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Chemical kinetics has a reputation for tables: units of k for each order, half-life formulas, graphs. Students memorise them and then mix them up under pressure. Almost all of it follows from two ideas: the rate law comes from experiment, and everything else comes from the rate law. This post shows how, with three worked examples.
Can you read the order from the balanced equation?
No, and this is the most common mistake in the chapter. The NCERT Class 12 chapter on Chemical Kinetics says the rate law "cannot be predicted by merely looking at the balanced chemical equation" and "must be determined experimentally". The coefficients tell you how much reacts, not how the rate depends on concentration.
The confusion is widespread. Cakmakci (2010) studied 191 secondary school and undergraduate students in Turkey and found that several alternative ideas about kinetics persisted from school into university, and that weak understanding of thermodynamics and equilibrium fed into them. A review by Bain and Towns (2016) notes that mathematics is often the main language used to teach kinetics, so it is easy to move symbols around without thinking about molecules. Keep asking what each equation says about them.
Both the JEE (Main) 2026 syllabus and the NEET (UG)-2026 syllabus list order, rate law, units of k, integrated forms of zero- and first-order reactions, half-lives and the Arrhenius equation. Check your own year's syllabus.
How do you find order from initial rates?
Change one concentration at a time and watch the rate.
Example 1. For A + 2B → products:
[A] = 0.10 M, [B] = 0.10 M: initial rate 3.0 × 10⁻⁴ M s⁻¹
[A] = 0.20 M, [B] = 0.10 M: initial rate 6.0 × 10⁻⁴ M s⁻¹
[A] = 0.10 M, [B] = 0.20 M: initial rate 3.0 × 10⁻⁴ M s⁻¹
Doubling [A] doubles the rate, so the order in A is 1.
Doubling [B] leaves the rate unchanged, so the order in B is 0, even though B's coefficient is 2.
Rate = k[A]. From the first row, k = 3.0 × 10⁻⁴ ÷ 0.10 = 3.0 × 10⁻³ s⁻¹.
If the rate had gone up four times when [A] doubled, the order would be 2, because 2² = 4. In general, if a concentration is multiplied by a and the rate by b, the order is log b ÷ log a.
What do the units of k tell you?
For a reaction of overall order n, rate = k[conc]ⁿ, so k has units of (mol L⁻¹)¹⁻ⁿ s⁻¹. That gives mol L⁻¹ s⁻¹ for zero order, s⁻¹ for first order and L mol⁻¹ s⁻¹ for second order. You can run it backwards: NCERT's example identifies k = 2.3 × 10⁻⁵ L mol⁻¹ s⁻¹ as a second-order rate constant from its units alone.
Where do half-life formulas come from?
Derive them; it takes one line each.
First order: k = (2.303/t) log([R]₀/[R]). Put [R] = [R]₀/2, and log 2 = 0.301, so t½ = 0.693/k. No concentration appears, so the half-life does not depend on how much you start with.
Zero order: [R] = [R]₀ − kt. Put [R] = [R]₀/2, so t½ = [R]₀/2k. Double the starting amount and the half-life doubles.
Example 2. For the reaction in Example 1, at fixed [B], k = 3.0 × 10⁻³ s⁻¹.
t½ = 0.693 ÷ (3.0 × 10⁻³) = 231 s.
Time to fall to one-eighth: three half-lives, 693 s.
Time for 90% to react: [R]₀/[R] = 10, so t = (2.303 ÷ 3.0 × 10⁻³) × log 10 ≈ 768 s. That is about 3.3 half-lives, not three.
How does temperature change the rate?
The Arrhenius equation in two-temperature form is log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ − 1/T₂). NCERT notes that for many reactions a 10° rise nearly doubles the rate constant.
Example 3. k doubles between 300 K and 310 K. Find Eₐ.
log 2 = 0.301.
1/T₁ − 1/T₂ = (310 − 300) ÷ (300 × 310) = 10 ÷ 93,000.
Eₐ = 2.303 × 8.314 × 0.301 × 93,000 ÷ 10 ≈ 53,600 J mol⁻¹, about 53.6 kJ mol⁻¹.
Use R = 8.314 J mol⁻¹ K⁻¹ and temperatures in kelvin, and convert to kJ only at the end.
Which mistakes should you watch for?
Orders from coefficients. Example 1 shows why this fails.
Natural log and log base 10 mixed. 2.303 converts between them; use it once, not twice.
Celsius in Arrhenius. Kelvin only.
Zero-order half-life treated as constant. It shrinks as the reaction proceeds.
Pseudo first order missed. When one reactant, such as water, is in large excess, its concentration barely changes and the reaction behaves as first order.
The same half-life algebra appears in physics; our guide to radioactive decay and half-life problems uses it.
A practice routine
Derive the zero- and first-order half-lives from memory every day for a week.
Solve five initial-rate tables, including at least one where the order differs from the coefficient.
Do four Arrhenius problems, writing units on every line.
Add the derivations, not the final formulas, to your own formula sheet.
When your method goes wrong and you cannot see where, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, take one kinetics question you got wrong and solve it again starting from the rate law.
