Answer to Question #24463 in Physical Chemistry for arundhati roy

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 ?
1
Expert's answer
2021-03-18T02:42:53-0400

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


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Comments

Assignment Expert
17.03.21, 14:31

Dear Candace Haudenschild, thank you very much

Candace Haudenschild
15.03.21, 23:13

I 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 Expert
27.05.20, 20:50

Dear 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.

upasana
20.05.20, 12:49

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 ?

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