The rate constant for the decomposition of $N_2O_5$ at various temperatures is given below: | T/°C | 0 | 20 | 40 | 60 | 80 | |---|---|---|---|---|---| | $10^5\times k$/$\text{s}^{-1}$ | 0.0787 | 1.70 | 25.7 | 178 | 2140 | Draw a graph between $\ln k$ and $1/T$ and calculate the values of $A$ and $E_a$. Predict the rate constant at 30° and 50°C.

The rate constant for the decomposition of N₂O₅ at – Class 12 Chemistry

Exercises · 3.22

The rate constant for the decomposition of $N_2O_5$ at various temperatures is given below:

T/°C	0	20	40	60	80
$10^5\times k$ / $\text{s}^{-1}$	0.0787	1.70	25.7	178	2140

Draw a graph between $\ln k$ and $1/T$ and calculate the values of $A$ and $E_a$ . Predict the rate constant at 30° and 50°C.

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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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The key idea is the Arrhenius equation in its linear form:

$\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$ .

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

Step 1 – Convert data

Convert $T$ to Kelvin: $T/K = t/°C + 273.15$ . The given $k$ values are $10^5 \times k$ , so the actual $k$ used for $\ln k$ is the table value $\times 10^{-5}$ .

$T/°C$	$T/K$	$10^3/T$ (K $^{-1}$ )	actual $k$ (s $^{-1}$ )	$\ln k$
0	273.15	3.661	$7.87\times10^{-7}$	$-14.055$
20	293.15	3.411	$1.70\times10^{-5}$	$-10.982$
40	313.15	3.193	$2.57\times10^{-4}$	$-8.266$
60	333.15	3.002	$1.78\times10^{-3}$	$-6.331$
80	353.15	2.832	$2.14\times10^{-2}$	$-3.844$

Step 2 – Plot and find slope

Plot $\ln k$ (y-axis) vs $10^3/T$ (x-axis). Using the first and last points:

Slope $= \frac{(-3.844) - (-14.055)}{(2.832 - 3.661)\times 10^{-3}} = \frac{10.211}{-0.829 \times 10^{-3}} \approx -1.232 \times 10^4$ K.

Step 3 – Calculate $E_a$ and $A$

$E_a = -\text{slope} \times R = 1.232 \times 10^4 \times 8.314 \approx 1.024 \times 10^5$ J/mol = 102.4 kJ/mol.

Intercept $\ln A = \ln k + \frac{E_a}{R}\cdot\frac{1}{T}$ . Using the (central) point at 40°C:

$\ln A = -8.266 + (1.232\times 10^4)(3.193\times 10^{-3}) = -8.266 + 39.34 = 31.07$

$A = e^{31.07} \approx 3.1 \times 10^{13}$ s $^{-1}$ . …

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Using the Arrhenius equation $\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$ , we plot $\ln k$ vs $1/T$ to get a straight line. From its slope ( $-E_a/R$ ) and intercept ( $\ln A$ ), we find $E_a \approx 102.4\ \text{kJ mol}^{-1}$ and $A \approx 3.1\times 10^{13}\ \text{s}^{-1}$ . Then we predict $k_{30^\circ\text{C}} \approx 7.0\times 10^{-5}\ \text{s}^{-1}$ and $k_{50^\circ\text{C}} \approx 8.6\times 10^{-4}\ \text{s}^{-1}$ .

The Arrhenius equation is the backbone of temperature-dependent kinetics. It tells us that the rate constant $k$ depends exponentially on temperature:

k = A e^{-E_a/RT}

Taking natural logs gives a linear form:

\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 $m = -E_a/R$ , and intercept $c = \ln A$ . So if we plot $\ln k$ against $1/T$ , we get a straight line — and from its slope and intercept we can extract both $E_a$ and $A$ .

Watch out

Temperature must be in kelvin when using $1/T$ in the Arrhenius plot. A common mistake is to use Celsius directly — that gives a completely wrong slope.

Let’s work through it step by step.

1. Convert temperatures to kelvin and compute $1/T$ and $\ln k$

$T$ (°C)	$T$ (K)	$1/T$ (K $^{-1}$ )	$k$ (s $^{-1}$ )	$\ln k$
0	273.15	$3.661 \times 10^{-3}$	$0.0787 \times 10^{-5}$	$-14.055$
20	293.15	$3.411 \times 10^{-3}$	$1.70 \times 10^{-5}$	$-10.982$
40	313.15	$3.193 \times 10^{-3}$	$25.7 \times 10^{-5}$	$-8.266$
60	333.15	$3.002 \times 10^{-3}$	$178 \times 10^{-5}$	$-6.331$
80	353.15	$2.832 \times 10^{-3}$	$2140 \times 10^{-5}$	$-3.844$

Notice that $k$ values are given as $10^5 \times k$ , so we divide by $10^5$ to get actual $k$ in s $^{-1}$ .

2. Plot $\ln k$ vs $1/T$

On a graph, the points fall beautifully on a straight line. The slope is negative (since $k$ increases with $T$ , $\ln k$ increases as $1/T$ decreases). We can calculate the slope using any two well-separated points, but for accuracy, use the first and last:

\text{slope} = \frac{\Delta (\ln k)}{\Delta (1/T)} = \frac{(-3.844) - (-14.055)}{(2.832 - 3.661) \times 10^{-3}} = \frac{10.211}{-0.829 \times 10^{-3}} \approx -1.232 \times 10^4\ \text{K}

Tip

Using the two extreme points gives a quick estimate. For exam problems, this is usually sufficient — but if you have time, a least-squares fit (or averaging slopes from multiple pairs) gives a more reliable result.

3. Calculate $E_a$ from the slope

Since slope $= -E_a/R$ , we have:

-\frac{E_a}{R} = -1.232 \times 10^4\ \text{K}

E_a = 1.232 \times 10^4 \times R = 1.232 \times 10^4 \times 8.314\ \text{J mol}^{-1}

E_a \approx 1.024 \times 10^5\ \text{J mol}^{-1} = 102.4\ \text{kJ mol}^{-1}

4. Calculate $A$ from the intercept

The intercept $c = \ln A$ . From the graph, the line crosses the $\ln k$ axis at $1/T = 0$ (theoretical). Using the point-slope form with any data point, say at $T = 40^\circ$ C:

\ln A = \ln k + \frac{E_a}{R} \cdot \frac{1}{T}

\ln A = -8.266 + (1.232 \times 10^4) \times (3.193 \times 10^{-3})

\ln A = -8.266 + 39.34 = 31.07

So:

$A = e^{31.07} \approx 3.1 \times 10^{13}\ \text{s}^{-1}$ …

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Method: Graphical Arrhenius Analysis (Two-Point & Linear Regression)

The Arrhenius equation in logarithmic form is:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

This is a straight line: $y = mx + c$ , where:

$y = \ln k$
$x = 1/T$ (in Kelvin)
Slope $m = -E_a/R$
Intercept $c = \ln A$

Step 1: Convert temperatures to Kelvin and compute $1/T$ and $\ln k$

The given $k$ values in the table are $10^5 \times k$ , so the actual $k$ used to compute $\ln k$ is the table value $\times 10^{-5}$ .

$T$ (°C)	$T$ (K)	$1/T$ (K $^{-1}$ )	$k \times 10^5$ (s $^{-1}$ )	actual $k$ (s $^{-1}$ )	$\ln k$
0	273.15	$3.661 \times 10^{-3}$	0.0787	$7.87\times10^{-7}$	$-14.055$
20	293.15	$3.411 \times 10^{-3}$	1.70	$1.70\times10^{-5}$	$-10.982$
40	313.15	$3.193 \times 10^{-3}$	25.7	$2.57\times10^{-4}$	$-8.266$
60	333.15	$3.002 \times 10^{-3}$	178	$1.78\times10^{-3}$	$-6.331$
80	353.15	$2.832 \times 10^{-3}$	2140	$2.14\times10^{-2}$	$-3.844$

Step 2: Plot $\ln k$ (y-axis) vs $1/T$ (x-axis)

You will get a straight line with a negative slope.

Step 3: Calculate slope from the graph

Using the first and last points (for a quick estimate):

\text{slope} = \frac{\Delta \ln k}{\Delta (1/T)} = \frac{-3.844 - (-14.055)}{(2.832 - 3.661) \times 10^{-3}}

= \frac{10.211}{-0.829 \times 10^{-3}} \approx -1.232 \times 10^4 \text{ K}

Step 4: Calculate activation energy $E_a$

From slope $m = -E_a/R$ :

E_a = -m \times R = 1.232 \times 10^4 \times 8.314

\boxed{E_a \approx 1.024 \times 10^5 \text{ J/mol} = 102.4 \text{ kJ/mol}}

Step 5: Calculate pre-exponential factor $A$

From intercept $c = \ln A$ :

Using the central point at $T = 313.15$ K ( $1/T = 3.193 \times 10^{-3}$ , $\ln k = -8.266$ ):

\ln A = \ln k + \frac{E_a}{R} \cdot \frac{1}{T}

\ln A = -8.266 + (1.232 \times 10^4)(3.193 \times 10^{-3})

\ln A = -8.266 + 39.34 = 31.07

\boxed{A = e^{31.07} \approx 3.1 \times 10^{13} \text{ s}^{-1}}

Step 6: Predict rate constants at 30°C and 50°C

Use the fitted line $\ln k = 31.07 - \dfrac{1.232 \times 10^4}{T}$ .

At 30°C (303.15 K):

$\ln k = 31.07 - (1.232 \times 10^4)(3.299 \times 10^{-3})$ …

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Common Mistakes & How to Avoid Them (Arrhenius Equation)

1. ✗ Forgetting to convert temperature to Kelvin

The Mistake: Students plot $1/T$ using °C values directly (e.g., $1/20$ instead of $1/293$ ).

Why it's wrong: The Arrhenius equation uses absolute temperature:

k = A e^{-E_a/RT}

$T$ must be in Kelvin ( $K = °C + 273$ ).

✓ How to avoid: Always write the conversion step explicitly:

$0°C = 273\,K$
$20°C = 293\,K$
$40°C = 313\,K$ , etc.

2. ✗ Using $k$ directly instead of $\ln k$

The Mistake: Plotting $k$ vs $1/T$ (a curve) instead of $\ln k$ vs $1/T$ (a straight line).

Why it's wrong: The Arrhenius equation in linear form is:

\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}

Only $\ln k$ vs $1/T$ gives a straight line with slope $= -E_a/R$ .

✓ How to avoid: Before plotting, compute $\ln k$ for each $k$ value. Use a table:

$T/K$	$10^5 k$	$\ln k$	$1/T$
273	0.0787	$\ln(0.0787 \times 10^{-5})$	0.00366
...	...	...	...

3. ✗ Mishandling the $10^5$ factor in $k$

The Mistake: Taking $\ln(0.0787)$ instead of $\ln(0.0787 \times 10^{-5})$ .

Why it's wrong: The given $k$ values are $10^5 \times k$ . So actual $k = (\text{table value}) \times 10^{-5}$ .

✓ How to avoid: Write clearly:

k_{\text{actual}} = (\text{table value}) \times 10^{-5}

Then take $\ln$ of this actual value.

4. ✗ Using $R$ in wrong units

The Mistake: Using $R = 0.0821$ (L·atm/mol·K) instead of $R = 8.314$ (J/mol·K).

Why it's wrong: $E_a$ is typically in J/mol or kJ/mol. The correct $R$ for energy calculations is:

R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}

✓ How to avoid: Remember:

For $E_a$ in J/mol: use $R = 8.314$
For $E_a$ in kJ/mol: use $R = 0.008314$

5. ✗ Confusing slope sign when finding $E_a$

The Mistake: Taking $E_a = \text{slope} \times R$ instead of $E_a = -\text{slope} \times R$ .

Why it's wrong: From $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$ , the slope is negative:

\text{slope} = -\frac{E_a}{R}

So $E_a = -\text{slope} \times R$ (which gives a positive value).

✓ How to avoid:

Plot the graph
Calculate slope $= \frac{\Delta \ln k}{\Delta (1/T)}$ (will be negative)
Then $E_a = -\text{slope} \times R$

6. ✗ Using wrong points for interpolation at 30°C and 50°C

The Mistake: Reading $k$ directly from the curved $k$ vs $T$ plot.

Why it's wrong: The relationship is linear only for $\ln k$ vs $1/T$ . …

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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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