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Most students remember that a reaction "doubles its rate for every 10 °C rise". It is a handy rule, and numericals use it, but it is only an approximation, and some questions are built to test whether you know when it breaks. This post defines the temperature coefficient, shows how it follows from the Arrhenius equation, works out what activation energy the doubling rule assumes, and tests it against half-life data printed in NCERT itself.
Where is this in NCERT and the syllabus?
The current NCERT Class 12 Chemistry Part I chapter Chemical Kinetics (2026-27 reprint) opens Section 3.4 on temperature dependence by noting that for a chemical reaction "with rise in temperature by 10°, the rate constant is nearly doubled", and then introduces the Arrhenius equation k = A e^(−Ea/RT) (Eq. 3.18) to explain it. The same section gives half-lives for N₂O₅ decomposition: 12 min at 50 °C, 5 h at 25 °C and 10 days at 0 °C. The JEE (Main) 2026 syllabus lists "the effect of temperature on the rate of reactions, Arrhenius theory, activation energy and its calculation", and the NEET (UG) 2026 syllabus has the same line. Check your exam year's syllabus.
For finding Ea from a graph of ln k against 1/T, see our guide to the Arrhenius equation and activation energy. This post is about the 10-degree ratio itself.
What is the temperature coefficient?
The temperature coefficient, often written μ, is the ratio of rate constants at two temperatures 10 degrees apart:
μ = k(T + 10)/k(T)
If μ is constant over a range, a rise of 10n degrees multiplies the rate constant by μⁿ. With μ = 2, a rise from 25 °C to 65 °C is four steps of 10, so k rises 2⁴ = 16 times. For a first-order reaction the half-life falls by the same factor, so a reaction that takes 80 min at 30 °C would take 80/2³ = 10 min at 60 °C.
What activation energy makes the rule exact?
From the Arrhenius equation, ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂). Set k₂/k₁ = 2 with T₁ = 298 K and T₂ = 308 K:
1/298 − 1/308 = 1.0895 × 10⁻⁴ K⁻¹.
ln 2 = 0.693.
Ea = 8.314 × 0.693 ÷ (1.0895 × 10⁻⁴) ≈ 5.29 × 10⁴ J/mol, about 52.9 kJ/mol.
This is the calculation in NCERT's intext question 3.8. So "rate doubles every 10 degrees" is really "this reaction has an Ea near 53 kJ/mol and we are near room temperature".
How does μ change with Ea and temperature?
Keep the range at 298 K to 308 K and vary Ea:
Ea = 30 kJ/mol: μ ≈ 1.48.
Ea = 52.9 kJ/mol: μ = 2.00.
Ea = 100 kJ/mol: μ ≈ 3.71.
Now keep Ea at 52.9 kJ/mol and move up the temperature scale:
298 K to 308 K: μ = 2.00.
348 K to 358 K: μ ≈ 1.67.
398 K to 408 K: μ ≈ 1.48.
Two patterns follow. A higher activation energy means stronger temperature sensitivity. And the same 10-degree step matters less at high temperature, because 1/T changes less between, say, 398 and 408 K than between 298 and 308 K.
Worked example: does N₂O₅ obey the rule?
Use NCERT's half-lives. N₂O₅ decomposition is first order, so k is proportional to 1/t½ and the ratio of rate constants equals the inverse ratio of half-lives.
From 25 °C to 50 °C. Half-life goes from 5 h (300 min) to 12 min, so k rises 300/12 = 25 times over 25 degrees, which is 2.5 steps of 10. Then μ = 25^(1/2.5) ≈ 3.6.
From 0 °C to 25 °C. Ten days is 14,400 min, so k rises 14,400/300 = 48 times over 2.5 steps. Then μ = 48^(1/2.5) ≈ 4.7.
Activation energy. For 25 °C to 50 °C: ln 25 = 3.22, and 1/298.15 − 1/323.15 = 2.595 × 10⁻⁴ K⁻¹, so Ea = 8.314 × 3.22 ÷ (2.595 × 10⁻⁴) ≈ 103 kJ/mol. The 0 °C to 25 °C data give about 105 kJ/mol. These two estimates agree, so the reaction follows Arrhenius behaviour; it simply has a high Ea, and the doubling rule badly underestimates it. The larger μ at the colder end is the temperature effect from the list above.
Why does a small rise make such a big difference?
NCERT explains this with the Maxwell-Boltzmann distribution: only molecules with energy at or above Ea react, and a 10-degree rise roughly doubles the fraction above that threshold for a typical reaction. The factor e^(−Ea/RT) in the Arrhenius equation is that fraction. Because it is exponential in Ea, a small change in T shifts it far more for a large Ea.
Which mistakes cost marks?
Adding instead of multiplying. Three 10-degree steps with μ = 2 give 2³ = 8, not 2 × 3 = 6.
Using °C in the Arrhenius equation. Convert to kelvin first; 1/25 − 1/35 is meaningless.
Assuming μ = 2 when data are given. If the question supplies two rate constants or half-lives, calculate μ from them.
Forgetting half-life goes the other way. A faster reaction has a shorter half-life. Divide, do not multiply.
Confusing ln and log. With log, the equation needs 2.303R.
If you keep losing the thread between μ, Ea and the Arrhenius equation, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
A practice routine
With μ = 3, find how much faster a reaction runs at 55 °C than at 25 °C. Check: 3³ = 27 times.
Find Ea if μ = 2 between 348 K and 358 K. Check: about 71.8 kJ/mol.
Repeat the N₂O₅ calculation using only the 0 °C and 50 °C data.
Explain in two lines why the same Ea gives a smaller μ at higher temperature.
Tonight, redo the 52.9 kJ/mol calculation from a blank page and keep every unit in place.
