JEE Mains Chemistry · Chemical Kinetics

Zero Order and Finding the Order

A zero-order reactant falls in a straight line and its half-life shrinks with the concentration; how the half-life depends on the starting concentration, or which plot is straight, tells you the order.

All Chemical Kinetics notes JEE Mains Chemistry strategy

Why this matters

Sixteen PYQs, seven of them multiple choice, and three from 2026. Six use the zero-order law, its half-life or a zero-order plot; six find the order from how the half-life changes with the starting concentration or pressure; four name the order from the shape of a graph. Three ideas cover the page.

Concept 1 of 3: The zero-order law and its half-life

In a zero-order reaction the rate does not depend on how much reactant is left — for example when a gas decomposes on a metal surface that is fully covered. The concentration falls by the same amount every minute, in a straight line. So the more there is, the longer half of it takes.

Definition

$r = k$ ; $[A] = [A]_0 - kt$ ; $k$ in mol L $^{-1}$ s $^{-1}$ .
$[A]$ against $t$ : a straight line, slope $-k$ , intercept $[A]_0$ .
$t_{1/2} = \dfrac{[A]_0}{2k}$ , proportional to $[A]_0$ .
Completion: $t_{100\%} = \dfrac{[A]_0}{k} = 2t_{1/2}$ . A first-order reaction, by contrast, never completes.
Each later half-life is half the one before, so the time to fall to $\tfrac14$ is $1.5\,t_{1/2}$ , not $2t_{1/2}$ .
Examples: $\mathrm{NH_3}$ decomposing on hot platinum at high pressure; $\mathrm{H_2 + Cl_2}$ in light.

Zero-order law

[A] = [A]_0 - kt\qquad t_{1/2} = \frac{[A]_0}{2k}

Worked example

A zero-order reaction has

t_{1/2} = 40

min when

[A]_0 = 0.80

M. Find

k

, and the time for

[A]

to fall from 0.30 M to 0.10 M.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q120Moderate

Example 1 · Chemical Kinetics · Zero Order and Finding the Order

Half life of zero order reaction

A \rightarrow

product is 1 hour, when initial concentration of reaction is

2.0\text{ }mol{\text{ }L}^{- 1}

. The time required to decrease concentration of A from 0.50 to

0.25\text{ }mol{\text{ }L}^{- 1}

is:

One-quarter taken as two half-lives

That holds only for first order. For zero order the second half-life is half the first, so the reactant reaches one-quarter at

1.5\,t_{1/2}

Zero-order half-life treated as fixed

t_{1/2} = \dfrac{[A]_0}{2k}

depends on the concentration you start from. Work out

k

from the stated half-life first, then find any later time from

[A] = [A]_0 - kt

Drill 5 more on the zero-order law and its half-life

Concept 2 of 3: Order from how the half-life depends on the starting concentration

For order

n

t_{1/2} \propto [A]_0^{\,1-n}

. So compare two half-lives measured from two starting concentrations (or pressures): the power that links them gives

1 - n

. A constant half-life means first order; one that grows with

[A]_0

means an order below 1.

Definition

$n = 0$ : $t_{1/2} \propto [A]_0$ . $n = \tfrac12$ : $t_{1/2} \propto \sqrt{[A]_0}$ . $n = 1$ : constant. $n = 2$ : $t_{1/2} \propto \dfrac{1}{[A]_0}$ .
$\dfrac{t_{1/2}'}{t_{1/2}} = \left(\dfrac{[A]_0'}{[A]_0}\right)^{1-n}$ . For a gas, initial pressure stands in for $[A]_0$ .
Put both half-lives in the same unit before dividing.
Time to $\tfrac14$ exactly twice the time to $\tfrac12$ means equal successive half-lives: first order.
The order of a reaction is fixed; it does not change when the starting concentration changes.

Half-life and order

t_{1/2} \propto [A]_0^{\,1-n}\quad\Rightarrow\quad \frac{t_{1/2}'}{t_{1/2}} = \left(\frac{[A]_0'}{[A]_0}\right)^{1-n}

Worked example

The half-life of a reaction is 120 s when

[A]_0 = 0.20

M and 30 s when

[A]_0 = 0.80

M. Find the order.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q32Moderate

Example 2 · Chemical Kinetics · Zero Order and Finding the Order

For a given reaction

R \rightarrow P,t_{1/2}

is related to

\lbrack A\rbrack_{0}

as given in table :

$\lbrack A\rbrack_{0}/molL^{- 1}$	$t_{1/2}/min$
0.100	200
0.025	100

Given :

log2 = 0.30

Which of the following is true? (A) The order of the reaction is

\frac{1}{2}

. (B) If

\lbrack A\rbrack_{0}

is 1 M , then

t_{1/2}

200\sqrt{10}\text{ }\min

t_{1/2}

is 800 min for

\lbrack A\rbrack_{0}= 1.6M

Choose the correct answer from the options given below :

Seconds compared with minutes

240 s at one pressure and 4.0 min at another are the SAME half-life, which means first order. Convert both to one unit before taking the ratio.

The order changing with concentration

A half-life table is used to find one fixed order. A statement that the order becomes 1 when the concentration is raised is false.

Drill 5 more on order from how the half-life depends on the starting concentration

Concept 3 of 3: Identifying zero and first order from the shape of a graph

Each order has one pair of axes that gives a straight line. Zero order is straight in

[A]

against

t

; first order is straight in

\ln[A]

against

t

. Read which plot is straight and which way it slopes.

Definition

Zero order: $[A]$ against $t$ is straight; the rate is constant, so rate against $t$ or against $[A]$ is horizontal.
First order: $\ln[A]$ or $\log\dfrac{[A]}{[A]_0}$ against $t$ is straight; rate against $[A]$ is a line through the origin; $t_{1/2}$ against $[A]_0$ is horizontal.
$\log\dfrac{[A]}{[A]_0}$ falls with time: its slope is $-\dfrac{k}{2.303}$ . Its reverse, $\log\dfrac{[A]_0}{[A]}$ , rises with slope $+\dfrac{k}{2.303}$ .

Plot	Zero order	First order
$[A]$ against $t$	Straight line, slope $-k$	Falling exponential curve that never reaches zero
$\ln[A]$ against $t$	Curve bending downward	Straight line, slope $-k$
$\log\dfrac{[A]}{[A]_0}$ against $t$	Curve bending downward	Straight line through the origin, slope $-\dfrac{k}{2.303}$
Rate against $t$	Horizontal line	Falling exponential curve
Rate against $[A]$	Horizontal line	Straight line through the origin, slope $k$
$t_{1/2}$ against $[A]_0$	Straight line through the origin	Horizontal line

Find the straight plot first; its slope then gives k.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q102Moderate

Example 3 · Chemical Kinetics · Zero Order and Finding the Order

Given below are two statements : Statement (I): The plot of

t_{1/2}

against

[R]_{0}

in the first figure is valid for a first order reaction. Statement (II): The plot of

\log\frac{[R]}{[R]_{0}}

against time in the second figure, with slope

=\frac{k}{2.303}

, is valid for a first order reaction. In the light of the above statements, choose the correct answer from the options given below :

The sign of the slope

\log\dfrac{[A]}{[A]_0}

falls with time, so its slope is

-\dfrac{k}{2.303}

. A statement giving the slope as

+\dfrac{k}{2.303}

for this plot is false.

A rate–time line read as a concentration–time line

A horizontal RATE against time line means zero order. A horizontal CONCENTRATION against time line would mean nothing is reacting. Read the axis label first.

Drill 3 more on identifying zero and first order from the shape of a graph

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

The zero-order law and its half-life
Zero-order law
$[A] = [A]_0 - kt\qquad t_{1/2} = \frac{[A]_0}{2k}$
Order from how the half-life depends on the starting concentration
Half-life and order
$t_{1/2} \propto [A]_0^{\,1-n}\quad\Rightarrow\quad \frac{t_{1/2}'}{t_{1/2}} = \left(\frac{[A]_0'}{[A]_0}\right)^{1-n}$

Reference tables (1)

Identifying zero and first order from the shape of a graph6 rows

Plot	Zero order	First order
$[A]$ against $t$	Straight line, slope $-k$	Falling exponential curve that never reaches zero
$\ln[A]$ against $t$	Curve bending downward	Straight line, slope $-k$
$\log\dfrac{[A]}{[A]_0}$ against $t$	Curve bending downward	Straight line through the origin, slope $-\dfrac{k}{2.303}$
Rate against $t$	Horizontal line	Falling exponential curve
Rate against $[A]$	Horizontal line	Straight line through the origin, slope $k$
$t_{1/2}$ against $[A]_0$	Straight line through the origin	Horizontal line

Find the straight plot first; its slope then gives k.

Watch out for (6)

One-quarter taken as two half-lives
→ The zero-order law and its half-life
Zero-order half-life treated as fixed
→ The zero-order law and its half-life
Seconds compared with minutes
→ Order from how the half-life depends on the starting concentration
The order changing with concentration
→ Order from how the half-life depends on the starting concentration
The sign of the slope
→ Identifying zero and first order from the shape of a graph
A rate–time line read as a concentration–time line
→ Identifying zero and first order from the shape of a graph

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