Rate Law and Order Determination from Experimental Data
You cannot read a rate law off a balanced equation. The orders in a rate law are experimental facts, and every kinetics question that gives you a table of data is really asking one thing: which method will get the order out of these numbers? There are four standard methods, and each suits a different kind of data. This page works all four, with the arithmetic done in full.
The rate law
Here k is the rate constant (its units depend on n), x is the order with respect to A and y the order with respect to B. Orders can be zero, whole numbers, fractions or even negative. They are found by experiment, never by inspection of the stoichiometry.
Method 1 — the initial-rate ratio method
Run the reaction several times, changing one concentration at a time, and measure the rate right at the start (before the concentrations have had time to fall). Then take ratios.
Data for A + B → products
| Experiment | [A]₀ / mol L⁻¹ | [B]₀ / mol L⁻¹ | Initial rate / mol L⁻¹ s⁻¹ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.20 | 1.6 × 10⁻³ |
Order in A: compare 1 and 2, where [B] is held constant.
[A] doubles; rate goes from 2.0 × 10⁻⁴ to 8.0 × 10⁻⁴, a factor of 4.
So 2x = 4, giving x = 2.
Order in B: compare 2 and 3, where [A] is held constant.
[B] doubles; rate goes from 8.0 × 10⁻⁴ to 1.6 × 10⁻³, a factor of 2.
So 2y = 2, giving y = 1.
Rate law: rate = k[A]²[B], overall order 3.
Now k, from experiment 1:
2.0 × 10⁻⁴ = k × (0.10)² × (0.10) = k × 1.0 × 10⁻³
k = (2.0 × 10⁻⁴) ÷ (1.0 × 10⁻³) = 0.20 L² mol⁻² s⁻¹
Check against experiment 3 — always use the run you did not use to find
k:
k[A]²[B] = 0.20 × (0.20)² × 0.20 = 0.20 × 0.040 × 0.20 = 1.6 × 10⁻³ ✔ matches the measured
rate.
Method 2 — the logarithm method, when the ratio is not neat
Real data rarely doubles conveniently. Taking logarithms turns the ratio method into simple division:
Experiment 1: [A] = 0.050 mol L⁻¹, rate = 1.5 × 10⁻³ mol L⁻¹ s⁻¹
Experiment 2: [A] = 0.125 mol L⁻¹, rate = 5.9 × 10⁻³ mol L⁻¹ s⁻¹
rate₂ ÷ rate₁ = 5.9 ÷ 1.5 = 3.9333
[A]₂ ÷ [A]₁ = 0.125 ÷ 0.050 = 2.5
ln 3.9333 = 1.3695 and ln 2.5 = 0.9163
x = 1.3695 ÷ 0.9163 = 1.49 ≈ 1.5
The order is three-halves. That is not a mistake in the data — fractional orders are genuine and are typical of chain reactions, where the observed rate depends on the square root of a radical concentration. Any base of logarithm works here, because the base cancels in the division; log₁₀ gives the same 1.49.
Method 3 — the integrated rate law: which graph is a straight line?
If you have concentration measured against time for a single run, you cannot use initial rates. Instead, plot the data three ways and see which one gives a straight line.
| Order | Integrated form | Plot that is linear | Slope | Units of k |
|---|---|---|---|---|
| 0 | [A] = [A]₀ − kt | [A] against t | −k | mol L⁻¹ s⁻¹ |
| 1 | ln[A] = ln[A]₀ − kt | ln[A] against t | −k | s⁻¹ |
| 2 | 1/[A] = 1/[A]₀ + kt | 1/[A] against t | +k | L mol⁻¹ s⁻¹ |
Data for the decomposition of A
| t / s | 0 | 100 | 200 | 300 | 400 |
|---|---|---|---|---|---|
| [A] / mol L⁻¹ | 0.1000 | 0.0779 | 0.0607 | 0.0472 | 0.0368 |
Test first order. Take natural logarithms:
ln 0.1000 = −2.3026
ln 0.0779 = −2.5523
ln 0.0607 = −2.8018
ln 0.0472 = −3.0533
ln 0.0368 = −3.3023
Successive differences: −0.2497, −0.2495, −0.2515, −0.2490. They are constant to within rounding, so ln[A] against t is a straight line and the reaction is first order.
Slope, using the two end points:
slope = (−3.3023 − (−2.3026)) ÷ (400 − 0) = −0.9997 ÷ 400 = −2.499 × 10⁻³ s⁻¹
k = 2.50 × 10⁻³ s⁻¹
Confirm it is not second order. Reciprocals: 10.00, 12.84, 16.47, 21.19, 27.17. Successive differences: 2.84, 3.63, 4.72, 5.98 — steadily increasing, so 1/[A] against t is a curve, not a line. Second order is ruled out.
Method 4 — the half-life test
Half-life depends on the starting concentration in a way that is different for every order, which makes it a quick order test in its own right:
zero order: t½ = [A]₀ ÷ 2k (halves when [A]₀ halves)
first order: t½ = 0.693 ÷ k (does not depend on [A]₀ at all)
second order: t½ = 1 ÷ (k[A]₀) (doubles when [A]₀ halves)
Apply it to the data above. If the reaction really is first order with
k = 2.50 × 10⁻³ s⁻¹, then
t½ = 0.693 ÷ (2.50 × 10⁻³) = 277 s
Does the data agree? [A] starts at 0.1000 and must fall to 0.0500. The table shows 0.0607 at 200 s and 0.0472 at 300 s, so it passes 0.0500 between those times — consistent with 277 s. Going on, [A] should reach 0.0250 at about 554 s, a further 277 s. A half-life that stays the same as the reaction proceeds is the signature of first order, and it is the fastest visual check you can run on a data table.
The isolation method — turning a hard rate law into an easy one
When two reactants change at once, flood the mixture with a large excess of one of them. If [B] is 100 times [A], then [B] barely changes during the run and can be absorbed into the constant:
The reaction now behaves as if it were order x only — this is what "pseudo-first-order" means. Acid-catalysed ester hydrolysis in dilute solution is the standard example: water is in vast excess, so a genuinely second-order reaction is measured as first order. Repeat the experiment at a different [B]₀ and you recover y from how k′ changes.
Common mistakes that cost marks
- Reading the order off the balanced equation. For 2N₂O₅ → 4NO₂ + O₂ the order is 1, not 2. Order equals the stoichiometric coefficient only for a single elementary step, and most reactions are not elementary.
- Changing two concentrations between the runs you compare. The ratio method only isolates one order if everything else is held constant.
- Using rates measured part-way through the reaction as "initial" rates. By then the concentrations are no longer the values in the table.
- Quoting k without units. The units follow from the overall order — mol L⁻¹ s⁻¹ for zero, s⁻¹ for first, L mol⁻¹ s⁻¹ for second, L² mol⁻² s⁻¹ for third. A wrong unit is a wrong answer, and the unit itself is a free clue to the order.
- Deciding "first order" from three points that look straightish. Test the other plots too, as in method 3. A short data set can look linear on more than one plot.
- Forgetting the minus sign in the slope. For zero and first order the slope is −k, so k is the magnitude of a negative slope.
Where this appears in exams
| Level | Typical question |
|---|---|
| CBSE/ICSE Class 12 | Find the order and k from a four-row initial-rate table; state units of k |
| JEE/NEET | Order from half-life behaviour, pseudo-first-order reasoning |
| IIT-JAM / CUET-PG | Choose the linear plot for a given data set; integrated rate law algebra |
| GATE / CSIR-NET | Fractional orders, isolation method, fitting ln[A] against t by least squares |
Method 3 is a straight-line fit, so let the calculator do it. Enter your (t, ln[A]) pairs in the Linear Regression tool, fit y = mx + c, and the slope m is −k directly — with the intercept c giving ln[A]₀ as a check that the fit is sensible.
Open the Linear Regression (y = mx + c) Calculator →Chemical kinetics is one of the highest-scoring chapters in Class 12 once the data-handling clicks. ABC Chemistry runs Class 11–12 chemistry at its Gurugram centre and in online batches across India, plus home tuition in Delhi-NCR — see abcchemistry.in.