A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentrati

Exercises · 3.6

Q. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i) doubled

(ii) reduced to half?

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Concept understanding — Effect of Concentration

Effect of Concentration: The Intuition

Imagine you're in a large, empty hall with just one other person. The two of you are trying to bump into each other accidentally. It'll take a while, right? Now imagine the same hall packed with a thousand people. Bumping into someone becomes almost certain within seconds.

That's the core idea behind the effect of concentration on reaction rates. Concentration simply means how much of a substance is packed into a given space. Higher concentration = more particles in the same volume.

When particles are more crowded, they collide more frequently. And since chemical reactions happen only when particles collide with enough energy and the right orientation, more collisions mean more reactions per second. The reaction speeds up.

The Precise Statement

For most chemical reactions, the rate of a reaction is directly proportional to the molar concentration of the reactants (raised to some power, which we'll get to).

\text{Rate} \propto [\text{Reactant}]^n

Here, $[ \ ]$ means "concentration in moles per litre" (mol/L or M), and $n$ is the order of reaction with respect to that reactant.

What does "order" mean?

For a simple reaction like $A \rightarrow \text{Products}$ :

First order ( $n=1$ ): Double the concentration of A → double the rate.
Second order ( $n=2$ ): Double the concentration of A → quadruple the rate ( $2^2 = 4$ ).
Zero order ( $n=0$ ): Changing concentration has no effect on the rate. This happens when the reaction is limited by something else (like a catalyst surface that's already fully covered).

Watch out

The order $n$ is not the same as the stoichiometric coefficient from the balanced equation. It must be determined experimentally. For example, the reaction $2A \rightarrow B$ could be first order in A, not second order.

Why does this happen? The collision theory

The rate depends on two things:

Collision frequency — how often particles meet.
Fraction of effective collisions — how many of those collisions have enough energy (activation energy) and the right orientation.

Doubling the concentration doubles the number of particles per unit volume. This roughly doubles the collision frequency. For a first-order reaction, that directly doubles the rate. For higher orders, the effect compounds because multiple reactant particles must meet simultaneously.

A concrete example

Consider the reaction between hydrochloric acid and sodium thiosulphate:

$\text{Na}_2\text{S}_2\text{O}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{S}(s) + \text{SO}_2(g) + \text{H}_2\text{O}(l)$ …

Why this formula?

Effect of Concentration on Reaction Rate — The Reasoning

The Effect of Concentration is rooted in collision theory. The key idea is simple:

More particles in the same volume → more frequent collisions → higher reaction rate.

Let's break down why the mathematical relationships hold.

1. The Rate Law — Why $r = k[A]^m[B]^n$ ?

This is not derived from theory alone — it is empirical (found experimentally). But the reasoning behind its form comes from collision probability.

For an elementary reaction (one step):

Consider:

A + B \rightarrow \text{products}

The rate depends on how often A and B molecules meet.
In a given volume, the number of A molecules is proportional to $[A]$ , and the number of B molecules is proportional to $[B]$ .
The number of A–B collisions per second is proportional to the product:

\text{Collision frequency} \propto [A] \times [B]

Therefore:

r \propto [A][B]

r = k[A][B]

For a reaction with coefficient $aA + bB$ :

If the reaction is elementary, the stoichiometric coefficients become the exponents:

r = k[A]^a[B]^b

Why? Because for $2A$ to react, two A molecules must collide simultaneously — the probability of that happening is proportional to $[A] \times [A] = [A]^2$ .

2. The Integrated Rate Laws — Why These Forms?

These come from solving the differential equation $r = -\frac{d[A]}{dt} = k[A]^n$ .

Zero-order ( $n=0$ ):

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

Reasoning: Rate is independent of concentration. This happens when the reaction is limited by something else (e.g., a saturated catalyst surface).
Integrate:

\int_{[A]_0}^{[A]} d[A] = -k \int_0^t dt

\Rightarrow [A] = [A]_0 - kt

First-order ( $n=1$ ):

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

Reasoning: Rate is directly proportional to $[A]$ . Each molecule has a constant probability of reacting per unit time (like radioactive decay).
Integrate:

\int_{[A]_0}^{[A]} \frac{d[A]}{[A]} = -k \int_0^t dt

\Rightarrow \ln[A] = \ln[A]_0 - kt

[A] = [A]_0 e^{-kt}

Second-order ( $n=2$ ):

-\frac{d[A]}{dt} = k[A]^2

Reasoning: Rate depends on two molecules of A colliding. Doubling $[A]$ quadruples the collision frequency.
Integrate:

\int_{[A]_0}^{[A]} \frac{d[A]}{[A]^2} = -k \int_0^t dt

\Rightarrow \frac{1}{[A]} = \frac{1}{[A]_0} + kt

3. The Half-Life — Why It Depends on Order …

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For a second-order reaction, rate ∝ [reactant]². Doubling the concentration quadruples the rate; halving it reduces the rate to one-fourth.

Why the rate changes this way

The order of a reaction tells you how the rate depends on the concentration of each reactant. For a reaction that is second order with respect to a single reactant A, the rate law is:

\text{Rate} = k[\text{A}]^2

Here $k$ is the rate constant (which does not change when you change concentration). The exponent 2 is the key: it means the rate is proportional to the square of the concentration.

If you change $[\text{A}]$ by some factor, the rate changes by the square of that factor. That is the entire physical idea — no more, no less.

Watch out

A common mistake is to think "second order means double the concentration → double the rate". That would be true only for a first-order reaction. For second order, the exponent 2 means the effect is amplified: a factor of 2 in concentration becomes a factor of $2^2 = 4$ in rate.

Step-by-step calculation

Let the initial concentration be $[\text{A}]_0$ and the initial rate be $r_0 = k[\text{A}]_0^2$ .

1. Concentration is doubled

New concentration: $[\text{A}]_1 = 2[\text{A}]_0$

New rate: $r_1 = k(2[\text{A}]_0)^2 = k \cdot 4[\text{A}]_0^2 = 4 \cdot k[\text{A}]_0^2 = 4r_0$

So the rate becomes four times the original rate.

2. Concentration is reduced to half

New concentration: $[\text{A}]_2 = \frac{1}{2}[\text{A}]_0$ …

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Here are the common mistakes students make when solving this type of question, along with how to avoid each.

✗ Mistake 1: Confusing order with the exponent

The error:

Students think "second order" means the rate is simply multiplied by 2 when concentration is doubled. They write:

Rate becomes $2 \times$ original (wrong)

Why it happens:

They confuse the order (which is an exponent) with a simple multiplication factor.

How to avoid:

Always write the rate law first. For a second-order reaction with respect to reactant $A$ :

\text{Rate} = k[A]^2

The exponent 2 tells you the rate depends on the square of concentration, not the concentration itself.

✗ Mistake 2: Incorrect factor for doubling concentration

The error:

When $[A]$ is doubled, students write:

New rate $= k(2[A]) = 2k[A]$ (wrong)

How to avoid:

Substitute correctly into the rate law:

\text{New rate} = k(2[A])^2 = k \cdot 4[A]^2 = 4 \times (\text{original rate})

Key result: Doubling concentration quadruples the rate.

✗ Mistake 3: Incorrect factor for halving concentration

The error:

When $[A]$ is halved, students write:

New rate $= k\left(\frac{[A]}{2}\right) = \frac{1}{2}k[A]$ (wrong)

How to avoid:

Again, use the rate law correctly:

\text{New rate} = k\left(\frac{[A]}{2}\right)^2 = k \cdot \frac{[A]^2}{4} = \frac{1}{4} \times (\text{original rate})

Key result: Halving concentration reduces the rate to one-fourth.

✗ Mistake 4: Forgetting to square the factor

The error:

Students apply the factor to $[A]$ but forget to square it. For example:

Doubling: factor = $2$ , but they forget $2^2 = 4$
Halving: factor = $\frac{1}{2}$ , but they forget $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$

How to avoid: …

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

CBSE 2026Set 56/3/11 markMCQ
$Q. Assertion (A) : Order of reaction is applicable to elementary as well as complex reactions. Reason (R) : Order of a reaction is an experimental quantity. Options : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
The assertion is false because order is defined only for simple rate laws (elementary steps or overall reactions with a simple rate expression), not for all complex reactions. The reason is true: order is experimental. So the correct option is (D).

Concept First: What "Order" Really Means

The order of a reaction is the sum of the exponents of concentration terms in the experimentally determined rate law. For an elementary reaction (a single step), the order equals the molecularity — that’s straightforward. But for a complex reaction (a sequence of steps), the overall rate law can be messy: it might involve fractional exponents, negative exponents, or even terms that don’t look like a simple power law at all. In such cases, the concept of "order" simply doesn’t apply in the usual sense.

The reason given is a fundamental truth: order is always found by experiment, never deduced from the balanced equation (except for elementary steps). That’s correct.

Now let’s examine the assertion carefully.

Step-by-Step Reasoning

What does "applicable" mean here?

The assertion says order is "applicable" to both elementary and complex reactions. If a reaction has a rate law of the form $r = k[\text{A}]^m[\text{B}]^n$ , then we can define order $= m+n$ . For an elementary reaction, this always works. For a complex reaction, it works only if the overall rate law happens to be a simple power law — which is not guaranteed.

Counterexample: a complex reaction where order is not defined

Consider the reaction $2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$ . Its mechanism involves a pre-equilibrium, and the experimental rate law is $r = k[\text{NO}]^2[\text{O}_2]$ . Here order = 3, so it is applicable. But take the decomposition of $\text{N}_2\text{O}_5$ : the rate law is $r = k[\text{N}_2\text{O}_5]$ , so order = 1 — again applicable.

However, consider a reaction like $\text{H}_2 + \text{Br}_2 \rightarrow 2\text{HBr}$ . The experimental rate law is:

$r = \frac{k[\text{H}_2][\text{Br}_2]^{1/2}}{1 + k'[\text{HBr}]/[\text{Br}_2]}$
This is not of the form $k[\text{A}]^m[\text{B}]^n$ — it has a denominator with a concentration term. You cannot assign a single "order" to this reaction. The concept of order is simply not applicable here. …
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CBSE 2025Set ANNUAL1 mark
$Q. The conversion of molecule X to Y follows third order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?$
›Reveal solutionSolution

For a third-order reaction, Rate ∝ [X]^3, so tripling [X] multiplies the rate by 3^3 = 27.

Rate law: Rate = k[X]^3 (third order in X).

Original rate: Rate1 = k[X]^3 …

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CBSE 2024Set D1 markMCQ
$Q. The rate of reaction of a substance depends upon$
(a) Atomic mass
(b) Equivalent mass
(c) Molecular mass
(d) Active mass
›Reveal solutionSolution

Rate depends on active mass (concentration), per the law of mass action.

The law of mass action states that the rate of a chemical reaction is proportional to the product of the ACTIVE MASSES (molar concentrations) of the reacting substances, each raised to a power. Atomic, equivalent and molecular masses are fixed properties of a substance and do not govern how fast a reaction …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A reaction in which reactant (R) are converted into products (P) follows second order kinetics. If concentration of reactant (R) is increased by 4 times, what will be the increase in rate of formation of P?$
(a) 9 times
(b) 4 times
(c) 16 times
(d) 8 times
›Reveal solutionSolution

Second order kinetics means Rate = k[R]^2; scaling the concentration by a factor scales the rate by that factor squared.

Rate = k[R]^2 (second order in R)

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Rate law for the reaction A + 2B \to C is found to be Rate = K[A][B]. Concentration of reactant B is doubled keeping the concentration of A constant, the value of rate constant will be :$
(a) the same
(b) doubled
(c) quadroupled
(d) halved
›Reveal solutionSolution

The rate constant $k$ is an intrinsic property of a reaction at a given temperature — it does NOT depend on the concentrations of reactants; only the rate itself changes when concentration changes.

For Rate $= k[A][B]$ , doubling $[B]$ while keeping $[A]$ constant doubles the RATE of the reaction (since rate $\propto [B]$ ), but $k$ itself is unaffected — $k$ is a constant determined only by the nature of the reaction, the temperature, and the …

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CBSE 2024Set ANNUAL1 mark
$Q. In general, what happens to the rate of reaction as the reaction progresses?$
›Reveal solutionSolution

As reactants are used up over time, their concentration drops, and since rate depends on concentration, the rate falls too.

For most reactions, the rate law has the form $Rate = k[\text{reactant}]^n$ (n > 0), i.e. the rate is directly related to the concentration of the reactant(s).

As a reaction proceeds, reactant molecules are continuously converted to products, so the concentration of the reactants keeps decreasing with time. Since the rate depends on this concentration, the rate of reaction generally decreases as the reaction progresses, being fastest at the very start (when reactant concentration is highest) and slowing down as the reaction approaches completion.

…

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CBSE 2023Set F1 markMCQ
$Q. The rate law equation of a chemical reaction is represented as Rate = K[A][B]^2. If the concentration of B is trebled keeping that of A constant, then rate becomes$
(a) double
(b) trebled
(c) quadrupled
(d) nine times
›Reveal solutionSolution

Rate depends on [B]^2, so tripling [B] increases the rate by a factor of 9.

Given Rate = K[A][B]^2 and [A] kept constant, only the [B]^2 term changes. If [B] is trebled (multiplied by 3):

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. A reaction is first order in A and second order in B. How is the rate affected when concentrations of both A and B are doubled?$
(a) It increases 4 times
(b) It increases 6 times
(c) It increases 8 times
(d) It reduces 8 times
›Reveal solutionSolution

Substituting the doubled concentrations into the rate law Rate = k[A]^1[B]^2 shows the rate scales by a factor of 8.

Given: order in A = 1, order in B = 2, so Rate = k[A][B]^2.

If [A] -> 2[A] and [B] -> 2[B]: …

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CBSE 2022Set HE2181 mark
$Q. Fill in the blank: The rate of reaction is ______ of concentration of reactant.$
›Reveal solutionSolution

The rate law states that the rate of a reaction is proportional to the concentration(s) of the reactant(s) raised to some power (the order), not necessarily equal to the stoichiometric coefficient.

For a general reaction A -> products, the experimentally determined rate law is written as:

Rate = k[A]^n

where k is the rate constant and n is the order of the reaction with respect to A (found experimentally, not simply read off the balanced equation). This shows the rate of reaction is a function of (depends on / is proportional to) the concentration of the reactant — as concentration increases, the frequency of effective molecular collisions increases, so the rat …

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CBSE 2020Set ANNUAL1 mark
$Q. State the rate law for chemical reaction.$
›Reveal solutionSolution

The rate law is the experimentally determined equation expressing reaction rate as proportional to reactant concentrations raised to their (experimental) orders.

For a general reaction, $aA + bB \rightarrow \text{Products}$ , the rate law (or rate equation) expresses how the rate of reaction depends on the concentration of each reactant:

$\text{Rate} = k[A]^x[B]^y$ …

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CBSE 2018Set ANNUAL1 mark
$Q. What is rate law ?$
›Reveal solutionSolution

The rate law connects reaction rate to reactant concentrations via experimentally-found exponents (orders).

For a reaction, the rate law (or rate equation) is the mathematical expression relating the instantaneous rate of the reaction to the molar concentrations of the reactants, each raised to a power (the order with respect to that reactant), determined experimentally (not necessarily from the stoichiometric coefficients):

$Rate = k[A]^x[B]^y$ …

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←Previous question3.5 — Mention the factors that affect the rate of a chemical reaction.→Next question3.7 — What is the effect of temperature on the rate constant of a reaction? …

Q.A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

Concept understanding — Effect of Concentration

Effect of Concentration: The Intuition

The Precise Statement

What does "order" mean?

Why does this happen? The collision theory

A concrete example

Why this formula?

Effect of Concentration on Reaction Rate — The Reasoning

1. The Rate Law — Why $r = k[A]^m[B]^n$ ?

For an elementary reaction (one step):

For a reaction with coefficient $aA + bB$ :

2. The Integrated Rate Laws — Why These Forms?

Zero-order ( $n=0$ ):

First-order ( $n=1$ ):

Second-order ( $n=2$ ):

3. The Half-Life — Why It Depends on Order …

Why the rate changes this way

Step-by-step calculation

Method: Rate Law Substitution Method

Steps

✗ Mistake 1: Confusing order with the exponent

✗ Mistake 2: Incorrect factor for doubling concentration

✗ Mistake 3: Incorrect factor for halving concentration

✗ Mistake 4: Forgetting to square the factor

Concept First: What "Order" Really Means

Step-by-Step Reasoning

Q.A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

Why the rate changes this way

Step-by-step calculation

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