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Two-temperature Arrhenius questions are common, but the equation is really about a graph. Exams also give a table of rate constants, or a plot of ln k against 1/T, and ask for the activation energy, the pre-exponential factor or the rate at a new temperature. This post works one full data set through every step and points out where graph questions go wrong.
Where does the equation come from?
The NCERT Class 12 Chemical Kinetics chapter states that the temperature dependence of rate was first proposed by J.H. van 't Hoff, and that Svante Arrhenius provided its physical justification and interpretation. Arrhenius's 1889 paper in the Zeitschrift für physikalische Chemie was a study of the rate at which acids invert cane sugar. His 1903 Nobel Prize in Chemistry was awarded for something else: "his electrolytic theory of dissociation".
NCERT gives the equation as k = A e^(−Ea/RT), where A is the pre-exponential or frequency factor. Taking natural logs gives ln k = −Ea/RT + ln A, so a plot of ln k against 1/T is a straight line with slope −Ea/R and intercept ln A. The JEE (Main) 2026 and NEET (UG) 2026 syllabi both list Arrhenius theory and "activation energy and its calculation". Check your own year's syllabus.
Why do graph questions trip students up?
Yan and Subramaniam gave a four-tier diagnostic test on reaction kinetics to 137 Grade 12 students and uncovered 25 alternative conceptions. Nearly 70% of them came from questions that featured graphs. The Arrhenius plot has two traps of its own. The x-axis is 1/T, so higher temperatures sit to the left. And the slope is negative, so a steeper line means a larger activation energy.
Worked example: activation energy from a table
A first-order reaction gives these rate constants. The data are illustrative.
300 K: k = 8.7 × 10⁻³ s⁻¹
320 K: k = 5.7 × 10⁻² s⁻¹
340 K: k = 0.30 s⁻¹
Step 1. Convert to plotting values.
300 K: 1/T = 3.333 × 10⁻³ K⁻¹, ln k = −4.744
320 K: 1/T = 3.125 × 10⁻³ K⁻¹, ln k = −2.865
340 K: 1/T = 2.941 × 10⁻³ K⁻¹, ln k = −1.204
Step 2. Find the slope from the end points.
Slope = (−1.204 − (−4.744)) ÷ (2.941 × 10⁻³ − 3.333 × 10⁻³) = 3.540 ÷ (−3.92 × 10⁻⁴) ≈ −9.03 × 10³ K.
Step 3. Activation energy.
Ea = −slope × R = 9.03 × 10³ × 8.314 ≈ 7.51 × 10⁴ J mol⁻¹, or 75 kJ mol⁻¹.
Step 4. Check with the middle point. Using 300 K and 320 K alone gives a slope of −9.02 × 10³ K, which agrees. If one point disagrees badly with the others, suspect a copying slip before you suspect the chemistry.
Step 5. Pre-exponential factor. ln A = ln k + Ea/RT = −4.744 + 75 100 ÷ (8.314 × 300) = −4.744 + 30.11 ≈ 25.4, so A ≈ 1.0 × 10¹¹ s⁻¹. A has the same units as k.
Step 6. Rate at a new temperature. Find k at 330 K from the 300 K value:
ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = 9030 × (1/300 − 1/330) = 9030 × 3.03 × 10⁻⁴ ≈ 2.74.
k₂/k₁ = e^2.74 ≈ 15.4, so k at 330 K ≈ 8.7 × 10⁻³ × 15.4 ≈ 0.134 s⁻¹. That sits between the 320 K and 340 K values, as it should.
What if the graph uses log k?
NCERT also writes the two-temperature form with base-10 logs: log(k₂/k₁) = [Ea ÷ (2.303R)] × [(T₂ − T₁) ÷ (T₁T₂)]. On a plot of log k against 1/T, the slope is −Ea/(2.303R). With the data above, log(0.30 ÷ 8.7 × 10⁻³) = 1.538, and Ea = 1.538 × 2.303 × 8.314 ÷ (3.92 × 10⁻⁴) ≈ 75 kJ mol⁻¹ again. Use the factor 2.303 once, with log, and not at all with ln.
Which mistakes should you avoid?
Plotting k against T, or ln k against T. Only ln k against 1/T is straight.
Dropping the minus sign in the slope and getting a negative Ea.
Using 2.303 with natural logs.
Using temperatures in Celsius.
Mixing kJ and J. Keep Ea in J mol⁻¹ until the final line.
Reading the slope from two points that are very close together, which magnifies rounding error.
The physical meaning of Ea, and why it controls the rate so strongly, is in our post on collision theory and catalysts. For order, half-life and the basic two-temperature form, see our chemical kinetics guide, and for slopes and intercepts in general, our post on reading physics graphs. When a slope calculation does not match the answer key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
A practice routine
Day 1: convert three data tables to 1/T and ln k, and sketch each plot by hand.
Day 2: four problems giving a graph, asking for Ea and A.
Day 3: four two-temperature problems, two with ln and two with log.
Day 4: a timed set of previous-year Arrhenius questions.
Today, plot the three points from the worked example on graph paper and measure the slope yourself.
