Rate Laws: Zero, First & Second Order
How fast a reaction runs depends on how much is left. In a zero-order reaction the rate stays the same, in a first-order reaction it falls in step with the concentration and in a second-order reaction with its square. Find the concentration after a time, the starting concentration, the rate constant or the time it takes, with the half-life — or find the order and k from your own measurements.
What Is a Rate Law?
The rate law rate = k[A]n says how the speed of a reaction depends on the concentration. The order n is found by experiment, not from the reaction equation. In a zero-order reaction the concentration falls in a straight line (for example when an enzyme or a catalyst surface is fully busy). In a first-order reaction it falls by the same fraction in every equal time step, like radioactive decay. In a second-order reaction it starts fast and then slows down a lot, because two particles have to meet.
Integrating the rate law gives the concentration over time. Each order has its own straight-line plot: [A], ln [A] or 1/[A] against t. The plot that comes out straightest tells you the order, and its slope gives k. The half-life is the time until half is left; only for first order is it the same at every concentration.
Concentration Over Time
[A]Worked Techniques
The same fraction disappears in every equal time step, whatever the concentration.
1/[A] grows in a straight line. Every next half-life is twice as long as the one before.
Make all three plots: the one that is a straight line gives the order, and its slope gives k (−k for the first two, +k for 1/[A]). With measurements over less than about one half-life the three are hard to tell apart.
Sources & Verification
- P. Atkins, J. de Paula & J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford University Press, 2018 — Topic 17B (Integrated rate laws).
- IUPAC Compendium of Chemical Terminology (Gold Book) — “order of reaction”, “rate constant”, “half-life”.
Frequently Asked Questions
How do I use the integrated rate law for a first-order reaction?
[A] = [A]₀·e^(−kt). Example: [A]₀ = 0.5 mol/L and k = 0.002 s⁻¹. How much is left after 10 minutes?
1. Use the same time unit as k: t = 600 s, so kt = 0.002 × 600 = 1.2.
2. e^(−1.2) = 0.301, so [A] = 0.5 × 0.301 = 0.151 mol/L: about 30% is left.
3. The half-life is ln 2 / k = 347 s.
How do I calculate the half-life of a reaction?
It depends on the order. Zero order: t½ = [A]₀ / 2k (2 mmol/L at k = 0.05 mmol/(L·min) gives 20 min). First order: t½ = ln 2 / k (k = 0.002 s⁻¹ gives 347 s), the same at every concentration. Second order: t½ = 1 / (k[A]₀) (0.01 mol/L at k = 0.5 L/(mol·s) gives 200 s), and every next half-life is twice as long. The calculator gives it with every result.
How do I find the order of a reaction from data?
Plot [A], ln [A] and 1/[A] against time: the plot that is a straight line gives the order (zero, first or second), and its slope gives k. Choose Measurements: find the order and enter one time and concentration per line.
For the first example data set, ln [A] against t gives R² = 0.9996, while the other two plots give 0.938 and 0.947: the reaction is first order with k = 0.0298 per minute. Measure over at least one or two half-lives, or the three plots look almost equally straight.
What are the units of the rate constant k?
They depend on the order, so that rate = k[A]ⁿ comes out in concentration per time. With mol/L and seconds: zero order mol/(L·s), first order 1/s, second order L/(mol·s). In general (mol/L)^(1 − n) per second. The calculator shows the unit that belongs to your order and units.
How long does it take until 90% has reacted?
For a first-order reaction t = ln 10 / k, about 3.32 half-lives, whatever the concentration: with k = 0.002 s⁻¹ that is 1151 s. For zero order it is 0.9·[A]₀/k and for second order 9/(k[A]₀). The calculator shows it as Time until 90% has reacted, or click Solve next to the time and enter 10% of the starting concentration.
Why does a zero-order reaction stop?
Its rate does not depend on the concentration, so [A] falls in a straight line and reaches 0 at t = [A]₀/k. Then the reaction simply stops. With 2 mmol/L and k = 0.05 mmol/(L·min), everything is used up after 40 minutes. Zero-order behaviour usually means something else limits the rate, such as an enzyme or a catalyst surface that is fully busy.
Can I tell the order from the reaction equation?
In general no: the order has to be found by experiment, because most reactions go through several steps. Only for a single elementary step does the order follow from the equation (for example, a step where two molecules collide is second order). Measuring the concentration over time and finding the straightest plot is the usual way.