Assertion: All collision of reactant molecules lead to product formation. Reason: Only those collisions in whi

NCERT Exemplar · Q65

Q. Assertion: All collision of reactant molecules lead to product formation. Reason: Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.

(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 assertion is FALSE — not every collision leads to product formation, only effective collisions do. The reason, however, correctly states the actual condition for a successful collision (correct orientation + sufficient kinetic energy) and is TRUE. This is an "Assertion incorrect, Reason correct" case, which none of the four listed options captures correctly — option (iv) as printed ("Both assertion and reason are incorrect") wrongly implies the reason is also false.

Assertion — "All collisions of reactant molecules lead to product formation" — is FALSE.

By collision theory, only a small fraction of collisions are effective. A collision yields product only when the molecules (i) carry kinetic energy at least equal to the activation energy $E_a$ , and (ii) are correctly oriented. The overwhelming majority of collisions are ineffective.

Reason — "Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation" — is TRUE. This is the standard statement of collision theory, and it directly explains why the assertion is false (it describes the actual, narrower condition, contradicting "all collisions"). …

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

Common Mistake 1: Misreading the Assertion as True

What students do:

They see the word “collision” and think of the collision theory of chemical reactions. They assume the assertion must be correct because reactions happen when molecules collide.

Why it’s wrong:

The assertion says all collisions lead to product formation. Collision theory clearly states that only a fraction of collisions — those with sufficient energy (≥ activation energy) and correct orientation — are effective. Most collisions are ineffective.

How to avoid:

Read the assertion literally — look for absolute words like “all,” “always,” “never.”
Recall the two conditions for an effective collision:
1. Sufficient kinetic energy (≥ activation energy)
2. Proper orientation
If either condition is missing, no product forms. So “all collisions” is false.

Common Mistake 2: Thinking the Reason is Incorrect

What students do:

They confuse the reason with the assertion. They think the reason says “only those collisions with correct orientation and sufficient energy lead to product formation” is wrong because they remember that sometimes even with correct orientation and energy, a reaction may not occur due to other factors (like steric hindrance).

Why it’s wrong:

The reason is actually correct — it is a standard statement of collision theory. The two conditions (energy + orientation) are necessary for a reaction to occur. The reason does not claim they are sufficient for every possible reaction, but they are the minimum requirements.

How to avoid:

Separate the assertion (a claim) from the reason (an explanation).
The reason is a textbook definition — it is factually correct.
Do not overcomplicate: if the reason matches the standard theory, mark it as correct.

Common Mistake 3: Choosing Option (i) — Both correct, reason explains assertion

What students do:

They see both statements as true and assume the reason explains the assertion. They think: “Since the reason says only some collisions work, that explains why all collisions don’t lead to product formation.”

Why it’s wrong:

The assertion is false (all collisions do not lead to product formation). The reason is true. So the reason cannot explain a false assertion. Option (i) requires both statements to be correct.

How to avoid:

First, decide if the assertion is true or false.
Then decide if the reason is true or false.
Only then check if the reason explains the assertion (only possible if both are true).
Here: Assertion = false, Reason = true → correct answer is (iv).

Common Mistake 4: Choosing Option (ii) — Both correct, reason does not explain

What students do: …

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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.Assertion: All collision of reactant molecules lead to product formation.
Reason: Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.

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: Collision Theory of Chemical Reactions

Method: Statement Analysis Method

Common Mistake 1: Misreading the Assertion as True

Common Mistake 2: Thinking the Reason is Incorrect

Common Mistake 3: Choosing Option (i) — Both correct, reason explains assertion

Common Mistake 4: Choosing Option (ii) — Both correct, reason does not explain

Q.Assertion: All collision of reactant molecules lead to product formation. Reason: Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.

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Q.Assertion: All collision of reactant molecules lead to product formation.
Reason: Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.

Unlock everything free for 14 days