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?

Andhra Pradesh BieapShort· 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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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 13 on this concept.

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\exp\left(-\dfrac{E_a}{RT}\right)$
›Reveal solutionSolution
This tests the meaning of each term in the Arrhenius equation; the exponential term represents the fraction of molecules with energy exceeding the activation energy.

Concept and Intuition

The Arrhenius equation, $k = A e^{-E_a/RT}$ , splits the rate constant into two physically distinct factors: $A$ (the frequency/pre-exponential factor, related to collision frequency and orientation) and $e^{-E_a/RT}$ (the fraction of molecular collisions that have enough energy to cross the activation energy barrier, derived from the Maxwell-Boltzmann energy distribution). Only molecules with energy $\geq E_a$ can react upon collision; the exponential term quantifies what fraction of the population meets this threshold at a given temperature.

Step-by-Step Solution

Write the Arrhenius equation: $k = A\,e^{-E_a/RT}$ .

Recognize $A$ is the frequency factor — related to the rate of collisions and their proper orientation, independent of energy considerations.

Recognize $e^{-E_a/RT}$ arises from integrating the Maxwell-Boltzmann distribution above the energy threshold $E_a$ — it represents the fraction of molecules whose energy exceeds $E_a$ at temperature $T$ . …
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AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$2.5 \times 10^{-4} s^{-1}$
›Reveal solutionSolution
This tests the Arrhenius equation relating rate constants at two temperatures; solving gives $T_2 \approx 625\,K$ .

Concept and Intuition

A faster half-life means a larger rate constant. The Arrhenius equation connects the ratio of two rate constants at two temperatures to the activation energy — a bigger $E_a$ needs a bigger temperature jump to boost $k$ by the same factor.

Step-by-Step Solution

$T_1 = 300°C = 573\,K$ , $k_1 = 2.5\times10^{-4}\,s^{-1}$ .

Desired half-life $t_{1/2}=138.6\,s$ for a first-order reaction: $k_2 = \dfrac{0.693}{138.6} = 5\times10^{-3}\,s^{-1}$ .

Ratio: $k_2/k_1 = \dfrac{5\times10^{-3}}{2.5\times10^{-4}} = 20$ .

Arrhenius (two-temperature form): $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

$\dfrac{E_a}{2.303R} = \dfrac{42000}{2.303\times2} = 9118.5$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$1.6\times10^{-5}s^{-1}$
›Reveal solutionSolution
The Arrhenius equation converts rate constants at two temperatures via activation energy; the given antilog hint (antilog(0.4771)=3.0) is a shortcut for the final power-of-ten step, giving $k_2=4.8\times10^{-3}\ s^{-1}$ .

Concept and Intuition

The two-point form of the Arrhenius equation lets us find how much a rate constant changes when temperature changes, purely from the activation energy — a higher $E_a$ makes the rate far more sensitive to temperature. The problem hands us the exact antilog we'll need at the end, which is a strong signal the intended computation lands exactly on $\log(k_2/k_1)=2.4771$ .

Step-by-Step Solution

Write the two-temperature Arrhenius relation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Compute $\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{600}-\dfrac{1}{700} = \dfrac{700-600}{600\times700} = \dfrac{100}{420000} = \dfrac{1}{4200}$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{198870}{2.303\times8.3} = \dfrac{198870}{19.1149} \approx 10403.9$ .

Multiply: $10403.9 \times \dfrac{1}{4200} \approx 2.4771$ . …
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AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$\to$
›Reveal solutionSolution
This tests the two-point form of the Arrhenius equation to find an unknown temperature given a rate constant ratio. The answer works out to T ≈ 311 K.

Concept and Intuition

The Arrhenius equation $k = A e^{-E_a/RT}$ can be converted, for two temperatures, into a linear relation between $\log(k_2/k_1)$ and $(1/T_1 - 1/T_2)$ , letting us solve for an unknown temperature when the rate constants and activation energy are known.

Step-by-Step Solution

Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303 R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Given $k_1 = 5\times10^3\ s^{-1}$ at $T_1 = 300\ K$ , and $k_2 = 1.0\times10^4\ s^{-1}$ at $T_2 = T$ (to find). Ratio $k_2/k_1 = \dfrac{1.0\times10^4}{5\times10^3} = 2$ .

Given $\log 2 = 0.3$ , so LHS $= 0.3$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{50000}{2.303 \times 8.3} = \dfrac{50000}{19.115} \approx 2615.7$ .

So $0.3 = 2615.7\left(\dfrac{1}{300} - \dfrac{1}{T}\right)$ , giving $\dfrac{1}{300} - \dfrac{1}{T} = \dfrac{0.3}{2615.7} \approx 1.1468\times10^{-4}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$\left(\dfrac{k_{610}}{k_{600}}\right)$
›Reveal solutionSolution
Same reaction time for 10% completion at 600K and 20% completion at 610K lets the time cancel, leaving $k_{610}/k_{600}=\log(1.25)/\log(1.111)\approx2.12$ .

Concept and Intuition

For a first-order reaction, $k=\dfrac{1}{t}\ln\dfrac{1}{1-x}$ where $x$ is the fraction reacted. If the same time $t$ produces different completions at two temperatures, that common $t$ cancels out when taking the ratio of the two rate constants, leaving a ratio of two logarithmic (completion) factors only.

Step-by-Step Solution

At 600 K, 10% complete in time $t$ : $k_{600} = \dfrac{1}{t}\ln\dfrac{1}{0.90} = \dfrac{2.303}{t}\log(1.111)$ .

At 610 K, 20% complete in the same time $t$ : $k_{610} = \dfrac{1}{t}\ln\dfrac{1}{0.80} = \dfrac{2.303}{t}\log(1.25)$ .

Ratio: $\dfrac{k_{610}}{k_{600}} = \dfrac{\log(1.25)}{\log(1.111)} = \dfrac{0.0969}{0.0457} \approx 2.12$ . …
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AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$E_a$
›Reveal solutionSolution
Statement I is correct because reaction orders are experimentally determined and can be zero, positive integers, or fractions. Statement II is incorrect because the frequency factor (pre-exponential factor) is not a fraction of molecules with energy above $E_a$ ; that fraction is given by the exponential term $e^{-E_a/RT}$ . Therefore, the correct option is (C).

The Arrhenius equation is the backbone of chemical kinetics when it comes to temperature dependence. It is written as:
$k = A e^{-E_a/RT}$
Here, $k$ is the rate constant, $A$ is the frequency factor (also called the pre-exponential factor), $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. The key intuition:

$A$ represents how often molecules collide in the correct orientation (a frequency, not a fraction).

The exponential term $e^{-E_a/RT}$ is the fraction of molecules that have energy equal to or greater than $E_a$ . Mixing these two up is a classic mistake.

Now, let’s evaluate each statement carefully.

Statement I: "Order of a reaction can be obtained from experiment and can have zero or positive integer or positive fraction values"

The order of a reaction is defined as the sum of the exponents of concentration terms in the experimentally determined rate law.

It is not derived from the stoichiometric coefficients (that would be molecularity, which applies only to elementary steps).

Orders can indeed be zero (e.g., decomposition of ammonia on a platinum surface), positive integers (most common), or positive fractions (e.g., the reaction $H_2 + Br_2 \rightarrow 2HBr$ has order 1.5).

Negative orders are also possible, but the statement only claims zero, positive integers, or positive fractions — all of which are valid.

Conclusion: Statement I is correct.

Statement II: "In the Arrhenius equation, the frequency factor is the fraction of molecules that can have energy higher than $E_a$ "

Let’s revisit the Arrhenius equation: $k = A e^{-E_a/RT}$ . …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$2\times10^{-5}\ \mathrm{s^{-1}}$
›Reveal solutionSolution
Plugging the two rate constants and temperatures into the two-point Arrhenius equation gives an activation energy of about 88 kJ/mol.

Concept and Intuition

The Arrhenius equation's two-temperature form lets you find activation energy directly from two rate constants at two known temperatures, without knowing the pre-exponential factor.

Step-by-Step Solution

Two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

$\dfrac{k_2}{k_1} = \dfrac{4\times10^{-3}}{2\times10^{-5}} = 200$ ; $\log 200 = \log 2 + \log 100 = 0.3 + 2 = 2.3$ .

$\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{400} - \dfrac{1}{500} = \dfrac{500-400}{400\times500} = \dfrac{100}{200000} = 5\times10^{-4}\ \mathrm{K^{-1}}$ .

$2.3 = \dfrac{E_a}{2.303 \times 8.3} \times 5\times10^{-4}$ .

$2.303 \times 8.3 = 19.115$ . …
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AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
$3.46 \times 10^{-2}\ s^{-1}$
›Reveal solutionSolution
This applies the two-temperature Arrhenius equation to find the rate constant at a higher temperature given the activation energy; the result is closest to $0.692\ s^{-1}$ .

Concept and Intuition

The Arrhenius equation, $k = Ae^{-E_a/RT}$ , shows that rate constants increase sharply with temperature for reactions with a sizeable activation energy. Taking the equation at two temperatures and dividing eliminates the pre-exponential factor $A$ , giving a convenient logarithmic two-point formula for finding a rate constant at a new temperature.

Step-by-Step Solution

Write the two-point Arrhenius equation: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$ .

Compute $\dfrac{1}{T_1} - \dfrac{1}{T_2} = \dfrac{1}{298} - \dfrac{1}{350} = \dfrac{350-298}{298\times350} = \dfrac{52}{104300} \approx 4.986\times10^{-4}\ \text{K}^{-1}$ .

Compute $\dfrac{E_a}{2.303R} = \dfrac{50100}{2.303\times8.314} = \dfrac{50100}{19.147} \approx 2616.6$ .

Multiply: $\log\dfrac{k_2}{k_1} \approx 2616.6 \times 4.986\times10^{-4} \approx 1.305$ .

So $\dfrac{k_2}{k_1} = 10^{1.305} = 10 \times 10^{0.305}$ . Using $\log 2 = 0.3010$ , $10^{0.305}$ is just slightly above 2, i.e. $\approx 2.02$ , giving $k_2/k_1 \approx 20.2$ . …
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AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
$\mathrm{kJ\,mol^{-1}}$
›Reveal solutionSolution
Plugging the rate-doubling data into the two-point Arrhenius equation gives $E_a \approx 53.33\ \mathrm{kJ\,mol^{-1}}$ — option (D).

Concept and Intuition

The Arrhenius equation, $k = Ae^{-E_a/RT}$ , links a reaction's rate constant to temperature through its activation energy. When we know the rate constant at two temperatures, we can eliminate the pre-exponential factor $A$ and solve directly for $E_a$ using the two-temperature form of the equation — this is the standard way "activation energy from a rate-doubling" problems are solved.

Step-by-Step Solution

Two-point Arrhenius form: $\log\dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

Here $k_2/k_1 = 2$ (rate doubled), so $\log 2 = 0.3$ (given).

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{300}-\dfrac{1}{310} = \dfrac{310-300}{300\times310} = \dfrac{10}{93000} = 1.0753\times10^{-4}\ \mathrm{K^{-1}}$ .

$0.3 = \dfrac{E_a}{2.303\times 8.3}\times 1.0753\times10^{-4}$ .

$2.303\times 8.3 = 19.115$ . …
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AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$k_1$
›Reveal solutionSolution
Plugging $E_a$ , $R$ , $T_1=300$ K and $T_2=400$ K into the two-point Arrhenius equation gives $\ln(k_2/k_1)\approx 4.166$ .

Concept and Intuition

The Arrhenius equation $k = A\,e^{-E_a/RT}$ shows that rate constants at two temperatures are related by $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac1{T_1}-\dfrac1{T_2}\right)$ , since the pre-exponential factor $A$ cancels out. A higher activation energy or a larger temperature gap gives a bigger ratio $k_2/k_1$ .

Step-by-Step Solution

Convert $E_a$ to joules: $41.5\ \mathrm{kJ/mol} = 41500\ \mathrm{J/mol}$ .

Compute $\dfrac{E_a}{R} = \dfrac{41500}{8.3} = 5000$ .

Compute $\dfrac1{T_1}-\dfrac1{T_2} = \dfrac1{300}-\dfrac1{400} = \dfrac{4-3}{1200} = \dfrac1{1200}$ . …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$\mathrm{C_2H_5I(g) \rightarrow C_2H_4(g) + HI(g)}$
›Reveal solutionSolution
Plugging the given rate-constant ratio and temperatures into the two-point Arrhenius equation gives an activation energy of 48.16 kJ/mol.

Concept and Intuition

The Arrhenius equation relates the rate constant at two different temperatures to the activation energy: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ . Given rate constants at two temperatures, we can solve directly for $E_a$ .

Step-by-Step Solution

$k_1 = x$ at $T_1=600$ K, $k_2=4x$ at $T_2=700$ K, so $k_2/k_1 = 4$ .

$\ln4 = 2.303\log4 = 2.303\times0.6 = 1.3818$ .

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac1{600}-\dfrac1{700} = \dfrac{700-600}{600\times700}=\dfrac{100}{420000}=2.381\times10^{-4}\ \text{K}^{-1}$ . …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
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›Reveal solutionSolution
Applying the two-point Arrhenius equation gives activation energy $\approx 33.45$ kJ/mol.

Concept and Intuition

The Arrhenius equation $k = A e^{-E_a/RT}$ links rate constant to temperature through the activation energy. Taking the ratio of rate constants at two temperatures eliminates the pre-exponential factor $A$ , letting us solve directly for $E_a$ .

Step-by-Step Solution

Write $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)$ .

$\dfrac{k_2}{k_1} = \dfrac{0.2}{0.02}=10$ , so $\ln 10 = 2.303$ .

$\dfrac{1}{T_1}-\dfrac{1}{T_2} = \dfrac{1}{500}-\dfrac{1}{700} = \dfrac{700-500}{500\times700} = \dfrac{200}{350000} = 5.714\times10^{-4}\ \text{K}^{-1}$ . …
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