Mark the incorrect statements. (Two or more than two options may be correct.) (i) Catalyst provides an alterna

NCERT Exemplar · Q55

Q. Mark the incorrect statements. (Two or more than two options may be correct.)

(i) Catalyst provides an alternative pathway to reaction mechanism.

(ii) Catalyst raises the activation energy.

(iii) Catalyst lowers the activation energy.

(iv) Catalyst alters enthalpy change of the reaction.

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Concept understanding — Arrhenius Equation Plot

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

You already know that heating things up makes reactions go faster. A cold chai takes forever to dissolve sugar; hot chai does it in seconds. But how much faster? And is there a pattern that holds for every reaction?

That pattern is the Arrhenius equation, and plotting it in a clever way reveals something fundamental about how molecules need to collide to react.

The core idea: an energy barrier

Imagine a ball sitting in a valley. To get to the next valley, it must first be pushed up over a hill. That hill is the activation energy ( $E_a$ ) — the minimum energy two molecules need to have when they collide, for the reaction to happen.

At a low temperature, most molecules move slowly. Only a tiny fraction have enough energy to climb that hill. Raise the temperature, and suddenly many more molecules have the required energy. The fraction of molecules with energy $\ge E_a$ is given by the Boltzmann distribution:

\text{fraction} = e^{-E_a / RT}

where $R$ is the gas constant and $T$ is the absolute temperature (in Kelvin). This exponential is the heart of the story.

The Arrhenius equation (precise statement)

The rate constant $k$ of a reaction depends on temperature as:

k = A \, e^{-E_a / RT}

$k$ = rate constant (how fast the reaction proceeds)
$A$ = pre-exponential factor (frequency of collisions, times a steric factor — how often molecules hit in the right orientation)
$E_a$ = activation energy (J/mol or kJ/mol)
$R$ = 8.314 J/(mol·K)
$T$ = temperature in Kelvin

Important

$k$ is not the reaction rate itself — it's the proportionality constant in the rate law. But for a fixed concentration, a larger $k$ means a faster reaction.

Why plot it? The linear trick

The equation $k = A e^{-E_a / RT}$ is exponential in $1/T$ . That's hard to eyeball. But take the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is the equation of a straight line:

y = c + m x

where:

$y = \ln k$
$x = 1/T$
slope $m = -E_a / R$
intercept $c = \ln A$

So if you measure $k$ at several temperatures and plot $\ln k$ versus $1/T$ , you get a straight line — provided the reaction follows Arrhenius behaviour (most do, over moderate temperature ranges).

Tip

Always use Kelvin for $T$ . Celsius will give you a curved mess because $1/T$ is not linear in Celsius.

What the plot tells you

From the slope, you get $E_a$ :

E_a = - (\text{slope}) \times R

A steep negative slope means a large $E_a$ — the reaction is very sensitive to temperature. A shallow slope means a small $E_a$ — temperature doesn't affect it much.

From the intercept, you get $A$ :

A = e^{\text{intercept}}

This tells you about the collision frequency and orientation factor. A high $A$ means molecules are colliding often and in the right geometry.

A typical Arrhenius plot looks like this

$T$ (K)	$k$ (s⁻¹)	$1/T$ (K⁻¹)	$\ln k$
300	0.0012	0.00333	-6.72
310	0.0028	0.00323	-5.88
320	0.0061	0.00313	-5.10
330	0.0125	0.00303	-4.38

Plot $\ln k$ (y-axis) vs $1/T$ (x-axis). The points fall on a straight line. Draw the best-fit line, measure its slope, and compute $E_a$ . …

Why this formula?

Arrhenius Equation Plot: Why It Holds

The Arrhenius equation is not a guess — it emerges from a deep physical picture of how molecules react. Let's build that understanding step by step.

The Core Idea: Molecules Need Energy to React

For a reaction to occur, molecules must collide with enough energy to break existing bonds and form new ones. This minimum energy is called the activation energy ( $E_a$ ).

But not all collisions succeed — only those with kinetic energy $\geq E_a$ lead to a reaction.

The Key Formula

The Arrhenius equation is:

k = A e^{-E_a / (RT)}

Where:

$k$ = rate constant
$A$ = pre-exponential factor (frequency of collisions with correct orientation)
$E_a$ = activation energy (J/mol)
$R$ = gas constant (8.314 J/mol·K)
$T$ = absolute temperature (K)

Why the Exponential Term Appears

Step 1: The Boltzmann Distribution

Molecules in a gas or liquid have a distribution of kinetic energies. The fraction of molecules with energy $\geq E$ is given by the Boltzmann factor:

\text{Fraction} = e^{-E / (k_B T)}

For molar quantities, replace $k_B$ with $R$ :

\text{Fraction} = e^{-E_a / (RT)}

This is not arbitrary — it comes from statistical mechanics. The exponential arises because the probability of a molecule having energy $E$ decreases exponentially as $E$ increases.

Step 2: Rate Depends on This Fraction

The rate constant $k$ is proportional to:

The collision frequency (how often molecules meet)
The fraction of collisions with energy $\geq E_a$

Thus:

k \propto \text{(collision frequency)} \times e^{-E_a / (RT)}

The collision frequency is captured by $A$ , giving:

k = A e^{-E_a / (RT)}

Why the Plot is Linear

Take the natural logarithm of both sides:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is of the form $y = mx + c$ , where:

$y = \ln k$
$x = 1/T$
Slope $m = -E_a / R$
Intercept $c = \ln A$

Thus, plotting $\ln k$ vs $1/T$ gives a straight line — this is the Arrhenius plot.

What the Slope Tells Us

From the slope: …

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A catalyst works by providing an alternative reaction pathway with a lower activation energy, without changing the enthalpy change of the reaction. The correct statements are (i) and (iii); (ii) and (iv) are incorrect.

Let’s understand why. The Arrhenius equation is the key here:

k = A e^{-E_a / RT}

It tells us that the rate constant $k$ depends exponentially on the activation energy $E_a$ . A catalyst speeds up a reaction by lowering $E_a$ — it offers a different path over the energy hill, not by pushing the reactants harder.

Now, evaluate each statement one by one.

Statement (i): Catalyst provides an alternative pathway to reaction mechanism.

This is true. A catalyst participates in the reaction, forming an intermediate, and is regenerated at the end. The new pathway has a different (lower) activation energy. This is the very definition of catalysis.
Statement (ii): Catalyst raises the activation energy.

This is false. Raising $E_a$ would decrease the rate (since $k$ drops exponentially). That’s the opposite of what a catalyst does. An inhibitor does this, not a catalyst.
Statement (iii): Catalyst lowers the activation energy.

This is true. By providing an alternative mechanism with a lower $E_a$ , the catalyst increases the fraction of molecules with energy above the barrier, speeding up the reaction.
Statement (iv): Catalyst alters enthalpy change of the reaction. …

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Here is a breakdown of the common mistakes students make on this specific question, along with the conceptual corrections.

The Core Concept: What a Catalyst Actually Does

A catalyst works by providing an alternative reaction pathway with a lower activation energy ( $E_a$ ). It does not change the thermodynamics of the reaction.

Enthalpy change ( $\Delta H$ ) is a state function (difference between products and reactants). A catalyst does not change the initial or final states, so $\Delta H$ remains unchanged.
Activation energy ( $E_a$ ) is the energy barrier. A catalyst lowers this barrier, allowing more molecules to have sufficient energy to react at a given temperature.

Common Mistake #1: Confusing "Alternative Pathway" with "Changing the Mechanism"

The Mistake: Students think statement (i) is incorrect because they believe a catalyst changes the steps of the reaction, not just the pathway.

Why it’s wrong: A catalyst does provide an alternative pathway. This is the definition of catalysis. The new pathway involves different elementary steps (e.g., forming an intermediate with the catalyst), but the overall reaction (reactants → products) remains the same.

How to Avoid: Remember: "Alternative pathway" = "Different route, same destination." The catalyst participates in the reaction but is regenerated. Statement (i) is correct.

Common Mistake #2: Misreading "Raises" vs. "Lowers"

The Mistake: Students see "activation energy" and automatically assume a catalyst lowers it, so they mark (ii) as incorrect and (iii) as correct. This is correct, but the trap is in the wording.

Why it’s a trap: Statement (ii) says "raises the activation energy." A catalyst lowers it. So (ii) is incorrect. Statement (iii) says "lowers the activation energy." This is correct.

How to Avoid: Read each statement independently. Don't assume a pattern. For every statement, ask: Does a catalyst do this? If yes, it's correct. If no, it's incorrect.

(ii) → No → Incorrect
(iii) → Yes → Correct

Common Mistake #3: Forgetting That $\Delta H$ is a State Function

The Mistake: Students think a catalyst can change the enthalpy change ( $\Delta H$ ) because it speeds up the reaction, or because it provides a different pathway. …

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

CBSE 2026Set 56/1/11 markMCQ
$E_a$
›Reveal solutionSolution
The fraction of molecules with energy equal to or greater than the activation energy $E_a$ is given by the Boltzmann factor $e^{-E_a/RT}$ , which appears directly in the Arrhenius equation. The correct option is (B).

The Arrhenius equation is the starting point here. It tells us that the rate constant $k$ depends on temperature as:
$k = A e^{-E_a/RT}$
where $A$ is the pre-exponential factor (related to collision frequency and orientation), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature.

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 — that is, molecules with kinetic energy equal to or greater than $E_a$ . This comes from the Maxwell–Boltzmann distribution of molecular energies: the fraction of molecules with energy $\geq E_a$ is proportional to $e^{-E_a/RT}$ .

So the question is simply asking: which of the given expressions matches this Boltzmann factor?

Let’s check each option:

Option (A): $+\dfrac{E_a}{RT}$

This is a positive number (since $E_a$ , $R$ , $T$ are all positive). It grows as temperature decreases or activation energy increases. But a fraction must lie between 0 and 1 — this expression can be much larger than 1, so it cannot represent a fraction of molecules. Discard.

Option (B): $e^{-E_a/RT}$

This is the Boltzmann factor. For typical values ( $E_a \sim 50$ kJ/mol, $T \sim 300$ K, $R = 8.314$ J/mol·K), $E_a/RT \approx 20$ , so $e^{-20} \approx 2 \times 10^{-9}$ — a tiny fraction, which makes sense: only a very small proportion of molecules have enough energy to react at room temperature. This matches the physical meaning exactly.

Option (C): $-\dfrac{E_a}{RT}$

This is negative. A fraction cannot be negative. Discard. …
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CBSE 2024Set 56/1/11 markMCQ
$\log k$
›Reveal solutionSolution
The Arrhenius equation $k = A e^{-E_a/(RT)}$ becomes linear when we take the natural log and plot $\ln k$ vs. $1/T$ . Converting to $\log_{10}$ gives slope $= -E_a/(2.303 R)$ and intercept $= \log A$ , so the correct option is (D).

The Arrhenius equation is one of the most elegant relationships in chemical kinetics — it connects the rate constant $k$ to temperature $T$ through two parameters: the activation energy $E_a$ and the pre-exponential factor $A$ . The equation is:
$k = A e^{-E_a/(RT)}$
If you plot $k$ directly against $T$ , you get a curve. But the trick is to take logarithms — that turns the exponential into a straight line. Why does that help? Because a straight line is easy to interpret: its slope and intercept give you $E_a$ and $A$ directly.

Let’s see how.

Take the natural logarithm of both sides

Starting from $k = A e^{-E_a/(RT)}$ , we get:

$\ln k = \ln A - \frac{E_a}{RT}$
This is already in the form $y = mx + c$ , where:

$y = \ln k$

$x = 1/T$

slope $m = -E_a/R$

intercept $c = \ln A$

So a plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ and intercept $\ln A$ .

But the question uses $\log k$ — that’s base 10

In many Indian exam contexts, $\log$ means $\log_{10}$ . To convert from natural log to base 10, use:

$\ln k = 2.303 \log_{10} k$
Substitute into the equation above:
$2.303 \log k = \ln A - \frac{E_a}{RT}$
Divide through by 2.303:
$\log k = \frac{\ln A}{2.303} - \frac{E_a}{2.303 \, RT}$

Identify slope and intercept

Now compare with $y = mx + c$ :

$y = \log k$

$x = 1/T$

slope $m = -\dfrac{E_a}{2.303 \, R}$

intercept $c = \dfrac{\ln A}{2.303} = \log A$

That’s exactly what option (D) says. …
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CBSE 2023Set 56/1/11 markMCQ
$\Delta H$
›Reveal solutionSolution
A catalyst provides an alternative reaction pathway with a lower activation energy ( $E_a$ ). It does not change the thermodynamic state functions $\Delta H$ , $\Delta G$ , or $\Delta S$ for the overall reaction. Therefore, the correct answer is (C).

The key to this question lies in distinguishing between kinetics (how fast a reaction occurs) and thermodynamics (whether a reaction is spontaneous and how much energy is exchanged). A catalyst is a kinetic tool — it speeds up a reaction without being consumed, but it never alters the starting point or the destination of the reaction.

Think of a mountain pass. The reactants are at the base of one side, the products at the base of the other. The height difference between them is $\Delta H$ (enthalpy change). The overall steepness and direction of the slope is $\Delta G$ (free energy change). The disorder along the path is $\Delta S$ (entropy change). A catalyst is like a tunnel through the mountain — it lowers the peak you have to climb over (the activation energy $E_a$ ), but the heights of the two bases and the distance between them remain exactly the same.

Let’s examine each option carefully.

$\Delta H$ (Enthalpy change)

$\Delta H$ is the difference in enthalpy between products and reactants. It depends only on the initial and final states of the system. A catalyst does not appear in the overall balanced equation and is regenerated at the end. Since the reactants and products are identical with or without the catalyst, $\Delta H$ is unchanged.
$\Delta H_{\text{catalysed}} = \Delta H_{\text{uncatalysed}}$

$\Delta G$ (Gibbs free energy change)

$\Delta G = \Delta H - T\Delta S$ is the thermodynamic driving force for a reaction. It determines spontaneity. A catalyst cannot make a non-spontaneous reaction spontaneous — it only accelerates a reaction that is already thermodynamically favourable. The equilibrium constant $K$ is related to $\Delta G$ by $\Delta G^\circ = -RT\ln K$ , and a catalyst does not shift equilibrium. So $\Delta G$ remains the same.

$E_a$ (Activation energy) …
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Q.Mark the incorrect statements. (Two or more than two options may be correct.)

Concept understanding — Arrhenius Equation Plot

The Arrhenius Equation Plot: Why Temperature Changes Reaction Speed

The core idea: an energy barrier

The Arrhenius equation (precise statement)

Why plot it? The linear trick

What the plot tells you

A typical Arrhenius plot looks like this

Why this formula?

Arrhenius Equation Plot: Why It Holds

The Core Idea: Molecules Need Energy to React

The Key Formula

Why the Exponential Term Appears

Step 1: The Boltzmann Distribution

Step 2: Rate Depends on This Fraction

Why the Plot is Linear

What the Slope Tells Us

Concept: Catalysis and Activation Energy

Method: Fact-Checking Against Catalyst Properties

Steps:

The Core Concept: What a Catalyst Actually Does

Common Mistake #1: Confusing "Alternative Pathway" with "Changing the Mechanism"

Common Mistake #2: Misreading "Raises" vs. "Lowers"

Common Mistake #3: Forgetting That $\Delta H$ is a State Function

Q.Mark the incorrect statements. (Two or more than two options may be correct.)

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