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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←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

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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