For a certain reaction large fraction of molecules has energy more than the threshold energy, yet the rate of

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Q.For a certain reaction large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow. Why?

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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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Even when many molecules have energy above the threshold, the reaction can be slow if the activation energy is high — because the Arrhenius equation shows that rate depends exponentially on the negative of activation energy, not just on the fraction above threshold. The key is that threshold energy and activation energy are related but not identical; a high activation energy means a tiny fraction of the already-energetic molecules actually have the correct orientation and sufficient energy to overcome the barrier in the transition state.

The question touches on a subtle but critical point in chemical kinetics. Let’s start with the core idea.

The Arrhenius equation is:

k = A e^{-E_a / RT}

Here, $k$ is the rate constant, $A$ is the frequency factor (related to collision frequency and orientation), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature. The fraction of molecules with energy greater than or equal to the activation energy is given by the Boltzmann factor $e^{-E_a / RT}$ . But the question says “large fraction of molecules has energy more than the threshold energy” — so why is the rate slow?

The trap is in the phrase “threshold energy.” In many textbooks, threshold energy ( $E_{th}$ ) is the minimum energy required for a reaction to occur. But the activation energy $E_a$ is the additional energy above the average energy of reactants that molecules must possess to reach the transition state. They are related by:

E_a = E_{th} - \text{(average energy of reactants)}

If a large fraction of molecules has energy above $E_{th}$ , that means the average energy of reactants is already high — perhaps because the temperature is high or the molecules are inherently energetic. But the rate constant $k$ depends on $e^{-E_a/RT}$ , and if $E_a$ itself is large, even a large fraction above threshold may correspond to a small fraction that actually surmounts the barrier from the average energy level.

Let’s break this down step by step.

Distinguish between threshold energy and activation energy.

Threshold energy $E_{th}$ is the absolute minimum energy a molecule must have to react. Activation energy $E_a$ is the energy barrier relative to the average energy of reactants. If the average energy of reactants is $\bar{E}$ , then $E_a = E_{th} - \bar{E}$ . So even if many molecules have energy above $E_{th}$ , the activation energy $E_a$ could still be large if $\bar{E}$ is also large — meaning the barrier from the average state is steep.
The Arrhenius exponential factor is about the fraction that overcomes the barrier from the average, not the fraction above threshold.

The Boltzmann factor $e^{-E_a/RT}$ gives the fraction of molecules with energy at least $E_a$ above the average. If $E_a$ is large, this fraction is tiny — even if the absolute number above $E_{th}$ is large. For example, suppose $E_{th} = 100$ kJ/mol, and the average energy $\bar{E} = 80$ kJ/mol, so $E_a = 20$ kJ/mol. At room temperature, $e^{-20000/(8.314 \times 298)} \approx e^{-8.07} \approx 0.0003$ , a very small fraction. But if many molecules have energy above 100 kJ/mol (say 40% of them), that doesn’t contradict the fact that only 0.03% have energy 20 kJ/mol above the average — because the average itself is high.
The frequency factor $A$ also matters.

Even if the exponential factor were large, a small $A$ (due to steric hindrance or orientation requirements) could make $k$ small. The question says “large fraction has energy more than threshold” — but orientation might be wrong. The Arrhenius equation’s $A$ includes the steric factor $p$ , which accounts for the fraction of collisions with proper orientation. If $p$ is very small (e.g., $10^{-6}$ ), the rate can be slow despite high energy.
A concrete analogy. …

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Here are the common mistakes students make when answering this question, along with how to avoid each.

Mistake 1: Forgetting the Role of Molecular Orientation

The Mistake: Students often assume that having enough energy (energy $\ge$ activation energy, $E_a$ ) is the only condition for a reaction to occur. They then get confused because the rate is slow despite a "large fraction" of molecules having sufficient energy.

Why it's Wrong: The Arrhenius equation ( $k = A e^{-E_a/RT}$ ) includes the pre-exponential factor $A$ , which accounts for the frequency of collisions and the steric factor (orientation). Even if molecules have enough kinetic energy, they must collide with the correct spatial orientation for bonds to break and form.

How to Avoid: Always remember the Collision Theory checklist:

Energy: Molecules must have $E \ge E_a$ .
Orientation: Molecules must hit each other at the reactive site (e.g., the correct end of a polar molecule).
Frequency: They must actually collide often enough.

The Answer: The reaction is slow because, despite sufficient energy, the molecules are not colliding with the correct orientation (or the collision frequency is low). The steric factor (part of $A$ ) is very small.

Mistake 2: Misinterpreting "Large Fraction"

The Mistake: Students think "large fraction" means "most molecules" or "almost all molecules." They then assume the rate must be fast.

Why it's Wrong: In chemical kinetics, "large fraction" is relative. For a reaction with a very high activation energy ( $E_a$ ), even a "large fraction" might be a tiny absolute number (e.g., 1 in $10^{15}$ ). The fraction is given by $e^{-E_a/RT}$ . If $E_a$ is huge, $e^{-E_a/RT}$ is still extremely small, even if it's "large" compared to another reaction.

How to Avoid: Always quantify the fraction mentally. Use the Boltzmann distribution:

Fraction $= e^{-E_a/RT}$
If $E_a$ is very high, this fraction is always tiny (e.g., $10^{-10}$ ).
"Large" is a comparative term, not an absolute one.

The Answer: The "large fraction" is still a very small absolute number because the activation energy is extremely high. The rate constant $k$ is proportional to this tiny fraction, so the rate is slow.

Mistake 3: Confusing Threshold Energy with Activation Energy

The Mistake: Students use the terms "threshold energy" and "activation energy" interchangeably without realizing the nuance. They then fail to explain why a high threshold energy slows the reaction.

Why it's Wrong: Threshold energy ( $E_{th}$ ) is the minimum total energy (kinetic + potential) required for a reaction. Activation energy ( $E_a$ ) is the additional energy the reactants need above their average energy to reach the transition state. The Arrhenius equation uses $E_a$ , not $E_{th}$ .

How to Avoid: Be precise:

$E_a = E_{th} - \text{(average energy of reactants)}$
A high $E_{th}$ means a high $E_a$ , which makes the exponential term $e^{-E_a/RT}$ very small.

The Answer: The reaction has a high activation energy ( $E_a$ ). Even though a "large fraction" of molecules have energy above the threshold, the exponential factor $e^{-E_a/RT}$ is still extremely small, leading to a small rate constant $k$ and a slow reaction.

Mistake 4: Ignoring the Pre-Exponential Factor ( $A$ ) Entirely

The Mistake: Students only discuss the exponential term ( $e^{-E_a/RT}$ ) and completely forget that $A$ (the frequency factor) can also be very small. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

COMEDK 2026Set 2026-A1 markMCQ
$50 \%$
›Reveal solutionSolution
For a first‑order reaction, the half‑life is inversely proportional to the rate constant. Using the two half‑lives at different temperatures in the Arrhenius equation gives the activation energy. The result is approximately 43.8 kJ mol⁻¹, which corresponds to option (B).

Concept & Intuition

For a first‑order reaction, the half‑life $t_{1/2}$ is related to the rate constant $k$ by
$t_{1/2} = \frac{\ln 2}{k}.$
Thus, $k \propto 1/t_{1/2}$ . When temperature changes, the rate constant changes according to the Arrhenius equation:
$\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right).$
Since $k \propto 1/t_{1/2}$ , we can replace the ratio of rate constants by the inverse ratio of half‑lives. This lets us find $E_a$ directly from the given half‑life data.

Step‑by‑step solution

Write the half‑life relation for a first‑order reaction

$t_{1/2} = \frac{\ln 2}{k} \quad \Rightarrow \quad k = \frac{\ln 2}{t_{1/2}}.$
So at temperature $T_1 = 300\ \text{K}$ , $t_{1/2}^{(1)} = 30\ \text{min}$ ; at $T_2 = 320\ \text{K}$ , $t_{1/2}^{(2)} = 10\ \text{min}$ .

Form the ratio of rate constants

$\frac{k_2}{k_1} = \frac{\ln 2 / t_{1/2}^{(2)}}{\ln 2 / t_{1/2}^{(1)}} = \frac{t_{1/2}^{(1)}}{t_{1/2}^{(2)}} = \frac{30}{10} = 3.$

Apply the Arrhenius equation in logarithmic form

$\ln \frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right).$
Substitute $\frac{k_2}{k_1} = 3$ and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ :
$\ln 3 = \frac{E_a}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right).$

Compute the temperature difference term

$\frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800}.$
So
$\ln 3 = \frac{E_a}{8.314} \times \frac{1}{4800}.$

Solve for $E_a$

$E_a = 8.314 \times 4800 \times \ln 3.$
Use $\ln 3 = 2.3026 \times \log_{10} 3$ (since $\ln x = 2.3026 \log_{10} x$ ). Given $\log 3 = 0.4771$ , …
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COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{k}_1$
›Reveal solutionSolution
The key idea is to set the two Arrhenius‑type rate constants equal and solve for temperature. The result is $T = \frac{1000}{2.303}$  K, which corresponds to option (C).

We are given two rate constants of the Arrhenius form:

$k_1 = 10^{16} e^{-2000/T}$ and $k_2 = 10^{15} e^{-1000/T}$ .

The problem asks for the temperature at which they are equal.

The natural approach is to set $k_1 = k_2$ and solve for $T$ . Because the expressions involve exponentials, we will use logarithms to bring the exponents down.

Set the two expressions equal

$10^{16} e^{-2000/T} = 10^{15} e^{-1000/T}$

Divide both sides by $10^{15}$ to simplify the pre‑exponential factors:

$\frac{10^{16}}{10^{15}} e^{-2000/T} = e^{-1000/T}$ $10 \, e^{-2000/T} = e^{-1000/T}$

Take the natural logarithm of both sides

$\ln(10) + \ln\!\left(e^{-2000/T}\right) = \ln\!\left(e^{-1000/T}\right)$
Since $\ln(e^u) = u$ , this becomes:
$\ln(10) - \frac{2000}{T} = -\frac{1000}{T}$

Collect the terms with $T$ Add $\frac{2000}{T}$ to both sides:

$\ln(10) = \frac{2000}{T} - \frac{1000}{T}$ $\ln(10) = \frac{1000}{T}$

Solve for $T$

$T = \frac{1000}{\ln(10)}$
Now recall that $\ln(10) \approx 2.302585$ , so we can write:

$T = \frac{1000}{2.303} \ \text{K}$ …
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COMEDK 2025Set 2025-A1 markMCQ
$20 \mathrm{~kJ} / \mathrm{mol}$
›Reveal solutionSolution
The key idea is that the catalyst keeps the rate constant the same at a lower temperature by reducing the activation energy. Using the Arrhenius equation in logarithmic form, we equate the rate constants at the two different conditions and solve for the original activation energy, which comes out to be $60\ \mathrm{kJ/mol}$ .

Concept and Intuition

The Arrhenius equation tells us that the rate constant $k$ depends on both the activation energy $E_a$ and the temperature $T$ :
$k = A e^{-E_a/(RT)}$
If a catalyst lowers $E_a$ by $\Delta E$ , the reaction can proceed at the same rate (same $k$ ) at a lower temperature. Here, the uncatalyzed reaction at 600 K has the same rate constant as the catalyzed reaction at 400 K, with the catalyst reducing $E_a$ by 20 kJ/mol. We set the two Arrhenius expressions equal and solve for the original $E_a$ .

Step-by-step solution

Write the Arrhenius equation for both cases. For the uncatalyzed reaction at $T_1 = 600\ \mathrm{K}$ :

$k = A e^{-E_a/(R T_1)}$
For the catalyzed reaction at $T_2 = 400\ \mathrm{K}$ , the activation energy is $E_a - 20$ (in kJ/mol):
$k = A e^{-(E_a - 20)/(R T_2)}$
The pre-exponential factor $A$ is assumed unchanged by the catalyst.

Equate the two expressions for $k$ . Since the rate constant is the same:

$A e^{-E_a/(R T_1)} = A e^{-(E_a - 20)/(R T_2)}$
Cancel $A$ (non-zero) and take natural logs:
$-\frac{E_a}{R T_1} = -\frac{E_a - 20}{R T_2}$
Multiply both sides by $-R$ :
$\frac{E_a}{T_1} = \frac{E_a - 20}{T_2}$

Substitute the temperatures (in Kelvin).
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COMEDK 2025Set 2025-A1 markMCQ
$Q. (i) and$
(ii) are 2 chemical reactions carried out at TK. (i). $\mathrm{X} \rightarrow \mathrm{Y}+\mathrm{W}$ with $\mathrm{k}_1$ as rate constant. (ii). $\mathrm{X} \rightarrow \mathrm{Z}+\mathrm{W}$ with $k_2$ as rate constant. The Activation energy for reaction
(ii) is 3 times that of reaction (i). What is the expression for $k_1$ ? (A) $k_1=-2 k_2 e^{2 E a_1 / R T}$ (B) $k_1=k_2 e^{2 E a_1 / R T}$ (C) $k_1=2 k_2 e^{E a_1 / R T}$ (D) $k_1=k_2 / 2 e^{-E a_1 / R T}$
›Reveal solutionSolution
The problem gives two parallel reactions from the same reactant X, with activation energies related by $E_{a2} = 3E_{a1}$ . Using the Arrhenius equation and the fact that the pre-exponential factors are equal (same reactant, similar reactions), we find $k_1 = k_2 e^{2E_{a1}/RT}$ , which corresponds to option (B).

The key idea here is the Arrhenius equation, which relates the rate constant $k$ to the activation energy $E_a$ and temperature $T$ :
$k = A e^{-E_a / RT}$
where $A$ is the pre-exponential factor (frequency factor). For two reactions starting from the same reactant and producing similar products, it is reasonable to assume the pre-exponential factors are equal: $A_1 = A_2$ . The problem gives $E_{a2} = 3E_{a1}$ . We can then write expressions for $k_1$ and $k_2$ and eliminate the common $A$ to relate them.

Write the Arrhenius equations for both reactions For reaction (i): $k_1 = A_1 e^{-E_{a1}/RT}$ For reaction (ii): $k_2 = A_2 e^{-E_{a2}/RT}$ Given $E_{a2} = 3E_{a1}$ and assuming $A_1 = A_2 = A$ , we have:

$k_1 = A e^{-E_{a1}/RT}, \quad k_2 = A e^{-3E_{a1}/RT}$

Divide the two equations to eliminate $A$

$\frac{k_1}{k_2} = \frac{A e^{-E_{a1}/RT}}{A e^{-3E_{a1}/RT}} = e^{(-E_{a1} + 3E_{a1})/RT} = e^{2E_{a1}/RT}$

Solve for $k_1$ Multiply both sides by $k_2$ : $k_1 = k_2 \, e^{2E_{a1}/RT}$ …
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COMEDK 2025Set 2025-E1 markMCQ
$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$
›Reveal solutionSolution

From the Arrhenius equation with $k_2/k_1=1.12$ between $300\ \text{K}$ and $310\ \text{K}$ , $E_a\approx 8.76\ \text{kJ/mol}$ — option (D).

A $12\%$ increase means $\dfrac{k_2}{k_1}=1.12$ . The two-temperature Arrhenius form:
$\ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right).$ $\frac{1}{T_1}-\frac{1}{T_2}=\frac{1}{300}-\frac{1}{310}=\frac{10}{93000}=1.075\times10^{-4}\ \text{K}^{-1},$ $\ln(1.12)=0.1133.$
Solving for $E_a$ (with $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$ ): …
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COMEDK 2025Set 2025-M1 markMCQ
$(\mathrm{k})$
›Reveal solutionSolution
The assertion that the rate constant roughly doubles for a 10 °C rise is empirically true for many reactions near room temperature, but the reason given — increased bimolecular collision frequency — is not the correct explanation; the real cause is the exponential increase in the fraction of molecules with energy above the activation barrier.

The key concept here is the Arrhenius equation, which separates the effect of temperature on reaction rates into two parts: the collision frequency and the fraction of collisions that are energetic enough to overcome the activation energy. While both factors increase with temperature, the dominant effect for a typical 10 °C rise is the exponential increase in the fraction of high-energy molecules, not the modest increase in collision frequency.

Understanding the Assertion (A) The statement that the rate constant $k$ nearly doubles for a 10 °C rise is a well-known rule of thumb for many reactions near room temperature. This comes from the Arrhenius equation:

$k = A e^{-E_a/(RT)}$
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.

For a typical activation energy of about 50 kJ/mol, a 10 °C rise from 300 K to 310 K gives:
$\frac{k_{310}}{k_{300}} = \exp\left[-\frac{E_a}{R}\left(\frac{1}{310} - \frac{1}{300}\right)\right] \approx \exp\left( \frac{50\,000}{8.314} \times \frac{10}{300 \times 310} \right) \approx e^{0.65} \approx 1.92$
So the assertion is correct for many common reactions.

Understanding the Reason (R) The reason states that the number of bimolecular collisions between reactant molecules increases with temperature. This is true — from kinetic molecular theory, the collision frequency is proportional to $\sqrt{T}$ . For a 10 °C rise from 300 K to 310 K, the increase is:

$\frac{\sqrt{310}}{\sqrt{300}} \approx 1.016$
That’s only about a 1.6% increase, far from the ~100% increase in the rate constant. So while the reason is factually correct, it is not sufficient to explain the near-doubling of $k$ .

Why the reason fails as an explanation …
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COMEDK 2025Set 2025-M1 markMCQ
$26.8 \mathrm{~kJ} / \mathrm{mol}$
›Reveal solutionSolution
The ratio of rate constants for two reactions with equal frequency factors depends only on the difference in activation energies via the Arrhenius equation. Using $\Delta E_a = 26.8\ \mathrm{kJ/mol}$ at $T=300\ \mathrm{K}$ , we find $x = e^{\Delta E_a/(RT)} \approx 4.631 \times 10^4$ , so option (A) is correct.

The key idea is the Arrhenius equation:

$k = A e^{-E_a/(RT)}$

When two reactions have the same frequency factor $A$ , the ratio of their rate constants depends only on the difference in activation energies. This lets us directly compute $x = k_2/k_1$ without needing absolute values.

Write the Arrhenius expressions For reaction 1: $k_1 = A e^{-E_{a1}/(RT)}$ For reaction 2: $k_2 = A e^{-E_{a2}/(RT)}$ Since $A$ is equal, the ratio is:

$\frac{k_2}{k_1} = e^{-(E_{a2} - E_{a1})/(RT)}$

Interpret the given difference The activation energies differ by $26.8\ \mathrm{kJ/mol}$ . We are not told which is larger, but the problem states $k_2 = x k_1$ . If $E_{a2} < E_{a1}$ , then $k_2 > k_1$ and $x > 1$ . Given the options, $x$ is large, so we take $E_{a1} - E_{a2} = 26.8\ \mathrm{kJ/mol}$ (i.e., reaction 2 has the lower activation energy). Thus:

$\frac{k_2}{k_1} = e^{(E_{a1} - E_{a2})/(RT)} = e^{\Delta E_a/(RT)}$

Plug in values with consistent units $\Delta E_a = 26.8\ \mathrm{kJ/mol} = 26800\ \mathrm{J/mol}$ $R = 8.314\ \mathrm{J/(mol \cdot K)}$ $T = 300\ \mathrm{K}$ So:
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KCET 2024Set B-21 markMCQ
$\log k = \log A - \dfrac{E_a}{2.303RT}$
›Reveal solutionSolution

Three options are algebraic rearrangements of $k = Ae^{-E_a/RT}$ ; the odd one out has lost the minus sign in the exponent.

Step 1 — The Arrhenius equation.
$k = A\,e^{-E_a/RT}$
The exponential factor $e^{-E_a/RT}$ is the Boltzmann fraction — the fraction of molecular collisions with energy at least $E_a$ . A fraction must be $\le 1$ , which forces the exponent to be negative.

Step 2 — Check each option.

(B) $k = Ae^{-E_a/RT}$ — this is the Arrhenius equation. ✓ (represents it)

(C) Take $\ln$ of (B):
$\ln k = \ln A + \ln\!\left(e^{-E_a/RT}\right) = \ln A - \frac{E_a}{RT} = -\frac{E_a}{RT} + \ln A. \;✓$
(A) Convert (C) to base-10 using $\ln y = 2.303\log y$ :
$2.303\log k = 2.303 \log A - \frac{E_a}{RT} \;\Rightarrow\; \log k = \log A - \frac{E_a}{2.303\,RT}. \;✓$
(D) $k = Ae^{+E_a/RT}$ — the sign is flipped. ✗ …
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COMEDK 2024Set 2024-E1 markMCQ
$27^{\circ} \mathrm{C}$
›Reveal solutionSolution

Lowering $E_a$ by 100 J/mol at 300 K raises the rate constant by a factor $e^{100/RT}=e^{0.0401}\approx1.04$ .

From the Arrhenius equation, at the same temperature:

$\frac{k_2}{k_1}=\frac{A e^{-E_{a2}/RT}}{A e^{-E_{a1}/RT}}=e^{(E_{a1}-E_{a2})/RT}=e^{\Delta E_a/RT}$ …
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COMEDK 2024Set 2024-E1 markMCQ
$A \rightarrow B+C$
›Reveal solutionSolution

$A/k=e^{E_a/RT}=e^{0.04606}\approx1.047$ .

Arrhenius equation: $k=Ae^{-E_a/RT}$ , so
$\frac{A}{k}=e^{E_a/RT}$
Here $E_a=0.04606\,RT$ , hence $E_a/RT=0.04606$ . …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}$
›Reveal solutionSolution

Apply the two-temperature Arrhenius equation with $k_1=0.004\,\text{s}^{-1}$ at $500\,\text{K}$ and $k_2=0.014\,\text{s}^{-1}$ ; solving gives $T_2\approx700\,\text{K}$ .

$\ln\frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$
With $E_a=18231\,\text{J mol}^{-1}$ , $R=8.314\,\text{J mol}^{-1}\text{K}^{-1}$ :

$\ln\frac{0.014}{0.004}=\ln(3.5)=1.253,\qquad \frac{E_a}{R}=\frac{18231}{8.314}=2193.$ …
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KCET 2023Set D-21 markMCQ
$500$
›Reveal solutionSolution
A catalyst does not change the equilibrium constant — it only speeds up the rate at which equilibrium is reached. So $K_c$ remains $2 \times 10^{-5}$ at the same temperature. The correct option is (D).

The key idea here is simple but often misunderstood: a catalyst affects the rate of a reaction, not the position of equilibrium. Let’s see why.

What a catalyst does

A catalyst provides an alternative reaction pathway with a lower activation energy. This means both the forward and reverse reactions are accelerated equally. Because it speeds up both directions by the same factor, the ratio of the forward rate constant to the reverse rate constant — which is exactly $K_c$ — stays unchanged.

Equilibrium constant depends only on temperature

For a given reaction, $K_c$ is a function of temperature alone. It is determined by the standard Gibbs free energy change:

$\Delta G^\circ = -RT \ln K_c$
A catalyst does not alter $\Delta G^\circ$ ; it only lowers the activation barrier. So at the same temperature, $K_c$ is fixed.

The “10 times faster” detail …
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Q.For a certain reaction large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow. Why?

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 Explanation: Arrhenius Equation & Activation Energy

Method: Energy Barrier + Orientation Factor Analysis

Step 1: Identify what "large fraction has energy > threshold" tells us

Step 2: Identify the missing factor — the frequency factor $A$

Step 3: Apply the logic

Mistake 1: Forgetting the Role of Molecular Orientation

Mistake 2: Misinterpreting "Large Fraction"

Mistake 3: Confusing Threshold Energy with Activation Energy

Mistake 4: Ignoring the Pre-Exponential Factor ( $A$ ) Entirely

Q.For a certain reaction large fraction of molecules has energy more than the threshold energy, yet the rate of reaction is very slow. Why?

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