Home / Environmental Engineering / Water Treatment / Biodegradation Half-Life

Formula & Calculator

Biodegradation Half-Life

When a pollutant or biodegradable material breaks down in the environment, it often follows first-order decay. The half-life is the time it takes for half of the material to disappear. It equals the natural log of 2 divided by the decay rate constant k. For example, if k = 0.1 per day, the half-life is about 6.9 days. It is used to assess how long pollutants like pesticides or oil will persist in soil or water, and to set clean-up targets.

Environmental Pollution Control Environmental Chemistry
VORZEP · ENVIRONMENTAL FATE & DECAY TOOLS

Biodegradation Half‑Life Calculator

Solve first-order decay in any direction — half-life from a rate constant, rate constant from half-life, or the time required to reach a target remaining fraction. Includes a decay curve, persistence gauge and companion DT50/DT90 metrics.

t½ = ln(2) / k
Visual relationship — t½ = ln(2) ÷ k
t½ half-life 10 d
=
ln(2) constant ≈ 0.6931 0.6931
÷
k decay constant 0.0693 /d
DECAY CURVE — C(t) / C₀
Solve for
Decay Rate Constant (k) First-order rate constant. Larger k = faster decay = shorter half-life. Units are reciprocal time.
Rate constant must be greater than zero.
Half-Life (t½) Time required for the substance to decay to 50% of its initial concentration.
days
Half-life must be greater than zero.
Target Remaining (%) Percentage of the initial concentration still remaining at time t. 50% corresponds to one half-life.
%
Target remaining must be between 0.1% and 99.9%.
Time to Reach Target (t) Time required for the concentration to fall to the target remaining percentage. t = ln(1/f) / k.
days
HALF-LIFE (t½)
10 days
1.000 × 10¹ days
t½ = ln(2) / k
Step-by-step solution
First-order decay curve
C(t)/C₀ remaining Half-life markers (t½, 2t½, …) Target time
Persistence classification
Sensitivity
Companion Decay Metrics
How k, t½ and time-to-target relate

Exponential decay

Concentration falls as C(t) = C₀·e^(−kt). Each half-life removes exactly half of whatever remains — so the curve never quite reaches zero.

k and t½ are reciprocal

Doubling the rate constant halves the half-life. The product k · t½ is a universal constant, ln(2) ≈ 0.6931, regardless of the substance.

Targets scale with half-life

Time to 10% remaining (DT90) is ≈ 3.32 × t½. Time to 1% remaining is ≈ 6.64 × t½. Every target fraction maps to a fixed multiple of the half-life.

First-order assumption

Real biodegradation only follows first-order kinetics at low substrate concentrations. Above ~1–10 mg/L, microbial saturation slows the decay and the simple model over-predicts loss.

How to Use This Calculator

The first-order decay relation links three quantities: the rate constant k, the half-life t½, and the time t required to reach a target remaining fraction. Pick which one you want the tool to solve, then fill in the others.

1. Choose what to solve for

Pick t½ (half-life from k), k (rate constant from t½), or t (time to a target remaining percentage) — the tool rearranges the formula automatically.

2. Enter the rate constant or half-life

Type k and pick its unit (1/day, 1/hour, 1/year), or type t½ directly in days. Quick presets cover common laboratory and field values.

3. Set the target remaining percentage

Use 50% for the half-life itself, 10% for DT90 (a common regulatory endpoint), or 1% for near-complete removal.

4. Read the result and steps

The result card shows the solved value with engineering notation, and the step-by-step panel shows exactly how the number was reached.

5. Check the curve and persistence gauge

The decay curve plots C(t)/C₀ with half-life markers, and the gauge places your substance on the very-fast → very-persistent scale.

6. Switch to Advanced for more depth

Advanced mode adds a 1% sensitivity check plus DT50, DT90, time-to-1% and cross-unit k values, alongside the parameter-relationship breakdown.

What This Tool Offers

Built around the standard first-order decay relation t½ = ln(2)/k, the workhorse equation behind every environmental persistence and biodegradation estimate.

Solve any variable

Go from rate constant to half-life, from half-life to rate constant, or work out how long it takes to reach any target remaining fraction.

Flexible k units

Enter the rate constant in 1/day, 1/hour or 1/year — the tool normalises internally and reports the result in whichever unit you chose.

Live parameter diagram

A visual t½ = ln(2) ÷ k diagram updates with your inputs, alongside a mini decay curve so the relationship is always visible.

Persistence gauge

A five-band very-fast → very-persistent scale (log-scaled) that instantly places your substance in a recognised environmental-fate category.

Built-in sensitivity check

Advanced mode shows exactly how much a 1% error in k, t½ or the target fraction propagates into the solved value.

Companion metrics

Every result is cross-reported as DT50, DT90, time to 1% remaining, and k in alternative units — no manual conversions needed.

Reference Tables & Conversions

A quick t½ ↔ k converter, plus lookup tables for decay-rate units, persistence classification, and time-to-target multiples.

Quick t½ ↔ k converter

days
1/day
= 0.0693 /day · DT90 = 33.2 d

Half-Life ↔ Rate Constant Reference

For a given half-life, the corresponding first-order rate constant in several common time units. The product k · t½ is always ln(2) ≈ 0.6931.

t½ (days)k (1/day)k (1/hour)k (1/year)Class
0.51.3860.0578506Very fast
10.6930.0289253Very fast
50.1390.0057750.6Readily biodegradable
100.06930.0028925.3Readily biodegradable
300.02310.0009638.43Inherently biodegradable
600.01160.0004814.22Persistent
1800.003850.0001601.41Very persistent
3650.001900.00007910.693Very persistent

Persistence Classification Bands

Half-life bands commonly used in environmental fate assessments. Exact thresholds vary by regulation; these follow the widely-cited OECD / REACH style guidance.

Half-life (t½)ClassInterpretation
< 2 daysVery fastRapid mineralisation; unlikely to persist or bioaccumulate
2 – 10 daysReadily biodegradableMeets the OECD 301 “readily biodegradable” 10-day window
10 – 60 daysInherently biodegradableDegrades but not at the rate required for “readily” classification
60 – 180 daysPersistentMeets the P criterion in most regulatory frameworks
> 180 daysVery persistentMeets the vP criterion; long-range transport concern

Time to Reach Target vs Half-Life Multiples

The time required to reach any remaining fraction is a fixed multiple of the half-life. These are the values behind the DT50, DT90 and DT99 endpoints.

Remaining %Fraction fMultiplier of t½Endpoint name
50%0.501.000 × t½DT50 (= t½)
25%0.252.000 × t½—
10%0.103.322 × t½DT90
5%0.054.322 × t½—
1%0.016.644 × t½DT99
0.1%0.0019.966 × t½—

What Does t½ = ln(2) / k Measure?

First-order decay describes any process where the rate of loss is proportional to the amount remaining. Radioactive decay, many chemical hydrolyses, and the biodegradation of dilute organic contaminants all follow this pattern. The half-life t½ is the time required for the concentration to fall to exactly half its starting value, and it is set entirely by the rate constant k.

The First-Order Decay Formula

The defining relationship is:

t½ = ln(2) / k

The same relation rearranged to solve for the other two variables:

k = ln(2) / t½      t = ln(1/f) / k

where f is the target remaining fraction (0 < f ≤ 1). Because the underlying concentration follows C(t) = C₀·e^(−kt), the half-life is independent of the starting concentration — the same t½ applies whether you begin at 1,000 mg/L or 1 mg/L.

Decay Calculation Variables

SymbolMeaningTypical values
kFirst-order decay rate constant0.001 – 1.5 per day
t½Half-life0.5 – 365+ days
tTime to reach a target remaining fraction0 – 2,000+ days
fTarget remaining fraction (C/C₀)0.001 – 0.999

Why the Half-Life Is Independent of Concentration

Because the rate of loss is proportional to the amount present, the fraction lost per unit time is constant. That is what makes the half-life such a useful number: 50% of whatever is there disappears in t½, then 50% of the remainder disappears in the next t½, and so on. The concentration never reaches exactly zero, which is why regulatory endpoints are defined at 90% or 99% removal rather than at “complete” disappearance.

How the Rate Constant Affects Half-Life

The relationship is exactly reciprocal. A substance with k = 0.0693 per day has t½ = 10 days; double k to 0.1386 per day and the half-life drops to 5 days. This reciprocity makes the product k · t½ a universal constant, ln(2) ≈ 0.6931, no matter what is decaying — which is precisely why the equation is so widely used across radiochemistry, pharmacology and environmental science.

Decay Calculation Examples

Given: k = 0.0693 /day — solve for t½
t½ = ln(2) / 0.0693 = 0.6931 / 0.0693
t½ = 10.0 days
Given: t½ = 30 days — solve for k
k = ln(2) / 30 = 0.6931 / 30
k = 0.0231 /day
Given: t½ = 10 days, target = 10% remaining — solve for t
t = ln(1/0.10) / k = ln(10) / 0.0693 = 2.3026 / 0.0693
t = 33.2 days (DT90)

Frequently Asked Questions

It is the time required for half of the initial amount of a substance to disappear, under the assumption that the loss rate is proportional to the amount present. The number ln(2) ≈ 0.6931 appears because solving C₀/2 = C₀·e^(−k·t½) for t½ gives t½ = ln(2)/k.

Yes — DT50 (“disappearance time to 50%”) is the environmental-fate term for the half-life. DT90 and DT99 are the times to 90% and 99% disappearance, and they are simply fixed multiples of DT50: 3.322 × DT50 and 6.644 × DT50.

At low substrate concentrations, microbial enzymes are not saturated, so the degradation rate is roughly proportional to the contaminant concentration — exactly the condition for first-order kinetics. At high concentrations the enzymes saturate and the reaction becomes zero-order (constant rate), which the simple model does not capture.

Under true first-order kinetics, no — the half-life is a constant for a given substance and environment. In practice, however, higher concentrations can inhibit or saturate the microbial community, which slows the apparent decay and makes the measured half-life concentration-dependent. Always check that your data actually follow a straight line on a log-concentration plot before trusting a single t½.

Most regulatory frameworks (REACH, Stockholm Convention, UNEP) treat a half-life above 60 days in soil or water as “persistent” (P) and above 180 days as “very persistent” (vP). Below 10 days is generally considered readily biodegradable. The exact thresholds vary by compartment (soil, water, sediment) and by regulation.

Mathematically, no — an exponential decay curve approaches zero asymptotically. In practice, once the remaining concentration falls below the analytical detection limit or below the threshold at which microbes can scavenge it, the substance is treated as effectively gone. That is why regulatory endpoints use 90% or 99% removal rather than 100%.

Calculating...

Variables

Symbol Quantity Unit
t1/2 Biodegradation half-life days
k First-order biodegradation rate constant 1/day

What it means

The half‑life is a measure of the persistence of a substance in the environment. For biodegradation, it is the time needed for the concentration of a chemical to decrease by half through microbial action. It is calculated as the natural logarithm of 2 divided by the first‑order degradation rate constant k. A short half‑life indicates rapid degradation, while a long half‑life suggests persistence, which is important for assessing the risk of pollutants. This concept is used in modelling contaminant transport, designing bioremediation systems, and setting regulatory limits. Example: If a pollutant has a first‑order degradation rate of 0.01 day⁻¹, its half‑life is ln(2)/0.01 ≈ 69.3 days. This helps in estimating how long it will take for the pollutant to decrease to safe levels in the environment.

Worked example

Biodegradation Half‑Life – Two Examples

Real‑World
Scenario: A chemical degrades with rate constant k = 0.05 day⁻¹. Calculate the half‑life.
ParameterValue
k0.05 day⁻¹
1t₁/₂ = ln(2)/0.05 = 0.693/0.05 = 13.86 days
Result 13.9 days ✓ Readily biodegradable
Scenario: A persistent pollutant has k = 0.02 day⁻¹. Find half‑life.
ParameterValue
k0.02 day⁻¹
1t₁/₂ = 0.693/0.02 = 34.65 days
Result 34.7 days ✓ Slower degradation
Environmental insight: Half‑life indicates persistence – shorter half‑life means faster breakdown in the environment.

Common mistakes

  • Rate constant k: The first‑order degradation rate constant – in units of 1/time (e.g., day⁻¹, year⁻¹).
  • ln(2): Natural logarithm of 2 (≈0.693).
  • Half‑life: The time required for the concentration of a substance to decrease by half – in the same time units as k.
  • Assumptions: First‑order kinetics (exponential decay) – applicable to many biodegradation and photolysis processes.
  • Environmental conditions: k depends on temperature, pH, moisture, and microbial activity – use site‑specific values.

Applications

Biodegradation half‑life (t₁/₂) is the time required for the concentration of a compound to be reduced by half through biological degradation, calculated as t₁/₂ = ln(2) / k, where k is the first‑order degradation rate constant. This parameter is essential for assessing the persistence of organic pollutants in the environment, for designing bioremediation strategies, and for regulatory risk assessment. Environmental chemists and engineers use half‑life to predict the fate of pesticides, petroleum hydrocarbons, and other contaminants in soil, water, and sediments. Shorter half‑lives indicate faster natural attenuation, which influences remediation feasibility and cleanup timeframes. By understanding half‑life, professionals can evaluate the effectiveness of bioremediation and set realistic cleanup goals.

  • Risk assessment for organic contaminants in soil and water
  • Design of bioremediation systems (e.g., bioventing, bioaugmentation)
  • Evaluation of natural attenuation for site closure
  • Predictive modelling of pollutant persistence in the environment
  • Regulatory decision‑making on chemical safety and persistence

Suggest an improvement

Found a mistake? Have a better example? Let us know.
You may send up to 3 suggestions per 24 hours.

Page details

2,005

Views
  • Published July 11, 2026
  • Last updated July 21, 2026