Foundations — Temperature dependence — Arrhenius equation k = A·e^(−Ea - RT)

This page assumes nothing. Before you can read the parent Arrhenius note, every letter in $k=A\cdot e^{-E_a/(RT)}$ must mean something to you as a picture, not just a squiggle. We build them one at a time, each resting on the last.

1. The rate constant $k$ — "how eager is this reaction?"

The picture. Imagine two identical beakers with the same amount of chemical. In one, the reaction fizzes away in seconds; in the other, it crawls for hours. Same amount of reactant, different speed — that difference lives entirely in $k$.

Why the topic needs it. The entire Arrhenius equation exists to answer one question: how does $k$ change when we heat things up? So $k$ is the star we are tracking. Everything else (temperature, energy) are the dials that move it.

To fully understand where $k$ sits inside a rate expression, see Rate Laws and Rate Constants.

2. Temperature $T$ and why it must be in Kelvin

The picture. Look at the figure below: cold molecules (left) drift slowly; hot molecules (right) zip and crash hard.

Why Kelvin, not Celsius? The energy of motion is proportional to $T$. Doubling the true jiggling energy means doubling $T$. On the Celsius scale, $0{}^{\circ }\text{C}$ is not zero energy — molecules are still moving fast at the freezing point of water. Only the Kelvin scale starts at true zero, so only Kelvin makes "energy $\propto T$" honest.

$T(\text{K})=T{(}^{\circ }\text{C})+273.15$

3. Energy $E$, and the special barrier energy $E_a$

The picture. Think of a hill between "reactants" valley and "products" valley. To roll the ball across, you must first push it up to the top of the hill. The height of that hilltop is $E_a$.

Why the topic needs it. The whole reason temperature matters is this barrier. If there were no hill ($E_a=0$), every collision would react and heating would barely matter. The taller the hill, the more the reaction cares about temperature. For the deeper story of what the hilltop actually is, see Activation Energy and Transition State Theory.

4. The Boltzmann factor $e^{-E_a/(RT)}$ — the fraction that makes it over

This is the heart of the equation, so we build it in three moves.

4a. What is $e$ and what is an exponential?

The picture. As shown below, $e^{-x}$ starts at $1$ (when $x=0$) and slides smoothly down toward — but never reaching — zero.

Why an exponential and not, say, a straight line? Nature distributes energy among molecules in an exponential way: a few molecules have lots of energy, many have little, and the count drops off exponentially as you demand more energy. This is the Maxwell-Boltzmann Distribution. An exponential is the only shape that matches how a gas actually shares energy — that is why it appears, not a straight line or a parabola.

4b. The gas constant $R$ — the exchange rate between temperature and energy

The picture. Think of $RT$ as the size of the "energy budget" the temperature hands to each mole of molecules. Warmer $T$ → bigger budget $RT$.

Why the topic needs it. We want to compare the barrier $E_a$ against how much energy the temperature can supply. You cannot compare an energy to a raw temperature directly — $R$ is the translator that puts them in the same units so the ratio makes sense.

4c. Putting it together: the ratio $\frac{E_a}{RT}$

Now the pieces click:

$\frac{E_a}{RT}=\frac{\text{energy needed (barrier height)}}{\text{energy available (thermal budget)}}$

Because top and bottom are both energies (in J/mol), the ratio is a pure number — no units. It answers: "How many times bigger is the wall than the energy we've got?"

• If the wall is huge compared to the budget → ratio is large → $e^{-\text{large}}$ is tiny → almost no molecules react.
• If the wall is small compared to the budget → ratio is small → $e^{-\text{small}}\approx 1$ → almost every collision reacts.

5. The pre-exponential factor $A$ — "how often, and how well-aimed"

The picture. Two cars can crash a lot (high frequency) but must also hit at the right angle to lock bumpers (orientation). $A$ counts the well-aimed collisions per second.

Why the topic needs it. Energy isn't the whole story: molecules also need to meet and be pointed correctly. $A$ carries all of that so the Boltzmann factor is free to carry only the energy story. The related idea of lowering the barrier without changing $A$ much is Catalysis.

6. Logarithms $\ln$ — the tool that straightens the curve

Why the topic needs it. The Arrhenius curve $k$ vs $T$ is a bent, hard-to-read exponential. Taking $\ln$ of both sides unbends it into a straight line, and a straight line's slope is easy to measure — that slope hands us $E_a$ directly. Using a log here is a deliberate choice: it is the one operation that peels the variable out of the exponent.

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

This is just $y=c+mx$ in disguise, with $y=\ln k$ and $x=1/T$.

How the foundations feed the topic

Equipment checklist

Test yourself — can you answer each before revealing?