Answer on the basis of the following reaction (species labelled (a)–(d) as printed in the Exemplar): $\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3CH(Cl)CH_2CH_3} \rightarrow \underset{(c)}{CH_3CH(OH)CH_2CH_3} + \underset{(d)}{Cl^-}}$ (2-chlorobutane; in the printed diagram the central carbon of (b) carries $\mathrm{CH_3}$ up, $\mathrm{C_2H_5}$ on a hashed bond to the left, $\mathrm{Cl}$ in plane to the right and H on a wedge — and the product (c) is drawn with exactly the same arrangement: $\mathrm{C_2H_5}$ still hashed, H still on the wedge, and $\mathrm{OH}$ in plane in the position $\mathrm{Cl}$ occupied, i.e. the drawn configuration is unchanged.) Which of the following statements are correct about the kinetics of this reaction? (Two or more than two options may be correct.) (i) The rate of reaction depends on the concentration of only (b). (ii) The rate of reaction depends on concentration of both (a) and (b). (iii) Molecularity of reaction is one. (iv) Molecularity of reaction is two.

Answer on the basis of the following reaction (species labelled (a)–(d) as printed in the Exemplar): mathrmund

NCERT Exemplar · Q36

\mathrm{\underset{(a)}{HO^-} + \underset{(b)}{CH_3CH(Cl)CH_2CH_3} \rightarrow \underset{(c)}{CH_3CH(OH)CH_2CH_3} + \underset{(d)}{Cl^-}}

(b) carries

\mathrm{CH_3}

up,

\mathrm{C_2H_5}

on a hashed bond to the left,

\mathrm{Cl}

in plane to the right and H on a wedge — and the product

\mathrm{C_2H_5}

still hashed, H still on the wedge, and

\mathrm{OH}

in plane in the position

\mathrm{Cl}

occupied, i.e. the drawn configuration is unchanged.)
Which of the following statements are correct about the kinetics of this reaction? (Two or more than two options may be correct.)

(i) The rate of reaction depends on the concentration of only (b).

(ii) The rate of reaction depends on concentration of both

(a) and (b).

(iii) Molecularity of reaction is one.

(iv) Molecularity of reaction is two.

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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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The drawn reaction of the Q.35/36 cluster proceeds by $\mathrm{S_N1}$ (the printed product keeps the substrate's configuration, which a concerted $\mathrm{S_N2}$ attack could never give). $\mathrm{S_N1}$ kinetics follow from that: the slow step involves only the substrate, so $\text{Rate} = k[\text{(b)}]$ and the molecularity is one. The correct options are (i) and (iii).

1. Carry the mechanism over from the drawing

This question shares its reaction — and its printed diagram — with the mechanism question just before it (the Exemplar's Q.35). The drawing shows the product (c) with exactly the same spatial arrangement as the substrate (b): $\mathrm{C_2H_5}$ on the hashed bond, H on the wedge, and $\mathrm{OH}$ in the in-plane position $\mathrm{Cl}$ occupied. A concerted $\mathrm{S_N2}$ attack must invert the carbon, so the drawn outcome rules $\mathrm{S_N2}$ out; the depicted pathway is $\mathrm{S_N1}$ , through a planar carbocation intermediate. The kinetics question must be answered for that mechanism.

2. Write the two steps and find the rate-determining step

\text{Step 1 (slow):}\quad \mathrm{CH_3CH(Cl)CH_2CH_3} \longrightarrow \mathrm{CH_3\overset{+}{C}HCH_2CH_3} + \mathrm{Cl^-}

\text{Step 2 (fast):}\quad \mathrm{CH_3\overset{+}{C}HCH_2CH_3} + \mathrm{OH^-} \longrightarrow \mathrm{CH_3CH(OH)CH_2CH_3}

The slow, rate-determining step is the ionisation of the C–Cl bond — and it involves only the substrate (b). The hydroxide ion (a) reacts after the bottleneck, so its concentration does not appear in the rate law.

\text{Rate} = k\,[\text{(b)}]

Watch out

Don't let " $\mathrm{OH^-}$ is a strong nucleophile" pull you toward a bimolecular rate law here. However strong the nucleophile is, in an $\mathrm{S_N1}$ reaction it only captures a carbocation that has already formed — speeding up a step that is not rate-determining changes nothing in the measured rate.

3. Molecularity …

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Common Mistakes Students Make on This Question

Mistake 1: Assuming $\mathrm{S_N2}$ because a strong nucleophile is present

The mistake: Students see $\mathrm{HO^-}$ + an alkyl halide and reflexively write bimolecular kinetics — rate depending on both (a) and (b), molecularity two.

Why it's wrong: The printed diagram for this reaction cluster shows the product with its configuration unchanged — the same wedge/hash arrangement as the substrate, with OH simply occupying Cl's in-plane position. A concerted $\mathrm{S_N2}$ attack must invert the carbon, so the reaction depicted is $\mathrm{S_N1}$ . In $\mathrm{S_N1}$ the nucleophile enters only after the slow ionisation step, so its concentration never appears in the rate law: $\text{Rate} = k[\text{(b)}]$ .

How to avoid: Decide the mechanism first — from the drawing, not from the reagent alone — and only then write the kinetics.

Mistake 2: Confusing molecularity with the overall stoichiometry

The mistake: The overall equation shows two reactants, so students conclude "molecularity = 2" and tick option (iv).

Why it's wrong: Molecularity is counted for the rate-determining elementary step, not the overall equation. In $\mathrm{S_N1}$ , the slow step is the ionisation of the substrate alone — one species — so the molecularity is one, even though the overall reaction consumes both (a) and (b).

How to avoid: Always split the mechanism into elementary steps, find the slow one, and count the species in that step only.

Mistake 3: Answering the mechanism and kinetics questions inconsistently

The mistake: Treating this kinetics question in isolation from its companion mechanism question about the same drawn reaction — e.g. calling the mechanism $\mathrm{S_N1}$ there but writing second-order kinetics here.

Why it's wrong: The two questions describe one reaction. If the mechanism is $\mathrm{S_N1}$ (carbocation intermediate), the kinetics must be unimolecular/first order: (i) and (iii). Mixed answers contradict themselves.

How to avoid: For paired questions on one reaction, fix the mechanism once and derive everything else from it.

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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

1. Carry the mechanism over from the drawing

2. Write the two steps and find the rate-determining step

3. Molecularity …

Method: Kinetics from the mechanism (not from the overall equation)

Step 1 – Fix the mechanism from the printed drawing

Step 2 – Identify the rate-determining step

Step 3 – Write the rate law

Step 4 – Count the molecularity …

Common Mistakes Students Make on This Question

Mistake 1: Assuming $\mathrm{S_N2}$ because a strong nucleophile is present

Mistake 2: Confusing molecularity with the overall stoichiometry

Mistake 3: Answering the mechanism and kinetics questions inconsistently

1. Carry the mechanism over from the drawing

2. Write the two steps and find the rate-determining step

3. Molecularity …

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