Question

A researcher is attempting to produce ethanol using an enzyme catalyzed batch reactor. The ethanol is produced from corn starch by first-order kinetics with a rate constant of 0.05 hr-1. Assuming the concentration of ethanol initially is 1 mg/L, what will be the concentration of ethanol (in mg/L) after 24 hours

Answer 1

The correct solution is "3.32 gm/L".

Step-by-step explanation:

Given:

Rate constant,

`K = 0.05 \ hr^(-1)`

Time,

`t = 24 \ hours`

Concentration of ethanol,

`C_o= 1 \ mg/L`

Now,

The concentration of ethanol after 24 hours will be:

⇒
`C_o=C* e^(-K* t)`

By putting the values, we get

`1=C* e^(-0.05* 24)`

`1=C* 0.30119`

`C= 3.32 \ gm/L`