The rate constant for the decomposition of hydrocarbons is $2.418\times10^{-5}\ \text{s}^{-1}$ at 546 K. If the energy of activation is $179.9\ \text{kJ/mol}$, what will be the value of pre-exponential factor?

> [!TLDR] > Using the Arrhenius equation $k = A e^{-E_a/(RT)}$, we solve for the pre-exponential factor $A$. With $k = 2.418 \times 10^{-5}\ \text{s}^{-1}$, $E_a = 179.9\ \text{kJ/mol}$, and $T = 546\ \text{K}$, the value of $A$ is approximately $3.9 \times 10^{12}\ \text{s}^{-1}$. The Arrhenius equation is the backbone of chemical kinetics when temperature dependence is involved. It tells us that the rate constant $k$ depends on two things: the fraction of molecules with enough energy to react (given by the exponential term $e^{-E_a/(RT)}$) and how often collisions happen in the right orientation (the pre-exponential factor $A$). Here, we know $k$, $E_a$, and $T$, and we need to find $A$ — so we simply rearrange the equation. > [!FORMULA] > $$k = A e^{-E_a/(RT)}$$ Where: - $k$ = rate constant ($\text{s}^{-1}$) - $A$ = pre-exponential factor (same units as $k$) - $E_a$ = activation energy (J/mol — careful with units!) - $R$ = gas constant ($8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$) - $T$ = temperature (K) 1. **Convert activation energy to consistent units.** The given $E_a = 179.9\ \text{kJ/mol}$. Since $R$ is in J/mol·K, we must convert: $$E_a = 179.9 \times 10^3\ \text{J/mol} = 1.799 \times 10^5\ \text{J/mol}$$ 2. **Compute the exponent $\frac{E_a}{RT}$.** First, $RT = 8.314 \times 546$. $$RT = 8.314 \times 546 = 4539.444\ \text{J/mol}$$ Then, $$\frac{E_a}{RT} = \frac{1.799 \times 10^5}{4539.444} \approx 39.63$$ 3. **Find $e^{-E_a/(RT)}$.** $$e^{-39.63} = \frac{1}{e^{39.63}}$$ This is a very small number. Using a calculator: $$e^{39.63} \approx 1.62 \times 10^{17}$$ So, $$e^{-39.63} \approx 6.17 \times 10^{-18}$$ 4. **Rearrange the Arrhenius equation to solve for $A$.** $$A = \frac{k}{e^{-E_a/(RT)}} = k \times e^{E_a/(RT)}$$ Substitute the values: $$A = (2.418 \times 10^{-5}) \times (1.62 \times 10^{17})$$ 5. **Calculate $A$.** $$A = 2.418 \times 1.62 \times 10^{12} \approx 3.917 \t

The rate constant for the decomposition of hydrocarbons is 2.418×10⁻⁵s⁻¹ at 546 K. If the energy of activation

Exercises · 3.23

2.418\times10^{-5}\ \text{s}^{-1}

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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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Using the Arrhenius equation $k = A e^{-E_a/(RT)}$ , we solve for the pre-exponential factor $A$ . With $k = 2.418 \times 10^{-5}\ \text{s}^{-1}$ , $E_a = 179.9\ \text{kJ/mol}$ , and $T = 546\ \text{K}$ , the value of $A$ is approximately $3.9 \times 10^{12}\ \text{s}^{-1}$ .

The Arrhenius equation is the backbone of chemical kinetics when temperature dependence is involved. It tells us that the rate constant $k$ depends on two things: the fraction of molecules with enough energy to react (given by the exponential term $e^{-E_a/(RT)}$ ) and how often collisions happen in the right orientation (the pre-exponential factor $A$ ). Here, we know $k$ , $E_a$ , and $T$ , and we need to find $A$ — so we simply rearrange the equation.

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

Where:

$k$ = rate constant ( $\text{s}^{-1}$ )
$A$ = pre-exponential factor (same units as $k$ )
$E_a$ = activation energy (J/mol — careful with units!)
$R$ = gas constant ( $8.314\ \text{J mol}^{-1}\ \text{K}^{-1}$ )
$T$ = temperature (K)

Convert activation energy to consistent units. The given $E_a = 179.9\ \text{kJ/mol}$ . Since $R$ is in J/mol·K, we must convert:

E_a = 179.9 \times 10^3\ \text{J/mol} = 1.799 \times 10^5\ \text{J/mol}

Compute the exponent $\frac{E_a}{RT}$ . First, $RT = 8.314 \times 546$ .

RT = 8.314 \times 546 = 4539.444\ \text{J/mol}

Then,

\frac{E_a}{RT} = \frac{1.799 \times 10^5}{4539.444} \approx 39.63

Find $e^{-E_a/(RT)}$ .

e^{-39.63} = \frac{1}{e^{39.63}}

This is a very small number. Using a calculator:

e^{39.63} \approx 1.62 \times 10^{17}

So,

e^{-39.63} \approx 6.17 \times 10^{-18}

Rearrange the Arrhenius equation to solve for $A$ .

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

Substitute the values:

A = (2.418 \times 10^{-5}) \times (1.62 \times 10^{17})

Calculate $A$ . $A = 2.418 \times 1.62 \times 10^{12} \approx 3.917 \times 10^{12}\ \text{s}^{-1}$ …

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Common Mistakes with the Arrhenius Equation (Pre-exponential Factor)

This is a classic "plug-and-chug" problem from Chemical Kinetics. While the calculation is straightforward, students often lose marks due to unit errors and logarithm mishandling.

Mistake #1: Forgetting to Convert Activation Energy to J/mol

The error:

The activation energy $E_a$ is given in kJ/mol (179.9 kJ/mol), but the gas constant $R$ is usually taken as $8.314\ \text{J mol}^{-1}\text{K}^{-1}$ . If you plug in $E_a$ in kJ without converting, the ratio $\frac{E_a}{RT}$ becomes 1000 times too small, giving a wildly wrong $A$ .

How to avoid:

Always convert $E_a$ to J/mol before using it in the Arrhenius equation.

E_a = 179.9 \times 10^3\ \text{J/mol}

Quick check: If your $\frac{E_a}{RT}$ value is less than about 10 (for typical exam problems), you likely forgot the conversion.

Mistake #2: Using the Wrong Form of the Arrhenius Equation

The error:

Students sometimes use the integrated form (for finding $k$ at two temperatures) when the problem only gives one temperature and asks for $A$ .

How to avoid:

Recognize the direct form:

k = A e^{-E_a / RT}

Here, you have $k$ , $E_a$ , and $T$ — solve directly for $A$ :

A = k \cdot e^{E_a / RT}

No need for two temperatures or the logarithmic form unless the problem asks for it.

Mistake #3: Mishandling the Natural Logarithm When Solving

The error:

When rearranging $k = A e^{-E_a/RT}$ , some students incorrectly write:

A = \frac{k}{e^{-E_a/RT}} \quad \text{(correct)}

but then compute $e^{-E_a/RT}$ instead of $e^{+E_a/RT}$ .

How to avoid:

Rewrite clearly:

A = k \times e^{+E_a/RT}

Remember: The negative sign in the exponent flips when you bring $e^{-E_a/RT}$ to the other side.

Mistake #4: Incorrectly Computing the Exponential

The error:

Students often compute $e^{E_a/RT}$ by first calculating $\frac{E_a}{RT}$ and then pressing the $e^x$ button — but they forget to use parentheses on their calculator, leading to:

e^{179900} / (8.314 \times 546) \quad \text{(wrong!)}

instead of:

e^{179900 / (8.314 \times 546)} \quad \text{(correct)}

How to avoid:

On a scientific calculator, type: …

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

GUJCET 2025Set 031 markMCQ
$Q. According to Arrhenius equation which of the following statement is correct? (A) Decrease in temperature or increase in Activation Energy will increase rate of reaction. (B) Increase in temperature and increase in Activation Energy will increase rate of reaction. (C) Increase in temperature or decrease in Activation Energy will increase rate of reaction. (D) Decrease in temperature and decrease in Activation Energy will increase rate of reaction.$
›Reveal solutionSolution
[!TLDR]

Higher temperature or lower activation energy makes the exponential term larger, so the rate increases.

Concept

The Arrhenius equation $k = A e^{-E_a/RT}$ links the rate constant to temperature $T$ and activation energy $E_a$ .

Solution

The factor $e^{-E_a/RT}$ controls the rate:

Increasing $T$ makes the exponent $-E_a/RT$ less negative, so $k$ increases.

Decreasing $E_a$ also makes the exponent less negative, so $k$ increases. …
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GUJCET 2024Set 131 markMCQ
$Q. Which of the following statements is incorrect for a reaction carried out in presence of catalyst? (A) Potential energy of reactants and products change (B) Equilibrium constant of the reaction does not change (C) There is no change in Gibbs energy of the reaction (D) The activation energy of the reaction decreases$
›Reveal solutionSolution
A catalyst only changes the pathway/ $E_a$ ; the energy levels of reactants and products (and hence $\Delta H$ , $\Delta G$ , $K$ ) are unchanged.

Concept: A catalyst provides an alternative path with lower activation energy. It does not alter the potential energies of the reactants or products, so $\Delta G$ , $\Delta H$ and the equilibrium constant $K$ remain the same.

(B) $K$ unchanged — correct. …
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←Previous question3.22 — The rate constant for the decomposition of N₂O₅ at various temperature…→Next question3.24 — Consider a certain reaction A arrow Products with k = 2.0×10⁻²s⁻¹. Cal…

Q.The rate constant for the decomposition of hydrocarbons is $2.418\times10^{-5}\ \text{s}^{-1}$ at 546 K. If the energy of activation is $179.9\ \text{kJ/mol}$ , what will be the value of pre-exponential factor?

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 Arrhenius Form (Solving for Pre-Exponential Factor)

Steps

Common Mistakes with the Arrhenius Equation (Pre-exponential Factor)

Mistake #1: Forgetting to Convert Activation Energy to J/mol

Mistake #2: Using the Wrong Form of the Arrhenius Equation

Mistake #3: Mishandling the Natural Logarithm When Solving

Mistake #4: Incorrectly Computing the Exponential

Concept

Solution

Q.The rate constant for the decomposition of hydrocarbons is 2.418×10−5 s−12.418\times10^{-5}\ \text{s}^{-1}2.418×10−5 s−1 at 546 K. If the energy of activation is 179.9 kJ/mol179.9\ \text{kJ/mol}179.9 kJ/mol, what will be the value of pre-exponential factor?

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Q.The rate constant for the decomposition of hydrocarbons is $2.418\times10^{-5}\ \text{s}^{-1}$ at 546 K. If the energy of activation is $179.9\ \text{kJ/mol}$ , what will be the value of pre-exponential factor?

Unlock everything free for 14 days