The value of rate constant of a pseudo first order reaction ______. (i) depends on the concentration of reacta

NCERT Exemplar · Q19

Q. The value of rate constant of a pseudo first order reaction ____________.

(i) depends on the concentration of reactants present in small amount.

(ii) depends on the concentration of reactants present in excess.

(iii) is independent of the concentration of reactants.

(iv) depends only on temperature.

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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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For a pseudo first-order reaction, the rate constant depends on the concentration of the reactant present in excess, because that concentration is absorbed into the observed constant.

The key to this question lies in understanding what a pseudo first-order reaction actually is. It’s a trick of kinetics: you have a reaction that is truly second-order (or higher), but you make one reactant’s concentration so large that it barely changes during the reaction. That effectively constant concentration gets lumped into the rate constant, making the reaction appear first-order in the other reactant.

Let’s break it down.

Start with the true rate law. Suppose you have a reaction: $A + B \rightarrow \text{products}$ , and the actual rate law is second-order:

\text{Rate} = k [A][B]

Here, $k$ is the true rate constant, which depends only on temperature (and the nature of the reaction).

Create the pseudo condition. Now, if you take $B$ in huge excess — say $[B]_0$ is 100 times $[A]_0$ — then as the reaction proceeds, $[B]$ changes so little that it’s essentially constant. You can write:

[B] \approx [B]_0

throughout the reaction.

Define the pseudo rate constant. Substitute this constant into the rate law:

\text{Rate} = k [A] [B]_0 = (k [B]_0) [A]

The product $k [B]_0$ is a new constant, called the pseudo first-order rate constant, often denoted $k'$ or $k_{\text{obs}}$ .

k_{\text{obs}} = k [B]_0

Notice: $k_{\text{obs}}$ now contains $[B]_0$ , the initial concentration of the reactant in excess.

What does this mean for the options?
- Option (i) says it depends on the concentration of reactants present in small amount. That’s false — the small amount ( $[A]$ ) determines how fast the reaction proceeds, but it’s not part of $k_{\text{obs}}$ .
- Option (ii) says it depends on the concentration of reactants present in excess. That’s exactly right: $k_{\text{obs}} = k [B]_0$ , so it depends on $[B]_0$ . …

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Here are the common mistakes students make with this specific question on the Arrhenius Equation and pseudo first order reactions, along with how to avoid each.

Mistake 1: Confusing "pseudo first order" with "first order"

The error: Students think that because a reaction is treated as first order, the rate constant $k$ behaves exactly like a true first order reaction (i.e., independent of all concentrations).

Why it’s wrong: In a pseudo first order reaction, one reactant is in large excess. The rate law is:

\text{Rate} = k[\text{A}][\text{B}]

If $[\text{B}] \gg [\text{A}]$ , then $[\text{B}]$ is nearly constant. The observed rate constant is:

k_{\text{obs}} = k[\text{B}]

So $k_{\text{obs}}$ depends on the concentration of the reactant in excess (ii), not on the limiting reactant (i).

How to avoid: Always identify which reactant is in excess. The rate constant of a pseudo first order reaction depends on the concentration of the excess reactant.

Mistake 2: Thinking the Arrhenius equation applies blindly without context

The error: Students recall that $k = A e^{-E_a/RT}$ and conclude that $k$ depends only on temperature (option D). They forget that in a pseudo first order reaction, $k_{\text{obs}}$ is a composite constant.

Why it’s wrong: The Arrhenius equation describes the true rate constant $k$ . But here, the question asks about the observed rate constant of a pseudo first order reaction, which includes the concentration of the excess reactant.

How to avoid: Read the question carefully. If it says "pseudo first order," the observed $k$ is not the same as the elementary rate constant. It includes a concentration factor.

Mistake 3: Choosing option (i) — "depends on concentration of reactants present in small amount"

The error: Students think the rate constant depends on the limiting reactant because the rate depends on it.

Why it’s wrong: The rate does depend on the small-concentration reactant, but the rate constant $k_{\text{obs}}$ does not — it is defined so that:

\text{Rate} = k_{\text{obs}} [\text{limiting reactant}]

Here $k_{\text{obs}}$ is constant for a fixed excess concentration. Changing the small reactant changes the rate, but not $k_{\text{obs}}$ .

How to avoid: Distinguish between rate (depends on concentration of limiting reactant) and rate constant (does not depend on limiting reactant concentration in pseudo first order).

Mistake 4: Ignoring the "pseudo" condition entirely

The error: Students treat it as a normal reaction and pick option (iii) — "independent of concentration of reactants" — which is true for elementary first order reactions, but not for pseudo first order. …

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

CBSE 2024Set 56/3/11 markMCQ
$Q. When a catalyst increases the rate of a chemical reaction, then the rate constant (k) : (A) remains constant (B) decreases (C) increases (D) may increase or decrease depending on the order of the reaction$
›Reveal solutionSolution
A catalyst lowers the activation energy, which directly increases the rate constant $k$ through the Arrhenius equation. The answer is (C).

The rate constant $k$ is not just a number we measure—it encodes how the molecular-scale energy barrier controls reaction speed. To see why a catalyst must increase $k$ , we need the Arrhenius equation.
$k = A e^{-E_a/RT}$
This equation tells us that $k$ depends exponentially on the activation energy. A catalyst works by providing an alternative reaction pathway with a lower $E_a$ —it doesn't change the thermodynamics (reactants and products stay the same), but it reduces the energy hill molecules must climb to react.

Step-by-step reasoning

What a catalyst does at the molecular level

A catalyst participates in the reaction mechanism but is regenerated at the end. It creates intermediate steps with lower energy barriers than the uncatalyzed path. The net effect: $E_a$ (catalyst) $< E_a$ (no catalyst).

Impact on the exponential term

When $E_a$ decreases, the exponent $-E_a/RT$ becomes less negative (closer to zero). Since $e^x$ is an increasing function, $e^{-E_a/RT}$ becomes larger.

The pre-exponential factor $A$

This factor relates to collision frequency and orientation. A catalyst typically doesn't change $A$ significantly—the main effect is on $E_a$ .

Independence from reaction order

The rate constant $k$ appears in the rate law (e.g., $\text{rate} = k[A]^n$ ), but its value is determined by the Arrhenius equation, not by the order $n$ . The order tells us how concentration affects rate; the activation energy tells us the intrinsic speed at given concentrations. A catalyst lowers $E_a$ regardless of whether the reaction is zeroth, first, second, or any other order.

Quantitative example

Suppose $E_a = 100 \, \text{kJ/mol}$ without catalyst and $E_a = 50 \, \text{kJ/mol}$ with catalyst at $T = 300 \, \text{K}$ (with $R = 8.314 \, \text{J/(mol·K)}$ ): …
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CBSE 2020Set 56/1/11 mark
$E_{act}$
›Reveal solutionSolution
When activation energy is zero, the Arrhenius equation reduces to $k = A$ , making the rate constant independent of temperature.

Why activation energy matters

The Arrhenius equation connects temperature to the rate constant through the activation energy—the minimum energy barrier reactants must overcome to transform into products. The equation captures a fundamental idea: higher temperatures give molecules more kinetic energy, increasing the fraction that can surmount the barrier.
$k = A e^{-E_a/RT}$
The exponential term $e^{-E_a/RT}$ embodies the temperature dependence. When $E_a$ is large, even small temperature changes dramatically alter $k$ . But what happens when there's no barrier at all?

The special case: $E_a = 0$

Substitute zero activation energy into the Arrhenius equation:

$k = A e^{-0/RT} = A e^0 = A \cdot 1 = A$

Interpret the result:

The rate constant collapses to just the pre-exponential factor $A$ . This factor represents the frequency of collisions with proper orientation—it depends on molecular properties and collision geometry, but crucially, it has no temperature dependence built into the exponential term.

Physical meaning:

A zero activation energy means every collision between properly oriented molecules leads to reaction, regardless of their kinetic energy. There's no energy threshold to cross. Temperature might still affect collision frequency slightly through changes in molecular speed, but the dominant exponential temperature dependence vanishes. …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. Arrhenius equation is$
(a) k = -Ae^(-Ea/RT)
(b) k = Ae^(Ea/RT)
(c) k = Ae^(-Ea/RT)
(d) k = -Ae^(Ea/RT)
›Reveal solutionSolution
The Arrhenius equation relating the rate constant to temperature is $k = A e^{-E_a/RT}$ .

The Arrhenius equation expresses how the rate constant $k$ of a reaction varies with absolute temperature $T$ :

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

where:

$A$ = the Arrhenius (pre-exponential/frequency) factor, related to the frequency of collisions with proper orientation

$E_a$ = activation energy of the reaction

$R$ = universal gas constant

$T$ = absolute temperature …
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CBSE 2017Set ANNUAL1 mark
$Q. Explain Arrhenius equation.$
›Reveal solutionSolution
The Arrhenius equation shows that a rate constant increases exponentially with temperature because more molecules acquire energy equal to or greater than the activation energy.

The Arrhenius equation is:

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

where:

$k$ = rate constant of the reaction

$A$ = pre-exponential (frequency) factor, related to the frequency of collisions with proper orientation

$E_a$ = activation energy of the reaction

$R$ = universal gas constant

$T$ = absolute temperature (K)

The equation shows that as temperature $T$ increases, the exponential term $e^{-E_a/RT}$ increases (since $-E_a/RT$ becomes less negative), so a larger fraction of reactant molecules possess energy equal to or greater than $E_a$ , and the rate constant $k$ increases — explaining why reaction rates generally rise sharply with temperature.

Taking the natural log of both sides gives the linear form: …
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Q.The value of rate constant of a pseudo first order reaction ____________.

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

Method: Conceptual Analysis of Pseudo First-Order Reactions

Step 1 – Understand the definition of a pseudo first-order reaction

Step 2 – Write the rate law

Step 3 – Identify what the pseudo rate constant depends on

Step 4 – Match with the given options

Mistake 1: Confusing "pseudo first order" with "first order"

Mistake 2: Thinking the Arrhenius equation applies blindly without context

Mistake 3: Choosing option (i) — "depends on concentration of reactants present in small amount"

Mistake 4: Ignoring the "pseudo" condition entirely

Step-by-step reasoning

Why activation energy matters

The special case: $E_a = 0$

Q.The value of rate constant of a pseudo first order reaction ____________.

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