This rate law calculator evaluates rate = k[A]ᵐ[B]ⁿ — the rate constant times each reactant concentration raised to its reaction order — and reports the overall order and the units k must carry. Orders are experimental facts, not balancing coefficients; this page computes with the orders you supply and explains how they’re found.

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Orders Are Measured, Never Read Off the Equation

The most-repeated sentence in kinetics exists because students keep violating it: the exponents m and n in rate = k[A]ᵐ[B]ⁿ are determined BY EXPERIMENT (the method of initial rates: vary one concentration, hold the rest, watch the rate’s response ratio) and only coincidentally match stoichiometric coefficients — true solely for elementary steps, which most overall reactions are not. The defaults carry a typical profile: k = 0.5, [A] = 0.2 M second-order, [B] = 0.3 M first-order → rate = 0.5 × 0.04 × 0.3 = 0.006 M/s, overall order 3, and k’s units (1/(M²·s)) are forced by the requirement that the right side land in M/s — the calculator prints them because a k quoted without units is a number without a meaning, and its units are a fingerprint of the overall order.

Where orders come from mechanistically: the rate-determining step’s molecularity writes the law for elementary reactions (unimolecular → first order, bimolecular → second), and multi-step mechanisms can produce the exotic cases — zero-order in a reactant present in saturating excess, fractional orders from pre-equilibria (the classic 1.5-order cases), negative orders from inhibition (product slowing its own formation, as in some catalytic cycles). Temperature enters through k alone (concentrations carry no T), via the Arrhenius equation — our Activation Energy Calculator works that side of the street: k = A·e^(−Ea/RT), and the two-point form extracts Ea from k at two temperatures.

Worked Examples (Verified)

  • Defaults: 0.5 × 0.2² × 0.3 = 6.00 × 10⁻³ M/s, overall order 3.
  • The doubling test: double [A] to 0.4 → rate = 0.5 × 0.16 × 0.3 = 0.024 M/s — quadrupled, the experimental signature of second order (this ratio test IS how orders are measured).
  • Zero-order plateau: ammonia decomposition on a saturated platinum surface runs rate = k regardless of [NH₃] — order 0, k in M/s; the surface, not the gas, is the bottleneck.

Rate Law Discipline

  • Never transcribe coefficients into orders unless the step is stated to be elementary — the single most penalized error in kinetics coursework.
  • Initial rates are measured before concentrations drift (≈ first 5–10% of reaction); a “rate” averaged over half the reaction is a different, usually meaningless, quantity.
  • k is temperature-specific: a k from a table at 25 °C misprices a 60 °C reactor by the Arrhenius factor, often an order of magnitude.
  • Units of k change with overall order — first check when comparing k values across papers: 0.5 1/s and 0.5 1/(M·s) describe incomparable worlds.
  • Rate laws describe the forward reaction’s speed at given concentrations; equilibrium (K) describes where it ends. The two meet in mechanisms, not in this formula.

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Frequently Asked Questions

How do you calculate the rate from a rate law?

Substitute: rate = k × [A]^order × [B]^order. With k = 0.5, [A] = 0.2 M (order 2), [B] = 0.3 M (order 1): 0.5 × 0.04 × 0.3 = 0.006 M/s. This calculator also reports the overall order (sum of orders) and the units k must have — 1/(M²·s) for an overall third-order law like this one.

How are reaction orders determined?

Experimentally, by initial rates: run the reaction at several starting concentrations, changing one reactant at a time. If doubling [A] doubles the rate, order 1; quadruples, order 2; no change, order 0; a 2.83× change (≈ 2^1.5) signals order 1.5. The orders are then inserted into the law, and k falls out of any run’s data. Integrated rate laws (plots of ln[A] or 1/[A] vs time) provide an independent second route.

Why don’t rate law exponents match the balanced equation?

Because balanced equations summarize bookkeeping (what’s consumed and produced overall), while rate laws describe the mechanism’s slow step(s). The famous case: H₂ + Br₂ → 2HBr has a complex law with fractional order and product inhibition, nothing like its tidy equation. Only for an ELEMENTARY step — a single molecular event — do orders equal the step’s own coefficients, which is why mechanism problems insist on the word “elementary.”

What are the units of k?

Whatever makes rate come out in M/s: overall order 0 → M/s; 1 → 1/s; 2 → 1/(M·s); 3 → 1/(M²·s); in general M^(1−order)/s. The units are diagnostic — seeing k in 1/s tells you the overall order is 1 before you read another word. This calculator states them on every result.

What does a zero-order reaction mean?

The rate is independent of that reactant’s concentration — it proceeds at rate = k until the reactant nearly runs out. Causes: saturated catalysts/enzymes (every active site busy), photolysis at fixed light intensity, evaporation from a fixed surface. The concentration falls in a straight line (constant rate), unlike first-order’s exponential decay — the shape of the concentration–time curve is the order’s signature.

How does temperature change the rate?

Entirely through k, via Arrhenius: k = A·e^(−Ea/RT). A useful rule of thumb — many reactions roughly double in rate per 10 °C near room temperature (that’s an Ea ≈ 50 kJ/mol talking). Concentrations in the law are instantaneous values; the law itself doesn’t age, but k must be quoted at its temperature. Our Activation Energy page inverts two k(T) points into Ea.

Rate law vs integrated rate law?

The (differential) rate law here gives instantaneous speed from current concentrations. The integrated forms give concentration as a function of TIME: [A] = [A]₀ − kt (zero order), [A] = [A]₀e^(−kt) (first), 1/[A] = 1/[A]₀ + kt (second) — plus their half-life formulas. Use the differential law for “how fast now/at these conditions,” the integrated for “how much remains at time t.”

Is this rate law calculator free?

Yes — free, instant, no sign-up. Enter k, two concentrations and their orders for the rate, overall order and k’s required units.