What happens to most probable kinetic energy and the energy of activation with increase in temperature?

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Q. What happens to most probable kinetic energy and the energy of activation with increase in temperature?

Arunachal CbseLong· 2mImportance★★★★★

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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 Arrhenius equation shows that both the most probable kinetic energy and the activation energy are independent of temperature — the former because it is a fixed property of the Maxwell–Boltzmann distribution shape, and the latter because it is an intrinsic energy barrier for a reaction.

Why this question is tricky

Many students instinctively think that raising the temperature "gives molecules more energy" and therefore must change the activation energy or the most probable kinetic energy. That is a misunderstanding. Temperature changes the distribution of molecular energies, not the fixed energy values that define the reaction.

Let’s separate the two quantities clearly.

1. Most probable kinetic energy ( $E_{mp}$ )

The most probable kinetic energy comes from the Maxwell–Boltzmann distribution of molecular speeds. For an ideal gas, the distribution of kinetic energies is given by:

f(E) = \frac{2}{\sqrt{\pi}} \left( \frac{1}{k_B T} \right)^{3/2} \sqrt{E} \, e^{-E/k_B T}

The most probable kinetic energy is the energy at which $f(E)$ is maximum. Differentiating and setting the derivative to zero gives:

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

This is a function of temperature — it increases linearly with $T$ . So the most probable kinetic energy does increase with temperature.

Watch out

A common mistake is to confuse "most probable kinetic energy" with "activation energy." They are completely different concepts. The most probable kinetic energy is a statistical property of the molecular population; activation energy is a fixed barrier for a specific reaction.

2. Energy of activation ( $E_a$ )

The activation energy is defined by the Arrhenius equation:

k = A e^{-E_a / RT}

Here $k$ is the rate constant, $A$ is the pre-exponential factor, $R$ is the gas constant, and $T$ is temperature. $E_a$ is the minimum energy that reactant molecules must possess for a collision to result in a reaction.

Crucially, $E_a$ is a constant for a given reaction under normal conditions. It does not change with temperature. What changes with temperature is the fraction of molecules that have energy $\geq E_a$ — that fraction increases as $T$ rises, which is why reaction rates increase.

E_a = \text{constant (for a given reaction)}

--- …

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Here are the common mistakes students make on this concept (Arrhenius Equation + Kinetic Energy) and how to avoid each.

Mistake 1: Confusing “Most Probable Kinetic Energy” with “Average Kinetic Energy”

What students do wrong:

They think both quantities behave identically. They often say “most probable kinetic energy increases by the same factor as average kinetic energy.”

Why it’s wrong:

Average kinetic energy ( $\overline{KE}$ ) is proportional to $T$ : $\overline{KE} = \frac{3}{2}kT$
Most probable kinetic energy ( $KE_{mp}$ ) is also proportional to $T$ : $KE_{mp} = \frac{1}{2}kT$

Both increase linearly with $T$ , but they are different values — the most probable is half the average. The mistake is treating them as identical.

How to avoid:

Memorise the two distinct formulas:
- $\overline{KE} \propto T$
- $KE_{mp} \propto T$
Remember: both increase with temperature, but the most probable is always smaller than the average.

Mistake 2: Thinking Activation Energy ( $E_a$ ) Changes with Temperature

What students do wrong:

They assume $E_a$ increases or decreases when temperature rises, because the reaction rate changes.

Why it’s wrong:

$E_a$ is an intrinsic property of the reaction — it depends on the nature of reactants and the transition state, not on temperature.
The Arrhenius equation: $k = A e^{-E_a / RT}$ shows that $E_a$ appears as a constant in the exponent. Only the fraction of molecules with energy $\geq E_a$ changes with $T$ .

How to avoid:

Treat $E_a$ like a fixed barrier height — like the height of a wall. Raising temperature doesn’t lower the wall; it gives more molecules the energy to jump over it.
Practice saying: “ $E_a$ is constant for a given reaction; temperature changes the number of molecules that can cross it.”

Mistake 3: Confusing “Most Probable Kinetic Energy” with “Energy of Activation”

What students do wrong:

They think $KE_{mp}$ and $E_a$ are the same thing, or that $E_a$ equals the most probable energy.

Why it’s wrong:

$KE_{mp}$ is the most common kinetic energy among molecules at a given temperature.
$E_a$ is the minimum energy required for a reaction to occur — it is usually much larger than $KE_{mp}$ .
They are different concepts and different values.

How to avoid:

Draw a Maxwell-Boltzmann distribution curve. Mark $KE_{mp}$ at the peak. Mark $E_a$ far to the right on the tail.
Remember: $KE_{mp}$ is about what most molecules have; $E_a$ is about what molecules need to react.

Mistake 4: Saying “Most Probable Kinetic Energy Decreases with Temperature”

What students do wrong: …

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Q.What happens to most probable kinetic energy and the energy of activation with increase in temperature?

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

Why this question is tricky

1. Most probable kinetic energy ( $E_{mp}$ )

2. Energy of activation ( $E_a$ )

Arrhenius Equation — Effect of Temperature on Energy Terms

Method: Kinetic Molecular Interpretation of Arrhenius Theory

Step 1: Recall the Arrhenius Equation

Step 2: Effect on Most Probable Kinetic Energy ( $E_{mp}$ )

Step 3: Effect on Activation Energy ( $E_a$ )

Mistake 1: Confusing “Most Probable Kinetic Energy” with “Average Kinetic Energy”

Mistake 2: Thinking Activation Energy ( $E_a$ ) Changes with Temperature

Mistake 3: Confusing “Most Probable Kinetic Energy” with “Energy of Activation”

Mistake 4: Saying “Most Probable Kinetic Energy Decreases with Temperature”

Q.What happens to most probable kinetic energy and the energy of activation with increase in temperature?

Why this question is tricky

1. Most probable kinetic energy (EmpE_{mp}Emp​)

2. Energy of activation (EaE_aEa​)

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1. Most probable kinetic energy ( $E_{mp}$ )

2. Energy of activation ( $E_a$ )

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