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Initial-rate tables are easy when doubling a concentration doubles or quadruples the rate. Exams rarely stop there. They give you tables where two concentrations change together, where the rate changes by 2.82 times, or where the order is not a whole number. This post works through those harder cases using NCERT's own data. For the basics of rate laws, half-lives and the Arrhenius equation, see our post on chemical kinetics and rate laws.
Why do we measure initial rates?
The NCERT Class 12 chapter Chemical Kinetics states that a rate law "cannot be predicted by merely looking at the balanced chemical equation" and "must be determined experimentally". Measuring the rate at the very start, before concentrations have changed much, lets you link each rate to a known set of starting concentrations. Order and rate law sit in both the JEE (Main) 2026 syllabus and the NEET (UG)-2026 syllabus.
For rate = k[A]^x[B]^y, the method is: find two experiments where only [A] changes, take the ratio of rates, and the unknown [B] terms cancel. In general, if [A] is multiplied by a and the rate by b, then a^x = b, so x = log b ÷ log a.
Worked example 1: NCERT's NO and O₂ table
NCERT Table 3.2 gives initial rates of formation of NO₂ for 2NO + O₂ → 2NO₂ (concentrations in mol L⁻¹, rates in mol L⁻¹ s⁻¹):
Exp 1: [NO] = 0.30, [O₂] = 0.30, rate = 0.096
Exp 2: [NO] = 0.60, [O₂] = 0.30, rate = 0.384
Exp 3: [NO] = 0.30, [O₂] = 0.60, rate = 0.192
Exp 4: [NO] = 0.60, [O₂] = 0.60, rate = 0.768
Exp 1 to 2: [NO] doubles, rate goes 0.096 → 0.384, four times, so order in NO = 2. Exp 1 to 3: [O₂] doubles, rate doubles, so order in O₂ = 1. Rate = k[NO]²[O₂], third order overall.
Now use the row nobody uses: Exp 4. Both concentrations doubled, so the prediction is 2² × 2 = 8 times Exp 1, that is 8 × 0.096 = 0.768. It matches. A spare row is a free check.
The rate constant from Exp 1: k = 0.096 ÷ (0.30² × 0.30) = 0.096 ÷ 0.027 ≈ 3.56 L² mol⁻² s⁻¹.
Worked example 2: a fractional order (NCERT exercise 3.10)
Exp 1: [A] = 0.20, [B] = 0.30, r₀ = 5.07 × 10⁻⁵
Exp 2: [A] = 0.20, [B] = 0.10, r₀ = 5.07 × 10⁻⁵
Exp 3: [A] = 0.40, [B] = 0.05, r₀ = 1.43 × 10⁻⁴
Exp 1 to 2: [A] fixed, [B] falls to a third, and the rate does not change. Order in B = 0.
Exp 2 to 3: [A] doubles while [B] halves. Because the order in B is 0, B's change has no effect, and the whole rate change is due to A. Rate ratio = 1.43 × 10⁻⁴ ÷ 5.07 × 10⁻⁵ ≈ 2.82. So 2^x = 2.82 and x = log 2.82 ÷ log 2 = 0.450 ÷ 0.301 ≈ 1.5.
Rate = k[A]^1.5. That 2.82 is 2√2, a clue worth recognising. NCERT notes that order "can be 0, 1, 2, 3 or even a fraction".
Worked example 3: when both concentrations change (NCERT exercise 3.11)
For 2A + B → C + D, rate of formation of D in mol L⁻¹ min⁻¹:
I: [A] = 0.1, [B] = 0.1, rate = 6.0 × 10⁻³
II: [A] = 0.3, [B] = 0.2, rate = 7.2 × 10⁻²
III: [A] = 0.3, [B] = 0.4, rate = 2.88 × 10⁻¹
IV: [A] = 0.4, [B] = 0.1, rate = 2.40 × 10⁻²
I to IV: [B] fixed, [A] × 4, rate × 4 (6.0 × 10⁻³ to 2.40 × 10⁻²). Order in A = 1. II to III: [A] fixed, [B] × 2, rate × 4 (7.2 × 10⁻² to 2.88 × 10⁻¹). Order in B = 2.
Rate = k[A][B]². Note that A's coefficient is 2 but its order is 1, and B's coefficient is 1 but its order is 2.
From I: k = 6.0 × 10⁻³ ÷ (0.1 × 0.1²) = 6.0 × 10⁻³ ÷ 1.0 × 10⁻³ = 6.0 L² mol⁻² min⁻¹. Check with II: 6.0 × 0.3 × 0.2² = 6.0 × 0.3 × 0.04 = 0.072 = 7.2 × 10⁻². It agrees.
Which mistakes cost marks?
Comparing rows where two things changed without first cancelling the known order.
Ratios upside down. Divide the larger concentration by the smaller and the matching rate by its partner; keep both in the same direction.
Rounding too early. 2.82 gives 1.5; rounding it to 3 would give 1.58 and tempt you to write 1.6.
Orders from coefficients. Example 3 shows they differ.
Units of k. For overall order n, k has units of (mol L⁻¹)^(1−n) × time⁻¹. Example 3 is third order, so k is in L² mol⁻² min⁻¹.
Skipping the check row. Always test your rate law on an unused experiment.
A practice routine
Redo the three examples above without looking, then check every ratio.
Solve NCERT exercise 3.12, which asks you to fill in missing concentrations and rates; it trains you to run the method backwards.
Make up one table with a fractional order, swap it with a friend, and solve each other's.
Memorise 2^1.5 ≈ 2.83, 2^0.5 ≈ 1.41, 3² = 9 and 4^1.5 = 8 so you can spot orders quickly.
If you can follow the solutions but cannot see where to start on a new table, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next study session, solve NCERT exercise 3.12 and check each answer against the rate law you were given.
