Activation Energy from an Arrhenius Graph
Measure a rate constant at five temperatures and you can find the activation energy of the reaction. The tool for it is a graph — the Arrhenius plot — and the whole answer lives in its slope. This page shows how to build the plot, how to extract the slope properly, and how to turn that slope into Ea in kJ mol⁻¹, with every number computed in full.
From the Arrhenius equation to a straight line
Compare that with y = mx + c:
| Straight line | Arrhenius plot | Meaning |
|---|---|---|
| y | ln k | natural log of the rate constant (k must be a pure number here) |
| x | 1/T | reciprocal of the absolute temperature, in K⁻¹ |
| m (slope) | −Ea/R | always negative for a normal reaction |
| c (intercept) | ln A | the value of ln k extrapolated to 1/T = 0 |
So Ea = −R × slope, with R = 8.314 J K⁻¹ mol⁻¹. Because the slope is negative, Ea comes out positive, and it arrives in joules per mole — divide by 1000 for kJ mol⁻¹.
The log₁₀ version — both appear in textbooks
Many Indian textbooks plot log₁₀k instead of ln k. The line is still straight, but the slope changes by the factor 2.303:
Neither version is "the correct one" — they are the same equation. What matters is that you match the constant to the axis you actually plotted. Using ln on the axis and 2.303 in the formula inflates Ea by 2.303 times, and that is a whole-question error.
Worked example 1 — the full plot
A first-order reaction gives these rate constants:
| T / K | 300 | 310 | 320 | 330 | 340 |
|---|---|---|---|---|---|
| k / s⁻¹ | 1.97 | 3.76 | 6.89 | 12.2 | 20.8 |
Step 1 — build the two columns you will actually plot.
| T / K | 1/T / 10⁻³ K⁻¹ | k / s⁻¹ | ln k |
|---|---|---|---|
| 300 | 3.3333 | 1.97 | 0.6780 |
| 310 | 3.2258 | 3.76 | 1.3244 |
| 320 | 3.1250 | 6.89 | 1.9301 |
| 330 | 3.0303 | 12.2 | 2.5014 |
| 340 | 2.9412 | 20.8 | 3.0350 |
Step 2 — the slope. The quickest honest estimate uses the two extreme
points, provided they sit on the line:
slope = (3.0350 − 0.6780) ÷ (2.9412 × 10⁻³ − 3.3333 × 10⁻³)
= 2.3570 ÷ (−3.9216 × 10⁻⁴) = −6010 K
Step 3 — activation energy.
Ea = −R × slope = −8.314 × (−6010) = 49 967 J mol⁻¹ = 50.0 kJ mol⁻¹
Step 4 — the least-squares check. Fitting all five points rather than
two gives slope = −6012 K, so
Ea = 8.314 × 6012 = 49 984 J mol⁻¹ = 50.0 kJ mol⁻¹
The two routes agree to three significant figures, which tells you the data really is
linear. If they had disagreed badly, one of the end points would be an outlier and the
fitted line would be the one to trust.
Worked example 2 — the two-point shortcut
If you are given only two rate constants, you do not need a graph at all. Subtracting the two logarithmic forms eliminates ln A:
Take k₁ = 1.97 s⁻¹ at T₁ = 300 K and k₂ = 6.89 s⁻¹ at T₂ = 320 K from the same data.
k₂ ÷ k₁ = 6.89 ÷ 1.97 = 3.4975, so ln(k₂/k₁) = 1.2520
1/T₁ − 1/T₂ = (1 ÷ 300) − (1 ÷ 320) = 3.33333 × 10⁻³ − 3.12500 × 10⁻³ = 2.0833 × 10⁻⁴ K⁻¹
Ea = R × 1.2520 ÷ (2.0833 × 10⁻⁴) = 8.314 × 6009.9 = 49 966 J mol⁻¹ = 50.0 kJ mol⁻¹
Same answer as the full plot — which is exactly what should happen when the data is clean. The value of the graph is not a better answer from good data; it is protection against bad data, because one wild point stands out visually and barely moves a five-point fit.
Worked example 3 — the intercept and the pre-exponential factor
Using the fitted line from example 1: slope = −6012 K, and the mean values are x̄ = 3.1311 × 10⁻³ K⁻¹ and ȳ = 1.8938.
Every least-squares line passes through (x̄, ȳ), so
intercept = ȳ − (slope × x̄) = 1.8938 − (−6012 × 3.1311 × 10⁻³)
= 1.8938 + 18.8253 = ln A = 20.719
A = e20.719 = 1.0 × 10⁹ s⁻¹
A carries the same units as k, so for this first-order reaction it is s⁻¹. Note how far the extrapolation runs: the data sits between 1/T = 2.94 × 10⁻³ and 3.33 × 10⁻³, but the intercept is read at 1/T = 0, which corresponds to infinite temperature. A small error in the slope therefore becomes a large error in A. Ea from an Arrhenius plot is reliable; A from the same plot is much less so, and that is worth saying in an answer.
Reading a slope off a drawn graph
In a practical exam you may have to measure the slope from the graph paper itself. Two rules make the difference between full marks and half:
- Take both points from the line you drew, not from two plotted crosses. The whole point of the best-fit line is that it averages the scatter; picking two raw points throws that away.
- Use a large triangle. Choose points near the two ends of the line so that Δy and Δx are big; a small triangle magnifies your reading error.
- Do not force the line through the origin. The intercept is a real quantity (ln A) and is almost never zero.
Also watch the axis scaling. If you label the x-axis in units of 10⁻³ K⁻¹, the slope you read is in those units too, and must be multiplied by 10³ before you use it. In example 1 the raw fit gave −6.012 per 10⁻³ K⁻¹, which is −6012 K.
Common mistakes that cost marks
- Using °C instead of K. 1/T must use absolute temperature. Adding 273.15 is the first line of the working, every time.
- Plotting T instead of 1/T. ln k against T is a curve and its slope means nothing.
- Losing the minus sign. The slope is −Ea/R, so Ea = −R × slope. A negative activation energy from a normal reaction is a sign error, not a discovery.
- Mixing ln and log₁₀. ln axis → Ea = −R × slope; log₁₀ axis → Ea = −2.303 × R × slope.
- Mixing J and kJ. With R = 8.314 J K⁻¹ mol⁻¹ the answer is in J mol⁻¹. Divide by 1000 at the end, not in the middle.
- Forgetting the 10⁻³ on the x-axis. This silently divides Ea by 1000 and produces an answer of about 50 J mol⁻¹ instead of 50 kJ mol⁻¹. A quick sanity check helps: most ordinary reactions sit roughly between 40 and 200 kJ mol⁻¹.
- Taking logs of a quantity with units. Strictly, ln k means ln of the numerical value of k in the stated units. Keep the units consistent across every temperature and this takes care of itself.
Where this appears in exams
| Level | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Two-point calculation of Ea; state what the slope of ln k against 1/T represents |
| School practical work | Plot the graph from tabulated k values and measure the slope |
| IIT-JAM / CUET-PG | Interpret a given Arrhenius plot; compare two reactions from their slopes |
| GATE / CSIR-NET | Least-squares fits, the intercept as ln A, and where the simple Arrhenius picture breaks down |
Check your slope before you trust your Ea. The Arrhenius calculator takes two rate constants and their temperatures and returns the activation energy, so you can verify the two-point result in seconds and compare it with the value your graph gave.
Open the Arrhenius Calculator →For the full five-point fit, the Linear Regression (y = mx + c) tool will fit your (1/T, ln k) pairs and hand you the slope and intercept directly.
Struggling to turn graphs into marks? ABC Chemistry teaches Class 11–12 chemistry at its Gurugram centre and in online batches across India, with home tuition available in Delhi-NCR — abcchemistry.in.