A plot between concentration of reactant [R] and time 't' is shown below. Which of the given order of reaction

Question

Q. A plot between concentration of reactant [R] and time 't' is shown below. Which of the given order of reaction is indicated by the graph ? (The graph shows [R] decreasing linearly with time — a straight line with negative slope; the graph is shown as a figure.) (A) Third order (B) Second order (C) First order (D) Zero order

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Zero Order Reaction

Zero Order Reaction – From Intuition to Precision

Imagine you are filling a bucket with water from a tap that runs at a constant speed. No matter how much water is already in the bucket, the tap adds the same amount every second. The rate at which the bucket fills does not depend on how full it already is.

Now flip that picture: a reaction where the rate of disappearance of a reactant is constant, regardless of how much reactant is left. That is a zero order reaction. The reaction does not slow down as the reactant gets used up — it keeps going at the same pace until the reactant is gone.

Note

The "order" of a reaction tells you how the rate depends on concentration. Zero order means the rate depends on concentration to the power zero — i.e., not at all.

The Precise Statement

For a zero order reaction involving a single reactant $A$ :

\text{Rate} = -\frac{d[A]}{dt} = k

Here $k$ is the rate constant (units: concentration/time, e.g., mol L $^{-1}$ s $^{-1}$ ). The negative sign indicates that $[A]$ decreases with time.

Integrating this differential equation gives the integrated rate law:

[A] = [A]_0 - kt

where $[A]_0$ is the initial concentration at $t=0$ .

[A] = [A]_0 - kt

This is a straight line with slope $-k$ when you plot $[A]$ versus $t$ . The concentration falls linearly until it hits zero at time $t = [A]_0 / k$ .

Key Features at a Glance

Property	Zero Order
Rate law	$\text{Rate} = k$
Units of $k$	concentration $\cdot$ time $^{-1}$
Plot for straight line	$[A]$ vs $t$
Slope of that plot	$-k$
Half-life	$t_{1/2} = \dfrac{[A]_0}{2k}$

Notice the half-life is not constant — it depends on the starting concentration. The more reactant you start with, the longer it takes to halve it. This is the opposite of first order reactions, where half-life is fixed.

Watch out

A common mistake: assuming half-life is always constant. For zero order, it changes with $[A]_0$ . Do not mix this up with first order behaviour.

Why Does a Reaction Become Zero Order? …

Start your 14-day free trial to unlock the full solution →

A straight-line plot of concentration $[R]$ vs. time $t$ means the rate is constant — this is the hallmark of a zero-order reaction. The correct option is (D).

The key to this question is recognising what each order of reaction looks like when you plot concentration against time. Many students memorise only the integrated rate laws, but the shape of the graph is what the exam directly tests.

For a zero-order reaction, the rate is independent of concentration:

\text{Rate} = -\frac{d[R]}{dt} = k

Integrating gives $[R] = [R]_0 - kt$ , which is a straight line with slope $-k$ and intercept $[R]_0$ . That is exactly what the graph shows — a linear decrease.

For first-order reactions, the plot of $\ln[R]$ vs $t$ is linear, not $[R]$ vs $t$ . For second-order, it's $1/[R]$ vs $t$ that gives a straight line. Third-order would involve $1/[R]^2$ vs $t$ .

Watch out

A common mistake is to see a decreasing line and think "first order" because first-order is the most familiar. But first-order gives an exponential decay curve in $[R]$ vs $t$ , not a straight line. The straight line is unique to zero order.

Let's walk through the reasoning step by step.

Identify what the graph shows. The figure plots $[R]$ on the y-axis and time $t$ on the x-axis. The line is straight and sloping downward — a linear relationship of the form $[R] = [R]_0 - mt$ , where $m$ is a positive constant (the slope magnitude).
Relate the graph to the rate law. The slope of this $[R]$ vs $t$ plot is $\frac{d[R]}{dt}$ , which is the rate of the reaction (negative because reactant is being consumed). Since the slope is constant (straight line), the rate is constant: $-\frac{d[R]}{dt} = \text{constant}$ .
Match constant rate to reaction order. A constant rate means the rate does not depend on $[R]$ . The general rate law is $\text{Rate} = k[R]^n$ . For the rate to be independent of $[R]$ , the exponent $n$ must be zero: $[R]^0 = 1$ , so $\text{Rate} = k$ , a constant.
Confirm with the integrated form. The zero-order integrated law is $[R] = [R]_0 - kt$ . This is a linear equation of the form $y = c + mx$ , with slope $= -k$ and intercept $= [R]_0$ . The graph matches perfectly. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 30 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The half life period of a zero order reaction is proportional to:$
(a) Initial concentration of the reactant
(b) (Initial concentration of the reactant)^1/2
(c) (Initial concentration of the reactant)^2
(d) (Initial concentration of the reactant)^3
›Reveal solutionSolution

For a zero order reaction, half-life is directly proportional to the initial concentration: $t_{1/2}\propto[A]_0$ .

For a zero order reaction $A\rightarrow$ products, the integrated rate law is $[A] = [A]_0 - kt$ . At $t=t_{1/2}$ , $[A]=\dfrac{[A]_0}{2}$ , so:

$\frac{[A]_0}{2}=[A]_0-kt_{1/2}$

$kt_{1/2}=[A]_0-\frac{[A]_0}{2}=\frac{[A]_0}{2}$

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

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The order of reaction for thermal decomposition of HI on gold surface is$
(a) 0
(b) 1
(c) 2
(d) 3
›Reveal solutionSolution

In a surface-catalysed (heterogeneous) reaction, once the catalyst surface is saturated with reactant, the rate becomes independent of the reactant's concentration - a zero-order reaction.

Thermal decomposition of HI on a gold surface is a classic textbook example of a zero-order reaction: at the working pressure the gold surface is completely covered by adsorbed HI molecules, so increasing the gas-phase HI concentration further …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The half-life period of a zero order reaction is equal to:$
(a) 0.693/K
(b) 2.303/K
(c) 2K/[A]0
(d) [A]0/2K
›Reveal solutionSolution

For a zero-order reaction, the integrated rate law gives $t_{1/2} = [A]_0/2K$ — half-life is directly proportional to the initial concentration.

For a zero-order reaction, rate $= K$ (constant, independent of concentration). The integrated rate law is:

$[A] = [A]_0 - Kt$

At $t = t_{1/2}$ , $[A] = [A]_0/2$ . Substituting:

$\dfrac{[A]_0}{2} = [A]_0 - Kt_{1/2}$

$Kt_{1/2} = [A]_0 - \dfrac{[A]_0}{2} = \dfrac{[A]_0}{2}$ …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Give one example of zero-order reaction.$
›Reveal solutionSolution

Decomposition of NH₃ on a hot Pt surface (2NH₃ → N₂ + 3H₂) is a classic zero-order reaction.

A zero-order reaction is one whose rate does not depend on the concentration of the reactant — rate = k[reactant]⁰ = k, a constant.

Example: the catalytic decomposition of ammonia gas on a hot platinum (Pt) surface at high temperature (around 1130 K):

$2NH_3(g) \xrightarrow[1130\,K]{Pt} N_2(g) + 3H_2(g)$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set 56/5/11 markMCQ
$Q. A plot between concentration of reactant [R] and time 't' is shown below. Which of the given order of reaction is indicated by the graph ? (The graph shows [R] decreasing linearly with time — a straight line with negative slope; the graph is shown as a figure.) (A) Third order (B) Second order (C) First order (D) Zero order$
›Reveal solutionSolution
A straight-line plot of concentration $[R]$ vs. time $t$ means the rate is constant — this is the hallmark of a zero-order reaction. The correct option is (D).

The key to this question is recognising what each order of reaction looks like when you plot concentration against time. Many students memorise only the integrated rate laws, but the shape of the graph is what the exam directly tests.

For a zero-order reaction, the rate is independent of concentration:
$\text{Rate} = -\frac{d[R]}{dt} = k$
Integrating gives $[R] = [R]_0 - kt$ , which is a straight line with slope $-k$ and intercept $[R]_0$ . That is exactly what the graph shows — a linear decrease.

For first-order reactions, the plot of $\ln[R]$ vs $t$ is linear, not $[R]$ vs $t$ . For second-order, it's $1/[R]$ vs $t$ that gives a straight line. Third-order would involve $1/[R]^2$ vs $t$ .

Watch out

A common mistake is to see a decreasing line and think "first order" because first-order is the most familiar. But first-order gives an exponential decay curve in $[R]$ vs $t$ , not a straight line. The straight line is unique to zero order.

Let's walk through the reasoning step by step.

Identify what the graph shows. The figure plots $[R]$ on the y-axis and time $t$ on the x-axis. The line is straight and sloping downward — a linear relationship of the form $[R] = [R]_0 - mt$ , where $m$ is a positive constant (the slope magnitude).

Relate the graph to the rate law. The slope of this $[R]$ vs $t$ plot is $\frac{d[R]}{dt}$ , which is the rate of the reaction (negative because reactant is being consumed). Since the slope is constant (straight line), the rate is constant: $-\frac{d[R]}{dt} = \text{constant}$ .

Match constant rate to reaction order. A constant rate means the rate does not depend on $[R]$ . The general rate law is $\text{Rate} = k[R]^n$ . For the rate to be independent of $[R]$ , the exponent $n$ must be zero: $[R]^0 = 1$ , so $\text{Rate} = k$ , a constant.

Confirm with the integrated form. The zero-order integrated law is $[R] = [R]_0 - kt$ . This is a linear equation of the form $y = c + mx$ , with slope $= -k$ and intercept $= [R]_0$ . The graph matches perfectly. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Fill in the blank: At high pressure, the decomposition of gaseous ammonia on a hot platinum surface is an example of ________ order reaction.$
›Reveal solutionSolution

At high pressure, decomposition of NH3 on a hot platinum surface is a classic example of a (pseudo) zero order reaction, since the catalyst surface is fully saturated with gas.

2NH3(g) --Pt catalyst, high pressure--> N2(g) + 3H2(g)

At high pressure, the platinum surface becomes completely covered (saturated) with adsorbed NH3 molecules. Beyond this saturation point, increasing the gas pressure (concentration) further does not increase the amount of NH3 actually reacting on the surface at any instant, since all active catalytic sites are …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. For a reaction, a plot between concentration of reactant, [R] and time, t is as shown below : The order of the reaction is _____.$
›Reveal solutionSolution

A straight-line plot of [R] versus t (concentration falling linearly with time) is the hallmark of a zero order reaction.

For a zero order reaction, R → P, the rate is independent of the concentration of reactant:

rate = k[R]^0 = k

Integrating -d[R]/dt = k gives the integrated rate law:

[R] = [R]0 - kt

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. For a zero order reaction, the integrated rate equation is :$
(a) [A] = -kt + [A]0
(b) kt = [A]/[A]0
(c) kt = [A] - [A]0
(d) [A] = kt - [A]0
›Reveal solutionSolution

For a zero order reaction the rate is independent of concentration, so [A] decreases linearly with time: [A] = [A]0 - kt.

For a zero order reaction, Rate = k[A]^0 = k (a constant, independent of concentration).

Rate = -d[A]/dt = k

Separating variables and integrating from [A]0 at t=0 to [A] at time t:

-d[A] = k dt => integrating both sides gives [A]0 - [A] = kt

Rearranging: [A] = [A]0 - kt, i.e. [A] = -kt + [A]0.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. The unit of rate constant for a zero order reaction is ______.$
›Reveal solutionSolution

For a zero order reaction, rate = k, so the rate constant k has the same units as rate, i.e. mol L^-1 s^-1 (or mol L^-1 time^-1).

In general, for a reaction of order n, the rate constant has units of (concentration)^(1-n) x (time)^-1, derived from Rate = k[A]^n.

For a zero order reaction, n = 0:

Rate = k[A]^0 = k

Units of Rate = units of k

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. The unit of rate constant for the zero order reaction is$
(a) L s^-1
(b) L mol^-1 s^-1
(c) mol L^-1 s^-1
(d) mol s^-1
›Reveal solutionSolution

For a zero order reaction, rate = k (independent of concentration), so k carries the same units as rate itself.

For a zero order reaction: Rate = k[A]^0 = k

Rate is always expressed in concentration/time = mol L^-1 s^-1 (or mol L^-1 time^-1). Since Rate = k directly here (no concentration term), k must carry exactly the same units as rate.

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$Q. What is the order of a reaction whose rate constant has same unit as the rate of reaction?$
›Reveal solutionSolution
The units of a rate constant depend on the overall order; only for zero order does $k$ carry the same units as rate itself, $mol\,L^{-1}\,s^{-1}$ .

For a reaction of general order $n$ , $\text{Rate} = k[A]^n$ , so:

$k = \dfrac{\text{Rate}}{[A]^n} = \dfrac{mol\,L^{-1}\,s^{-1}}{(mol\,L^{-1})^n}$

The units of $k$ for common orders:

Zero order ( $n=0$ ): $k$ has units $mol\,L^{-1}\,s^{-1}$ — identical to the units of rate.

First order ( $n=1$ ): $k$ has units $s^{-1}$ .

Second order ( $n=2$ ): $k$ has units $L\,mol^{-1}\,s^{-1}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. When initial concentration of the reactant is doubled, the half-life period of a zero order reaction is ......................................$
(a) Halved
(b) Doubled
(c) Tripled
(d) Same
›Reveal solutionSolution

For a zero order reaction, half-life is directly proportional to initial concentration, so doubling it doubles the half-life.

For a zero order reaction, rate $= k$ (independent of concentration), and the half-life is $t_{1/2} = \frac{[A]_0}{2k}$ . Since $k$ is a constant for a given reaction and temperature, $t_{1/2} \propto [A]_0$ . So if the initial concentration $[A]_0$ is doubled, the half-life also doubles. (Contrast this with a first order reaction, …

Start your 14-day free trial to see the full solution →

🎓Unlock everything free for 14 days

Unlock everything free for 14 days