The decomposition of A into product has value of $k$ as $4.5\times10^{3}\ \text{s}^{-1}$ at 10°C and energy of activation $60\ \text{kJ mol}^{-1}$. At what temperature would $k$ be $1.5\times10^{4}\ \text{s}^{-1}$?

The decomposition of A into product has value of k as 4.5×10³s⁻¹ at 10°C and energy of activation 60kJ mol⁻¹.

Exercises · 3.28

k

CBSENCERTSubjective· 3mImportance★★★★★

40% · 47/117 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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

Start your 14-day free trial to unlock the full solution →

Using the Arrhenius equation in its two-point form, the temperature at which the rate constant rises to $1.5 \times 10^{4}\ \text{s}^{-1}$ is approximately $297\ \text{K}$ (or $24^\circ\text{C}$ ).

The key to this problem is the Arrhenius equation, which tells us how the rate constant $k$ depends on temperature and activation energy. The equation is:

k = A e^{-E_a / RT}

where $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the absolute temperature. When we have two different temperatures and their corresponding rate constants, we can eliminate $A$ by taking a ratio. This gives the very useful two-point form:

\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

This is the formula we'll use. Notice the order: $k_2$ is the rate constant at the unknown temperature $T_2$ , and $k_1$ is the known rate constant at $T_1$ . The activation energy must be in the same energy units as $R$ , so we'll use $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ and convert $E_a$ from kJ to J.

Let's work through it step by step.

Convert all data to consistent units.

The activation energy is $60\ \text{kJ mol}^{-1} = 60\,000\ \text{J mol}^{-1}$ .

The first temperature is $10^\circ\text{C} = 283\ \text{K}$ (since $T(\text{K}) = T(^\circ\text{C}) + 273$ ).

The first rate constant is $k_1 = 4.5 \times 10^{3}\ \text{s}^{-1}$ .

The second rate constant is $k_2 = 1.5 \times 10^{4}\ \text{s}^{-1}$ .

We need to find $T_2$ .
Write the two-point Arrhenius equation with the known values.

\ln\left(\frac{1.5 \times 10^{4}}{4.5 \times 10^{3}}\right) = \frac{60\,000}{8.314}\left(\frac{1}{283} - \frac{1}{T_2}\right)

Simplify the left-hand side.

\frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} = \frac{15}{4.5} = \frac{10}{3} \approx 3.333

So $\ln(3.333) \approx 1.204$ (using a calculator or knowing $\ln(10/3) = \ln 10 - \ln 3 \approx 2.303 - 1.099 = 1.204$ ).

Simplify the right-hand side constant.

\frac{E_a}{R} = \frac{60\,000}{8.314} \approx 7217\ \text{K}

So the equation becomes:

1.204 = 7217 \left(\frac{1}{283} - \frac{1}{T_2}\right)

Solve for the bracket. …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make when solving this Arrhenius-type problem, along with how to avoid each.

1. Forgetting to Convert Celsius to Kelvin

The Mistake:

Plugging $T_1 = 10^\circ\text{C}$ directly into the Arrhenius equation.

The equation uses absolute temperature (Kelvin), not Celsius.

How to Avoid:

Always convert:

T_1 = 10 + 273 = 283\ \text{K}

Write this step explicitly before substituting.

2. Using the Wrong Form of the Arrhenius Equation

The Mistake:

Using $k = A e^{-E_a/RT}$ directly without realising you need the two-point form to compare two temperatures.

How to Avoid:

When you have $k$ at two different $T$ , always use:

\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)

This avoids needing the pre-exponential factor $A$ .

3. Mixing Up $k_1$ and $k_2$ (or $T_1$ and $T_2$ )

The Mistake:

Assigning $k_1 = 1.5 \times 10^4$ and $k_2 = 4.5 \times 10^3$ , then getting a negative or illogical temperature.

How to Avoid:

Label clearly:

$k_1 = 4.5 \times 10^3\ \text{s}^{-1}$ at $T_1 = 283\ \text{K}$
$k_2 = 1.5 \times 10^4\ \text{s}^{-1}$ at $T_2 = ?$

Since $k_2 > k_1$ , $T_2$ must be greater than $T_1$ . Check your final answer for sense.

4. Unit Errors in $E_a$ and $R$

The Mistake:

Using $E_a = 60\ \text{kJ mol}^{-1}$ but $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ without converting.

How to Avoid:

Convert $E_a$ to J/mol:

E_a = 60 \times 10^3 = 60\,000\ \text{J mol}^{-1}

Then $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ works directly.

5. Arithmetic Errors in the Fraction $\frac{1}{T_1} - \frac{1}{T_2}$

The Mistake:

Computing $\frac{1}{283} - \frac{1}{T_2}$ incorrectly, or rounding too early.

How to Avoid:

Keep at least 4–5 decimal places during calculation.

Example: …

Start your 14-day free trial to unlock the full solution →

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 $R = 8.314 \, \text{J/(mol·K)}$ ): …
Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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: …
Start your 14-day free trial to see the full solution →

←Previous question3.27 — The rate constant for the first order decomposition of H₂O₂ is given b…→Next question3.29 — The time required for 10% completion of a first order reaction at 298 …

Q.The decomposition of A into product has value of $k$ as $4.5\times10^{3}\ \text{s}^{-1}$ at 10°C and energy of activation $60\ \text{kJ mol}^{-1}$ . At what temperature would $k$ be $1.5\times10^{4}\ \text{s}^{-1}$ ?

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: Two-Point Form of the Arrhenius Equation

Steps

1. Forgetting to Convert Celsius to Kelvin

2. Using the Wrong Form of the Arrhenius Equation

3. Mixing Up $k_1$ and $k_2$ (or $T_1$ and $T_2$ )

4. Unit Errors in $E_a$ and $R$

5. Arithmetic Errors in the Fraction $\frac{1}{T_1} - \frac{1}{T_2}$

Step-by-step reasoning

Why activation energy matters

The special case: $E_a = 0$

Q.The decomposition of A into product has value of kkk as 4.5×103 s−14.5\times10^{3}\ \text{s}^{-1}4.5×103 s−1 at 10°C and energy of activation 60 kJ mol−160\ \text{kJ mol}^{-1}60 kJ mol−1. At what temperature would kkk be 1.5×104 s−11.5\times10^{4}\ \text{s}^{-1}1.5×104 s−1?

🎓Unlock everything free for 14 days

Q.The decomposition of A into product has value of $k$ as $4.5\times10^{3}\ \text{s}^{-1}$ at 10°C and energy of activation $60\ \text{kJ mol}^{-1}$ . At what temperature would $k$ be $1.5\times10^{4}\ \text{s}^{-1}$ ?

Unlock everything free for 14 days