Rate law for the reaction $A + 2B \rightarrow C$ 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 ______. (i) the same (ii) doubled (iii) quadrupled (iv) halved

Rate law for the reaction A + 2B arrow C is found to be Rate = k[A][B] Concentration of reactant 'B' is double

NCERT Exemplar · Q14

A + 2B \rightarrow C

(i) the same

(ii) doubled

(iii) quadrupled

(iv) halved

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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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The rate constant $k$ is an intrinsic property of the reaction at a given temperature — it does not change when you change reactant concentrations. Doubling $[B]$ changes the rate, not $k$ . So the answer is (i) the same.

Why this question trips students up

The trap here is subtle. The problem gives you the rate law:

\text{Rate} = k[A][B]

It then asks: If you double $[B]$ (keeping $[A]$ constant), what happens to the rate constant $k$ ?

Many students see “doubled” and “quadrupled” in the options and think: “Well, the rate law says rate depends on $[B]$ , so if $[B]$ doubles, the rate doubles — so maybe $k$ doubles too?” That’s wrong, but it’s a very natural mistake.

The key distinction: rate and rate constant are not the same thing. The rate constant $k$ is a proportionality factor that links concentrations to the rate. It depends only on temperature (and, for some reactions, on the presence of a catalyst or the nature of the solvent). It does not depend on how much reactant you put in.

Step-by-step reasoning

Recall the definition of a rate constant.

In any rate law of the form $\text{Rate} = k[\text{A}]^m[\text{B}]^n$ , the constant $k$ is called the rate constant or specific rate constant. Its value is fixed for a given reaction at a given temperature. Changing the concentration of a reactant changes the rate, but $k$ stays the same — it’s like the “sensitivity” of the reaction to concentration.
Look at what the question actually asks.

The question says: “the value of rate constant will be ______.” It does not ask what happens to the rate. It asks about $k$ itself. Since $k$ is independent of concentration, it remains unchanged.
Check the effect of doubling $[B]$ on the rate (for completeness).

If you double $[B]$ , the rate becomes:

$\text{Rate}_{\text{new}} = k[A](2[B]) = 2k[A][B] = 2 \times \text{Rate}_{\text{old}}$ …

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Q.Rate law for the reaction $A + 2B \rightarrow C$ 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 ______.

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 this question trips students up

Step-by-step reasoning

Method: Direct Substitution in Rate Law

Steps:

Common Mistakes & How to Avoid Them

Mistake 1: Confusing "rate constant" with "rate of reaction"

Mistake 2: Thinking $k$ depends on the rate law expression

Mistake 3: Misreading "doubled" as "quadrupled" due to stoichiometry

Q.Rate law for the reaction A+2B→CA + 2B \rightarrow CA+2B→C is found to be Rate =k[A][B]= k[A][B]=k[A][B] Concentration of reactant 'B' is doubled, keeping the concentration of 'A' constant, the value of rate constant will be ______.

Why this question trips students up

Step-by-step reasoning

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Q.Rate law for the reaction $A + 2B \rightarrow C$ 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 ______.

Unlock everything free for 14 days