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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

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Cricetus

Answer:

The correct solution is "3.32 gm/L".

Explanation:

Given:

Rate constant,

[tex]K = 0.05 \ hr^{-1}[/tex]

Time,

[tex]t = 24 \ hours[/tex]

Concentration of ethanol,

[tex]C_o= 1 \ mg/L[/tex]

Now,

The concentration of ethanol after 24 hours will be:

⇒ [tex]C_o=C\times e^{-K\times t}[/tex]

By putting the values, we get

    [tex]1=C\times e^{-0.05\times 24}[/tex]

    [tex]1=C\times 0.30119[/tex]

    [tex]C= 3.32 \ gm/L[/tex]

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