In the Arrhenius equation, when $\log k$ is plotted against $1/T$, a straight line is obtained whose: (A) slope is $\frac{A}{R}$ and intercept is $E_a$. (B) slope is $A$ and intercept is $\frac{-E_a}{R}$. (C) slope is $\frac{-E_a}{RT}$ and intercept is $\log A$. (D) slope is $\frac{-E_a}{2 \cdot 303 R}$ and intercept is $\log A$.

> [!TLDR] > The Arrhenius equation $k = A e^{-E_a/(RT)}$ becomes linear when we take the natural log and plot $\ln k$ vs. $1/T$. Converting to $\log_{10}$ gives slope $= -E_a/(2.303 R)$ and intercept $= \log A$, so the correct option is **(D)**. The Arrhenius equation is one of the most elegant relationships in chemical kinetics — it connects the rate constant $k$ to temperature $T$ through two parameters: the activation energy $E_a$ and the pre-exponential factor $A$. The equation is: $$k = A e^{-E_a/(RT)}$$ If you plot $k$ directly against $T$, you get a curve. But the trick is to take logarithms — that turns the exponential into a straight line. Why does that help? Because a straight line is easy to interpret: its slope and intercept give you $E_a$ and $A$ directly. Let’s see how. --- 1. **Take the natural logarithm of both sides** Starting from $k = A e^{-E_a/(RT)}$, we get: $$\ln k = \ln A - \frac{E_a}{RT}$$ This is already in the form $y = mx + c$, where: - $y = \ln k$ - $x = 1/T$ - slope $m = -E_a/R$ - intercept $c = \ln A$ So a plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ and intercept $\ln A$. 2. **But the question uses $\log k$ — that’s base 10** In many Indian exam contexts, $\log$ means $\log_{10}$. To convert from natural log to base 10, use: $$\ln k = 2.303 \log_{10} k$$ Substitute into the equation above: $$2.303 \log k = \ln A - \frac{E_a}{RT}$$ Divide through by 2.303: $$\log k = \frac{\ln A}{2.303} - \frac{E_a}{2.303 \, RT}$$ 3. **Identify slope and intercept** Now compare with $y = mx + c$: - $y = \log k$ - $x = 1/T$ - slope $m = -\dfrac{E_a}{2.303 \, R}$ - intercept $c = \dfrac{\ln A}{2.303} = \log A$ That’s exactly what option **(D)** says. …

In the Arrhenius equation, when log k is plotted against 1/T, a straight line is obtained whose: (A) slope is

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\log k

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

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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 $k = A e^{-E_a/(RT)}$ becomes linear when we take the natural log and plot $\ln k$ vs. $1/T$ . Converting to $\log_{10}$ gives slope $= -E_a/(2.303 R)$ and intercept $= \log A$ , so the correct option is (D).

The Arrhenius equation is one of the most elegant relationships in chemical kinetics — it connects the rate constant $k$ to temperature $T$ through two parameters: the activation energy $E_a$ and the pre-exponential factor $A$ . The equation is:

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

If you plot $k$ directly against $T$ , you get a curve. But the trick is to take logarithms — that turns the exponential into a straight line. Why does that help? Because a straight line is easy to interpret: its slope and intercept give you $E_a$ and $A$ directly.

Let’s see how.

Take the natural logarithm of both sides

Starting from $k = A e^{-E_a/(RT)}$ , we get:

\ln k = \ln A - \frac{E_a}{RT}

This is already in the form $y = mx + c$ , where:

$y = \ln k$
$x = 1/T$
slope $m = -E_a/R$
intercept $c = \ln A$

So a plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ and intercept $\ln A$ .

But the question uses $\log k$ — that’s base 10

In many Indian exam contexts, $\log$ means $\log_{10}$ . To convert from natural log to base 10, use:

\ln k = 2.303 \log_{10} k

Substitute into the equation above:

2.303 \log k = \ln A - \frac{E_a}{RT}

Divide through by 2.303:

\log k = \frac{\ln A}{2.303} - \frac{E_a}{2.303 \, RT}

Identify slope and intercept

Now compare with $y = mx + c$ :
- $y = \log k$
- $x = 1/T$
- slope $m = -\dfrac{E_a}{2.303 \, R}$
- intercept $c = \dfrac{\ln A}{2.303} = \log A$
That’s exactly what option (D) says. …

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

CBSE 2026Set 56/1/11 markMCQ
$E_a$
›Reveal solutionSolution
The fraction of molecules with energy equal to or greater than the activation energy $E_a$ is given by the Boltzmann factor $e^{-E_a/RT}$ , which appears directly in the Arrhenius equation. The correct option is (B).

The Arrhenius equation is the starting point here. It tells us that the rate constant $k$ depends on temperature as:
$k = A e^{-E_a/RT}$
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.

The exponential term $e^{-E_a/RT}$ is the key. It represents the fraction of molecules that have enough energy to overcome the activation barrier — that is, molecules with kinetic energy equal to or greater than $E_a$ . This comes from the Maxwell–Boltzmann distribution of molecular energies: the fraction of molecules with energy $\geq E_a$ is proportional to $e^{-E_a/RT}$ .

So the question is simply asking: which of the given expressions matches this Boltzmann factor?

Let’s check each option:

Option (A): $+\dfrac{E_a}{RT}$

This is a positive number (since $E_a$ , $R$ , $T$ are all positive). It grows as temperature decreases or activation energy increases. But a fraction must lie between 0 and 1 — this expression can be much larger than 1, so it cannot represent a fraction of molecules. Discard.

Option (B): $e^{-E_a/RT}$

This is the Boltzmann factor. For typical values ( $E_a \sim 50$ kJ/mol, $T \sim 300$ K, $R = 8.314$ J/mol·K), $E_a/RT \approx 20$ , so $e^{-20} \approx 2 \times 10^{-9}$ — a tiny fraction, which makes sense: only a very small proportion of molecules have enough energy to react at room temperature. This matches the physical meaning exactly.

Option (C): $-\dfrac{E_a}{RT}$

This is negative. A fraction cannot be negative. Discard. …
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CBSE 2024Set 56/1/11 markMCQ
$\log k$
›Reveal solutionSolution
The Arrhenius equation $k = A e^{-E_a/(RT)}$ becomes linear when we take the natural log and plot $\ln k$ vs. $1/T$ . Converting to $\log_{10}$ gives slope $= -E_a/(2.303 R)$ and intercept $= \log A$ , so the correct option is (D).

The Arrhenius equation is one of the most elegant relationships in chemical kinetics — it connects the rate constant $k$ to temperature $T$ through two parameters: the activation energy $E_a$ and the pre-exponential factor $A$ . The equation is:
$k = A e^{-E_a/(RT)}$
If you plot $k$ directly against $T$ , you get a curve. But the trick is to take logarithms — that turns the exponential into a straight line. Why does that help? Because a straight line is easy to interpret: its slope and intercept give you $E_a$ and $A$ directly.

Let’s see how.

Take the natural logarithm of both sides

Starting from $k = A e^{-E_a/(RT)}$ , we get:

$\ln k = \ln A - \frac{E_a}{RT}$
This is already in the form $y = mx + c$ , where:

$y = \ln k$

$x = 1/T$

slope $m = -E_a/R$

intercept $c = \ln A$

So a plot of $\ln k$ vs $1/T$ gives a straight line with slope $-E_a/R$ and intercept $\ln A$ .

But the question uses $\log k$ — that’s base 10

In many Indian exam contexts, $\log$ means $\log_{10}$ . To convert from natural log to base 10, use:

$\ln k = 2.303 \log_{10} k$
Substitute into the equation above:
$2.303 \log k = \ln A - \frac{E_a}{RT}$
Divide through by 2.303:
$\log k = \frac{\ln A}{2.303} - \frac{E_a}{2.303 \, RT}$

Identify slope and intercept

Now compare with $y = mx + c$ :

$y = \log k$

$x = 1/T$

slope $m = -\dfrac{E_a}{2.303 \, R}$

intercept $c = \dfrac{\ln A}{2.303} = \log A$

That’s exactly what option (D) says. …
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CBSE 2023Set 56/1/11 markMCQ
$\Delta H$
›Reveal solutionSolution
A catalyst provides an alternative reaction pathway with a lower activation energy ( $E_a$ ). It does not change the thermodynamic state functions $\Delta H$ , $\Delta G$ , or $\Delta S$ for the overall reaction. Therefore, the correct answer is (C).

The key to this question lies in distinguishing between kinetics (how fast a reaction occurs) and thermodynamics (whether a reaction is spontaneous and how much energy is exchanged). A catalyst is a kinetic tool — it speeds up a reaction without being consumed, but it never alters the starting point or the destination of the reaction.

Think of a mountain pass. The reactants are at the base of one side, the products at the base of the other. The height difference between them is $\Delta H$ (enthalpy change). The overall steepness and direction of the slope is $\Delta G$ (free energy change). The disorder along the path is $\Delta S$ (entropy change). A catalyst is like a tunnel through the mountain — it lowers the peak you have to climb over (the activation energy $E_a$ ), but the heights of the two bases and the distance between them remain exactly the same.

Let’s examine each option carefully.

$\Delta H$ (Enthalpy change)

$\Delta H$ is the difference in enthalpy between products and reactants. It depends only on the initial and final states of the system. A catalyst does not appear in the overall balanced equation and is regenerated at the end. Since the reactants and products are identical with or without the catalyst, $\Delta H$ is unchanged.
$\Delta H_{\text{catalysed}} = \Delta H_{\text{uncatalysed}}$

$\Delta G$ (Gibbs free energy change)

$\Delta G = \Delta H - T\Delta S$ is the thermodynamic driving force for a reaction. It determines spontaneity. A catalyst cannot make a non-spontaneous reaction spontaneous — it only accelerates a reaction that is already thermodynamically favourable. The equilibrium constant $K$ is related to $\Delta G$ by $\Delta G^\circ = -RT\ln K$ , and a catalyst does not shift equilibrium. So $\Delta G$ remains the same.

$E_a$ (Activation energy) …
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