INTEGRATED RATE LAW CALCULATOR

Integrated rate law calculator

Zero, first or second order: concentration, time, half-life or k.

s⁻¹
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What are the integrated rate laws?

For a reaction whose rate depends on one reactant A (OpenStax Chemistry 2e, Table 12.2): zero order [A] = [A]₀ − kt; first order ln[A] = ln[A]₀ − kt, or [A] = [A]₀e^(−kt); second order 1/[A] = 1/[A]₀ + kt. The units of k follow the order: M s⁻¹, s⁻¹ and M⁻¹ s⁻¹ (with whatever time unit you use).

How do I calculate the half-life?

Zero order t½ = [A]₀/(2k), first order t½ = ln 2/k ≈ 0.693/k, second order t½ = 1/(k[A]₀). Only the first-order half-life is independent of concentration. OpenStax Example 12.11: hydrogen peroxide decomposing at 40 °C has t½ = 2.16 × 10⁴ s, so k = 0.693 ÷ 2.16 × 10⁴ s = 3.21 × 10⁻⁵ s⁻¹. Iodine-131 with k = 0.138 d⁻¹ has t½ = 5.02 days.

Example: a first-order reaction

Cyclobutane decomposes at 500 °C with k = 9.2 × 10⁻³ s⁻¹. How long until 80.0% has reacted? 20.0% is left, so t = ln(1 ÷ 0.200) ÷ k ≈ 1.7 × 10² s (OpenStax Example 12.6). Tip: for percentages, enter [A]₀ = 1 and [A] = the fraction left.

Example: a second-order reaction

Butadiene dimerizes with k = 5.76 × 10⁻² L mol⁻¹ min⁻¹. Starting at 0.200 M, after 10.0 min 1/[A] = 0.576 + 5.00 = 5.58 L/mol, so [A] = 0.179 M (Example 12.8), and the half-life is 1 ÷ (0.0576 × 0.200) = 86.8 min (Example 12.12). Each later half-life is twice as long as the one before.

Example: a zero-order reaction, and how to tell the order

Ammonia decomposing on hot tungsten is zero order with k ≈ 1.3 × 10⁻⁶ M s⁻¹ (Example 12.10). From 0.0028 M its half-life is about 18 min, and it reaches 0.0001 M after about 35 min. To find the order from data, plot [A], ln[A] and 1/[A] against time: the straight one gives the order, and k is minus the slope (plus the slope for 1/[A]).

Why do the units of k matter?

Because k carries the units that make the rate come out in concentration per time: M s⁻¹ for zero order, s⁻¹ for first order and M⁻¹ s⁻¹ for second order. The time you enter must be in the same unit as the time in k. If k is per minute and you enter seconds, the answer is off by a factor of 60.

Can the concentration become negative in a zero-order reaction?

The formula [A] = [A]₀ − kt would go below zero after t = [A]₀ ÷ k, but a concentration cannot. The reactant is used up at that time, so the calculator reports zero from then on, and a rate constant cannot be found from a data point with [A] = 0.

Comparing the three orders

Take [A]₀ = 1.00 M and a rate constant of numerical value 0.020 in the units each order needs, and look at t = 20 time units. Zero order leaves 0.600 M, first order 0.670 M and second order 0.714 M. The half-lives differ as well: 25.0 for zero order, 34.7 for first order (ln 2 ÷ k) and 50.0 for second order. The zero-order reactant is gone completely at t = 50, while first- and second-order reactions only approach zero. This is why a curve of concentration against time looks different for each order.

Finding k from two measurements

For a first-order reaction with [A]₀ = 0.100 M and [A] = 0.060 M after 30 s: k = ln(0.100 ÷ 0.060) ÷ 30 s = 0.0170 s⁻¹. The calculator’s data mode rearranges the integrated law in the same way for zero and second order. A single pair of points assumes you already know the order; if you do not, use several points and see which of [A], ln[A] or 1/[A] against time lies on a straight line.

Things to watch

  • These laws apply to a rate that depends on one reactant. A reaction with several reactants, or one that runs in both directions, needs a different treatment.
  • Only the first-order half-life is constant. For zero and second order it depends on [A]₀, so the half-life you compute applies to the starting concentration you entered.
  • Natural logarithm, not base 10. Using log₁₀ in the first-order equation changes k by a factor of 2.303.
  • Percent conversions: use fractions of the starting amount ([A]₀ = 1 and [A] = fraction left) when you do not know the molarity, which works for first order because the ratio is all that matters.

Sources: OpenStax Chemistry 2e, section 12.4 (Integrated Rate Laws) and Table 12.2, with its worked examples. Related tools: half-life calculator, activation energy calculator.