A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is th

Exercises · 3.9

Q. A reaction is first order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

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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 reaction first order in A and second order in B, the rate law is $r = k[A]^1[B]^2$ . Tripling [B] multiplies the rate by 9; doubling both [A] and [B] multiplies the rate by 8.

The core idea: the rate law is a product of concentration terms, each raised to its order.

The rate of a reaction tells us how fast reactants are used up or products form. For a simple reaction, the rate depends on the concentrations of the reactants, each raised to a power (the order). The overall rate law is:

r = k [A]^m [B]^n

where $m$ is the order in A, $n$ is the order in B, and $k$ is the rate constant (which depends only on temperature, not on concentration). Here, we are told the reaction is first order in A ( $m = 1$ ) and second order in B ( $n = 2$ ).

Step-by-step solution

1. Write the differential rate equation.

The differential rate equation directly expresses the instantaneous rate in terms of concentrations. Using the given orders:

r = k [A]^1 [B]^2

That’s it. The exponent on [A] is 1 (often omitted), and on [B] it’s 2. This is the complete rate law.

r = k [A][B]^2

2. Effect of tripling the concentration of B.

We want to see what happens to the rate when only [B] changes. Let the initial rate be $r_1 = k [A][B]^2$ .

Now, increase [B] to three times its original value: $[B]_{\text{new}} = 3[B]$ . The concentration of A stays the same. The new rate $r_2$ is:

r_2 = k [A] (3[B])^2 = k [A] \cdot 9 [B]^2 = 9 \cdot k [A][B]^2 = 9 r_1

So the rate becomes 9 times the original rate.

Why 9? Because the order in B is 2 — the rate is proportional to $[B]^2$ . Tripling B means squaring the factor: $3^2 = 9$ .

Watch out

A common mistake is to multiply the factor by the order (e.g., $3 \times 2 = 6$ ). That’s wrong. The order is an exponent, not a multiplier. Always raise the concentration factor to the power of the order.

3. Effect of doubling both [A] and [B]. …

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Here are the common mistakes students make on this exact question, along with how to avoid each one.

(i) Writing the Differential Rate Equation

Mistake 1: Writing the wrong order for B

Students often write:

Rate = $k[A]^1[B]^1$

Why this happens: They confuse the stoichiometric coefficient (from a balanced equation) with the order of reaction. Here, no balanced equation is given — only the order is stated.

How to avoid:

The order is an experimentally determined number.
If the reaction is first order in A → exponent on $[A]$ is 1.
If second order in B → exponent on $[B]$ is 2.

✓ Correct equation:

\text{Rate} = k[A]^1[B]^2

Mistake 2: Forgetting the proportionality constant $k$

Some students write:

Rate = $[A][B]^2$

Why this happens: They omit the rate constant, thinking it’s optional.

How to avoid:

The rate law always includes $k$ .
$k$ is the proportionality constant that makes the equation an equality.

✓ Always write:

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

(ii) Effect of Tripling the Concentration of B

Mistake 3: Thinking the rate triples

Students say:

Rate becomes 3 times

Why this happens: They apply a linear (first-order) logic to a second-order reactant.

How to avoid:

For a second-order reactant, if concentration is multiplied by $x$ , the rate is multiplied by $x^2$ .
Here, $x = 3$ , so rate multiplies by $3^2 = 9$ .

✓ Correct reasoning:

\text{New rate} = k[A][3B]^2 = k[A] \cdot 9[B]^2 = 9 \times (\text{original rate})

Key result: Rate becomes 9 times the original.

(iii) Effect of Doubling Both A and B

Mistake 4: Adding the effects instead of multiplying

Students say:

Rate becomes $2 + 4 = 6$ times

Why this happens: They treat the effect as additive rather than multiplicative.

How to avoid:

The rate law is multiplicative.
Doubling $[A]$ (order 1) multiplies rate by $2^1 = 2$ . …

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Q.A reaction is first order in A and second order in B.

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 …

The core idea: the rate law is a product of concentration terms, each raised to its order.

Step-by-step solution

Method: Rate Law Analysis Using Order of Reaction

(i) Write the differential rate equation

(ii) Effect when concentration of B is increased three times

(iii) Effect when concentrations of both A and B are doubled

(i) Writing the Differential Rate Equation

Mistake 1: Writing the wrong order for B

Mistake 2: Forgetting the proportionality constant $k$

(ii) Effect of Tripling the Concentration of B

Mistake 3: Thinking the rate triples

(iii) Effect of Doubling Both A and B

Mistake 4: Adding the effects instead of multiplying

Q.A reaction is first order in A and second order in B.

The core idea: the rate law is a product of concentration terms, each raised to its order.

Step-by-step solution

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