Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex

NCERT Exemplar · Q66

Q. Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules. Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

(i) Both assertion and reason are correct and the reason is correct explanation of assertion.

(ii) Both assertion and reason are correct but reason does not explain assertion.

(iii) Assertion is correct but reason is incorrect.

(iv) Both assertion and reason are incorrect.

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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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The Arrhenius equation gives accurate rate constants for simple reactions, but for complex molecules, steric factors and orientation matter — so the assertion is false. The reason is also false because molecules must have proper orientation for a reaction to occur. Both statements are incorrect.

The key here is to understand what the Arrhenius equation actually models and where it falls short. The equation $k = A e^{-E_a/RT}$ assumes that every collision with sufficient energy leads to a reaction — but that's only true for simple, small molecules. For complex molecules, the orientation during collision becomes critical, and the Arrhenius equation overestimates the rate unless corrected by a steric factor.

Let's break down each statement.

Assertion: "Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules."

This is incorrect. For simple molecules (like two atoms colliding), the Arrhenius equation works well because almost every energetic collision leads to reaction. But for complex molecules (large organic compounds, for instance), the molecule must hit the reactive site in the correct orientation. The Arrhenius equation ignores this — it assumes all collisions with enough energy are effective. That's why we introduce the steric factor $P$ in collision theory: $k = P Z e^{-E_a/RT}$ , where $P$ is often much less than 1 for complex molecules. So the assertion is false.
Reason: "Reactant molecules undergo chemical change irrespective of their orientation during collision." …

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Here’s a breakdown of the common mistakes students make on this specific assertion-reason question about the Arrhenius equation, along with how to avoid each.

Mistake 1: Confusing “Accuracy” with “Universality”

The Error: Students think the Arrhenius equation works perfectly for all reactions, including complex ones. They then mark the assertion as correct without checking the nuance.
Why It’s Wrong: The Arrhenius equation is empirical and works well for simple, elementary reactions. For complex molecules or multi-step reactions, the rate constant often deviates because the equation assumes a single activation energy barrier, which isn’t true for complex pathways.
How to Avoid: Remember: Arrhenius is accurate for simple, single-step reactions. For complex reactions, the plot of $\ln k$ vs. $1/T$ may be curved, not linear. The assertion says “fairly accurate for simple as well as complex molecules” — this is incorrect because it overstates the equation’s range.

Mistake 2: Misinterpreting the Reason (Orientation Factor)

The Error: Students think the reason sounds scientific and matches the topic of collision theory, so they assume it must be correct.
Why It’s Wrong: The reason states: “Reactant molecules undergo chemical change irrespective of their orientation during collision.” This is false. In reality, proper orientation is critical for effective collisions (the steric factor in collision theory). If orientation is wrong, no reaction occurs even if energy is sufficient.
How to Avoid: Link the reason to collision theory:
- Effective collision = sufficient energy (Arrhenius) + proper orientation.
- The reason denies the orientation requirement, which is a classic mistake. Always check if a statement contradicts basic collision theory.

Mistake 3: Assuming “Both Correct” Without Checking the Link

The Error: Students see two statements that both sound plausible and pick option (i) or (ii) without verifying if the reason actually explains the assertion.
Why It’s Wrong: Even if both were correct (they aren’t here), the reason (orientation is irrelevant) does not explain why the Arrhenius equation is accurate. The accuracy of Arrhenius depends on the reaction being elementary, not on orientation.
How to Avoid: For assertion-reason questions, always ask: “Does the reason directly cause or justify the assertion?” Here, the reason is about orientation during collision, while the assertion is about the equation’s accuracy — they are unrelated concepts.

Mistake 4: Forgetting the “Complex Molecules” Trap …

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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.Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

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

Method: Conceptual Evaluation of Assertion-Reason Statements

Step 1: Evaluate the Assertion

Step 2: Evaluate the Reason

Mistake 1: Confusing “Accuracy” with “Universality”

Mistake 2: Misinterpreting the Reason (Orientation Factor)

Mistake 3: Assuming “Both Correct” Without Checking the Link

Mistake 4: Forgetting the “Complex Molecules” Trap …

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.Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules. Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

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Q.Assertion: Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.
Reason: Reactant molecules undergo chemical change irrespective of their orientation during collision.

Unlock everything free for 14 days