About the Rate Law Calculator
This rate law calculator applies the method of initial rates. Enter the starting concentrations and measured initial rates from two experiments (one reactant) or three experiments (two reactants), and it solves for the order with respect to each reactant, the overall order, and the rate constant k with its correct units. It then writes out the rate law, rate = k[A]ᵐ[B]ⁿ, and predicts the initial rate for any new pair of concentrations you enter.
It is designed for general chemistry kinetics problems and lab reports, where a table of experiments is given and you need to work out how the rate responds to doubling or tripling each concentration. Unlike the usual by-hand method, the experiments do not have to hold one concentration constant: the calculator solves the logarithmic equations simultaneously, so any three independent experiments work.
Orders are first calculated exactly, then rounded to the nearest whole or half number when the data are close, because real data contain measurement noise. The rate constant is averaged over all experiments using the rounded orders. Concentrations are in mol/L and rates in M per time unit.
How to use the rate law calculator
- 1Choose one reactant (two experiments) or two reactants (three experiments).
- 2Enter the initial concentrations and measured initial rate for each experiment.
- 3Pick the time unit your rates use.
- 4Read the rate law, the orders and k with units.
- 5Enter new concentrations to predict the initial rate.
Formula and method
The rate law expresses the initial rate as a power of each reactant concentration. Taking logarithms turns it into a linear equation in the unknowns ln k, m and n, so three experiments give three equations that the calculator solves exactly (Cramer’s rule). When two experiments share the same [B], this reduces to the familiar ratio method: m = ln(rate ratio) ÷ ln(concentration ratio).
The orders are then rounded to the nearest integer or half-integer when within measurement error, and k is the average of rate ÷ ([A]ᵐ[B]ⁿ) across experiments. The units of k are M^(1−overall order) per time unit, which keeps the rate in M per time.
- k
- Rate constant
- m, n
- Reaction orders in A and B
- [A], [B]
- Initial molar concentrations
- r
- Initial rate (M per time unit)
Worked examples
Second order in A, first order in B
Doubling [A] (exp 1→2) quadruples the rate, so m = 2; doubling [B] (exp 1→3) doubles it, so n = 1. k = 0.002 ÷ (0.10² × 0.10) = 2.0 M⁻²·s⁻¹, and at 0.15 M each the rate is 2 × 0.15² × 0.15 = 0.00675 M/s.
Single reactant, first order
Doubling [A] doubles the rate, so the reaction is first order: m = ln 2 ÷ ln 2 = 1. k = 0.00015 ÷ 0.05 = 0.003 s⁻¹, predicting 0.003 × 0.15 = 0.00045 M/s.
Zero order in B
Tripling [B] leaves the rate unchanged, so n = 0, while doubling [A] doubles it, so m = 1. k = 0.006 ÷ 0.2 = 0.03 s⁻¹ and the rate law is simply rate = k[A].
Frequently asked questions
How do you find the order of a reaction from initial rates?+
Compare two experiments in which only one concentration changes. If doubling it doubles the rate the order is 1; if it quadruples the rate the order is 2; if the rate is unchanged the order is 0. In general, order = ln(rate ratio) ÷ ln(concentration ratio).
Can I get the rate law from the balanced equation?+
No. Orders must be measured experimentally because they reflect the reaction mechanism, not the stoichiometry. They only match the coefficients for elementary steps.
What are the units of k?+
They depend on the overall order: M·s⁻¹ for zero order, s⁻¹ for first order, M⁻¹·s⁻¹ for second order and M⁻²·s⁻¹ for third order. In general k has units of M^(1−n)·time⁻¹.
Why do my calculated orders come out as 1.96 or 0.48?+
Experimental rates have measurement error, so orders rarely come out exactly whole. The calculator rounds to the nearest integer or half-integer when close and shows the raw values so you can judge.
Can reaction orders be fractional or negative?+
Yes. Fractional orders such as ½ or 3/2 occur in chain and surface reactions, and negative orders appear when a species inhibits the reaction. They indicate a multistep mechanism.