Q.iii. The energy of activation for a first order reaction is 104 kJ/mol. The rate constant at 25 $^0$C is 3.7 $\times$ 10$^{-5}$ s$^{-1}$. What is the rate constant at 30$^0$C? (R = 8.314 J/K mol) (7.4 $\times$ 10$^{-5}$)

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:

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:

• $k$ = 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

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:

This is the equation of a straight line:

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).

What the plot tells you

From the slope, you get $E_a$:

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 $A$:

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

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 matters for exams

You will be asked to:

1. Calculate $E_a$ from two data points using the two-point form:

2. Predict $k$ at a new temperature if $E_a$ and $A$ are known.
3. Interpret a plot: steeper slope = higher $E_a$ = more temperature-sensitive reaction.
4. Identify the axes: always $\ln k$ vs $1/T$, never $k$ vs $T$.

The big picture

The Arrhenius plot is a tool to linearise an exponential relationship. It turns a messy curve into a clean straight line, letting you extract two fundamental properties of a reaction: how high the energy barrier is ($E_a$) and how often molecules try to cross it ($A$).

Once you see that $\ln k$ vs $1/T$ is a straight line, you've understood the core idea. Everything else — calculations, interpretations, exam problems — follows from that single linear relationship.

The Arrhenius equation is one of the most exam-relevant formulas in the NCERT/CBSE Class 12 Chemistry Chemical Kinetics chapter, and ‘Arrhenius equation derivation’ or ‘Arrhenius equation important questions’ are searched heavily by students preparing for board exams, JEE Main and NEET. This equation connects activation energy, temperature and the rate constant — a relationship tested across nearly every kinetics numerical in competitive chemistry exams.