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.

AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$E_a$
›Reveal solutionSolution
The fraction of molecules exceeding activation energy is the Boltzmann factor $e^{-E_a/RT}$ , which evaluates to $\exp(-2.4)$ here.

Concept and Intuition

The Arrhenius equation $k = Ae^{-E_a/RT}$ splits rate into a collision-frequency factor $A$ and an exponential Boltzmann factor $e^{-E_a/RT}$ , which physically represents the fraction of molecular collisions/molecules possessing energy at least $E_a$ . This fraction is exactly what the question calls $X$ .

Step-by-Step Solution

Fraction with energy $\geq E_a$ : $X = e^{-E_a/RT}$ .

Compute the exponent: $\dfrac{E_a}{RT} = \dfrac{10^4\ \text{J/mol}}{8.3\ \text{J mol}^{-1}\text{K}^{-1}\times500\ \text{K}} = \dfrac{10000}{4150} \approx 2.41 \approx 2.4$ . …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$\log k = 14 - \dfrac{1.25\times10^4 K}{T}$
›Reveal solutionSolution
Match the given empirical rate-constant equation to the Arrhenius equation in log form to extract $E_a$ and $A$ . Answer: $E_a=238.93$ kJ/mol, $A=10^{14}\ s^{-1}$ .

Concept and Intuition

The Arrhenius equation $k=Ae^{-E_a/RT}$ , written in base-10 log form, is $\log k = \log A - \dfrac{E_a}{2.303RT}$ — a straight line when $\log k$ is plotted against $1/T$ , with intercept $\log A$ and slope $-E_a/2.303R$ . Any experimentally fitted equation of this form can be matched term-by-term to read off $A$ and $E_a$ directly.

Step-by-Step Solution

Given: $\log k = 14 - \dfrac{1.25\times10^4}{T}$ .

Arrhenius form: $\log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}$ .

Matching the constant term: $\log A = 14 \Rightarrow A = 10^{14}\ s^{-1}$ .

Matching the $1/T$ coefficient: $\dfrac{E_a}{2.303R} = 1.25\times10^4\text{ K}$ .

$E_a = 1.25\times10^4 \times 2.303 \times 8.3\ \text{J/mol}$ . Compute: $1.25\times10^4\times2.303 = 28{,}787.5$ ; $\times 8.3 = 238{,}936.25$ J/mol $\approx 238.93$ kJ/mol. …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{kJ\, mol^{-1}}$
›Reveal solutionSolution
A textbook comparison of activation energies for acid- vs enzyme-catalysed sucrose hydrolysis: $X=6.22$ , $Y=2.15\ \mathrm{kJ\,mol^{-1}}$ .

Concept and Intuition

A catalyst speeds up a reaction by providing an alternate pathway with lower activation energy, without changing $\Delta H$ of the reaction. Enzymes are exceptionally efficient catalysts, so an enzyme-catalysed pathway typically has a much lower $E_a$ than the corresponding acid-catalysed (or uncatalysed) pathway for the same reaction, since $k = A e^{-E_a/RT}$ — a smaller $E_a$ gives a dramatically larger rate constant at the same temperature.

Step-by-Step Solution

Identify the reaction: hydrolysis of sucrose to glucose + fructose, run two ways — acid-catalysed and sucrase(enzyme)-catalysed.

Recall the standard reported values for this reaction: acid catalysis, $E_a = 6.22\ \mathrm{kJ\,mol^{-1}}$ ; sucrase catalysis, $E_a = 2.15\ \mathrm{kJ\,mol^{-1}}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\log\dfrac{k}{A} = -\dfrac{x}{T}$
›Reveal solutionSolution
Matching the given rate-law equation to the Arrhenius equation in logarithmic form directly identifies the activation energy as $E_a = 2.303\,x\,R$ .

Concept and Intuition

The Arrhenius equation, $k = A e^{-E_a/RT}$ , in logarithmic (base 10) form becomes:
$\log k = \log A - \frac{E_a}{2.303\,R\,T} \quad\Rightarrow\quad \log\frac{k}{A} = -\frac{E_a}{2.303\,R\,T}$
Comparing coefficients with a given empirical equation of the same form directly reveals $E_a$ .

Step-by-Step Solution

Standard Arrhenius log form: $\log\dfrac{k}{A} = -\dfrac{E_a}{2.303\,R\,T}$ .

Given equation: $\log\dfrac{k}{A} = -\dfrac{x}{T}$ .

Equate the coefficients of $\dfrac{1}{T}$ : $\dfrac{E_a}{2.303\,R} = x$ . …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\log_{10}\dfrac{k}{A} = 0.00174$
›Reveal solutionSolution
This tests applying the Arrhenius equation in logarithmic form to extract activation energy. The answer is 10.0 J/mol.

Concept and Intuition

The Arrhenius equation $k=Ae^{-E_a/RT}$ relates the rate constant to activation energy $E_a$ and temperature $T$ . Taking natural log and converting to base-10: $\log_{10}\dfrac{k}{A} = -\dfrac{E_a}{2.303RT}$ . So the magnitude of $\log_{10}(k/A)$ scales directly with activation energy at a given temperature — a small log ratio corresponds to a small (near-zero) activation energy.

Step-by-Step Solution

Write the relation: $\left|\log_{10}\dfrac{k}{A}\right| = \dfrac{E_a}{2.303RT}$ .

Rearranging: $E_a = 2.303\,R\,T\times\left|\log_{10}\dfrac{k}{A}\right|$ .

Substitute values: $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , $T=300$ K, $\left|\log_{10}(k/A)\right|=0.00174$ . …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\rightarrow$
›Reveal solutionSolution
On a reaction energy diagram, $E_a$ is the gap from reactant level to the peak, and $|\Delta H|$ is the gap between reactant and product levels — here $E_a=15$ kJ/mol and $|\Delta H|=5$ kJ/mol.

Concept and Intuition

A potential-energy vs reaction-coordinate diagram encodes both kinetics and thermodynamics in one picture: the height of the barrier above the reactants is the activation energy (how hard it is to get started), while the difference between the final resting level of products and the starting level of reactants is the heat of reaction (whether the overall process releases or absorbs energy).

Step-by-Step Solution

From the graph: reactant A is at the dashed gridline $y=10$ kJ/mol.

The curve rises through the hump to its highest point, which lines up with the $y=25$ kJ/mol gridline — this is the transition state / activated complex energy.

Activation energy $E_a = E_{\text{peak}} - E_{A} = 25 - 10 = 15$ kJ/mol.

The curve then falls to product P, at the $y=5$ kJ/mol gridline. …
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AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$\ln k$
›Reveal solutionSolution
Straightforward application of the Arrhenius equation's linear form; the slope directly gives $E_a$ after multiplying by $-R$ .

Concept and Intuition

The Arrhenius equation $k = Ae^{-E_a/RT}$ becomes linear on taking logarithms:
$\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$
Plotting $\ln k$ (y-axis) against $1/T$ (x-axis) gives a straight line of slope $-E_a/R$ and intercept $\ln A$ .

Step-by-Step Solution

Given slope $= -2\times10^4$ K.

Since slope $= -E_a/R$ : $E_a = -(\text{slope})\times R = (2\times10^4\ \text{K})(8.3\ \text{J K}^{-1}\text{mol}^{-1})$ . …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$Q. Given below are two statements Assertion (A): A catalyst, generally increases the rate of a reaction Reason (R): It lowers the activation energy of a reaction by providing a new path The correct answer is (A) Both (A) and (R) are correct and (R) is the correct explanation of (A) (B) Both (A) and (R) are correct and (R) is not the correct explanation of (A) (C) (A) is correct but (R) is not correct (D) (A) is not correct but (R) is correct$
›Reveal solutionSolution
The assertion and reason are both true, and the reason correctly explains why the assertion is true. The correct option is (A).

Understanding the Assertion (A):

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. This is a fundamental fact in chemistry — catalysts speed up both forward and reverse reactions, allowing equilibrium to be reached faster. So (A) is correct.

Understanding the Reason (R):

The reason states that a catalyst lowers the activation energy by providing an alternative reaction pathway. This is the standard explanation from the Arrhenius equation:

$k = A e^{-E_a/(RT)}$
Lowering $E_a$ (activation energy) increases the rate constant $k$ , and thus the reaction rate. The catalyst does not change the overall thermodynamics (ΔH or ΔG) — it only reduces the energy barrier. So (R) is also correct.

Checking if (R) explains (A): …
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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

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

Concept and Intuition

Step-by-Step Solution

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

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