coordinate: reactants rise over an activation-energy barrier through the activated complex to products, with a lower dashed catalysed path and the reaction enthalpy delta H — Reactants climb $E_{a}$ to the activated complex, then fall to products; $Δ H$ = net energy change.

Effective collisions

▸ $Threshold energy$ $=$ activation energy $+$ energy already possessed by the molecules.
▸A collision is $effective$ only if it has (i) energy $\geq$ threshold $and$ (ii) the correct orientation.
▸The $activated complex$ at the barrier top is an unstable arrangement that decays to products.

⚠️ Watch out · Limits of the model

Treating molecules as structureless hard spheres ignores their internal structure, so simple collision theory over-predicts rates for complex molecules — hence the steric factor

P

is needed to account for orientation.

🚫 Examiner Trap · Examiner traps

(1) An effective collision needs BOTH enough energy (

\geq

threshold) AND correct

orientation

(steric factor P). (2) Threshold energy

= E_{a} +

energy already in the molecules. (3) The

activated complex

sits at the barrier TOP (max PE), not at the products. (4) A catalyst lowers

E_{a}

but leaves

Δ H

unchanged.

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Frequently Asked Questions

What are the most important Chemical Kinetics formulas for JEE Main?

This Chemical Kinetics formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Rate of a chemical reaction, Factors affecting rate: concentration, temperature, pressure, catalyst, Elementary and complex reactions, Order and molecularity of reactions, Rate law and rate constant with units — each shown with the key result and, where useful, a worked example.

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How should I revise Chemical Kinetics formulas?

Blurt the Chemical Kinetics formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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