Concentration-time relationships and half-lives

Define the differential equation first

For this lesson, k belongs to the species depletion equation −d[A]/dt = k[A], at constant volume and constant k in an irreversible first-order model. Its solution is [A]t = [A]₀ exp(−kt). The exponent kt is dimensionless, and k has units s⁻¹ when t is in seconds.

This definition avoids losing a stoichiometric factor. If a different k is defined through a normalized reaction rate for a coefficient other than 1, the depletion equation must be converted before using an integrated formula. Reversible or competing reactions may need different equations.

Equal time intervals remove equal fractions

An original hypothetical model starts at [A]₀ = 0.800 mol L⁻¹ and sets k = ln(2)/(20.0 s). Treat the chosen model parameters as exact for this exercise. At 20, 40 and 60 s, concentrations are 0.400, 0.200 and 0.100 mol L⁻¹. Each 20 s interval halves the amount remaining.

The curve approaches zero without becoming negative. The removed absolute concentration gets smaller in successive intervals even though the fraction removed is the same. Plotted markers and the calculated curve show this stipulated equation; they are not experimental measurements.

Only this first-order half-life ignores the starting concentration

Half-life t₁/₂ is the time for concentration to reach half its initial value in the specified model. For first-order depletion t₁/₂ = ln(2)/k. For zero-order depletion −d[A]/dt = k, it is [A]₀/(2k); for second-order depletion −d[A]/dt = k[A]², it is 1/(k[A]₀).

At a fixed k within each separate model, doubling [A]₀ doubles the zero-order half-life, leaves the first-order value unchanged and halves the second-order value. The k units differ between these laws. A zero-order linear decline applies only while its regime holds and cannot be extended into negative concentration.

Use the fraction remaining to find time

Rearrange the first-order solution to ln([A]t/[A]₀) = −kt. The concentration ratio is dimensionless. Removing 90% leaves a fraction 0.10, so t = ln(10)/k. With the exact hypothetical k = ln(2)/(20.0 s), this gives about 66.4 s.

A linearized plot can help assess a candidate model, but a few points that look straight do not uniquely prove a mechanism. Compare the assumptions and residuals against the actual observations, and retain uncertainty in a fitted k and predicted time.

Worked examples

Try it yourself

1. What concentration remains at 40 s in the plotted model?

   Show the answer and reasoning

   0.200 mol L⁻¹.

   Two 20 s half-lives reduce 0.800 to 0.400 and then to 0.200 mol L⁻¹.

2. Does each first-order half-life remove the same absolute amount?

   Show the answer and reasoning

   No.

   It removes the same fraction of what remains; the absolute decrease becomes smaller.

3. Why write ln([A]t/[A]₀) rather than a logarithm of a bare dimensional concentration?

   Show the answer and reasoning

   The ratio is dimensionless.

   A logarithm’s argument needs a dimensionless convention.

4. Within −d[A]/dt = k[A]² at fixed k, what happens to half-life when [A]₀ doubles?

   Show the answer and reasoning

   It halves.

   t₁/₂ = 1/(k[A]₀) for this defined second-order depletion model.

What to carry forward

• Specify whether k describes depletion or a normalized reaction rate.
• First-order half-life is independent of starting concentration within its model.
• Use dimensionless concentration ratios in logarithms.
• Do not extend a depletion model beyond its physical or tested range.