How Many Half-Lives Equal a 1/8 Concentration Reduction?

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To determine how long it takes for a reactant concentration to drop to 1/8 of its initial value in a first-order reaction with a rate constant of 6.70×10−3, the half-life is calculated to be approximately 103.43 units of time. It is established that three half-lives are required to reach this concentration, as each half-life reduces the concentration by half: after one half-life, 1/2 remains; after two, 1/4; and after three, 1/8. The discussion confirms that the correct answer is indeed three half-lives. This understanding is crucial for solving similar problems in first-order kinetics. The clarity on half-lives aids in accurately predicting concentration changes over time.
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Homework Statement


A certain first-order reaction has a rate constant of 6.70×10−3 . How long will it take for the reactant concentration to drop to 1/8 of its initial value?

Homework Equations



half life =0.693/k

The Attempt at a Solution



half life = 0.693/(6.70×10−3) = 103.43


but after that I'm confused how many half lives does 1/8 refer to?
I tried doing it by half life =4 but the answer came wrong.
any help would be appreaciated.
 
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Since this is a first order reaction, the function at hand is exponential, meaning the half life is constant. This means that after 1 half life, 1/2 of the reactants is left, after 2 half lives, 1/4 is left and so on. Can you do it now?
 
Werg22 said:
Since this is a first order reaction, the function at hand is exponential, meaning the half life is constant. This means that after 1 half life, 1/2 of the reactants is left, after 2 half lives, 1/4 is left and so on. Can you do it now?

so its 3 half lives .. right ?
 
Yes it is.
 
Yup;

half of a half of a half = 1/2 x 1/2 x 1/2 = 1/8

Best of health.

Steve
 
Werg22 said:
Yes it is.

Thanks a lot Werg!
 
Smith4046 said:
Yup;

half of a half of a half = 1/2 x 1/2 x 1/2 = 1/8

Best of health.

Steve

Thanks a lot Smith !
 
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