The big idea: At HL we put a number on the link between temperature and rate. The Arrhenius equation says how the rate constant k depends on temperature:
k = A·e−E_{a/RT}
The rate constant grows when the temperature T rises, when the activation energy Ea is low, and when the Arrhenius factor A (collision frequency and orientation) is large.
- the rate constant
- the Arrhenius (frequency) factor — collision frequency × correct orientation (steric factor)
- the activation energy (J mol⁻¹)
- the gas constant = 8.31 J K⁻¹ mol⁻¹
- the absolute temperature (K)
Define your terms: - A — the Arrhenius (frequency) factor. It accounts for how often molecules collide and whether they collide with the right orientation (the steric factor). It has the same units as k. - Ea — the activation energy. The minimum energy a collision must have to react. It sits in the exponent, so it has a huge effect on k. - R — the gas constant, 8.31 J K⁻¹ mol⁻¹. - T — the absolute temperature in kelvin (always convert °C → K by adding 273).
Why k rises so STEEPLY with temperature: Ea is negative in the exponent and T is in the denominator. So a small rise in T makes the exponent less negative, and because it is an exponential, k can change a lot — a 10 K rise near room temperature often roughly doubles the rate constant.
This is the maths behind the Maxwell–Boltzmann picture: a higher T puts a much larger fraction of molecules above Ea.
Why k climbs so steeply with T: raising the temperature shifts the energy distribution right, so a much larger fraction of molecules has energy ≥ Eₐ. That growing fraction is exactly what the e−Eₐ/RT term in the Arrhenius equation captures.
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Reading the Arrhenius equation tells you, at a glance, what makes k bigger or smaller:
| Change | Effect on the exponent −Eₐ/RT | Effect on k |
|---|---|---|
| raise T | the exponent becomes less negative (T is in the denominator) | k increases — steeply (often roughly doubles per 10 K) |
| lower Eₐ (e.g. a catalyst) | the exponent becomes less negative | k increases |
| raise Eₐ | the exponent becomes more negative | k decreases |
| larger A (more frequent, better-oriented collisions) | no change to the exponent | k increases (k ∝ A) |
Take logs to get a straight line: The exponential form is awkward to handle. Taking the natural log of both sides turns it into the equation of a straight line:
- plotted on the y-axis
- plotted on the x-axis (T in K)
- the gradient of the straight line
- the y-intercept (where 1/T = 0)
Match it to y = mx + c: Compare ln k = ln A − (Ea/R)(1/T) with y = mx + c:
- y = ln k (vertical axis) - x = 1/T (horizontal axis) - gradient m = −Ea/R (always negative, because Ea > 0) - intercept c = ln A (the value of ln k when 1/T = 0)
So plotting ln k against 1/T gives a straight line whose slope gives Ea and whose intercept gives A.
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The rate constant of a reaction was measured at five temperatures. The values of ln k and 1/T are tabulated below ready to plot an Arrhenius plot (ln k on the y-axis, 1/T on the x-axis):
An Arrhenius plot: ln k against 1/T gives a straight line with gradient = −Eₐ/R and y-intercept = ln A. The steeper the (negative) gradient, the larger the activation energy.
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| T / K | 1/T / 10⁻³ K⁻¹ | k / s⁻¹ | ln k |
|---|---|---|---|
| 333 | 3.00 | 9.7 × 10⁻² | −2.33 |
| 323 | 3.10 | 5.9 × 10⁻² | −2.83 |
| 313 | 3.19 | 3.5 × 10⁻² | −3.35 |
| 303 | 3.30 | 2.0 × 10⁻² | −3.91 |
| 294 | 3.40 | 1.3 × 10⁻² | −4.34 |
Reading the gradient: The points lie on a straight line. Take two well-separated points on the line (the first and last rows) to find the gradient — gradient = Δ(ln k) ÷ Δ(1/T). Remember 1/T values are in 10⁻³ K⁻¹, so 3.00 means 3.00 × 10⁻³ K⁻¹.
Worked example — activation energy from the gradient
Use the first and last data points to find the gradient of the ln k vs 1/T line, then calculate the activation energy Eₐ in kJ mol⁻¹. (R = 8.31 J K⁻¹ mol⁻¹.)
Solution
- Formula first — the linear Arrhenius form tells us the gradient:
- Find the gradient from the two end points (1/T in K⁻¹, so 3.00 × 10⁻³ and 3.40 × 10⁻³):
- Work out top and bottom:
- Rearrange the relationship for Ea:
- Substitute (the two minus signs cancel):
- Convert J → kJ by dividing by 1000:
Final answer
gradient = −5.0 × 10³ K; Eₐ = −gradient × R = 41.6 kJ mol⁻¹.
Worked example — the Arrhenius factor A from the intercept
The y-intercept of the same ln k vs 1/T line (where 1/T = 0) is ln A = 12.7. Calculate the Arrhenius factor A.
Solution
- Formula first — the intercept of the linear form is ln A:
- Undo the natural log by raising e to that power:
- Work it out (same units as k, here s⁻¹):
Final answer
A = e¹²·⁷ = 3.3 × 10⁵ s⁻¹ (A has the same units as k).
You don't always have a full graph. If you know the rate constant at two temperatures, the two-point form of the Arrhenius equation gives Ea directly:
- rate constants at temperatures T₁ and T₂
- the two absolute temperatures (K)
- the activation energy (J mol⁻¹ — ÷ 1000 for kJ mol⁻¹)
- the gas constant = 8.31 J K⁻¹ mol⁻¹
Worked example — Eₐ from two rate constants
For a reaction, k = 1.0 × 10⁻³ s⁻¹ at 300 K (T₁) and k = 4.0 × 10⁻³ s⁻¹ at 320 K (T₂). Calculate the activation energy Eₐ in kJ mol⁻¹. (R = 8.31 J K⁻¹ mol⁻¹.)
Solution
- Formula first — the two-point Arrhenius form:
- Left side — the ratio of rate constants:
- The temperature bracket (T₂ = 320 K, T₁ = 300 K):
- Rearrange for Ea:
- Substitute (the two minus signs make Ea positive):
- Work it out, then convert J → kJ:
Final answer
Eₐ = 55.3 kJ mol⁻¹.
How this is tested: This is a classic HL Paper 2 skill.
- Graph version: you are given (or must plot) ln k against 1/T, read the gradient, then use Ea = −gradient × R and convert to kJ mol⁻¹. - Two-point version: you are given k at two temperatures and use the two-point form.
Both routes give the same Ea; examiners may also ask you to explain the shape of the graph (straight line, negative gradient) and to identify the intercept as ln A.
The traps that cost marks: (1) Always work in kelvin (add 273 to °C). (2) The gradient is −Ea/R, so Ea = −gradient × R — never drop the minus sign. (3) R is in J K⁻¹ mol⁻¹, so your raw Ea is in J mol⁻¹ — ÷ 1000 for kJ mol⁻¹.