During decomposition of an activated complex (Two or more than two options may be correct.) (i) energy is alwa

NCERT Exemplar · Q51

Q. During decomposition of an activated complex (Two or more than two options may be correct.)

(i) energy is always released

(ii) energy is always absorbed

(iii) energy does not change

(iv) reactants may be formed

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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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An activated complex (transition state) sits at the highest-energy point along the reaction coordinate — above BOTH the reactants and the products. Whichever way it decomposes (forward to products, or backward to reactants), the system moves downhill from that peak, so energy is always released. The complex can equally well fall back to re-form the original reactants. The correct options are (i) and (iv).

1. What is an activated complex?

In transition state theory, the activated complex is the fleeting, high-energy arrangement of atoms that exists at the top of the potential energy barrier along the reaction coordinate. It is not a stable species — old bonds are partly broken and new bonds are partly formed at the same time. Crucially, it sits at a potential energy MAXIMUM relative to both reactants and products on either side of it.

2. What happens when it decomposes?

"Decomposition" here means the complex breaks apart and the system moves away from that energy peak — either forward, converting to products, or backward, reverting to reactants.

Moving forward (complex → products): the system descends from the peak to the product energy level, so energy is released.
Moving backward (complex → reactants): the system descends from the same peak back down to the reactant energy level, so energy is released here too.

Because the activated complex is higher in energy than both reactants and products, any direction it decomposes in involves a drop in potential energy — so energy is always released during decomposition. This is why option (i) is correct, and why options (ii) "energy is always absorbed" and (iii) "energy does not change" are both wrong: energy absorption happens while the reactants are climbing UP to form the complex (that absorbed energy is the activation energy), not while the complex is decomposing. …

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Here’s a breakdown of the common mistakes students make on this question, along with how to avoid each.

The Core Concept: The Activated Complex (Transition State)

The activated complex (or transition state) is a high-energy, unstable arrangement of atoms that exists at the peak of the potential energy barrier in a chemical reaction. It is not a stable molecule; it exists only for an instant.

It can decompose forward to form products.
It can decompose backward to reform reactants.

The key is that the activated complex has higher energy than both the reactants and the products. Therefore, when it decomposes in either direction, energy is released as it moves down the energy hill.

Common Mistake #1: Assuming energy is always absorbed

The Error: Students think that because the activated complex is "activated" (high energy), breaking it down must require more energy input. They confuse the formation of the complex (which requires energy) with its decomposition.

Why it’s wrong: The activated complex is at a local energy maximum. Any movement away from this peak — whether toward reactants or products — involves a decrease in potential energy. A decrease in potential energy means energy is released (usually as heat or kinetic energy of the molecules).

How to Avoid: Draw the energy profile diagram. Mark the activated complex at the top. Ask yourself: "If I am at the top of a hill, which way do I go to lose energy?" The answer is any direction. Decomposition always releases energy.

Common Mistake #2: Thinking energy does not change

The Error: Students assume that because the complex is unstable and short-lived, its decomposition is "instantaneous" and involves no energy change.

Why it’s wrong: The very definition of the activated complex is that it sits at a potential energy maximum. Moving away from that maximum must involve a change in energy (a decrease). "No change" would imply it is at a flat plateau, which is not the case for a true transition state. …

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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.During decomposition of an activated complex (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

1. What is an activated complex?

2. What happens when it decomposes?

Concept: Activated Complex (Transition State) in Chemical Reactions

Method: Energy Profile Analysis

Steps:

The Core Concept: The Activated Complex (Transition State)

Common Mistake #1: Assuming energy is always absorbed

Common Mistake #2: Thinking energy does not change

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.During decomposition of an activated complex (Two or more than two options may be correct.)

1. What is an activated complex?

2. What happens when it decomposes?

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