Which of the following statement is not correct for the catalyst? (i) It catalyses the forward and backward re

NCERT Exemplar · Q18

Q. Which of the following statement is not correct for the catalyst?

(i) It catalyses the forward and backward reaction to the same extent.

(ii) It alters

\Delta G

of the reaction.

(iii) It is a substance that does not change the equilibrium constant of a reaction.

(iv) It provides an alternate mechanism by reducing activation energy between reactants and products.

Karnataka PUCMCQ· 1mImportance★★★★★

57% · 67/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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):

$\text{Average rate} = \frac{1}{\nu_X} \frac{\Delta [X]}{\Delta t}$ …

Start your 14-day free trial to unlock the full solution →

A catalyst speeds up both forward and backward reactions equally, does not change $\Delta G$ or the equilibrium constant, and works by lowering activation energy. The incorrect statement is (ii) — a catalyst never alters $\Delta G$ .

The question asks which statement about a catalyst is not correct. To answer this, you need a clear picture of what a catalyst actually does — and what it does not do.

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. It works by providing an alternative reaction pathway with a lower activation energy. This is a kinetic effect — it changes how fast equilibrium is reached, not where equilibrium lies.

Now, let’s examine each option carefully.

Option (i): "It catalyses the forward and backward reaction to the same extent."

This is correct. A catalyst lowers the activation energy for both the forward and reverse reactions by the same amount. Why? Because the transition state is the same for both directions — lowering its energy speeds up both paths equally. As a result, the equilibrium constant $K$ remains unchanged, and equilibrium is reached faster.
Option (ii): "It alters $\Delta G$ of the reaction."

This is the incorrect statement. $\Delta G$ (Gibbs free energy change) is a thermodynamic quantity — it depends only on the initial and final states of the reactants and products, not on the path taken. A catalyst changes the path (the mechanism) but not the start and end points. So $\Delta G$ stays exactly the same.

Watch out

A common mistake is to think that because a catalyst speeds up a reaction, it must change the energy difference. It does not — it only lowers the activation barrier, not the overall free energy change.
Option (iii): "It is a substance that does not change the equilibrium constant of a reaction." …

Start your 14-day free trial to unlock the full solution →

Concept: Catalysis and Thermodynamics

A catalyst is a substance that increases the rate of a reaction without being consumed. The key is understanding what a catalyst can and cannot change.

Method: Elimination by Thermodynamic Principles

Step 1: Recall the fundamental properties of a catalyst

A catalyst lowers the activation energy ( $E_a$ ) by providing an alternative pathway.
It affects both forward and backward reactions equally.
It does not change the equilibrium constant ( $K_{eq}$ ).
It does not change the thermodynamic state functions like $\Delta G$ , $\Delta H$ , or $\Delta S$ .

Step 2: Evaluate each option

Option	Statement	Correct?
(i)	Catalyses forward and backward reaction equally	✓ Correct — catalyst lowers $E_a$ for both directions by same amount
(ii)	Alters $\Delta G$ of the reaction	✗ Incorrect — $\Delta G$ is a state function; catalyst does not change it
(iii)	Does not change equilibrium constant	✓ Correct — $K_{eq}$ depends only on $\Delta G^\circ$ , not on catalyst

Start your 14-day free trial to unlock the full solution →

Let’s break down the concept of a catalyst first, then look at the common mistakes students make with this exact type of question.

Core Concept: What a Catalyst Does (and Doesn’t Do)

A catalyst:

Speeds up both forward and backward reactions equally.
Lowers activation energy by providing an alternate pathway.
Does not change the equilibrium constant ( $K_{eq}$ ).
Does not change $\Delta G$ (Gibbs free energy change) of the reaction.

The Question

Which statement is not correct for the catalyst?

Options:

(i) It catalyses the forward and backward reaction to the same extent.
(ii) It alters $\Delta G$ of the reaction.
(iii) It is a substance that does not change the equilibrium constant of a reaction.
(iv) It provides an alternate mechanism by reducing activation energy between reactants and products.

Correct answer: (ii) — because a catalyst does not alter $\Delta G$ .

Common Mistakes & How to Avoid Them

✗ Mistake 1: Thinking a catalyst changes $\Delta G$

Why it happens:

Students confuse activation energy ( $E_a$ ) with Gibbs free energy change ( $\Delta G$ ). Both involve “energy,” so they assume the catalyst changes both.

How to avoid:

Remember:

$\Delta G$ depends only on initial and final states (reactants and products).
A catalyst does not change reactants or products — it only changes the path between them.
So $\Delta G$ remains unchanged.

✓ Tip: Draw a reaction energy profile. The catalyst lowers the peak (activation energy), but the start and end heights (reactants and products) stay the same.

✗ Mistake 2: Thinking a catalyst affects only the forward reaction

Why it happens:

Students focus on “speeding up the reaction” and forget the reverse direction.

How to avoid:

A catalyst lowers $E_a$ for both forward and backward reactions equally. That’s why equilibrium is reached faster, but $K_{eq}$ stays the same.

✓ Tip: If a catalyst only helped the forward reaction, it would shift equilibrium — which it doesn’t. So it must help both sides equally.

✗ Mistake 3: Confusing “catalyst changes equilibrium constant” with “catalyst helps reach equilibrium faster”

Why it happens:

Students see “catalyst speeds up reaction” and incorrectly think it changes the equilibrium position.

How to avoid:

$K_{eq}$ depends only on temperature (for a given reaction).
A catalyst does not change temperature or the nature of reactants/products.
So $K_{eq}$ is unchanged. …

Start your 14-day free trial to unlock the full solution →

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

KCET 2025Set D-41 markMCQ
$E_a$
›Reveal solutionSolution
On an energy profile the forward and backward barriers are both measured to the same peak, so $E_a(\text{reverse}) = E_a(\text{forward}) - \Delta H = 215 - 90 = 125$ kJ.

Step 1 — Read the graph.

From the described energy profile:

The reactants sit at the lower level.

The curve rises to a peak (the activated complex / transition state).

The products sit at a higher level than the reactants — so the reaction is endothermic.

$E_a(\text{forward}) = 215\ \mathrm{kJ}$ — measured from reactants up to the peak.

$\Delta H = +90\ \mathrm{kJ}$ — measured from reactants up to products.

Step 2 — Set an energy scale.

Put the reactants at zero:
$E(\text{reactants}) = 0\ \mathrm{kJ}$ $E(\text{peak}) = 0 + E_a(f) = 215\ \mathrm{kJ}$ $E(\text{products}) = 0 + \Delta H = 90\ \mathrm{kJ}$
Step 3 — The reverse activation energy.

Running the reaction backwards, the molecules start at the products and must climb to the same peak (the transition state is common to both directions — this is the principle of microscopic reversibility):
$E_a(\text{reverse}) = E(\text{peak}) - E(\text{products}) = 215 - 90 = 125\ \mathrm{kJ}$
Step 4 — Verify with the general relation.

For any reaction, …
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{A}+3 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}$
›Reveal solutionSolution

Rate of reaction $=4.17\times10^{-5}$ , and rate of formation of $\text{C}$ is twice this $=8.32\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$ . Official key: (B).

Working

For $\text{A} + 3\text{B} \rightarrow 2\text{C} + \text{D}$ , the rate of reaction is
$\text{Rate} = -\frac{d[\text{A}]}{dt} = +\frac{1}{2}\frac{d[\text{C}]}{dt}$
The change in $[\text{A}]$ over $1$ minute ( $=60\ \text{s}$ ):
$-\frac{\Delta[\text{A}]}{\Delta t} = \frac{0.0150 - 0.0125}{60} = \frac{0.0025}{60} = 4.17\times10^{-5}\ \text{mol L}^{-1}\text{s}^{-1}$
Since $\text{C}$ has a stoichiometric coefficient of $2$ , its rate of formation is twice the rate of reaction: …
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{BrO}_3^{-} \text {(aq) }+5 \mathrm{Br}^{-} \text {(aq) }+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_2(\mathrm{l})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
›Reveal solutionSolution
The rate of appearance of a product and the rate of disappearance of a reactant are linked by stoichiometric coefficients, with opposite signs. For the given reaction, the correct relation is $\frac{d[\mathrm{Br}_2]}{dt} = -\frac{3}{5} \frac{d[\mathrm{Br}^-]}{dt}$ , which corresponds to option (D).

The key idea is that for a chemical reaction, the rate can be expressed in terms of any reactant or product, but we must account for stoichiometry and sign conventions. The rate of disappearance of a reactant is negative (its concentration decreases), while the rate of appearance of a product is positive (its concentration increases). To relate them, we divide each rate by its stoichiometric coefficient and set them equal in magnitude, then adjust signs.

Let’s work through it step by step.

Write the general rate expression. For a reaction $aA + bB \rightarrow cC + dD$ , the rate is defined as:

$\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}$
The negative signs for reactants ensure the rate is positive (since $\frac{d[\text{reactant}]}{dt}$ is negative).

Identify the species and coefficients. In the given reaction:

$\mathrm{BrO}_3^- + 5\mathrm{Br}^- + 6\mathrm{H}^+ \rightarrow 3\mathrm{Br}_2 + 3\mathrm{H}_2\mathrm{O}$
We care about $\mathrm{Br}^-$ (reactant, coefficient 5) and $\mathrm{Br}_2$ (product, coefficient 3).

Set up the equality from the general rate. Using the definition:

$-\frac{1}{5}\frac{d[\mathrm{Br}^-]}{dt} = \frac{1}{3}\frac{d[\mathrm{Br}_2]}{dt}$
The left side has a negative sign because $\mathrm{Br}^-$ is a reactant; the right side has no negative sign because $\mathrm{Br}_2$ is a product.

Solve for the desired relation. Multiply both sides by 3:
Start your 14-day free trial to see the full solution →
COMEDK 2024Set 2024-E1 markMCQ
$5 X+Y \rightarrow 3 Z$
›Reveal solutionSolution
The key idea is that reaction rates for different species are linked by stoichiometric coefficients. For the reaction $5X + Y \rightarrow 3Z$ , the rate of disappearance of X is $\frac{5}{3}$ times the rate of formation of Z, giving $4.0 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}$ .

The core concept here is stoichiometric rate relationships. In a chemical reaction, the rate at which reactants disappear and products appear are not independent — they are tied together by the coefficients in the balanced equation. This is because for every 5 molecules of X that react, 3 molecules of Z are formed. So the rate of change of concentration for each species must be scaled by its coefficient to give the same "overall reaction rate."

Why this works:

If we define the rate of reaction as $r = -\frac{1}{5}\frac{d[X]}{dt} = -\frac{1}{1}\frac{d[Y]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$ , then knowing any one of these rates lets us find the others by simple multiplication or division by the appropriate coefficient ratio.

Write the general rate relationship For the reaction $5X + Y \rightarrow 3Z$ , the rate of reaction $r$ is:

$r = -\frac{1}{5}\frac{d[X]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$
The negative sign indicates disappearance (decrease in concentration), while positive indicates formation.

Identify what we know We are given the rate of formation of Z:

$\frac{d[Z]}{dt} = 2.4 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1}$

Relate the rate of disappearance of X to the rate of formation of Z From the equality above:

$-\frac{1}{5}\frac{d[X]}{dt} = \frac{1}{3}\frac{d[Z]}{dt}$
Multiply both sides by $-5$ :

$\frac{d[X]}{dt} = -\frac{5}{3}\frac{d[Z]}{dt}$ …
Start your 14-day free trial to see the full solution →
COMEDK 2021Set 2021-B1 markMCQ
$Q. The equation showing the decomposition of Dinitrogen pentoxide is as follows : N2O5(g) -> N2O4(g) + 1/2 O2(g) If the initial pressure is 250 mm and after half an hour of the reaction the total pressure of the gaseous mixture is 296 mm, what is the average rate of the reaction ? (A) 4.04 * 10^-3 atm min^-1 (B) -2.02 * 10^-3 atm min^-1 (C) 2.19 * 10^-2 atm min^-1 (D) 1.09 * 10^-2 atm min^-1$
›Reveal solutionSolution

The pressure rise fixes the N $_2$ O $_5$ consumed at 92 mm; converting to atm over 30 min gives $4.04\times10^{-3}$ atm min $^{-1}$ .

Pressure bookkeeping ( $\text{N}_2\text{O}_5 \to \text{N}_2\text{O}_4 + \tfrac12\text{O}_2$ ), let $x$ = mm of N $_2$ O $_5$ reacted:
$P_{total} = (250-x) + x + \tfrac{x}{2} = 250 + \tfrac{x}{2}$
$250 + \tfrac{x}{2} = 296 \Rightarrow x = 92\ \text{mm}$ …
Start your 14-day free trial to see the full solution →
KCET 2020Set A-11 markMCQ
$k = PZe^{-E_a/RT}$
›Reveal solutionSolution
The rate constant $k$ increases when the activation energy $E_a$ is decreased, because $k$ depends exponentially on $-E_a/RT$ ; the correct answer is (D).

The Arrhenius equation $k = A e^{-E_a/RT}$ (where $A = PZ$ in the given notation) tells us exactly how temperature and activation energy control reaction speed. The exponential term $e^{-E_a/RT}$ is the key: it represents the fraction of molecules that have enough energy to overcome the activation barrier. To speed up a reaction, we want $k$ to be larger — so we need that exponential term to be as large as possible.

Since the exponent is negative, making $E_a$ smaller makes the exponent less negative, which increases $e^{-E_a/RT}$ . Making $T$ larger also makes the exponent less negative (because $RT$ is in the denominator), so increasing temperature speeds up the reaction. But the question asks which factor has to be decreased to speed up the reaction.

Let’s check each option carefully.

Option (A): $T$ — Temperature appears in the denominator of the exponent. Decreasing $T$ makes $E_a/RT$ larger, so the exponent becomes more negative, and $k$ gets smaller. That slows the reaction down, not speeds it up. So this is wrong.

Option (B): $Z$ — $Z$ is the collision frequency factor (part of the pre-exponential term $PZ$ ). It multiplies the exponential directly: $k = (PZ) e^{-E_a/RT}$ . Decreasing $Z$ reduces the pre-factor, which reduces $k$ . That also slows the reaction. So this is wrong too.

Option (C): Both $Z$ and $T$ — Since decreasing either one slows the reaction, decreasing both certainly won’t speed it up. This is incorrect. …
Start your 14-day free trial to see the full solution →
KCET 2018Set A-11 markMCQ
$30^\circ \mathrm{C}$
›Reveal solutionSolution

Rate ratio $= (\text{temperature coefficient})^{\Delta T/10} = 2^{60/10} = 2^{6} = 64$ .

Step 1 — Meaning of the temperature coefficient.

The temperature coefficient $\mu$ of a reaction is defined as the ratio of the rate constants for a $10^\circ$ C rise:
$\mu = \frac{k_{T+10}}{k_{T}}$
For most reactions $\mu \approx 2$ – $3$ (the familiar 'rate roughly doubles for every $10^\circ$ C rise' rule). Here $\mu = 2$ .

Step 2 — Count the $10^\circ$ intervals.
$\Delta T = 90^\circ\mathrm{C} - 30^\circ\mathrm{C} = 60^\circ\mathrm{C}$ $n = \frac{\Delta T}{10} = \frac{60}{10} = 6$
Step 3 — Compound the factor.

The effect is multiplicative, not additive — each successive $10^\circ$ step multiplies the rate by 2 again: …
Start your 14-day free trial to see the full solution →

←Previous questionQ17 — Compounds 'A' and 'B' react according to the following chemical equati…→Next questionQ19 — The value of rate constant of a pseudo first order reaction ______. (i…

Q.Which of the following statement is not correct for the catalyst?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days