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.

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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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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." …

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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

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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. …

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