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

Answer:

The correct solution is "3.32 gm/L".

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:

⇒ $C_o=C\times e^{-K\times t}$

By putting the values, we get

    $1=C\times e^{-0.05\times 24}$

    $1=C\times 0.30119$

    $C= 3.32 \ gm/L$