Question

Alcohol dehydrogenase (ADH) is a liver enzyme that catalyzes the oxidation of ethanol to acetaldehyde. The catalyzed reaction in the liver is zero-order with a rate constant of 0.17 molh−1. Halogenated hydrocarbons act as inhibitors to the ADH by binding strongly to the enzyme. The halogenated hydrocarbons can persist in the liver for days following exposure. Suppose that a worker who is exposed to halogenated hydrocarbons leaves work with 40 % of their ADH inhibited. The worker decides to have a few beers which are 341 mL and 5% alcohol by volume.

How many bottles of beer can the worker drink and be fully free of ethanol by the start of the next morning in 10 hours later?

Please show calculations.

Answer 1

The initial concentration of alcohol in liver after taking beer = 0.05 X 341 mL = 17.05 mL

Density of methanol = 0.79 g / mL

MAss of methanol = Density X volume = 0.79 X 17.05 = 13.47 g

Moles = Mass / mol wt = 13.47 / 32 = 0.42 moles / bottle

time = 10 hours

Rate constant = 0.17 mol / hour

it is zero order reaction so

[A] = -Kt + [A0]

for fully free from ethanol it means

[A] = 0 [after time t (10 hours)]

0= - 0.17 X 10 + [A]0

[A]0 = 1.7 moles

Number of bottles = Moles / moles in a bottle = 1.7 / 0.42 = 4 bottles