Which of the following statements are applicable to a balanced chemical equation of an elementary reaction? (T

NCERT Exemplar · Q47

Q. Which of the following statements are applicable to a balanced chemical equation of an elementary reaction? (Two or more than two options may be correct.)

(i) Order is same as molecularity.

(ii) Order is less than the molecularity.

(iii) Order is greater than the molecularity.

(iv) Molecularity can never be zero.

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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.

--- …

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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For an elementary reaction, the rate law is directly given by the stoichiometric coefficients, so order equals molecularity. Molecularity is always a positive integer (≥1) and can never be zero. Therefore, statements (i) and (iv) are correct.

Let’s first get the core idea straight. In chemical kinetics, the order of a reaction is an experimentally determined number — it tells you how the rate depends on the concentration of each reactant. The molecularity of a reaction, on the other hand, is a theoretical concept: it’s the number of molecules (or atoms, ions) that must collide simultaneously for an elementary step to occur. For an elementary reaction (a single-step process), the rate law is written directly from the balanced equation — the exponents are the stoichiometric coefficients. That means order and molecularity are the same number. Also, since molecularity counts actual colliding particles, it can only be a positive integer (1, 2, 3…). Zero makes no physical sense — you can’t have a reaction with zero molecules colliding.

Now let’s examine each option carefully.

Option (i): Order is same as molecularity. For an elementary reaction, the rate law is:

\text{Rate} = k [A]^a [B]^b \dots

where $a, b, \dots$ are the stoichiometric coefficients of the reactants in the balanced elementary equation. The sum $a + b + \dots$ is the molecularity. Since the exponents in the rate law are exactly those coefficients, the order (sum of exponents) equals the molecularity. This is a fundamental property of elementary reactions. So (i) is correct.

Option (ii): Order is less than the molecularity.

This would happen if the rate law had fractional or zero exponents, which is typical for complex (multi-step) reactions — but not for elementary ones. In an elementary step, every reactant molecule participates in the collision, so no exponent can be less than the coefficient. Hence (ii) is false.
Option (iii): Order is greater than the molecularity. …

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Method: Conceptual Analysis of Elementary Reactions

For an elementary reaction (a reaction that occurs in a single step), the molecularity and order have a direct relationship. Let's break this down step-by-step.

Step 1: Understand the definitions

Molecularity: The number of molecules (atoms, ions) that collide in an elementary step. It is always a positive integer (1, 2, 3…). It cannot be zero or fractional.
Order of reaction: The sum of the exponents of concentration terms in the rate law. For an elementary reaction, the rate law is written directly from the balanced equation — the exponents equal the stoichiometric coefficients.

Step 2: Apply the rule for elementary reactions

For an elementary reaction:

Order = Molecularity

This is because the rate law is derived from the molecular collision picture.

Step 3: Evaluate each statement

Statement	Analysis	Correct?
(i) Order is same as molecularity	True for elementary reactions	✓

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

Mistake 1: Assuming order always equals molecularity for elementary reactions

Why it happens:

Many students memorise “for elementary reactions, order = molecularity” as a blanket rule. They pick only option (i) and stop.

How to avoid:

Understand the definition of molecularity: it is the number of molecules (or atoms) that collide in the rate-determining step of an elementary reaction.

For a unimolecular elementary step (e.g., $A \rightarrow \text{products}$ ), molecularity = 1, order = 1.
For a bimolecular step (e.g., $A + B \rightarrow \text{products}$ ), molecularity = 2, order = 2.
For a termolecular step (e.g., $A + B + C \rightarrow \text{products}$ ), molecularity = 3, order = 3.

So order = molecularity is always true for elementary reactions.

But molecularity can never be zero — a reaction step must involve at least one molecule. So option (iv) is also correct.

Key takeaway: For elementary reactions, order = molecularity always, and molecularity ≥ 1. So both (i) and (iv) are correct.

Mistake 2: Thinking order can be less than or greater than molecularity for elementary reactions

Why it happens:

Students confuse elementary reactions with complex (multi-step) reactions.

In complex reactions, the observed order can be fractional, zero, or different from molecularity of the slow step.
But the question explicitly says “balanced chemical equation of an elementary reaction” — so it’s a single step.

How to avoid:

Read the question carefully. If it says “elementary reaction”, then:

Order = molecularity (always).
Molecularity is a positive integer (1, 2, or rarely 3).
Options (ii) and (iii) are never true for elementary reactions.

Memory aid:

“Elementary = one step → order = molecularity. No exceptions.”

Mistake 3: Forgetting that molecularity cannot be zero

Why it happens:

Some students think “zero-order reactions exist, so molecularity can be zero.”

But zero-order is an observed rate law (often due to a catalyst surface saturation), not a molecularity.

How to avoid:

Molecularity is a theoretical concept: the number of molecules that must collide in a single step.
You cannot have a reaction step with zero molecules colliding — that would mean no reaction.
So molecularity is always ≥ 1.

Key fact: …

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Q.Which of the following statements are applicable to a balanced chemical equation of an elementary reaction? (Two or more than two options may be correct.)

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