In any unimolecular reaction _______. (Two or more than two options may be correct.) (i) only one reacting spe

NCERT Exemplar · Q48

Q. In any unimolecular reaction ______________. (Two or more than two options may be correct.)

(i) only one reacting species is involved in the rate determining step.

(ii) the order and the molecularity of slowest step are equal to one.

(iii) the molecularity of the reaction is one and order is zero.

(iv) both molecularity and order of the reaction are one.

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Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

Imagine you are watching a simple reaction:

$2 \text{NO}_2 \rightarrow 2 \text{NO} + \text{O}_2$

As NO₂ disappears, NO appears twice as fast as O₂ appears. Why? Because the balanced equation says: for every 2 molecules of NO₂ that break apart, you get 2 molecules of NO and 1 molecule of O₂. The numbers in front of the species — the stoichiometric coefficients — tell you the relative speeds at which reactants vanish and products appear.

That is the core idea: reaction rate stoichiometry is the relationship between the rates of change of different species in a chemical reaction, dictated by their coefficients in the balanced equation.

The Intuitive Picture

Think of a factory assembly line. The balanced equation is like a recipe:

2 units of raw material A → 2 units of product B + 1 unit of byproduct C

If the line runs steadily, every time 2 units of A are consumed, 2 units of B are produced and 1 unit of C is produced. So the rate at which A disappears must be twice the rate at which C appears. The rate at which B appears equals the rate at which A disappears (both have coefficient 2).

Note

The stoichiometric coefficients are not speeds themselves — they are scaling factors that connect the speeds of different species.

The Precise Statement

For a general reaction:

aA + bB \rightarrow cC + dD

The rate of reaction (often called the rate of the process, $r$ ) is defined as:

r = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Here:

$\frac{d[X]}{dt}$ is the instantaneous rate of change of concentration of species X (in mol L⁻¹ s⁻¹).
The minus sign is used for reactants (their concentration decreases with time).
The plus sign is used for products (their concentration increases with time).
Dividing by the coefficient normalises the rate — so $r$ is the same number for every species.

r = -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{c}\frac{d[C]}{dt}

This single value $r$ is the intrinsic rate of the reaction, independent of which species you measure.

How to Use It (Step by Step)

Suppose you are given:

$2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}$

And you measure that $\frac{d[\text{H}_2]}{dt} = -0.040 \text{ M/s}$ (negative because H₂ is being consumed).

Step 1: Write the rate relation:

r = -\frac{1}{2}\frac{d[\text{H}_2]}{dt} = -\frac{1}{1}\frac{d[\text{O}_2]}{dt} = \frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt}

Step 2: Plug in the known value:

r = -\frac{1}{2}(-0.040) = +0.020 \text{ M/s}

Step 3: Find the rate for O₂:

-\frac{d[\text{O}_2]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{O}_2]}{dt} = -0.020 \text{ M/s}

Step 4: Find the rate for H₂O:

\frac{1}{2}\frac{d[\text{H}_2\text{O}]}{dt} = r \quad \Rightarrow \quad \frac{d[\text{H}_2\text{O}]}{dt} = 2r = 0.040 \text{ M/s}

Tip

A quick check: the coefficients tell you the relative rates. Here, H₂ disappears twice as fast as O₂, and H₂O appears at the same rate as H₂ disappears (both coefficient 2). Always verify your numbers match the coefficient ratios.

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Why this formula?

Average Rate of Reaction — Why the Formula Holds

Let’s build this from the ground up. The goal is to understand why the average rate formula looks the way it does — not just memorise it.

1. What does "rate of reaction" mean physically?

A chemical reaction changes the concentration of reactants (decreasing) and products (increasing) over time.

Rate = how fast this change happens.
If you measure the change over a finite time interval, you get the average rate.

2. The core idea: change per unit time

For any quantity $X$ that changes from $X_1$ to $X_2$ over time $t_1$ to $t_2$ :

\text{Average rate of change of } X = \frac{\Delta X}{\Delta t} = \frac{X_2 - X_1}{t_2 - t_1}

This is just the slope of the straight line connecting the two points on a concentration vs. time graph.

3. Applying this to a reaction

Consider a simple reaction:

A \rightarrow B

Reactant A is consumed: $[A]$ decreases.
Product B is formed: $[B]$ increases.

For reactant A (disappearing):

\text{Average rate} = -\frac{\Delta [A]}{\Delta t}

Why the minus sign?

Because $\Delta [A] = [A]_2 - [A]_1$ is negative (concentration drops). The rate itself must be positive (speed is never negative). So we multiply by $-1$ .

For product B (appearing):

\text{Average rate} = +\frac{\Delta [B]}{\Delta t}

Here $\Delta [B]$ is positive, so no minus sign needed.

4. The general formula for any reaction

For a balanced reaction:

aA + bB \rightarrow cC + dD

The average rate is defined per mole of reaction — so it’s the same number regardless of which species you track.

We divide each $\frac{\Delta [\text{species}]}{\Delta t}$ by its stoichiometric coefficient:

\text{Average rate} = -\frac{1}{a} \frac{\Delta [A]}{\Delta t} = -\frac{1}{b} \frac{\Delta [B]}{\Delta t} = +\frac{1}{c} \frac{\Delta [C]}{\Delta t} = +\frac{1}{d} \frac{\Delta [D]}{\Delta t}

Why divide by the coefficient?

If 2 moles of A disappear for every 1 mole of C formed, then $\frac{\Delta [A]}{\Delta t}$ is twice as large as $\frac{\Delta [C]}{\Delta t}$ . Dividing by the coefficient normalises them to the same "per mole of reaction" rate.

5. Key exam point: the formula in one line

For any species $X$ with stoichiometric coefficient $\nu_X$ (negative for reactants, positive for products):

$\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}$ …

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A unimolecular reaction involves only one molecule in the elementary step, so its molecularity is 1. For such an elementary step, the rate law follows directly from its molecularity, making the order also 1. Options (i), (ii), and (iv) are correct; (iii) is wrong because order cannot be zero for a unimolecular elementary step.

The key to this question lies in understanding two distinct but related ideas: molecularity and order. Molecularity is a property of an elementary step — it counts how many molecules must collide to make that step happen. Order is an experimental quantity, the sum of exponents in the rate law. For an elementary step, the order equals the molecularity because the rate law is written directly from the stoichiometry of that step.

A unimolecular reaction means exactly one molecule undergoes a change (like a decomposition or rearrangement). Think of a radioactive decay or the isomerisation of cyclopropane to propene — only one molecule is involved in the event. That single molecule is the only reacting species in that step.

Now let’s walk through each option carefully.

Option (i): "only one reacting species is involved in the rate determining step."

The rate-determining step is the slowest elementary step in a mechanism. If the overall reaction is unimolecular, the slowest step must itself be unimolecular — meaning exactly one molecule participates in it. So this statement is correct. It doesn’t say the whole reaction has only one reactant; it says the rate-determining step involves only one reacting species. That is the very definition of a unimolecular step.
Option (ii): "the order and the molecularity of slowest step are equal to one."

For an elementary step, molecularity equals the number of molecules in that step. For a unimolecular step, molecularity = 1. The rate law for that step is $\text{rate} = k[\text{A}]$ , so the order is also 1. Hence both are equal to one. This is a direct consequence of the definition of an elementary reaction.

Tip

A common shortcut: for any elementary step, order = molecularity. So if molecularity is 1, order is automatically 1 — no need to derive.
Option (iii): "the molecularity of the reaction is one and order is zero." …

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The Correct Answer

For a unimolecular reaction (like radioactive decay or isomerisation):

Molecularity = 1 (only one molecule decomposes or rearranges).
Order is usually 1 (rate ∝ concentration of that one species).

So the correct options are:

(i) only one reacting species is involved in the rate determining step.
(ii) the order and the molecularity of slowest step are equal to one.
(iv) both molecularity and order of the reaction are one.

Option (iii) is wrong — molecularity cannot be one and order zero for a unimolecular reaction.

Common Mistakes & How to Avoid Them

✗ Mistake 1: Confusing molecularity with order

What students do:

Think “unimolecular” always means order = 1. That’s usually true, but they forget that molecularity is a theoretical concept (number of molecules in the rate-determining step), while order is experimental.

How to avoid:

Molecularity = count of molecules in the slowest step (always a small integer, never zero).
Order = sum of exponents in the rate law (can be fractional, zero, or negative).
For a unimolecular elementary step, both are 1 — but if the reaction is complex, order may differ.

✗ Mistake 2: Thinking molecularity can be zero

What students do:

Pick option (iii) because they’ve seen zero-order reactions (e.g., decomposition on a surface). They wrongly assume a unimolecular reaction can have order zero.

How to avoid:

Molecularity is never zero — it’s the number of molecules colliding in the slow step.
Zero-order means rate is independent of concentration — that happens in catalysed or surface reactions, not in a simple unimolecular step.

✗ Mistake 3: Forgetting that “unimolecular” refers to the rate-determining step

What students do:

Assume the whole reaction is unimolecular, even if the slow step involves more than one molecule.

How to avoid:

A reaction is called unimolecular only if its rate-determining step involves one molecule.
Always check: “Which step is slowest? How many molecules are in it?”

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Q.In any unimolecular reaction ______________. (Two or more than two options may be correct.)

Concept understanding — Average Rate Of Reaction

Reaction Rate Stoichiometry – From Intuition to Precision

The Intuitive Picture

The Precise Statement

How to Use It (Step by Step)

Why this formula?

Average Rate of Reaction — Why the Formula Holds

1. What does "rate of reaction" mean physically?

2. The core idea: change per unit time

3. Applying this to a reaction

For reactant A (disappearing):

For product B (appearing):

4. The general formula for any reaction

5. Key exam point: the formula in one line

Method: Conceptual Elimination Using Definitions of Molecularity and Order

Step 1 — Recall the definitions

Step 2 — Apply to each option

The Correct Answer

Common Mistakes & How to Avoid Them

✗ Mistake 1: Confusing molecularity with order

✗ Mistake 2: Thinking molecularity can be zero

✗ Mistake 3: Forgetting that “unimolecular” refers to the rate-determining step

Q.In any unimolecular reaction ______________. (Two or more than two options may be correct.)

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