According to Maxwell Boltzmann distribution of energy, _____. (Two or more than two options may be correct.) (

NCERT Exemplar · Q52

Q. According to Maxwell Boltzmann distribution of energy, __________. (Two or more than two options may be correct.)

(i) the fraction of molecules with most probable kinetic energy decreases at higher temperatures.

(ii) the fraction of molecules with most probable kinetic energy increases at higher temperatures.

(iii) most probable kinetic energy increases at higher temperatures.

(iv) most probable kinetic energy decreases at higher temperatures.

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Concept understanding — Arrhenius Equation

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: 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 picture step by step.

1. The Core Question

Why does reaction rate increase dramatically with temperature?

For many reactions, a 10 °C rise can double or triple the rate. This cannot be explained by simple kinetic energy arguments alone — the relationship is exponential.

2. The Key Insight: An Energy Barrier

Before molecules can react, they must collide — but not all collisions lead to products.

There is a minimum energy threshold called activation energy ( $E_a$ ). Only collisions with energy ≥  $E_a$ can break old bonds and form new ones.

Think of it like pushing a boulder over a hill:

The hilltop is the transition state (activated complex).
$E_a$ is the height of that hill from the reactant valley.

3. The Boltzmann Factor — The "Why" of Exponential Dependence

At temperature $T$ , the fraction of molecules with energy ≥  $E_a$ is given by the Boltzmann distribution:

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

where:

$R$ = universal gas constant (8.314 J mol⁻¹ K⁻¹)
$T$ = absolute temperature (Kelvin)

Why this form?

The Boltzmann distribution tells us that the probability of a molecule having energy $E$ is proportional to $e^{-E/(k_B T)}$ .
For 1 mole of molecules, $k_B$ (Boltzmann constant) becomes $R$ via $R = N_A k_B$ .
So the fraction with energy ≥ $E_a$ is the integral of that distribution from $E_a$ to ∞, which yields $e^{-E_a/(RT)}$ .

Key takeaway: This exponential factor is not arbitrary — it comes directly from statistical mechanics.

4. The Pre-Exponential Factor ( $A$ )

Even if a collision has enough energy, it must also:

Have the correct orientation (steric factor)
Occur with sufficient collision frequency

These are bundled into the pre-exponential factor $A$ (also called the frequency factor):

A = \text{(collision frequency)} \times \text{(orientation factor)}

For simple gas-phase reactions, collision frequency can be calculated from kinetic molecular theory — it's on the order of $10^{10}$  L mol⁻¹ s⁻¹.

5. Putting It Together: The Arrhenius Equation

The rate constant $k$ is proportional to:

The number of effective collisions per second (given by $A$ )
The fraction of collisions with sufficient energy (given by $e^{-E_a/(RT)}$ )

Thus:

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

This is the Arrhenius equation.

6. Why It Works — The Physical Logic

Component	Physical meaning	Why it's there
$A$	Maximum possible rate if every collision worked	Accounts for collision frequency & geometry

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The Maxwell-Boltzmann distribution shifts and flattens at higher temperatures: the most probable kinetic energy increases, but the fraction of molecules at that exact energy decreases. So options (i) and (iii) are correct.

The Maxwell-Boltzmann distribution describes how molecular speeds (and therefore kinetic energies) are spread among molecules in a gas at a given temperature. The key idea is that temperature is a measure of the average kinetic energy — so when you heat a gas, the entire distribution shifts to the right (higher energies) and also broadens out.

Let’s unpack what happens to the most probable kinetic energy and the fraction of molecules at that energy.

What is the most probable kinetic energy? The most probable kinetic energy, $E_{mp}$ , is the energy at which the distribution curve peaks. From kinetic theory, for an ideal gas, the most probable kinetic energy is directly proportional to temperature:

E_{mp} = \frac{1}{2} k_B T

(where $k_B$ is Boltzmann’s constant). This is because the peak of the speed distribution shifts to higher speeds as $T$ increases, and kinetic energy goes as $v^2$ . So as temperature rises, $E_{mp}$ increases. That immediately tells us option (iii) is correct and (iv) is wrong.

What happens to the fraction of molecules at $E_{mp}$ ? The distribution function $f(E)$ gives the probability density for a molecule to have energy $E$ . At the peak, $f(E_{mp})$ is the maximum value of the curve. But here’s the crucial point: as temperature increases, the curve becomes broader and flatter — the peak height actually decreases. Why? Because the total area under the curve must remain 1 (all molecules are accounted for). If the curve spreads out to higher energies, the peak must come down to keep the area constant. So the fraction of molecules with exactly the most probable kinetic energy decreases at higher temperatures. This makes option (i) correct and (ii) wrong. …

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Let’s break this down. The question tests your understanding of the Maxwell-Boltzmann distribution of molecular kinetic energies — specifically, how temperature affects the most probable kinetic energy and the fraction of molecules having that energy.

✓ Correct Answer

(i) the fraction of molecules with most probable kinetic energy decreases at higher temperatures.
(iii) most probable kinetic energy increases at higher temperatures.

So the correct options are A and C.

✗ Common Mistakes & How to Avoid Them

1. Confusing “most probable kinetic energy” with “fraction of molecules at that energy”

Mistake: Students think that if the most probable kinetic energy increases, the fraction of molecules with that energy must also increase.

Why it’s wrong:

At higher temperature, the distribution curve flattens and broadens. The peak shifts to the right (higher energy), but the height of the peak decreases. So the fraction of molecules at the most probable energy actually goes down.

How to avoid:

Visualise the curve:

At low $T$ : tall, narrow peak.
At high $T$ : shorter, wider peak, shifted right. The peak height = fraction at most probable energy. It decreases with temperature.

2. Thinking “most probable kinetic energy” is constant

Mistake: Choosing option (iv) — that most probable kinetic energy decreases at higher temperature.

Why it’s wrong:

From kinetic theory, the most probable kinetic energy is proportional to $kT$ :

E_{\text{mp}} = \frac{1}{2} kT

So as $T$ increases, $E_{\text{mp}}$ increases.

How to avoid:

Remember the formula:

Most probable speed: $v_{\text{mp}} = \sqrt{\frac{2kT}{m}}$
Most probable kinetic energy: $\frac{1}{2} m v_{\text{mp}}^2 = kT$ (for 1D) or $\frac{1}{2}kT$ per degree of freedom. Both increase linearly with $T$ .

3. Misreading “fraction of molecules” as “number of molecules”

Mistake: Thinking that because the total number of molecules is fixed, the fraction at the most probable energy must stay the same.

Why it’s wrong: …

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Q.According to Maxwell Boltzmann distribution of energy, __________. (Two or more than two options may be correct.)

Concept understanding — Arrhenius Equation

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: Why It Holds

1. The Core Question

2. The Key Insight: An Energy Barrier

3. The Boltzmann Factor — The "Why" of Exponential Dependence

4. The Pre-Exponential Factor ( $A$ )

5. Putting It Together: The Arrhenius Equation

6. Why It Works — The Physical Logic

Concept: Maxwell-Boltzmann Distribution of Molecular Energies

Method: Graphical Analysis of Distribution Curves

Steps:

✓ Correct Answer

✗ Common Mistakes & How to Avoid Them

1. Confusing “most probable kinetic energy” with “fraction of molecules at that energy”

2. Thinking “most probable kinetic energy” is constant

3. Misreading “fraction of molecules” as “number of molecules”

Q.According to Maxwell Boltzmann distribution of energy, __________. (Two or more than two options may be correct.)

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