Match the items of Column I and Column II. Column I: (i) Diamond (ii) Instantaneous rate (iii) Average rate Co

NCERT Exemplar · Q60

Q. Match the items of Column I and Column II. Column I:

(i) Diamond

(ii) Instantaneous rate

(iii) Average rate
Column II:

(a) short interval of time

(b) ordinarily rate of conversion is imperceptible

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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 key idea is to match each term in Column I with its most natural description in Column II based on the time interval over which the rate is measured. Diamond reacts so slowly that its rate is imperceptible; instantaneous rate is measured over a very short interval; average rate is measured over a long duration. The correct matches are: (i) → (b), (ii) → (a), (iii) → (c).

The question asks you to match items from Column I (three terms related to reaction rates) with their appropriate descriptions in Column II. This is a conceptual matching problem, not a calculation. The core idea is understanding what each term means in the context of chemical kinetics.

Average rate of reaction is the change in concentration of a reactant or product over a finite, measurable time interval — typically a long duration, like minutes or hours. It gives you a single, overall speed for that entire period.

Instantaneous rate of reaction is the rate at a specific moment in time. To find it, you need to measure the change in concentration over an extremely short interval — so short that the rate is essentially constant during that interval. In practice, this is done by drawing a tangent to the concentration vs. time curve at the point of interest.

Diamond is a special case. Diamond is an allotrope of carbon that is thermodynamically unstable under standard conditions — it should slowly convert to graphite. However, the rate of this conversion is so incredibly slow at room temperature that it is effectively zero on human timescales. You cannot measure any change in concentration over any practical time interval. The rate is imperceptible.

Now, let's match them step by step.

Match (i) Diamond.

Diamond's conversion to graphite is so slow that no change is observed over any ordinary time period. The description that fits this is "ordinarily rate of conversion is imperceptible."

So, (i) matches with (b).
Match (ii) Instantaneous rate. …

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Here is a breakdown of the common mistakes students make when matching these items, along with the correct reasoning.

The Core Concept: Rate of Reaction

This question tests your understanding of reaction kinetics, specifically the difference between average rate and instantaneous rate, and how the physical state of a reactant (like a solid diamond) affects the observable rate.

Average Rate: The change in concentration over a long duration of time (e.g., the entire reaction).
Instantaneous Rate: The rate at a specific moment, measured over a very short interval of time (approaching zero).
Diamond: A solid with a very rigid, covalent structure. Its reaction rate (e.g., combustion) is imperceptible under normal conditions.

Common Mistake #1: Confusing "Short Interval" with "Instantaneous"

The Mistake:

Students often match (i) Diamond with (a) short interval of time or (c) long duration of time, thinking that because a diamond is "hard" or "takes a long time to form," it must be matched with a time interval.

Why it's wrong:

The column items (a), (b), and (c) describe the method of measurement or the observable nature of the rate, not the physical properties of the substance. A diamond's rate is not measured over a short or long interval; it is simply not measurable at room temperature.

How to Avoid:

Focus on the verb: The key phrase is "rate of conversion is imperceptible." This means you cannot see or measure any change. This is a direct property of diamond's stability.
Think about measurement: Ask yourself: "Can I measure the rate of diamond turning into graphite at room temperature?" The answer is no. Therefore, it matches with (b).

Correct Match: (i) → (b)

Common Mistake #2: Swapping "Average Rate" and "Instantaneous Rate"

The Mistake:

Students match (ii) Instantaneous rate with (c) long duration of time and (iii) Average rate with (a) short interval of time.

Why it's wrong:

This is a direct reversal of the definitions.

Average rate is calculated over the entire reaction or a long duration (e.g., total change in concentration divided by total time).
Instantaneous rate is the slope of the tangent at a single point, which requires a very short interval (theoretically, $\Delta t \to 0$ ).

How to Avoid:

Memorize the definitions with a time anchor:
- Average = Long duration (think: "average over the whole journey").
- Instantaneous = Short interval (think: "speedometer reading at one instant").
Use the formula: The instantaneous rate is $-\frac{d[R]}{dt}$ , where $dt$ is an infinitesimally small time interval. The average rate is $-\frac{\Delta[R]}{\Delta t}$ , where $\Delta t$ is a large, measurable time.

Correct Match:

(ii) → (a)
(iii) → (c)

Common Mistake #3: Ignoring the "Ordinarily" Clause

The Mistake: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$
›Reveal solutionSolution
This tests the definition of "rate of reaction" that is normalised by stoichiometric coefficients so it gives the same value regardless of which species you track. The rate works out to $2.5\times10^{-3}\ mol\,L^{-1}min^{-1}$ .

Concept and Intuition

For a general reaction $aA \rightarrow bB + cC$ , different species are consumed/produced at different numerical speeds proportional to their coefficients. To get one unambiguous rate of reaction, each species' rate of change is divided by its own coefficient (with a minus sign for reactants, since their concentration falls): $\text{Rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = \dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt}$ .

Step-by-Step Solution

Reaction: $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ , so the coefficient of $N_2O_5$ is 2.

Concentration of $N_2O_5$ changes from $2$ to $1.5\ mol\,L^{-1}$ , so $\Delta[N_2O_5] = 1.5 - 2 = -0.5\ mol\,L^{-1}$ , over $\Delta t = 100$ min.

Average rate of disappearance of $N_2O_5 = -\dfrac{\Delta[N_2O_5]}{\Delta t} = -\dfrac{-0.5}{100} = 5\times10^{-3}\ mol\,L^{-1}min^{-1}$ . …
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AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
$\mathrm{A \rightarrow B}$
›Reveal solutionSolution
This tests the basic definition of average rate of reaction as the change in concentration of a reactant over the time interval.

Concept and Intuition

The average rate of a reaction over a time interval is simply how much the concentration of a reactant (or product) changes, divided by how long it took — with a sign convention so the rate is always reported as a positive quantity.

Step-by-Step Solution

Average rate of disappearance of A $= -\dfrac{\Delta[A]}{\Delta t} = -\dfrac{[A]_{final} - [A]_{initial}}{t}$ .

Here $[A]$ decreases from $x$ to $y$ over $t = 100$ min, so $\Delta[A] = y - x$ (negative, since $y < x$ ).

Average rate $= -\dfrac{y-x}{100} = \dfrac{x-y}{100}$ , and writing it generally (regardless of which is larger) as a positive quantity: $\dfrac{|x-y|}{100}$ . …
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AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
$\mathrm{R \longrightarrow P}$
›Reveal solutionSolution
Average rate is simply the change in concentration of reactant divided by the time elapsed (converted to seconds); the 'first order' label is not even needed for this particular calculation. Answer: (B).

Concept and Intuition

The average rate of a reaction over a finite time interval is defined purely from the measured change in concentration and elapsed time — it does not require knowing the rate law or order of the reaction (order only matters for finding the instantaneous rate constant $k$ via the integrated rate law). Here we are only asked for the average velocity, so a direct $\Delta[\text{conc}]/\Delta t$ calculation suffices, with careful unit conversion from minutes to seconds since the answer must be in $\mathrm{mol\,L^{-1}\,s^{-1}}$ .

Step-by-Step Solution

Change in concentration of R: $\Delta[\mathrm{R}]=0.04-0.03=0.01\ \mathrm{mol\,L^{-1}}$ (a decrease, since R is being consumed).

Time elapsed: $40$ min $=40\times 60=2400$ s.

Average rate $=\dfrac{-\Delta[\mathrm{R}]}{\Delta t}=\dfrac{0.01}{2400}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ . …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$L^{-1}$
›Reveal solutionSolution
Rate is simply the drop in concentration divided by the elapsed time (converted to seconds) — $6.667\times10^{-6}\ \mathrm{mol\,L^{-1}s^{-1}}$ , answer (A).

Concept and Intuition

The (average) rate of a reaction over a time interval is defined as the change in concentration of a reactant divided by the time elapsed (with a negative sign convention for reactants, but the magnitude is what's asked here). Care must be taken to convert all times to consistent units (here, minutes to seconds, since the rate is required in $\mathrm{s^{-1}}$ ).

Step-by-Step Solution

Concentration drop: $\Delta[\text{reactant}] = 0.03 - 0.02 = 0.01\ \mathrm{mol\,L^{-1}}$ .

Time elapsed: $25\ \mathrm{min} = 25\times 60 = 1500\ \mathrm s$ .

Average rate $= \dfrac{0.01}{1500} = 6.667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ . …
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AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
$\mathrm{A} \rightarrow \mathrm{P}$
›Reveal solutionSolution
On a plain concentration-vs-time graph, the instantaneous rate at any point is just the magnitude of the tangent's slope there — here that is simply $m$ , regardless of the reaction being first order.

Concept and Intuition

The instantaneous rate of a reaction $A \to P$ is defined purely graphically as $\text{Rate} = -\dfrac{d[A]}{dt}$ , which is exactly the negative of the slope of the concentration-vs-time curve at that instant. This definition holds for a reaction of ANY order — it is a direct consequence of the definition of rate, not something that depends on the rate law. The factor of $2.303$ only enters when working with $\log_{10}[A]$ vs $t$ plots (used to extract the first-order rate constant $k$ from the slope $-k/2.303$ ) — it is irrelevant here since the axes are plain concentration and time, not logarithmic concentration.

Step-by-Step Solution

The graph plots $[A]$ (concentration of A) on the y-axis against time on the x-axis — a decaying curve, as expected since A is being consumed.

At point C, a tangent line is drawn with slope magnitude $m$ (the curve is decreasing, so the true signed slope is $-m$ , but this magnitude labelled $m$ represents how fast concentration is falling at that instant).

By definition, instantaneous rate $= -\dfrac{d[A]}{dt}\Big|_C = m$ (taking the tangent's slope magnitude as the rate). …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$M\ s^{-1}$
›Reveal solutionSolution
This tests the general rule "unit of $k$ = $(mol\,L^{-1})^{1-n} s^{-1}$ " for a reaction of order $n$ . The false statement is (D): first order $k$ has units $s^{-1}$ , not $M\,s^{-1}$ .

Concept and Intuition

For a reaction of order $n$ : $\text{rate} = k[A]^n$ , and rate always has units $M\,s^{-1}$ . Rearranging for $k$ :
$k = \dfrac{\text{rate}}{[A]^n} \;\Rightarrow\; \text{units of }k = (M)^{1-n} s^{-1}$
So the units of $k$ change with order — this is exactly why chemists use the units of an experimentally measured $k$ to identify the reaction order.

Step-by-Step Solution

Rate of reaction / rate of disappearance of reactant: both are $\dfrac{\Delta[\text{conc}]}{\Delta t}$ , units $M\,s^{-1}$ — (A), (B) correct.

Units of $k$ depend on order: for order $n$ , units are $M^{1-n}s^{-1}$ — (C) correct.

For $n=1$ (first order): units of $k = M^{1-1}s^{-1} = M^0 s^{-1} = s^{-1}$ , not $M\,s^{-1}$ . …
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Q.Match the items of Column I and Column II.
Column I:

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

Concept: Rates of Chemical Reactions

Method: Conceptual Matching by Time Interval

Steps

The Core Concept: Rate of Reaction

Common Mistake #1: Confusing "Short Interval" with "Instantaneous"

Common Mistake #2: Swapping "Average Rate" and "Instantaneous Rate"

Common Mistake #3: Ignoring the "Ordinarily" Clause

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Match the items of Column I and Column II. Column I:

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Q.Match the items of Column I and Column II.
Column I:

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