Blood alcohol content (BAC) is a measure of the amount of alcohol in the blood s

ID: 483144 • Letter: B

Question

Blood alcohol content (BAC) is a measure of the amount of alcohol in the blood stream of a person. In Pennsylvania, drivers of cars are required to have a BAC lower than 0.08 %. This corresponds to 80 mg of alcohol for every 100 mL of blood. The following data shows the BAC as a function of time for an individual. A typical adult would have this starting concentration of alcohol if about three drinks are consumed in an hour. No additional drinks are consumed after testing began. values, actual rates of alcohol metabolism depend on a variety of factors. What is the reaction order for metabolism of alcohol in the human body? Prepare three graphs (0th, 1st, and 2nd order) using the above data to support your answer. What is the rate constant for metabolism of alcohol using the above data? Impaired functioning has been shown for BAC of 0.02 %. How much time will it take for the BAC level in the person above to be less than 0.02%?

Explanation / Answer

Prepare the following three plots:

BAC% vs t (to check for 0 order reaction)

ln (BAC%) vs t (to check for 1st order reaction)

1/(BAC%) vs t (to check for 2nd order reaction)

Time (hours)

BAC%

ln (BAC%)

1/(BAC%)

0.12

-2.120

8.333

1.5

0.097

-2.333

10.309

2.5

0.083

-2.489

12.048

3.5

0.067

-2.703

14.925

5.0

0.045

-3.101

22.222

6.0

0.030

-3.506

33.333

Plot of BAC% vs t

Plot of ln (BAC%) vs t

Plot of 1/(BAC%) vs t

The plot is linear for the zero order reaction and hence the metabolism of alcohol is zero order (ans).

b) The rate constant can be obtained from the slope of the linear plot. From the linear equation, we note that slope = -0.015 and hence the rate constant has the value of -0.015. The unit for the rate constant is % h-1 and the rate constant is -0.015% h-1 (ans).

c) For this zero order reaction, we must have

-d(BAC%)/dt = k where k = -0.015% s-1

===> -d(BAC%) = k.dt

Integrate and obtain

(BAC%)0t = k.(t)0t

===> (BAC%)t – (BAC%)0 = k.t

We are given (BAC%)t = 0.02% and (BAC%)0 = 0.12. Put k = -0.015% h-1 and obtain t.

(0.02%) – (0.12%) = (-0.015% h-1).t

===> t = (-0.10%)/(0.015% h-1) = 6.647 h (ans).