Consider the Arrhenius equation given below and mark the correct option. k = A e-Ea/RT (i) Rate constant incre

NCERT Exemplar · Q7

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

(i) Rate constant increases exponentially with increasing activation energy and decreasing temperature.

(ii) Rate constant decreases exponentially with increasing activation energy and decreasing temperature.

(iii) Rate constant increases exponentially with decreasing activation energy and decreasing temperature.

(iv) Rate constant increases exponentially with decreasing activation energy and increasing temperature.

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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 Arrhenius equation shows that the rate constant $k$ depends on activation energy $E_a$ and temperature $T$ through an exponential factor $e^{-E_a/RT}$ . A lower $E_a$ and a higher $T$ both make the exponent less negative, so $k$ increases. The correct option is (iv).

The Arrhenius equation is one of the most elegant relationships in chemical kinetics. It tells us how the rate constant $k$ — the speedometer of a reaction — changes with two key factors: the activation energy barrier $E_a$ and the temperature $T$ . The equation is:

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

Here, $A$ is the pre-exponential factor (a constant for a given reaction), $R$ is the gas constant, and the exponential term $e^{-E_a/RT}$ is the fraction of molecules that have enough energy to overcome the barrier.

The core idea: the exponent is negative, so anything that makes $-\frac{E_a}{RT}$ less negative (i.e., closer to zero) will make $e^{\text{that}}$ larger, and hence $k$ larger. Let’s see how $E_a$ and $T$ affect this.

Effect of activation energy $E_a$

The exponent is $-\frac{E_a}{RT}$ . If $E_a$ is large, the negative number is more negative, so $e^{-E_a/RT}$ is very small — $k$ is small. If $E_a$ is small, the negative number is less negative, so $e^{-E_a/RT}$ is larger — $k$ is larger.

So: decreasing $E_a$ increases $k$ (exponentially, because $E_a$ sits in the exponent).
Effect of temperature $T$

$T$ appears in the denominator of the exponent. If $T$ increases, the fraction $\frac{E_a}{RT}$ becomes smaller, so $-\frac{E_a}{RT}$ becomes less negative, and $e^{-E_a/RT}$ increases — $k$ increases.

So: increasing $T$ increases $k$ (again, exponentially).
Putting it together

The rate constant $k$ increases exponentially when:
- Activation energy $E_a$ decreases
- Temperature $T$ increases …

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Common Mistakes on the Arrhenius Equation Plot Question

Mistake 1: Confusing the Exponential Relationship

The error: Students think that a negative exponent means the function decreases when the quantity inside increases — but they forget the sign of the exponent.

Why it happens: The equation is $k = A e^{-E_a/RT}$ . The exponent is $-E_a/RT$ . Many students see " $-E_a$ " and assume that increasing $E_a$ makes $k$ larger.

How to avoid:

Remember: A negative exponent means the function decreases as the exponent's magnitude increases.
Think of $e^{-x}$ : as $x$ gets bigger, $e^{-x}$ gets smaller.
Here $x = E_a/RT$ $x = E_{a} / R T$ , so:
- Higher $E_a$ → larger $x$ → smaller $k$
- Lower $T$ → larger $x$ (since $T$ is in denominator) → smaller $k$

Mistake 2: Misinterpreting "Exponentially"

The error: Students think "exponentially" always means "increases rapidly" — but it can also mean "decreases rapidly."

Why it happens: In everyday language, "exponential growth" is common, but "exponential decay" is equally valid in science.

How to avoid:

In the Arrhenius equation, $k$ follows an exponential decay with respect to $E_a/RT$ .
The phrase "increases exponentially" in options refers to the direction of change in $k$ , not the mathematical form.
Always check: does the quantity inside the exponent increase or decrease?

Mistake 3: Forgetting Temperature is in the Denominator

The error: Students think decreasing temperature always decreases the exponent's magnitude.

Why it happens: They see $T$ in the denominator but treat it like a numerator.

How to avoid:

Write it clearly: $-\frac{E_a}{R} \cdot \frac{1}{T}$
Lower $T$ → larger $1/T$ → larger $E_a/RT$ → more negative exponent → smaller $k$
Higher $T$ → smaller $1/T$ → smaller $E_a/RT$ → less negative exponent → larger $k$

Mistake 4: Mixing Up Activation Energy and Temperature Effects

The error: Students think both $E_a$ and $T$ affect $k$ in the same direction.

Why it happens: They memorize "higher temperature = faster reaction" but don't connect it to the equation.

How to avoid:

Make a quick table: …

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

AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$E_a$
›Reveal solutionSolution
The fraction of molecules exceeding activation energy is the Boltzmann factor $e^{-E_a/RT}$ , which evaluates to $\exp(-2.4)$ here.

Concept and Intuition

The Arrhenius equation $k = Ae^{-E_a/RT}$ splits rate into a collision-frequency factor $A$ and an exponential Boltzmann factor $e^{-E_a/RT}$ , which physically represents the fraction of molecular collisions/molecules possessing energy at least $E_a$ . This fraction is exactly what the question calls $X$ .

Step-by-Step Solution

Fraction with energy $\geq E_a$ : $X = e^{-E_a/RT}$ .

Compute the exponent: $\dfrac{E_a}{RT} = \dfrac{10^4\ \text{J/mol}}{8.3\ \text{J mol}^{-1}\text{K}^{-1}\times500\ \text{K}} = \dfrac{10000}{4150} \approx 2.41 \approx 2.4$ . …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$\log k = 14 - \dfrac{1.25\times10^4 K}{T}$
›Reveal solutionSolution
Match the given empirical rate-constant equation to the Arrhenius equation in log form to extract $E_a$ and $A$ . Answer: $E_a=238.93$ kJ/mol, $A=10^{14}\ s^{-1}$ .

Concept and Intuition

The Arrhenius equation $k=Ae^{-E_a/RT}$ , written in base-10 log form, is $\log k = \log A - \dfrac{E_a}{2.303RT}$ — a straight line when $\log k$ is plotted against $1/T$ , with intercept $\log A$ and slope $-E_a/2.303R$ . Any experimentally fitted equation of this form can be matched term-by-term to read off $A$ and $E_a$ directly.

Step-by-Step Solution

Given: $\log k = 14 - \dfrac{1.25\times10^4}{T}$ .

Arrhenius form: $\log k = \log A - \dfrac{E_a}{2.303R}\cdot\dfrac{1}{T}$ .

Matching the constant term: $\log A = 14 \Rightarrow A = 10^{14}\ s^{-1}$ .

Matching the $1/T$ coefficient: $\dfrac{E_a}{2.303R} = 1.25\times10^4\text{ K}$ .

$E_a = 1.25\times10^4 \times 2.303 \times 8.3\ \text{J/mol}$ . Compute: $1.25\times10^4\times2.303 = 28{,}787.5$ ; $\times 8.3 = 238{,}936.25$ J/mol $\approx 238.93$ kJ/mol. …
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AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$\mathrm{kJ\, mol^{-1}}$
›Reveal solutionSolution
A textbook comparison of activation energies for acid- vs enzyme-catalysed sucrose hydrolysis: $X=6.22$ , $Y=2.15\ \mathrm{kJ\,mol^{-1}}$ .

Concept and Intuition

A catalyst speeds up a reaction by providing an alternate pathway with lower activation energy, without changing $\Delta H$ of the reaction. Enzymes are exceptionally efficient catalysts, so an enzyme-catalysed pathway typically has a much lower $E_a$ than the corresponding acid-catalysed (or uncatalysed) pathway for the same reaction, since $k = A e^{-E_a/RT}$ — a smaller $E_a$ gives a dramatically larger rate constant at the same temperature.

Step-by-Step Solution

Identify the reaction: hydrolysis of sucrose to glucose + fructose, run two ways — acid-catalysed and sucrase(enzyme)-catalysed.

Recall the standard reported values for this reaction: acid catalysis, $E_a = 6.22\ \mathrm{kJ\,mol^{-1}}$ ; sucrase catalysis, $E_a = 2.15\ \mathrm{kJ\,mol^{-1}}$ . …
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AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$\log\dfrac{k}{A} = -\dfrac{x}{T}$
›Reveal solutionSolution
Matching the given rate-law equation to the Arrhenius equation in logarithmic form directly identifies the activation energy as $E_a = 2.303\,x\,R$ .

Concept and Intuition

The Arrhenius equation, $k = A e^{-E_a/RT}$ , in logarithmic (base 10) form becomes:
$\log k = \log A - \frac{E_a}{2.303\,R\,T} \quad\Rightarrow\quad \log\frac{k}{A} = -\frac{E_a}{2.303\,R\,T}$
Comparing coefficients with a given empirical equation of the same form directly reveals $E_a$ .

Step-by-Step Solution

Standard Arrhenius log form: $\log\dfrac{k}{A} = -\dfrac{E_a}{2.303\,R\,T}$ .

Given equation: $\log\dfrac{k}{A} = -\dfrac{x}{T}$ .

Equate the coefficients of $\dfrac{1}{T}$ : $\dfrac{E_a}{2.303\,R} = x$ . …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\log_{10}\dfrac{k}{A} = 0.00174$
›Reveal solutionSolution
This tests applying the Arrhenius equation in logarithmic form to extract activation energy. The answer is 10.0 J/mol.

Concept and Intuition

The Arrhenius equation $k=Ae^{-E_a/RT}$ relates the rate constant to activation energy $E_a$ and temperature $T$ . Taking natural log and converting to base-10: $\log_{10}\dfrac{k}{A} = -\dfrac{E_a}{2.303RT}$ . So the magnitude of $\log_{10}(k/A)$ scales directly with activation energy at a given temperature — a small log ratio corresponds to a small (near-zero) activation energy.

Step-by-Step Solution

Write the relation: $\left|\log_{10}\dfrac{k}{A}\right| = \dfrac{E_a}{2.303RT}$ .

Rearranging: $E_a = 2.303\,R\,T\times\left|\log_{10}\dfrac{k}{A}\right|$ .

Substitute values: $R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ , $T=300$ K, $\left|\log_{10}(k/A)\right|=0.00174$ . …
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AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
$\rightarrow$
›Reveal solutionSolution
On a reaction energy diagram, $E_a$ is the gap from reactant level to the peak, and $|\Delta H|$ is the gap between reactant and product levels — here $E_a=15$ kJ/mol and $|\Delta H|=5$ kJ/mol.

Concept and Intuition

A potential-energy vs reaction-coordinate diagram encodes both kinetics and thermodynamics in one picture: the height of the barrier above the reactants is the activation energy (how hard it is to get started), while the difference between the final resting level of products and the starting level of reactants is the heat of reaction (whether the overall process releases or absorbs energy).

Step-by-Step Solution

From the graph: reactant A is at the dashed gridline $y=10$ kJ/mol.

The curve rises through the hump to its highest point, which lines up with the $y=25$ kJ/mol gridline — this is the transition state / activated complex energy.

Activation energy $E_a = E_{\text{peak}} - E_{A} = 25 - 10 = 15$ kJ/mol.

The curve then falls to product P, at the $y=5$ kJ/mol gridline. …
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AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
$\ln k$
›Reveal solutionSolution
Straightforward application of the Arrhenius equation's linear form; the slope directly gives $E_a$ after multiplying by $-R$ .

Concept and Intuition

The Arrhenius equation $k = Ae^{-E_a/RT}$ becomes linear on taking logarithms:
$\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$
Plotting $\ln k$ (y-axis) against $1/T$ (x-axis) gives a straight line of slope $-E_a/R$ and intercept $\ln A$ .

Step-by-Step Solution

Given slope $= -2\times10^4$ K.

Since slope $= -E_a/R$ : $E_a = -(\text{slope})\times R = (2\times10^4\ \text{K})(8.3\ \text{J K}^{-1}\text{mol}^{-1})$ . …
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AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
$Q. Given below are two statements Assertion (A): A catalyst, generally increases the rate of a reaction Reason (R): It lowers the activation energy of a reaction by providing a new path The correct answer is (A) Both (A) and (R) are correct and (R) is the correct explanation of (A) (B) Both (A) and (R) are correct and (R) is not the correct explanation of (A) (C) (A) is correct but (R) is not correct (D) (A) is not correct but (R) is correct$
›Reveal solutionSolution
The assertion and reason are both true, and the reason correctly explains why the assertion is true. The correct option is (A).

Understanding the Assertion (A):

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. This is a fundamental fact in chemistry — catalysts speed up both forward and reverse reactions, allowing equilibrium to be reached faster. So (A) is correct.

Understanding the Reason (R):

The reason states that a catalyst lowers the activation energy by providing an alternative reaction pathway. This is the standard explanation from the Arrhenius equation:

$k = A e^{-E_a/(RT)}$
Lowering $E_a$ (activation energy) increases the rate constant $k$ , and thus the reaction rate. The catalyst does not change the overall thermodynamics (ΔH or ΔG) — it only reduces the energy barrier. So (R) is also correct.

Checking if (R) explains (A): …
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Q.Consider the Arrhenius equation given below and mark the correct option.
$k = A\, e^{-E_a/RT}$

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

Method: Arrhenius Equation Analysis — Direct Interpretation of the Exponential Form

Step 1: Write the Arrhenius equation in its standard form

Step 2: Identify the key relationship from the exponent

Step 3: Combine the effects …

Common Mistakes on the Arrhenius Equation Plot Question

Mistake 1: Confusing the Exponential Relationship

Mistake 2: Misinterpreting "Exponentially"

Mistake 3: Forgetting Temperature is in the Denominator

Mistake 4: Mixing Up Activation Energy and Temperature Effects

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Consider the Arrhenius equation given below and mark the correct option. k=A e−Ea/RTk = A\, e^{-E_a/RT}k=Ae−Ea​/RT

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Q.Consider the Arrhenius equation given below and mark the correct option.
$k = A\, e^{-E_a/RT}$

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