General Chemistry II · Chemical Kinetics

The Arrhenius Equation

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Study tools
Sources & references

In 30 seconds

The Arrhenius equation is the quantitative version of collision theory. It expresses the rate constant k in terms of three quantities: a frequency (pre-exponential) factor A, the activation energy Ea, and the absolute temperature T:

$k = A e -E a /RT$

The exponential term is the fraction of collisions with sufficient energy; A collects everything else (collision frequency and orientation). Because Ea sits in the exponent, k is exquisitely sensitive to temperature.

Why this matters

The Arrhenius equation is the single most important quantitative tool in kinetics for temperature effects. It underpins shelf-life prediction (accelerated aging tests), the design of reactors that run hot to maximize throughput, enzyme denaturation studies, and any calculation of how a reaction's speed changes with heat. Understanding its exponential sensitivity is key to interpreting why reactions that are slow at room temperature become fast in an oven.

The college version

Core Concept

$k = A e -E a /RT$

Key Ideas

k increases as T increases and as Ea decreases.
A is essentially constant over modest temperature ranges.
The two-point (integrated) form lets you find Ea from two rate constants, or predict k at a new temperature.
T must be in kelvin; Ea and RT must be in the same energy units.

Equations and Variables

Linear form (a straight line in a plot of ln k vs 1/T):

$ln k = ln A - E a R \cdot 1 T$

slope = −Ea/R
y-intercept = ln A

Two-point form (for two temperatures T₁, T₂ with rate constants k₁, k₂):

$ln k 2 k 1 = E a R (1 T 1 - 1 T 2)$

k = rate constant (same units on both sides)
A = frequency factor (same units as k)
Ea = activation energy (J/mol)
R = 8.314 J·mol⁻¹·K⁻¹
T = absolute temperature (K)

How It Works

Take natural logarithms of the Arrhenius equation and it becomes a line: ln k = ln A − (Ea/R)(1/T). This is why chemists plot ln k against 1/T to measure Ea — the slope is −Ea/R. A steep (more negative) slope means a large activation energy and a rate constant that changes dramatically with temperature.

The two-point form is the practical shortcut. Because ln A cancels when you subtract the equation at two temperatures, you can determine Ea from just two (k, T) pairs — or, knowing Ea, predict how k changes when the temperature moves.

A useful rule of thumb emerges: for a typical Ea near 50–100 kJ/mol, a 10 °C rise in temperature roughly doubles to triples the rate constant.

Worked Example

A reaction has Ea = 75.0 kJ/mol. Its rate constant at 298 K is k₁ = 1.00 × 10⁻³ s⁻¹. Find k₂ at 308 K.

Step 1 — Convert Ea to J/mol: 75.0 kJ/mol = 7.50 × 10⁴ J/mol.

Step 2 — Compute Ea/R:

$E a R = 7.50 \times 104 J mol -1 8.314 J mol -1 K -1 = 9020 K$

Step 3 — Compute (1/T₁ − 1/T₂):

$1298 - 1308 = 0.003356 - 0.003247 = 1.09 \times 10 -4 K -1$

Step 4 — Apply the two-point form:

$ln k 2 k 1 = 9020 K \times 1.09 \times 10 -4 K -1 = 0.983$

$k 2 k 1 = e 0.983 = 2.67$

$k 2 = 2.67 \times 1.00 \times 10 -3 s -1 = 2.67 \times 10 -3 s -1$

A 10 K rise more than doubles the rate constant — consistent with the rule of thumb.

How it works

A useful rule of thumb emerges: for a typical Ea near 50–100 kJ/mol, a 10 °C rise in temperature roughly doubles to triples the rate constant.

Common confusions

Using °C instead of K. 1/T is meaningless in Celsius and can even be negative or infinite.
Mixing units for Ea and R. If Ea is in kJ/mol, convert to J/mol (or use R = 8.314 × 10⁻³ kJ·mol⁻¹·K⁻¹).
Forgetting that A also carries units. A has the same units as k, whatever they are.
Misreading the sign. Higher T always raises k for a given Ea; the negative exponent guarantees it.

Quick review

Write the Arrhenius equation and name each symbol.
What is the slope of a plot of ln k vs 1/T?
Write the two-point form of the Arrhenius equation.
Why must temperature be in kelvin?
If Ea is doubled (at fixed T), does k increase or decrease?

Eli explains

The same idea, in plain words

Explain it like I’m 10

The Arrhenius equation is the math that says "heat speeds things up — a lot." The activation energy Ea is the hill the reaction must climb; only molecules with enough energy make it over. When you raise the temperature, you hand more molecules enough energy to clear the hill, and because that "enough energy" condition is exponential, a little heat goes a long way. The two-point form is just a clever trick: by measuring the speed at two different temperatures, you can measure the height of the hill itself — no direct view of the molecules needed.

Worked example

Worked Example

A reaction has Ea = 75.0 kJ/mol. Its rate constant at 298 K is k₁ = 1.00 × 10⁻³ s⁻¹. Find k₂ at 308 K.

Step 1 — Convert Ea to J/mol: 75.0 kJ/mol = 7.50 × 10⁴ J/mol.

Step 2 — Compute Ea/R:

$E a R = 7.50 \times 104 J mol -1 8.314 J mol -1 K -1 = 9020 K$

Step 3 — Compute (1/T₁ − 1/T₂):

$1298 - 1308 = 0.003356 - 0.003247 = 1.09 \times 10 -4 K -1$

Step 4 — Apply the two-point form:

$ln k 2 k 1 = 9020 K \times 1.09 \times 10 -4 K -1 = 0.983$

$k 2 k 1 = e 0.983 = 2.67$

$k 2 = 2.67 \times 1.00 \times 10 -3 s -1 = 2.67 \times 10 -3 s -1$

A 10 K rise more than doubles the rate constant — consistent with the rule of thumb.

Key takeaways

### High-Yield Facts
k = A·e^(−Ea/RT); larger T or smaller Ea → larger k.
ln k vs 1/T is linear with slope −Ea/R.
Two-point form: ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
Use kelvin, and keep Ea in J/mol to match R = 8.314 J·mol⁻¹·K⁻¹.
~10 °C rise ≈ 2–3× rate increase for typical Ea values.

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Practice General Chemistry II

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Study tools & related lessonsYou’ll learn to · Related

You’ll learn to

State the Arrhenius equation and identify each term.
Explain how the rate constant k depends on temperature and activation energy.
Use the two-point form to calculate Ea or a new k at a different temperature.
Recall that temperature must be in kelvin.

Sources & references

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In 30 seconds

Why this matters

The college version

Core Concept

Key Ideas

Equations and Variables

How It Works

Worked Example

How it works

Common confusions

Quick review

The same idea, in plain words

Worked example

Worked Example

Key takeaways

Keep learning

You’ll learn to

Related lessons

Sources & references