Reaction Order & Integrated Rate Laws
Watch [A] fall for zero, first and second order — and how the half-lives change.
About this simulation
The integrated rate laws made visible. Students pick zero-, first- or second-order kinetics and vary the rate constant k and initial concentration [A]₀; the [A]-versus-time curve is drawn with successive half-life markers, whose spacing is the tell: equal for first order (t½ = 0.693/k, independent of concentration), shrinking for zero order and stretching for second order. A one-tap 'Linearized' view switches the y-axis to [A], ln[A] or 1/[A] so the data straightens only for the matching order — exactly how order is found from experimental data. Half-life, the units of k and the straight-line plot update live.
How to use it
- Choose 'Reaction order' First and look at the orange dashed half-life markers on the [A] curve: they are evenly spaced.
- Switch to Zero and then Second and watch the spacing between successive half-life markers shrink or stretch.
- Change 'Initial concentration' and watch the half-life at the top of the stage: it stays put for First, while for Zero and Second both the first and the next t½ change.
- Set 'Plot' to Linearized: the y-axis becomes [A], ln [A] or 1 / [A] so the graph turns into a straight line for the chosen order.
- Increase 'Rate constant' and watch the 't½ =' label fall and the time axis shorten as every half-life gets quicker.
What to notice
- For a first-order reaction every half-life is the same length, so the dashed markers are equally spaced and t½ does not change with [A]₀.
- For zero order each half-life is shorter than the one before and the concentration falls in a straight line to zero.
- For second order each half-life is longer than the last, giving a long, slowly falling tail.
- The order of a reaction can be found from data by finding which plot is straight: [A] for zero, ln[A] for first, 1/[A] for second order.
The formulas
- Zero order: [A] = [A]₀ − kt, t½ = [A]₀ / 2k
- Concentration falls linearly with time t at rate constant k; half-life is proportional to the initial concentration [A]₀.
- First order: ln[A] = ln[A]₀ − kt, t½ = 0.693 / k
- The log of concentration falls linearly; half-life depends only on the rate constant k.
- Second order: 1/[A] = 1/[A]₀ + kt, t½ = 1 / (k[A]₀)
- The reciprocal of concentration rises linearly; half-life is inversely proportional to [A]₀.
Questions
Why is the half-life of a first-order reaction independent of initial concentration?
In a first-order reaction the rate is proportional to the concentration, so the same fraction of reactant reacts in each equal interval of time. Halving takes the same time whether you start with 2 M or 0.2 M, giving t½ = 0.693/k. Choose First and change the initial concentration: the 't½ =' label on the graph does not change.
How can you find the order of a reaction from a graph?
Plot the concentration data in three ways against time. If [A] versus t is a straight line the reaction is zero order; if ln[A] versus t is straight it is first order; if 1/[A] versus t is straight it is second order. The slope gives k. In the simulation, switch Plot to Linearized for each order to see the straight line.
What are the units of the rate constant for zero, first and second order reactions?
The units of k depend on the order n as (mol L⁻¹)¹⁻ⁿ s⁻¹. For zero order, k has the units of rate, mol L⁻¹ s⁻¹; for first order, s⁻¹; for second order, L mol⁻¹ s⁻¹. So the units of k tell you the order of the reaction, and a rate constant quoted in s⁻¹ always belongs to a first-order reaction.
How does the half-life of a zero-order reaction change as the reaction proceeds?
For a zero-order reaction the rate stays constant, so the concentration falls by the same amount in each second. The half-life is [A]₀/2k, proportional to the concentration at the start of that half-life. As the concentration falls, each half-life is shorter than the one before. Select Zero and watch the dashed markers bunch together towards the end.
How much time does a first-order reaction take to reach 99.9% completion?
Using t = (2.303/k) log([A]₀/[A]), for 99.9% completion [A]₀/[A] = 1000, so t = (2.303 × 3)/k = 6.909/k. Since t½ = 0.693/k, this is about ten half-lives. Similarly, 99% completion takes about 6.6 half-lives. Each half-life removes half of what is left, so the reactant never quite reaches zero.
