Rate law calculator (method of initial rates)
Enter the initial concentrations and the measured initial rate of each experiment. With one reactant two runs are enough; with two reactants enter three runs. The calculator solves for the orders, rounds them to the nearest half, and gives the rate constant with its units.
Runs in your browser: nothing you enter is sent anywhere.
How to use it
- Choose one reactant (two runs) or two reactants (three runs).
- Enter each run’s concentrations in mol/L and its initial rate in mol/(L·s).
- Read the orders, the overall order and the rate constant with its units.
How it works
The rate law rate = k × [A]^m × [B]^n becomes linear in logarithms:
ln rate = ln k + m × ln[A] + n × ln[B]
Three runs give three linear equations in ln k, m and n, solved by Cramer’s rule. With one reactant, two runs give m = ln(r₂ ÷ r₁) ÷ ln([A]₂ ÷ [A]₁). The orders are rounded to the nearest half and k is the mean of rate ÷ ([A]^m × [B]^n) over the runs.
Worked example
| Run |
[A] |
[B] |
Rate |
| 1 |
0.1 |
0.1 |
0.002 |
| 2 |
0.2 |
0.1 |
0.008 |
| 3 |
0.1 |
0.2 |
0.004 |
Runs 1 → 2: [A] doubles, the rate quadruples, so m = 2. Runs 1 → 3: [B] doubles, the rate doubles, so n = 1. Overall order 3.
k = 0.002 ÷ (0.1² × 0.1) = 2 L²/(mol²·s).
Tips
- Change one thing at a time. It makes the orders readable at a glance and the calculation robust.
- Measure early. The initial rate is taken before concentrations change noticeably, usually within the first few percent of reaction.
- Order is experimental. It cannot be read from the coefficients of the balanced equation, except for an elementary step.
Frequently asked questions
How does the method of initial rates work?
Run the reaction several times, changing one concentration at a time, and measure the rate right at the start. If doubling [A] quadruples the rate, 2^m = 4 and m = 2. In general m = ln(rate₂ ÷ rate₁) ÷ ln([A]₂ ÷ [A]₁).
What if my runs change both concentrations?
That is fine as long as the three runs are independent. The calculator takes logarithms, ln rate = ln k + m·ln[A] + n·ln[B], and solves the three equations together. It reports an error only when the runs cannot tell A and B apart, for example when [A] and [B] always change by the same factor.
Why are the orders rounded?
Real measurements are noisy, so an order may come out as 1.04 or 1.97. Orders are usually whole numbers or halves, so the calculator rounds to the nearest 0.5, shows the unrounded value next to it, and uses the rounded orders to compute k as the mean over all runs.
What are the units of the rate constant?
They depend on the overall order: mol/(L·s) for zero order, 1/s for first, L/(mol·s) for second and L²/(mol²·s) for third order, assuming concentrations in mol/L and the rate in mol/(L·s). With other time units, replace s.
Can the order be negative or zero?
Yes. A zero order means the rate does not depend on that reactant; a negative order (an inhibitor) means the rate falls as its concentration rises. Both come out of the same calculation.