How it Works
Chemists rarely know a reaction's rate law in advance — it has to be measured. The method of initial rates does this by running several trials, each starting from a different concentration [A]₀, and clocking only the instantaneous rate right at t=0 of each one, before any product buildup complicates the picture. Because rate₀ = k[A]₀ⁿ, plotting ln(rate₀) against ln([A]₀) turns the exponent n — the reaction order — into the slope of a straight line. Run two or more trials at different starting concentrations and switch the view to "Method of initial rates" to watch that slope emerge from the point cloud.
A single trial's full concentration-versus-time decay carries the same information in a different form. Integrating the rate law −d[A]/dt = k[A]ⁿ for each candidate order predicts a different quantity that should vary linearly with time: [A] itself for zero order, ln[A] for first order, or 1/[A] for second order. Toggle the linearization view to try each transformation of the same trial's decay data — only the correct one will straighten into a line, and its R² will sit close to 1.000 while the other two curve visibly away from their fitted lines.
The stats panel also tracks half-life — the time for a trial's concentration to fall to half its starting value — computed directly from the hidden rate law. Run several trials at different [A]₀ and compare the half-lives in the trial log: constant across trials points to first order, a half-life that grows with [A]₀ points to zero order, and one that shrinks as [A]₀ increases points to second order.
Method of initial rates: ln(rate₀) = ln k + n·ln[A]₀ (slope = n)
Integrated (linear) forms — n=0: [A]=[A]₀−kt · n=1: ln[A]=ln[A]₀−kt · n=2: 1/[A]=1/[A]₀+kt
Half-life — n=0: t½=[A]₀/2k · n=1: t½=ln2/k (constant) · n=2: t½=1/(k[A]₀)
Frequently Asked Questions
What is the "order" of a chemical reaction?
The order of a reaction with respect to a reactant is the exponent on that reactant's concentration in the experimentally determined rate law, rate = k[A]ⁿ. The overall order is the sum of all such exponents. Order is not read off the balanced equation — it must be measured, which is exactly what the method of initial rates and integrated rate laws are for.
What is the method of initial rates?
The method of initial rates measures the instantaneous rate at the very start of several separate trials, each begun at a different initial concentration [A]₀, before any product buildup or reverse reaction can complicate things. Plotting ln(rate₀) against ln([A]₀) gives a straight line whose slope is the reaction order n, since rate₀ = k[A]₀ⁿ implies ln(rate₀) = ln k + n·ln[A]₀.
How do you find the reaction order from initial-rate data at two different concentrations?
Divide the rate law written for two trials: rate₂/rate₁ = ([A]₀,₂/[A]₀,₁)ⁿ. Taking the ratio of measured rates and the ratio of starting concentrations and solving n = log(rate₂/rate₁) / log([A]₀,₂/[A]₀,₁) gives the order directly, without needing a full linear regression — though more trials make the estimate far more reliable.
What is an integrated rate law and why is it useful for finding order?
An integrated rate law expresses concentration as an explicit function of time, obtained by integrating the differential rate law rate = −d[A]/dt = k[A]ⁿ. Each order predicts a different function of concentration that varies linearly with time — [A], ln[A], or 1/[A] — so plotting the right one against t and checking for a straight line confirms the order from a single trial's full decay curve.
Why does plotting ln[A] vs time give a straight line only for a first-order reaction?
Integrating −d[A]/dt = k[A] gives ln[A] = ln[A]₀ − kt, which is linear in t with slope −k. For a zero-order or second-order reaction, ln[A] is a curved (concave) function of time, so only the truly first-order case produces a straight line on this particular axis pairing.
Why is the half-life of a first-order reaction independent of the starting concentration?
For a first-order reaction t½ = ln2/k, a formula that contains no [A]₀ term at all — halving the amount always takes the same amount of time no matter how much you started with, because the rate itself scales down proportionally as concentration drops. This constant half-life is a hallmark used to identify first-order kinetics experimentally.
How does half-life behave for zero-order and second-order reactions?
For zero-order kinetics t½ = [A]₀/2k, so a larger starting concentration takes proportionally longer to halve because the rate stays constant regardless of how much reactant remains. For second-order kinetics t½ = 1/(k[A]₀), so a larger starting concentration actually halves faster, since the rate falls off with the square of concentration.
What are the units of the rate constant k for zero-, first-, and second-order reactions?
Units of k always make rate come out in concentration per time. For zero order, k has units of M·s⁻¹ (rate itself); for first order, k has units of s⁻¹ (a frequency, independent of concentration); for second order, k has units of M⁻¹·s⁻¹. Checking which units make a fitted k dimensionally sensible is a quick sanity check on a proposed order.
What is R² (coefficient of determination) and why do chemists use it to identify reaction order?
R² measures how closely a set of points follows a straight line, ranging from 0 (no linear relationship) to 1 (a perfect line). Because real data always carries measurement noise, chemists fit all three integrated-rate-law transformations to the same decay data and pick the order whose plot gives the R² closest to 1 as the most likely true order.
Can a reaction have a fractional or negative order?
Yes — reaction order is an empirical quantity, not something guessable from stoichiometry, so multistep or catalytic mechanisms can produce fractional orders (like 1.5) or even negative orders (rate decreases as a species' concentration increases, often seen with inhibitors). This simulator only models the three classic integer cases (0, 1, 2) that appear in most introductory kinetics courses.