How to Calculate Rate Law
Why you can’t just read the rate law off the equation
Here’s the hard fact that bites every general-chemistry student: a rate law cannot be predicted from a balanced equation. The reaction 2 NO₂ → 2 NO + O₂ is second-order in NO₂. The reaction H₂ + I₂ → 2 HI is first-order in H₂ and first-order in I₂. The reaction H₂ + Br₂ → 2 HBr has a rate law involving a square root and a denominator term — nothing you’d ever guess from the stoichiometry. Orders come from the slow step of the mechanism, and the slow step is something you have to discover experimentally.
That’s what the method of initial rates is for. Run the reaction at several different starting concentrations, measure how fast it begins, and back out the exponents. Once you have the orders, the rate constant k falls out of any single experiment. With k and the orders you can predict the rate at any future condition — which is what kinetics is actually for: designing reactors, forecasting drug shelf life, modeling atmospheric chemistry.
The general form
For aA + bB → products:
Rate = k[A]^m [B]^n
m and n are the reaction orders (almost always 0, 1, or 2; occasionally fractional). Their sum is the overall order. Units of k depend on overall order — track them carefully or your rate constant will mean nothing.
The method of initial rates, walked through
You need at least one more experiment than you have reactants. Vary one concentration at a time, hold the others constant, and look at the rate ratio.
| Exp | [A] (M) | [B] (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 4.0 × 10⁻³ |
Order in A (compare experiments 1 and 2 — only [A] changes):
Rate₂ / Rate₁ = ([A]₂ / [A]₁)^m 8.0 × 10⁻³ / 2.0 × 10⁻³ = (0.20 / 0.10)^m 4 = 2^m → m = 2
Order in B (compare 1 and 3 — only [B] changes):
4.0 × 10⁻³ / 2.0 × 10⁻³ = (0.20 / 0.10)^n 2 = 2^n → n = 1
Rate law: Rate = k[A]²[B], overall third-order.
Rate constant (plug experiment 1 in):
2.0 × 10⁻³ = k(0.10)²(0.10) = k(1.0 × 10⁻³) k = 2.0 M⁻² s⁻¹
Always confirm by computing k from a second experiment — if you get the same number, you’re done. If not, your orders are wrong.
When the ratios aren’t powers of two
| Exp | [X] (M) | [Y] (M) | Rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 1.5 × 10⁻⁴ |
| 2 | 0.30 | 0.10 | 1.35 × 10⁻³ |
| 3 | 0.10 | 0.30 | 4.5 × 10⁻⁴ |
For X: 1.35 × 10⁻³ / 1.5 × 10⁻⁴ = 9.0; concentration tripled (3.0). 9 = 3^m → m = 2.
For Y: 4.5 × 10⁻⁴ / 1.5 × 10⁻⁴ = 3.0; concentration tripled. 3 = 3^n → n = 1.
When the ratio doesn’t pop out cleanly, take logs: m = log(rate ratio) / log(concentration ratio). For 9 and 3: log 9 / log 3 = 2.000 exactly.
Rate = k[X]²[Y]; k = 1.5 × 10⁻⁴ / (0.10² × 0.10) = 0.15 M⁻² s⁻¹.
Spotting a zero-order reactant
| Exp | [P] (M) | [Q] (M) | Rate (M/s) |
|---|---|---|---|
| 1 | 0.20 | 0.10 | 5.0 × 10⁻³ |
| 2 | 0.40 | 0.10 | 1.0 × 10⁻² |
| 3 | 0.20 | 0.30 | 5.0 × 10⁻³ |
P: rate doubles when [P] doubles → first order. Q: rate doesn’t budge when [Q] triples → zero order.
Rate = k[P], k = 0.025 s⁻¹.
A zero-order reactant means changes in its concentration don’t affect the rate at all. This shows up when one reactant is in large excess (so its concentration is effectively constant), or when it appears only in a fast step after the rate-determining one. Enzyme kinetics at saturating substrate is the textbook example.
Units of k — they encode the overall order
| Overall order | Units of k |
|---|---|
| 0 | M s⁻¹ |
| 1 | s⁻¹ |
| 2 | M⁻¹ s⁻¹ |
| 3 | M⁻² s⁻¹ |
General rule: k has units M^(1−n) s⁻¹. If your computed k comes out without units, you’ve skipped a step — assign units from the rate law form.
Once you have the orders, integrated rate laws unlock half-lives
| Order | Linear plot | Integrated form | Half-life |
|---|---|---|---|
| 0 | [A] vs t | [A] = [A]₀ − kt | [A]₀ / (2k) |
| 1 | ln[A] vs t | ln[A] = ln[A]₀ − kt | 0.693 / k |
| 2 | 1/[A] vs t | 1/[A] = 1/[A]₀ + kt | 1 / (k[A]₀) |
The first-order half-life is concentration-independent — that’s why radioactive decay (always first-order) gives clean half-life behavior, and why pharmacokinetics gets so much mileage out of “drug X has a half-life of 6 hours” without specifying dose.
Traps
Don’t assume order = stoichiometric coefficient. They sometimes match by coincidence and you’ll get rewarded for the wrong reasoning. The method of initial rates is the only way to know.
Hold everything else constant. If two concentrations change between experiments, you can’t isolate either order. Pick experiment pairs surgically.
Logs over guessing. If 5 = 2^m, m isn’t 2.5 — m = log 5 / log 2 = 2.32. Don’t round to a “nice” integer until you’ve checked.
Track units religiously. A rate constant without units is useless for prediction. Every time you write k, write its units.
Practice
Check your work with the Equilibrium Calculator:
- One reactant: Exp 1 [A] = 0.10, rate = 3.0×10⁻³; Exp 2 [A] = 0.20, rate = 1.2×10⁻²; Exp 3 [A] = 0.30, rate = 2.7×10⁻². Find the rate law.
- 2 NO + Cl₂ → 2 NOCl: Exp 1 [NO] = 0.10, [Cl₂] = 0.10, rate = 1.2×10⁻³; Exp 2 [NO] = 0.20, [Cl₂] = 0.10, rate = 4.8×10⁻³; Exp 3 [NO] = 0.10, [Cl₂] = 0.20, rate = 2.4×10⁻³. Determine orders and k.
- First-order, k = 0.045 s⁻¹. Time for 90% decomposition?
- Second-order, k = 0.50 M⁻¹ s⁻¹, [A]₀ = 0.80 M. Half-life?
- First-order half-life = 25 min. Fraction left after 75 min?
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