Chick’s Law

\frac{N}{N _{0}} = e^{- k t}

First-order die-off of microorganisms under constant disinfectant concentration. N/N0 is the surviving fraction; k is the inactivation rate. A 1-log (90%) reduction needs kt = 2.303; 2-log (99%) needs 4.605; 3-log (99.9%) needs 6.91. Chick–Watson multiplies k by C^n.

$N$: Surviving organism count
$N_{0}$: Initial count
$k$ (1/min): Inactivation rate constant
$t$ (min): Contact time

Step-by-step solved example

Need 99% (2-log) inactivation. k = 0.46 min⁻¹. Find t.

Disinfection: N/N0 = e^{-kt} in a contact tank.

1. Log inactivation
N/N0 = 0.01 = e^{−kt} → kt = −ln(0.01) = 4.605.
2. Time
t = 4.605/0.46 = 10.0 min.
$t = 10.0 min$

Answer: t = 10.0 min

10 practice questions

0/10 correct

1.A 1-log reduction means N/N0 =

2.kt for 99% inactivation is

3.t = 3 min, k = 0.50 min⁻¹. N/N0 =

4.Chick–Watson extends Chick’s law by writing

5.EPA CT tables are a practical form of

6.If k doubles, time for the same log kill

7.3-log inactivation is a surviving fraction of

8.N0 = 10^6, k = 0.23 min⁻¹, t = 10 min. N ≈

9.Chick’s law assumes disinfectant concentration is

10.Percent kill when N/N0 = e^{−2.303} is