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Chemical Kinetics: one-page cheat sheet

NEET Chemistry · Unit 8 · Physical Chemistry

ScoreChemby CSJ Sir
Weightage: 2–3 questions · 8–12 marks

1 · Rate of reaction

a A + b B \to c C + d D : rate = - \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}

Rate law

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

; overall order

n = x + y

Instantaneousslope of the tangent on the

[R]

t

curve

Order from datavary one reactant; if rate

\times 2^{x}

when its concentration is doubled, order

= x

Zero orderheterogeneous catalysis on a saturated surface, e.g. ammonia decomposition on Pt

2 · Units of $k$

unit of k = (mol L^{- 1})^{1 - n} s^{- 1}

Order	Unit of $k$
0	mol L $^{- 1}$ s $^{- 1}$
1	s $^{- 1}$
2	L mol $^{- 1}$ s $^{- 1}$
3	L $^{2}$ mol $^{- 2}$ s $^{- 1}$

3 · Order vs molecularity

	Order	Molecularity
From	experiment	mechanism
Values	0, fractional, integer	1, 2, 3 only
Applies to	any reaction	elementary steps only

Overall order matches the molecularity of the slowest step.

4 · Integrated rate laws

	Zero order	First order
Rate law	$k$	$k [R]$
Integrated	$[R] = [R]_{0} - k t$	$ln [R] = ln [R]_{0} - k t$
$k$	$\frac{[ R ] _{0} - [ R ]}{t}$	$\frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ]}$
Plot, slope	$[R]$ vs $t$ , $- k$	$ln [R]$ vs $t$ , $- k$
$t_{1/2}$	$[R]_{0} /2 k$	$0.693/ k$

Half-lifefirst order: independent of

[R]_{0}

; generally

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

Gas phaseA(g)

\to

B(g) + C(g):

k = \frac{2.303}{t} lo g \frac{p _{i}}{2 p _{i} - p _{t}}

Doubling

[R]_{0}

doubles

t_{1/2}

: zero order. It halves

t_{1/2}

: second order. No change: first order.

5 · Pseudo first order

Ideasecond-order reaction with one reactant in large excess

Ester hydrolysisCH

_{3}

COOC

_{2}

_{5}

+ H

_{2}

H^{+}

acid + alcohol; rate

= k [ester]

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

Cane sugarinversion to glucose + fructose; rate

= k [sucrose]

6 · Arrhenius equation

k = A e^{- E_{a} / R T}, lo g k = lo g A - \frac{E _{a}}{2.303 R T}

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

Plot

ln k

1/ T

: slope

= - E_{a} / R

, intercept

= ln A

Rule of thumbrate constant roughly doubles per

1 0^{\circ}

C rise

E_{a}

needsrate constants at two temperatures

7 · Catalysis & collision theory

Catalystlowers

E_{a}

by a new path; speeds forward and backward equally;

Δ H

Δ G

and

K

unchanged

rate = P \cdot Z_{A B} \cdot e^{- E_{a} / R T}

Effective collisionenergy

\geq

threshold (

E_{a}

+ reactant energy) and correct orientation (

P

)

8 · First-order shortcuts

Completion	Time
50%	$t_{1/2}$
75%	$2 t_{1/2}$
87.5%	$3 t_{1/2}$
99.9%	$\approx 10 t_{1/2}$

Mechanismthe slowest step fixes the rate; intermediates cancel in the overall equation

Seen in NEET 2023

Rate $= k [A]^{2} [B]$ . $[A]$ is tripled, $[B]$ constant. New rate?

rate^{'} = k (3 [A])^{2} [B] = 9 \times rate

Seen in NEET 2025

First order, $t_{1/2} = 1$ min. Time for 99.9% completion?

t = \frac{2.303}{k} lo g 1 0^{3} = \frac{6.909}{k} = \frac{6.909}{0.693} t_{1/2} \approx 10 min

Seen in NEET 2024

Rate constant quadruples from $2 7^{\circ}$ C to $5 7^{\circ}$ C. Find $E_{a}$ .

lo g 4 = \frac{E _{a}}{2.303 \times 8.314} \cdot \frac{30}{300 \times 330} \Rightarrow E_{a} = 38.04 kJ mol^{- 1}

Six traps to remember

Unit of $k$ is $(mol L^{- 1})^{1 - n} s^{- 1}$ ; first order is s $^{- 1}$ .
Molecularity is never zero or fractional; order can be.
First-order $t_{1/2} = 0.693/ k$ does not depend on concentration.
Zero-order $t_{1/2}$ is proportional to $[R]_{0}$ .
$E_{a}$ comes from $k$ at two temperatures, via the slope of $ln k$ vs $1/ T$ .
A catalyst lowers $E_{a}$ but leaves $Δ H$ and $K$ unchanged.

Chemical Kinetics: free one-page cheat sheet