Integrated Rate Laws: Concentration, Time and Half-Life

Turning a rate law into concentration against time

A differential rate law, such as rate = k[A], links the rate to the concentration at one moment. As the reaction runs, [A] falls, so the rate falls too. Integrating the rate law over time, which is a calculus step you do not need to repeat in each problem, gives an equation for [A] at any time t in terms of the starting concentration [A]₀ and the rate constant k.

The equations below apply to reactions whose rate depends on the concentration of one reactant, A, at constant temperature. They also apply when other reactants are present in such large excess that their concentrations barely change. That situation is called pseudo-first-order (or pseudo-zero-order) behavior.

Three equations to keep side by side

Each integrated rate law can be rearranged into the form of a straight line, y = mx + b, with time on the x-axis. The half-life, t½, is the time it takes for [A] to fall to half of its current value. It comes from setting [A] = ½[A]₀ in each equation.

Only the first-order half-life is independent of concentration. That single fact explains why first-order processes, including radioactive decay, are so often described by their half-lives.

• Zero order: [A] = [A]₀ − kt. A plot of [A] against t is a straight line with slope −k. Half-life: t½ = [A]₀/(2k), so it gets shorter as the reaction proceeds.
• First order: ln[A] = ln[A]₀ − kt, or [A] = [A]₀e⁻ᵏᵗ. A plot of ln[A] against t is a straight line with slope −k. Half-life: t½ = ln 2 / k = 0.693/k, the same at every stage.
• Second order: 1/[A] = 1/[A]₀ + kt. A plot of 1/[A] against t is a straight line with slope +k. Half-life: t½ = 1/(k[A]₀), so each half-life is twice as long as the one before.

Worked example: cyclobutane, a first-order decomposition

OpenStax section 12.4 gives the first-order rate constant for the decomposition of cyclobutane, C₄H₈ → 2 C₂H₄, as 9.2 × 10⁻³ s⁻¹ at 500 °C. The half-life is t½ = 0.693 / (9.2 × 10⁻³ s⁻¹) = 75 s, to two significant figures to match k.

What fraction of the cyclobutane is left after 3.00 min? Convert the time to seconds first, because k is in s⁻¹: t = 180 s. The exponent is kt = (9.2 × 10⁻³ s⁻¹)(180 s) = 1.656, so [A]/[A]₀ = e⁻ᵏᵗ = e raised to the power −1.656 = 0.19, and 19 percent remains. If the starting concentration was 0.0800 mol/L, then [A] = 0.0800 × 0.19 = 0.015 mol/L. You did not need the starting concentration to find the fraction. That is true only of first-order reactions.

How long does it take to fall to 5.0 percent of the original amount? Rearrange to t = ln([A]₀/[A]) / k = ln(1/0.050) / (9.2 × 10⁻³ s⁻¹) = 2.996 / (9.2 × 10⁻³ s⁻¹) = 3.3 × 10² s, about 5.4 min. As a check, that is about 4.3 half-lives, and (½)⁴·³ ≈ 0.05.

Worked example: butadiene dimerization, a second-order reaction

Butadiene molecules combine in pairs, 2 C₄H₆ → C₈H₁₂, in a reaction that is second order in butadiene. OpenStax section 12.4 quotes k = 5.76 × 10⁻² L mol⁻¹ min⁻¹ under the conditions it describes. Start with [C₄H₆]₀ = 0.0500 mol/L.

The first half-life is t½ = 1/(k[A]₀) = 1 / (5.76 × 10⁻² L mol⁻¹ min⁻¹ × 0.0500 mol/L) = 347 min. The second half-life starts at 0.0250 mol/L and lasts 1/(5.76 × 10⁻² × 0.0250) = 694 min, twice as long. A half-life that grows as the concentration falls is the signature of second-order kinetics.

To find the concentration after 2.00 h, use minutes so that the units match k: t = 120 min. 1/[A] = 1/0.0500 + (5.76 × 10⁻²)(120) = 20.0 + 6.912 = 26.91 L/mol, so [A] = 0.0372 mol/L. The answer is less than the starting value and more than half of it, as it must be at a time shorter than the first half-life.

Counting half-lives in first-order problems

For a first-order reaction, the fraction left after n half-lives is (½)ⁿ. After one half-life, 50 percent remains; after two, 25 percent; after three, 12.5 percent. When the target concentration is a whole number of halvings away, you can count instead of using logarithms.

Worked example. OpenStax section 12.4 reports that hydrogen peroxide decomposes by first-order kinetics at 40 °C with a half-life of 2.16 × 10⁴ s, which is 6.00 h. The rate constant is k = 0.693 / (2.16 × 10⁴ s) = 3.21 × 10⁻⁵ s⁻¹. Going from 0.800 mol/L to 0.100 mol/L is three halvings (0.800 → 0.400 → 0.200 → 0.100), so it takes 3 × 6.00 h = 18.0 h.

For 0.800 mol/L to 0.150 mol/L the count is not a whole number, so use the equation: t = ln(0.800/0.150) / k = 1.674 / (3.21 × 10⁻⁵ s⁻¹) = 5.22 × 10⁴ s, or 14.5 h. That falls between two half-lives (12.0 h) and three (18.0 h), as expected.

Choosing the order from a concentration–time table

If you have concentrations measured at several times, test each linear form. Calculate [A], ln[A] and 1/[A] for every point and see which one changes by the same amount in each equal time step. With real, noisy data, plot all three and choose the straightest line, or compare the correlation of linear fits.

Worked example. The following data were constructed for practice. At t = 0, 50, 100, 150 and 200 s, [A] = 0.1000, 0.0667, 0.0500, 0.0400 and 0.0333 mol/L. The drops in [A] over each 50 s step are 0.0333, 0.0167, 0.0100 and 0.0067 mol/L, which are not constant, so the reaction is not zero order. The changes in ln[A] are −0.405, −0.288, −0.223 and −0.183, which are not constant either, so it is not first order. The values of 1/[A] are 10.0, 15.0, 20.0, 25.0 and 30.0 L/mol, rising by 5.0 L/mol every 50 s. The reaction is second order, and the slope gives k = 5.0 / 50 s = 0.100 L mol⁻¹ s⁻¹.

The half-lives confirm the order. [A] fell from 0.100 to 0.0500 mol/L in 100 s, and the second-order formula predicts the next halving, to 0.0250 mol/L, will take 1/(0.100 × 0.0500) = 200 s.

Where integrated rate law answers go wrong

Most mistakes come from mixing up equations or units. A quick check is that every concentration you calculate must be smaller than [A]₀ and larger than zero.

• Using a first-order equation for a reaction whose order you have not established.
• Mixing time units: k in s⁻¹ with t in minutes gives a value of kt that is off by a factor of 60.
• Writing ln([A]/[A]₀) = kt and losing the minus sign, which makes the concentration grow with time.
• Using log (base 10) where the equation needs ln (base e).
• Applying t½ = 0.693/k to a second-order reaction, whose half-life depends on concentration.
• Taking the slope of a 1/[A] plot as −k. For second order the slope is +k.