IB Chemistry HLTopic 5 — How Fast? The Rate of ReactionPaper 1 & 2HL only | Practical skill~11 min read
Determining Ea and the Arrhenius Factor
Nobody measures activation energy directly. What you can measure is the rate constant at a few different temperatures — and with one clever rearrangement, those readings fall onto a straight line whose gradient hands you Ea. This is one of the most predictable long questions in the whole HL course, so it is worth drilling until it is automatic.
📘 What you need to know
Rearranged, the Arrhenius equation is ln k = –(Ea/R) × (1/T) + ln A.
That matches y = mx + c, so plot ln k on the y-axis against 1/T on the x-axis.
The graph is a straight line with a negative gradient.
gradient = –Ea/R, so Ea = –gradient × R. The two minus signs cancel, so Ea comes out positive.
intercept = ln A, so A = eintercept.
You can only read the intercept off the graph if the x-axis starts at zero. It almost never does.
When it does not, substitute a point from the line back into the equation to find ln A.
T must be in kelvin, and Ea comes out in J mol–1.
Turning a curve into a straight line
Plot k against T and you get a curve, which is useless for pulling numbers out. Take natural logs of both sides and the same data becomes a straight line, because 1/T and ln k are linked linearly.
The equation, written to look like a straight line
ln k = –EaR × 1T + ln A
In y = mx + c
In the Arrhenius equation
What you do with it
y
ln k
Take ln of every rate constant
x
1/T
Take the reciprocal of every temperature in kelvin
m, the gradient
–Ea/R
Multiply by –R to get Ea
c, the intercept
ln A
Take ec to get A
Notice which variable goes on which axis. It is ln k against 1/T, not k against T and not the other way round. Getting the axes the wrong way up turns the gradient upside down and wrecks every number after it.
What the graph looks like
Take the two ends of your line of best fit rather than two data points. Using the line averages out the scatter, and a big triangle keeps the percentage error in the gradient small.
From gradient to activation energy
The gradient is negative, and there is a minus sign in the formula. The two cancel, which is exactly why activation energy always comes out positive.
The two results you need
Ea = –gradient × R and A = eintercept
🤔 Why is the line straight, and why does it slope down?
Straight, because ln k depends on 1/T through a single multiplication by the fixed number –Ea/R — nothing is squared or curved. Downwards, because a bigger value of 1/T means a lower temperature, and a lower temperature means a smaller k. So moving right along the x-axis is moving to colder conditions, and the line has to fall.
The intercept trap
Textbook versions of this graph show the line crossing the y-axis at ln A. Real ones almost never do, because 1/T values for sensible temperatures are all crammed between about 0.002 and 0.004. Drawing an axis from zero would squash every point into a corner, so exam graphs start part way along — and that means the point where your line meets the left-hand edge is not the intercept.
🧩 Finding A when the axis does not start at zero
Work out the gradient and the activation energy first.
Pick any clear point on the line of best fit and read off its 1/T and ln k.
Substitute into ln k = –(Ea/R)(1/T) + ln A.
Rearrange for ln A and work it out.
Take eln A to get A, and give it the same units as k.
Faster route in the calculation: since gradient = –Ea/R, you can put the gradient straight into ln A = ln k – (gradient × 1/T) without converting to Ea and back. Fewer steps, fewer rounding errors.
Worked examples
WORKED EXAMPLE
The full graph method
The rate constant of a first order reaction was measured at five temperatures.
T / K
(1/T) / K–1
k / s–1
ln k
300
3.33 × 10–3
1.75 × 10–2
–4.05
320
3.13 × 10–3
6.43 × 10–2
–2.74
340
2.94 × 10–3
2.03 × 10–1
–1.59
360
2.78 × 10–3
5.65 × 10–1
–0.57
380
2.63 × 10–3
1.41
0.34
A graph of ln k against 1/T is a straight line. Calculate (a) the activation energy in kJ mol–1 and (b) the Arrhenius factor.
(a) Step 1: Gradient, using the two ends of the linegradient = 0.34 – (–4.05)(2.63 – 3.33) × 10–3= 4.39–0.70 × 10–3 = –6271 KStep 2: Turn the gradient into EaEa = –gradient × R = –(–6271) × 8.31= 52112 J mol–1Ea = 52.1 kJ mol–1 (3 s.f.)(b) Step 3: The x-axis does not start at zero, so substitute a pointUse the point (2.63 × 10⁻³, 0.34) from the line.ln A = ln k – (gradient × 1/T)= 0.34 – (–6271 × 2.63 × 10–3)= 0.34 + 16.49 = 16.83Step 4: Undo the logA = e16.83A = 2.04 × 107 s–1A takes the units of k, so a first order reaction gives A in s⁻¹
WORKED EXAMPLE
Two readings, no graph
The same reaction has k = 1.75 × 10–2 s–1 at 300 K and k = 1.41 s–1 at 380 K. Calculate Ea without plotting anything.
Step 1: The gradient is just rise over run for these two pointsln (1.41) = 0.344 and ln (1.75 × 10–2) = –4.045Δ(ln k) = 0.344 – (–4.045) = 4.389Step 2: Work out the change in 1/T1380 = 2.6316 × 10–3 and 1300 = 3.3333 × 10–3Δ(1/T) = –7.018 × 10–4Step 3: Gradient, then Eagradient = 4.389–7.018 × 10–4 = –6254 KEa = 6254 × 8.31 = 51970 J mol–1Ea = 52.0 kJ mol–1 (3 s.f.)same answer as the graph method – the graph just uses all five points instead of two
WORKED EXAMPLE
Comparing two lines on the same axes
Two reactions are plotted as ln k against 1/T on the same axes. Line P has a gradient of –3.10 × 103 K and line Q has a gradient of –9.40 × 103 K.
(a) Calculate both activation energies. (b) State which reaction is more sensitive to a change in temperature, and explain why.
(a) Multiply each gradient by –RP: Ea = 3100 × 8.31 = 25761 J mol–1P: Ea = 25.8 kJ mol–1Q: Ea = 9400 × 8.31 = 78114 J mol–1Q: Ea = 78.1 kJ mol–1(b) Which one cares more about temperatureQ has the steeper line, so its ln k changes more for the same change in 1/T.Q — a higher activation energy means the rate constant changes more sharply with temperaturesteeper down means bigger Eₐ – one of the quickest marks in this topic
🧠 The four-line summary
Plot ln k against 1/T. Gradient × –R gives Ea. Intercept in the ex button gives A. And if the axis does not start at zero, substitute a point instead of reading the intercept. Learn those four lines and this whole page fits on a revision card.
💡 Exam tip
Make the gradient triangle big. A small triangle magnifies your reading errors, and examiners look for a large one.
Take the triangle corners from the line of best fit, not from two plotted points.
The gradient has units of K, so Ea comes out in J mol–1. Divide by 1000 at the end.
Check the sign: a negative gradient must give a positive Ea. If yours is negative, you have dropped a minus.
When completing a table, keep 1/T to 3 significant figures and ln k to 2 decimal places unless told otherwise.
Look at where the x-axis starts before you go anywhere near the intercept.
Give A the same units as k, and expect a large number — often 107 or bigger.
⚠ Common mix-up
Plotting k against T. That gives a curve you cannot get a gradient from.
Reading the intercept off a truncated axis. The commonest error on this whole page.
Forgetting the minus in Ea = –gradient × R. It is what makes the answer positive.
Leaving the answer in J mol–1. Questions almost always want kJ mol–1.
Stopping at ln A. That is not A until you have pressed ex.
Using °C when working out 1/T. Kelvin only.
Reading a steeper line as a faster reaction. A steeper line means a bigger activation energy, which usually means a slower one.
That is the end of How Fast? The Rate of Reaction. Up next: Features of Dynamic Equilibrium, where the question changes from how fast a reaction goes to how far it gets.
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