Reaction Rate Law & Half-Life Calculator

A reaction rate law calculator only gives useful numbers if you pick the right order first. Students routinely plug data into the first-order equation by default, get a rate constant, and never check whether the fit actually holds. Order isn't a guess—it comes from experimental data or a proposed mechanism, and choosing wrong means every downstream number (half-life, time to reach a target concentration, shelf-life estimate) is garbage.

Zero-order reactions have a rate independent of concentration: rate = k. Enzyme-catalyzed reactions at substrate saturation behave this way—the enzyme is maxed out, so adding more substrate doesn't speed things up. First-order reactions have rate = k[A]: radioactive decay, many drug eliminations, and unimolecular decompositions follow this pattern. Second-order reactions have rate = k[A]² (one reactant) or rate = k[A][B] (two reactants): bimolecular collisions in the gas phase or solution are typical examples.

The practical difference matters immediately. A first-order reaction's half-life is constant regardless of starting concentration. A second-order reaction's half-life doubles every time you halve the concentration—it takes longer and longer to decay. Zero-order half-life shortens as concentration drops because the rate stays flat until the reactant is consumed. If you're using this calculator to estimate how long a reagent lasts on a shelf, picking the wrong order can be off by a factor of ten.

The differential rate law tells you the instantaneous rate. The integrated rate law tells you what the concentration is at time t—which is usually what you actually need. Zero-order: [A] = [A]₀ − kt. First-order: ln[A] = ln[A]₀ − kt. Second-order: 1/[A] = 1/[A]₀ + kt.

Each form is a straight-line equation (y = mx + b) when you plot the right transformation. Zero-order: [A] vs. t is linear with slope −k. First-order: ln[A] vs. t is linear with slope −k. Second-order: 1/[A] vs. t is linear with slope +k. Whichever plot gives you a straight line tells you the order. If all three look curved, your reaction might be fractional order or follow a more complex mechanism.

A common mistake: applying the first-order integrated law but forgetting to take the natural log. If you plot [A] vs. t for a first-order reaction, the curve looks exponential—it bends. Students see the bend, panic, and assume the data is bad. No—you just plotted the wrong thing. Take ln[A] and replot. Also, make sure you use natural log (ln), not log₁₀. The slope of a log₁₀ plot gives k/2.303, not k.

Half-life (t₁/₂) is the time for concentration to drop to half its current value. The formula depends entirely on order. Zero-order: t₁/₂ = [A]₀ / (2k). First-order: t₁/₂ = ln 2 / k ≈ 0.693 / k. Second-order: t₁/₂ = 1 / (k[A]₀).

The first-order result is the famous one: half-life is constant, independent of concentration. Carbon-14 decays with t₁/₂ = 5730 years whether you start with a microgram or a kilogram. This is unique to first order. For zero order, t₁/₂ depends on initial concentration—halve the starting amount and the half-life halves too. For second order, t₁/₂ also depends on [A]₀ but inversely—lower concentration means longer half-life. Each successive half-life takes twice as long as the previous one.

Test question trap: "A drug has a half-life of 4 hours. After 12 hours, what fraction remains?" This only works cleanly if the elimination is first-order (it usually is for drugs). Then 12 hours = 3 half-lives, so (1/2)³ = 1/8 remains. If someone assumes zero-order kinetics for a first-order process, they get a linear decrease and a completely different answer.

Two main methods: the graphical method and the method of initial rates. For graphical analysis, you plot [A] vs. t, ln[A] vs. t, and 1/[A] vs. t. Whichever gives a straight line reveals the order. For initial rates, you compare experiments where one reactant's concentration changes while others stay fixed. If doubling [A] doubles the rate, it's first order in A. If doubling [A] quadruples the rate, it's second order.

The initial rates method uses the ratio: rate₂/rate₁ = ([A]₂/[A]₁)ⁿ. Solve for n. If rate₂/rate₁ = 2 and [A]₂/[A]₁ = 2, then 2 = 2ⁿ, so n = 1. If rate₂/rate₁ = 4 and [A]₂/[A]₁ = 2, then 4 = 2ⁿ, so n = 2. Watch for cases where the ratio isn't a clean power of 2—take the log: n = log(rate ratio) / log(concentration ratio).

The units of k change with reaction order, and getting them wrong is a dead giveaway that you picked the wrong order or made an algebra error. Zero-order: k has units of M/s (or M·s⁻¹). First-order: k has units of s⁻¹ (concentration cancels). Second-order: k has units of M⁻¹s⁻¹ (or L·mol⁻¹·s⁻¹).

Quick dimensional check: rate always has units of M/s. The rate law is rate = k[A]ⁿ. So k = rate / [A]ⁿ = (M/s) / Mⁿ = M¹⁻ⁿ·s⁻¹. Plug in n = 0, 1, 2 and you recover the units above. If your calculated k comes out with units that don't match the expected order, recheck your algebra. This is the fastest sanity check available.

Can a reaction be fractional order? Yes. Fractional orders (like 1.5) arise from complex mechanisms—chain reactions, for instance. The rate law is determined experimentally, not from the balanced equation. This calculator handles integer orders (0, 1, 2), which cover the vast majority of textbook and practical cases.

Does temperature affect the rate constant? Absolutely. The Arrhenius equation k = A·e^−Eₐ/RT shows k increases exponentially with temperature. A 10 °C rise roughly doubles k for many reactions. The order doesn't change with temperature (for elementary reactions), but k does—sometimes dramatically.

Why does the balanced equation not tell me the order? Because the rate law reflects the mechanism, not the stoichiometry. 2NO₂ → 2NO + O₂ is second order experimentally, matching the single-step collision of two NO₂ molecules. But 2H₂O₂ → 2H₂O + O₂ is first order because it proceeds through a multi-step mechanism where the rate-determining step involves only one H₂O₂ molecule.

What if I only have two data points? Two points can give you k for an assumed order, but you can't determine the order from just two points—any order will produce a line through two points. You need at least three data points at different times (graphical method) or three experiments at different concentrations (initial rates method) to distinguish orders.

Problem: A first-order reaction has k = 0.0462 s⁻¹ and starts at [A]₀ = 0.800 M. Find the half-life, the concentration after 30 s, and the time to reach 0.100 M.

Notice how every answer cross-checks with the half-life count. At 30 s (2 half-lives), concentration is 1/4 of the start: 0.800 / 4 = 0.200 M. At 45 s (3 half-lives), it's 1/8: 0.800 / 8 = 0.100 M. If your integrated-law calculation doesn't agree with the half-life shortcut, you have an arithmetic error somewhere. Always cross-check.