In the presence of a catalyst, the heat evolved or absorbed during the reaction ______. (i) increases. (ii) de

NCERT Exemplar · Q2

Q. In the presence of a catalyst, the heat evolved or absorbed during the reaction ___________.

(i) increases.

(ii) decreases.

(iii) remains unchanged.

(iv) may increase or decrease.

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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.

Common Pitfall to Avoid

Watch out

Do not write $\frac{d[\text{reactant}]}{dt}$ as a positive number and then forget the minus sign. The rate of change of a reactant is negative (concentration falls). The minus sign in the definition flips it to a positive $r$ . If you skip the sign, you will get the wrong magnitude for other species.

Why This Matters

In exams (JEE, NEET, etc.), you will often be given the rate for one species and asked to find the rate for another. The stoichiometric relation is the only tool you need — no extra formulas. It also appears in more advanced topics like the rate law (where the exponents are not the coefficients) — but that is a separate concept. Reaction rate stoichiometry is purely about the definition of the reaction rate itself.

Final takeaway: The coefficients in the balanced equation are the conversion factors between the rates of different species. Always normalise by dividing by the coefficient to get the universal reaction rate $r$ .

Average rate of reaction is one of the first ideas introduced in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘average rate of reaction formula’ or ‘average vs instantaneous rate’ are common important-question topics in board exams and JEE Main chemistry. A solid grasp of this basic definition is also assumed in nearly every subsequent kinetics numerical asked in NEET and state CETs.

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}

$\nu_X$ is negative for reactants → the minus sign is already built in.
$\nu_X$ is positive for products.

6. Why this is the average rate (not instantaneous)

Average rate uses a finite $\Delta t$ — it’s the slope of the chord between two points.
Instantaneous rate uses $\Delta t \rightarrow 0$ — it’s the slope of the tangent at a single point.

The average rate formula is just the discrete version of the derivative:

\text{Instantaneous rate} = \frac{1}{\nu_X} \frac{d[X]}{dt}

Summary — the "why" in one sentence

The average rate formula holds because it measures change in concentration per unit time, uses a minus sign to keep rates positive for reactants, and divides by stoichiometric coefficients to give a single, comparable value for the whole reaction.

Always remember:

$\Delta [\text{reactant}]$ is negative → minus sign makes it positive.
$\Delta [\text{product}]$ is positive → no minus sign.
Divide by coefficient → normalise to "per mole of reaction".

A catalyst speeds up a reaction by lowering the activation energy, but it does not change the overall enthalpy change ( $\Delta H$ ) of the reaction. The heat evolved or absorbed remains unchanged.

The key idea here is simple but often misunderstood: a catalyst affects the path of a reaction, not its destination. Let’s unpack why.

The Concept: What a Catalyst Actually Does

A catalyst provides an alternative reaction pathway with a lower activation energy. This means more reactant molecules have enough energy to cross the energy barrier per unit time, so the reaction rate increases.

But here’s the crucial point: the initial and final states of the reaction — the reactants and products — are exactly the same with or without the catalyst. The catalyst is not consumed and does not appear in the overall balanced equation.

The heat evolved or absorbed in a reaction is the enthalpy change, $\Delta H = H_{\text{products}} - H_{\text{reactants}}$ . Since the reactants and products are identical in both the catalysed and uncatalysed reactions, $\Delta H$ must be the same.

Watch out

A common mistake is to think that because a catalyst lowers the activation energy, it also changes the heat released or absorbed. That’s false — activation energy and enthalpy change are completely different quantities. Activation energy is the barrier height; enthalpy change is the net energy difference between start and finish.

Step-by-Step Reasoning

Identify what the question asks

The “heat evolved or absorbed” is the enthalpy change of the reaction, $\Delta H$ . This is a thermodynamic property, not a kinetic one.
Recall the role of a catalyst

A catalyst speeds up the reaction by lowering the activation energy ( $E_a$ ) for both the forward and reverse reactions equally. It does not alter the energies of the reactants or products themselves.
Visualise the energy profile

Draw an energy diagram: the reactants are at some energy level, the products at another. The catalyst lowers the peak (the transition state) but leaves the two flat ends exactly where they were. The vertical drop (or rise) from reactants to products — that’s $\Delta H$ — stays the same.
Apply the principle

Since $\Delta H$ depends only on the initial and final states, and those are unchanged, the heat evolved or absorbed remains unchanged.

Tip

Think of a catalyst as a tunnel through a mountain instead of a path over the top. The tunnel is faster, but you start and end at the same two towns — the altitude difference between the towns hasn’t changed.

✓Final answer

The heat evolved or absorbed during the reaction remains unchanged . The correct option is (iii) .

Here is the breakdown of the common mistakes students make on this concept, along with how to avoid them.

The Core Concept

A catalyst provides an alternative pathway (mechanism) for the reaction. This new pathway has a lower activation energy ( $E_a$ ), which is why the reaction speeds up.

Crucially, a catalyst does not change the initial and final states of the reactants and products. Since the enthalpy change ( $\Delta H$ ) depends only on the difference in energy between these initial and final states (Hess's Law), the catalyst cannot change $\Delta H$ .

Therefore, the correct answer is (iii) remains unchanged.

Common Mistake #1: Confusing Rate with Energy Change

The Mistake: Students think that because a catalyst makes a reaction happen faster, it must also change how much heat is released or absorbed. They assume "faster" means "more" or "less" heat.
Why it happens: The word "catalyst" is often associated with "speed" in everyday language. Students fail to separate the kinetics (how fast) from the thermodynamics (how much energy).
How to Avoid:
- Draw the Energy Profile Diagram: Always sketch the reaction coordinate diagram. Draw the curve for the uncatalyzed reaction (high peak) and the catalyzed reaction (lower peak). Notice that the starting point (reactants) and ending point (products) are at the exact same energy levels on both curves.
- Remember the Definition: $\Delta H = H_{products} - H_{reactants}$ . The catalyst does not change $H_{products}$ or $H_{reactants}$ .
- Key Phrase: "A catalyst affects the path, not the destination."

Common Mistake #2: Thinking a Catalyst Absorbs or Releases Heat

The Mistake: Students believe the catalyst itself participates in the reaction by absorbing heat (making it "less exothermic") or releasing heat (making it "more exothermic").
Why it happens: Some students know that catalysts can be involved in the reaction mechanism (e.g., forming an intermediate) and assume this involvement changes the overall energy balance.
How to Avoid:
- The "Regeneration" Rule: A catalyst is chemically unchanged at the end of the reaction. If it were to absorb or release a net amount of heat, its own chemical structure or energy state would have to change permanently. Since it is regenerated, its net energy contribution to the system is zero.
- Think of a "Middleman": A catalyst is like a middleman who helps two people trade goods. The middleman facilitates the trade but doesn't keep any of the goods or money. The net value of the trade is the same whether the middleman is there or not.

Common Mistake #3: Confusing Catalyst with an "Initiator" or "Igniter"

The Mistake: Students think a catalyst "starts" a reaction that otherwise wouldn't happen, and therefore must supply the initial energy (heat) to get it going.
Why it happens: This is a confusion between a catalyst and an initiator (like a spark plug or a match). An initiator provides the initial activation energy and is consumed in the process.
How to Avoid:
- Compare and Contrast:
  - Initiator: Provides energy, gets consumed, changes $\Delta H$ of the overall process (e.g., burning a match to start a fire adds the match's energy to the system).
  - Catalyst: Lowers the energy barrier, is not consumed, does not change $\Delta H$ .
- The "Free Pass" Analogy: A catalyst is like a "free pass" that lowers the entrance fee to a concert. It doesn't change the price of the ticket (the $\Delta H$ ), it just makes it easier to get in (lowers $E_a$ ).

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 $N_2O_5 = -\dfrac{\Delta[N_2O_5]}{\Delta t} = -\dfrac{-0.5}{100} = 5\times10^{-3}\ mol\,L^{-1}min^{-1}$ .

Rate of the reaction $= \dfrac{1}{2}\times(\text{rate of disappearance of } N_2O_5) = \dfrac{1}{2}\times 5\times10^{-3} = 2.5\times10^{-3}\ mol\,L^{-1}min^{-1}$ .

Common Mistakes

Reporting the raw rate of disappearance of $N_2O_5$ ( $5\times10^{-3}$ ) without dividing by its coefficient 2 -- that gives option (A), a common trap.

Sign errors: forgetting the negative sign convention for a reactant, which doesn't change the magnitude here but can confuse students on which direction is positive.

✓Final answer
$2.5\times10^{-3}$

ANSWER: B
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: $\dfrac{|x-y|}{100}$ .

This has correct units of concentration/time (mol L $^{-1}$ min $^{-1}$ ), unlike the other options which either invert the ratio or square a term.

Common Mistakes

Inverting the fraction (time over concentration change) — this gives the wrong units.

Squaring the concentration difference, which is dimensionally and physically meaningless here.

✓Final answer
$\dfrac{|x-y|}{100}$

ANSWER: A
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 $=\dfrac{-\Delta[\mathrm{R}]}{\Delta t}=\dfrac{0.01}{2400}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

$\dfrac{0.01}{2400}=4.1667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

This matches option (B).

Common Mistakes

Leaving time in minutes instead of converting to seconds, which would give $2.5\times10^{-4}$ (option A) — a common careless-unit trap built into the distractors.

Forgetting the negative sign convention or misplacing a decimal, landing on $2.5\times10^{-5}$ (option D) instead of the correctly converted value.

✓Final answer
$4.167\times10^{-6}$

ANSWER: B
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 $= \dfrac{0.01}{1500} = 6.667\times10^{-6}\ \mathrm{mol\,L^{-1}\,s^{-1}}$ .

Common Mistakes

Forgetting to convert minutes to seconds (would give $4\times10^{-4}$ , a distractor option).

Confusing this simple average-rate calculation with computing the first-order rate constant $k$ (which needs the log form) — the question only asks for the rate, not $k$ .

✓Final answer
$6.667\times10^{-6}$

ANSWER: A
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).

The mention of 'first order' is a distractor here — the rate-from-tangent relationship on a $[A]$ vs $t$ graph does not depend on reaction order; that information would only matter if the graph or question asked for the rate constant $k$ .

So the instantaneous rate at C is simply $m$ , with no $2.303$ factor needed.

Common Mistakes

Applying the $2.303$ conversion factor here — that factor belongs to $\log[A]$ vs $t$ plots used for finding $k$ in first-order kinetics, not to reading the rate off a plain $[A]$ vs $t$ curve.

Being distracted by 'first order' into thinking the answer must involve the rate constant relation $k = \dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ , which is unrelated to reading a tangent slope.

✓Final answer
$The correct option is (B) — m.$

ANSWER: B
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 $M\,s^{-1}$ .

So statement (D), which claims $M\,s^{-1}$ for a first-order $k$ , is false — that unit actually belongs to a zero-order reaction ( $n=0 \Rightarrow M^{1}s^{-1}$ ).

Common Mistakes

Confusing the zero-order rate-constant unit ( $M\,s^{-1}$ ) with the first-order one ( $s^{-1}$ ).

Thinking rate constant units are always the same as rate units.

✓Final answer
$k$

ANSWER: D

←Previous questionQ1 — The role of a catalyst is to change _______. (i) gibbs energy of react…→Next questionQ3 — Activation energy of a chemical reaction can be determined by _______.…

Q.In the presence of a catalyst, the heat evolved or absorbed during the reaction ___________.

The Concept: What a Catalyst Actually Does

Step-by-Step Reasoning

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