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

AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ

\mathrm{S_2O_8^{2-} + 3I^- \rightarrow 2SO_4^{2-} + I_3^-}

Ex. No.	$[S_2O_8^{2-}]$	$[I^-]$	Initial rate (M/s)
1.	0.080	0.034	$2.2 \times 10^{-4}$
2.	0.080	0.017	$1.1 \times 10^{-4}$
3.	0.160	0.017	$2.2 \times 10^{-4}$

›Reveal solution

Tests deriving a rate law and rate constant from tabulated initial-rate data; the reaction is

first order in each reactant (second order overall), and $k \approx 8.0\times10^{-2}\ \mathrm{M^{-1}s^{-1}}$ .

Concept and Intuition

The method of initial rates finds the order in each reactant by comparing pairs of experiments in

which only one concentration changes while the other is held fixed — the ratio of rates then

directly reveals the order with respect to the varied species. Once both orders are known, the

rate constant is obtained by substituting any one experiment's data into the full rate law.

Step-by-Step Solution

Compare Exp 1 and Exp 2: $[\mathrm{S_2O_8^{2-}}]$ is unchanged (0.080), while $[\mathrm{I^-}]$ halves from 0.034 to 0.017. The rate also halves, from $2.2\times10^{-4}$ to $1.1\times10^{-4}$ . So rate $\propto [\mathrm{I^-}]^1$ .
Compare Exp 2 and Exp 3: $[\mathrm{I^-}]$ is unchanged (0.017), while $[\mathrm{S_2O_8^{2-}}]$ doubles from 0.080 to 0.160. The rate also doubles, from $1.1\times10^{-4}$ to $2.2\times10^{-4}$ . So rate $\propto [\mathrm{S_2O_8^{2-}}]^1$ .
Hence rate $= k[\mathrm{S_2O_8^{2-}}][\mathrm{I^-}]$ (overall second order).
Substitute Exp 1 data: $2.2\times10^{-4} = k \times 0.080 \times 0.034 = k \times 2.72\times10^{-3}$ . …

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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
Q.For the reaction $\mathrm{A + B \rightarrow C}$ $A + B \to C$ , the following data were obtained Expt. [A] M [B] M Initial rate ( $\mathrm{M\,min^{-1}}$ $M mi n^{- 1}$ )
1. 0.1 0.1 $1.0\times10^{-4}$
2. 0.1 0.3 $9.0\times10^{-4}$
3. 0.3 0.3 $2.7\times10^{-3}$
The order of reaction with respect to A and B are respectively (A) 1, 2 (B) 2, 1 (C) 1.5, 1.5 (D) 0, 3
›Reveal solutionSolution
Use the method of initial rates, comparing pairs of experiments where only one concentration changes, to find the order in each reactant. Answer: order in A = 1, order in B = 2.

Concept and Intuition

For a rate law $\text{rate} = k[A]^m[B]^n$ , comparing two experiments where only one concentration changes isolates that reactant's order: the ratio of rates equals the ratio of concentrations raised to that reactant's order.

Step-by-Step Solution

Compare Expt 1 and Expt 2: [A] stays at 0.1, [B] goes $0.1\to0.3$ (×3); rate goes $1.0\times10^{-4}\to9.0\times10^{-4}$ (×9).

So $3^n = 9 \Rightarrow n = 2$ (order in B). …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$
›Reveal solutionSolution
Adding an inert gas at constant volume doesn't change the concentrations of the reacting species, so it leaves the equilibrium (with equal moles of gas on each side) undisturbed.

Concept and Intuition

Le Chatelier's principle: a system at equilibrium responds to changes that alter concentration/partial pressure of the species actually involved in the equilibrium. Adding an inert gas at constant volume doesn't change any reactant/product concentration (each species still occupies the same volume), so at constant volume it has no effect on the position of equilibrium — especially clear here since $\Delta n_g = 0$ for this reaction anyway.

Step-by-Step Solution

Adding $H_2$ increases $[H_2]$ , shifting equilibrium forward (toward HI) — this does disturb equilibrium.

Removing $HI$ decreases $[HI]$ , shifting equilibrium forward to replace it — disturbs equilibrium. …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ

\mathrm{A(g) + B(g) \rightarrow C(g)}

Experiment	[A] / mol L $^{-1}$	[B] / mol L $^{-1}$	Initial rate / mol L $^{-1}$ s $^{-1}$
1	0.05	0.05	R
2	x	0.05	2R
3	0.20	0.10	z
(A) x = 0.10, z = 8R
(B) x = 0.05, z = 4R
(C) x = 0.10, z = 16R
(D) x = 0.20, z = 16R

›Reveal solution

This tests using a given rate law $r=k[A][B]^2$ to back-calculate an unknown concentration and an unknown rate from a data table. Answer: x = 0.10, z = 16R.

Concept and Intuition

The rate law tells us the reaction is first order in A and second order in B. Since [B] is unchanged between experiments 1 and 2, any change in rate there is due to [A] alone (first-order, so rate scales linearly with [A]). Between experiments 1 and 3, both concentrations change, so we substitute directly using $k$ eliminated as a common factor.

Step-by-Step Solution

Write the rate law: $r=k[A]^1[B]^2$ .
Experiment 1: $R=k(0.05)(0.05)^2=k(0.05)(0.0025)=1.25\times10^{-4}k$ .
Experiment 2: $[B]$ is still $0.05$ , only $[A]=x$ changes. $2R=k\,x\,(0.0025)$ .
Divide: $\dfrac{2R}{R}=\dfrac{k\,x(0.0025)}{k(0.05)(0.0025)}=\dfrac{x}{0.05}$ , so $2=\dfrac{x}{0.05}\Rightarrow x=0.10$ . …

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AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQ
$2A + B_2 \longrightarrow 2C$
›Reveal solutionSolution
A classic method-of-initial-rates problem: isolate each concentration's effect by comparing pairs of experiments where only one concentration changes. Overall order = 1 (answer D).

Concept and Intuition

For a rate law $Rate = k[A]^x[B]^y$ , we can find $x$ and $y$ separately by picking pairs of experiments where only one reactant's concentration changes while the other stays fixed — any change in rate must then be due to that one reactant alone.

Step-by-Step Solution

Find order in B: Compare Experiment 1 ( $[A]=0.5,[B]=0.5$ , rate $=1\times10^{-4}$ ) and Experiment 2 ( $[A]=0.5,[B]=1.0$ , rate $=2\times10^{-4}$ ). Here $[A]$ is unchanged, $[B]$ doubles, and rate doubles. Since $2^y = 2 \Rightarrow y=1$ . …
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AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$R_o = k[A_o]^2[B_o]$
›Reveal solutionSolution
Plugging the new concentrations into the rate law $R=k[A]^2[B]$ : the rate

increases by a factor of $(1.5)^2 \times 3 = 6.75$ .

Concept and Intuition

When a rate law has the form $R=k[A]^m[B]^n$ , scaling each concentration by a

factor scales the rate by that factor raised to its respective order — and

since these effects multiply independently, the overall factor is the product

of each individual scaling factor raised to its order.

Step-by-Step Solution

Given rate law: $R_o = k[A_o]^2[B_o]^1$ .

New concentrations: $[A] = 1.5[A_o]$ , $[B] = 3[B_o]$ .

New rate: $R = k(1.5[A_o])^2(3[B_o]) = k \times (1.5)^2[A_o]^2 \times 3[B_o]$ . …
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AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
$X$
›Reveal solutionSolution
Tripling a reactant raised to the power $n$ multiplies the rate by $3^n$ ;

tripling both X (order 3) and Y (order 4) multiplies the rate by

$3^3 \times 3^4 = 3^7$ .

Concept and Intuition

The rate law expresses rate as $k[X]^a[Y]^b$ where $a,b$ are the individual

orders. If a concentration is scaled by a factor $f$ , the rate contribution from

that species scales by $f^{\text{order}}$ ; when several concentrations change

simultaneously, their scaling factors multiply.

Step-by-Step Solution

Rate law: $Rate = k[X]^3[Y]^4$ (orders given as 3 and 4 respectively).

Tripling $[X]$ : contributes a factor of $3^3$ .

Tripling $[Y]$ : contributes a factor of $3^4$ . …
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AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$A + B \to$
›Reveal solutionSolution
The reaction is third order overall ( $1+2$ ), so rearranging the rate law for $K$ gives units $mol^{-2}\ L^2\ s^{-1}$ .

Concept and Intuition

The rate constant's units always depend on the overall order of the reaction, since $Rate = K \times (\text{concentration})^n$ must be dimensionally consistent, with Rate always in $mol\ L^{-1}\ s^{-1}$ (or $mol\ L^{-1}\ time^{-1}$ ). Solving for $K$ 's units directly from the given rate law is the safest approach — no need to memorize a general formula, just do the algebra.

Step-by-Step Solution

Given: $Rate = K[A][B]^2$ , with $Rate$ in $mol\ L^{-1}\ s^{-1}$ , and $[A],[B]$ in $mol\ L^{-1}$ .

Rearranging: $K = \dfrac{Rate}{[A][B]^2}$ .

Units: $K = \dfrac{mol\ L^{-1}\ s^{-1}}{(mol\ L^{-1})\times(mol\ L^{-1})^2} = \dfrac{mol\ L^{-1}\ s^{-1}}{mol^3\ L^{-3}}$ . …
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AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
›Reveal solutionSolution
Adding extra $H_2$ to the exothermic ammonia equilibrium (constant volume, closed container) shifts the reaction forward by Le Chatelier's principle, and since the forward direction releases heat, the system's temperature rises.

Concept and Intuition

Le Chatelier's principle says a system at equilibrium responds to a disturbance (here, adding more $H_2$ ) by shifting in the direction that partially counteracts the change — i.e. consuming the added $H_2$ by shifting the $N_2+3H_2\rightleftharpoons2NH_3$ equilibrium to the RIGHT (forward). Because the forward reaction is stated to be exothermic, this net forward conversion of reactants to products liberates heat. In a closed container (not necessarily in contact with a large external heat sink that instantly removes this heat), that liberated heat raises the temperature of the system.

Step-by-Step Solution

Adding extra $H_2$ increases $[H_2]$ , disturbing the existing equilibrium.

By Le Chatelier's principle, the equilibrium shifts in the direction that consumes some of the added $H_2$ — i.e. the FORWARD direction (toward $NH_3$ ).

Volume is held constant while more gas ( $H_2$ ) has been added, so total pressure necessarily increases initially — ruling out option (A).

The equilibrium position (concentrations) DOES change as the system re-equilibrates — ruling out option (B). …
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AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ

A

Exp No	Initial [A]	Initial [B]	Initial rate of formation of C
1	0.1	0.1	$6\times10^{-3}$
2	0.3	0.2	$7.2\times10^{-2}$
3	0.3	0.4	$2.88\times10^{-1}$
4	0.4	0.1	$2.4\times10^{-2}$

›Reveal solution

Tests determining reaction order from experimental rate data; the answer is (B), $Rate = K[A][B]^2$ .

Concept and Intuition

To find the order with respect to each reactant, compare pairs of experiments where only ONE concentration changes while the other is held constant, then see how the rate scales.

Step-by-Step Solution

Order in A: Compare Exp 1 ([A]=0.1, [B]=0.1, rate= $6\times10^{-3}$ ) and Exp 4 ([A]=0.4, [B]=0.1, rate= $2.4\times10^{-2}$ ) — [B] is constant, [A] increases by a factor of 4. Rate ratio $= \dfrac{2.4\times10^{-2}}{6\times10^{-3}} = 4$ . If order in A is $x$ : $4^x = 4 \Rightarrow x = 1$ .
Order in B: Compare Exp 2 ([A]=0.3, [B]=0.2, rate= $7.2\times10^{-2}$ ) and Exp 3 ([A]=0.3, [B]=0.4, rate= $2.88\times10^{-1}$ ) — [A] is constant, [B] doubles. Rate ratio $= \dfrac{2.88\times10^{-1}}{7.2\times10^{-2}} = 4$ . If order in B is $y$ : $2^y = 4 \Rightarrow y = 2$ . …

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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 ______.

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