VARIATION OF THE CONCENTRATION OF A POISON WITH TIMEThe zeroth order kinetics of ethanol
decay
Ethanol breaks down into ethanal (CH 3 CHO) in the body. Ethanol (common alco-
hol) does not follow first order kinetics. Instead, the variation of blood alcohol con-
centration (symbolized ‘BAC’) with time follows zeroth order kineticsin which the
reaction rate is independent of the concentration of A:
[A] = [A]o–kt (23.2)giving the concentration/time curve shown in Figure 23.3(b).
Calculations using equation 23.2 are very straightforward. BAC is commonly
expressed in units of milligrams of ethanol per 100 cm^3 of blood (mg/100 cm^3 ). At
room temperature kis approximately 15 milligrams of ethanol per 100 cm^3 of blood
per hourand we will use this value as applying to all individuals irrespective of gen-
der and health. Remember though, that if we use kin these units, tmust be in units
ofhours.
23.5
425Example 23.1
A person ingests a morphine overdose. The half-life of morphine
in blood is about 3.0 h. If the blood plasma concentration 5.0 h
after ingestion was 6.0 mg dm–3of blood, estimate the peak
concentration of morphine at the time of ingestion.Answer
Rearranging Equation 23.2 gives[A] 0 = [A]t/ e–kt (1)The first-order rate constant kis calculated using the equation:k= 0.693/t (^1) / 2
Ast (^1) / 2 =3.0 h, k= 0.693/3 = 0.23 h–1(the units are ‘hours to the minus one’).
Substitutingkandt(= 5.0 h) into equation (1) yields
[A] 0 = 2.0 / e–(0.235.0)
We start by working out 0.23 5.0 = 1.15. The quantity e–1.15is evaluated using
the exfunction on a calculator, giving e–1.15= 0.317. Finally
[A] 0 = 6.0 / 0.317 = 18.9 mg dm–3
In conclusion, we have estimated the peak concentration of morphine in blood as
about 19 mg dm–3.
The drug paracetamol
decays by first-order kinet-
ics in the body with a half-
life of 4.0 h. A person’s
blood is found to contain 50
mg dm–3of paracetamol
8.0 h after ingestion of the
drug.
Estimate the peak blood
concentration of drug.
Exercise 23A