Assertion: Order and molecularity are same. Reason: Order is determined experimentally and molecularity is the

NCERT Exemplar · Q63

Q. Assertion: Order and molecularity are same. Reason: Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.

(i) Both assertion and reason are correct and the reason is correct explanation of assertion.

(ii) Both assertion and reason are correct but reason does not explain assertion.

(iii) Assertion is correct but reason is incorrect.

(iv) Both assertion and reason are incorrect.

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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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The assertion is FALSE — order and molecularity are not, in general, the same. The reason, however, is actually a TRUE statement (it correctly defines order as experimental and molecularity as the elementary-step coefficient sum). This is an "Assertion incorrect, Reason correct" case, which is not represented by any of the four listed options exactly as worded — option (iv) as printed ("Both assertion and reason are incorrect") is not correct, because the reason itself is true. This mismatch needs review of the option text (the standard fourth choice for this assertion–reason format is normally "Assertion is incorrect but reason is correct").

Assertion — "Order and molecularity are same" — is FALSE.

Order is an experimental quantity: the sum of the powers of the concentration terms in the observed rate law. It may be zero, a whole number, or fractional, and depends on the mechanism.
Molecularity is a theoretical quantity defined only for an elementary step: the number of species that come together in that step. It is always a small positive integer (1, 2, or rarely 3).

Order and molecularity coincide only for a single-step (elementary) reaction; in general they differ (e.g. acid-catalysed ester hydrolysis is experimentally first order overall, even though the mechanism is multi-step). So the assertion is wrong. …

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Here’s a breakdown of the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Thinking Order and Molecularity Are Always the Same

The error: Students assume the assertion is correct because both terms describe the number of molecules involved in a reaction. They confuse the definition of molecularity with the experimental nature of order.
Why it’s wrong: Order can be fractional, zero, or negative — molecularity is always a positive integer (1, 2, 3). They are not the same concept.
How to avoid: Memorise the key difference:
- Order: Sum of exponents of concentration terms in the experimentally determined rate law.
- Molecularity: Number of molecules colliding in an elementary step (only for elementary reactions).
- Example: For $2NO + O_2 \rightarrow 2NO_2$ , the order can be 2 (experimentally), but molecularity of the rate-determining step might be 3 (if it’s a single step). They can match only for elementary reactions — not always.

Mistake 2: Misinterpreting the Reason Statement

The error: Students think the reason is correct because “molecularity is the sum of stoichiometric coefficients of the rate-determining step” sounds familiar. They forget that molecularity is defined only for an elementary step, not for the overall reaction.
Why it’s wrong: The reason says “sum of stoichiometric coefficients of rate determining elementary step” — this is correct for molecularity. But the assertion is false, so the reason cannot explain it.
How to avoid: Always check the truth value of each statement separately before linking them. Write:
- Assertion: False (order ≠ molecularity in general).
- Reason: True (molecularity = sum of coefficients in the rate-determining step).
- Then decide: Since assertion is false, options (i), (ii), (iii) are all invalid. Only (iv) fits.

Mistake 3: Confusing “Correct Explanation” with “Correct Statement”

The error: Students see both statements as individually correct and automatically pick (i) or (ii). They forget to check whether the reason actually explains the assertion.
Why it’s wrong: Even if both were true (which they aren’t here), the reason does not explain why order and molecularity are the same — it actually highlights a difference (experimental vs theoretical).
How to avoid: After verifying truth values, ask: “Does the reason logically lead to the assertion?” If not, even if both are true, the answer is (ii), not (i).

Mistake 4: Forgetting That Order Is Experimental, Molecularity Is Theoretical

The error: Students treat both as theoretical concepts derived from the balanced equation.
Why it’s wrong: Order comes from experimental data (e.g., initial rate method). Molecularity comes from mechanism (theoretical).
How to avoid: Remember the mnemonic: Order = Observed (experiment), Molecularity = Mechanism (theory). If a question says “order is determined experimentally,” that part is always correct. …

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Q.Assertion: Order and molecularity are same.
Reason: Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.

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: Assertion–Reason Analysis

Steps

Mistake 1: Thinking Order and Molecularity Are Always the Same

Mistake 2: Misinterpreting the Reason Statement

Mistake 3: Confusing “Correct Explanation” with “Correct Statement”

Mistake 4: Forgetting That Order Is Experimental, Molecularity Is Theoretical

Q.Assertion: Order and molecularity are same. Reason: Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.

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Q.Assertion: Order and molecularity are same.
Reason: Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.

Unlock everything free for 14 days