Question #24463

The half-life for a first-order reaction is 32 sec.What was the original concentration if,after 2.0 minuites, the concentration is 0.062 M ?

Expert's answer

The half life is the time required for the concentration to drop to one-half its original value. 2 minutes is 2*60= 120 sec.

If you have 120 sec and time of half life, you can calculate number of concentration decreasing to one-half of its previous value.

120 sec / 32 sec = 3.75 and it means that you need divide original concentration on 2 3.75 times to get 0.062 M, so

[C]/2^3.75 = 0.062 M

[C]/13.45 = 0.062 M

[C] = 0.062 M*13.75

[C] = 0.8525 M

Learn more about our help with Assignments: Physical Chemistry

## Comments

Assignment Expert17.03.21, 14:31Dear Candace Haudenschild, thank you very much

Candace Haudenschild15.03.21, 23:13I think that 2 x 60 is 120. So you would have less than 5 half lives. More like 3.75. So you could use the integrated rate law for a first order reaction.

Assignment Expert27.05.20, 20:50Dear Upasana, Questions in this section are answered for free. We can't fulfill them all and there is no guarantee of answering certain question but we are doing our best. And if answer is published it means it was attentively checked by experts. You can try it yourself by publishing your question. Although if you have serious assignment that requires large amount of work and hence cannot be done for free you can submit it as assignment and our experts will surely assist you.

upasana20.05.20, 12:49The half-life for a first-order reaction is 32 sec.What was the original concentration if,after 2.0 minuites, the concentration is 0.062 M ?

## Leave a comment