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 ?

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

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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

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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 ?

Answer to Question #24463 in Physical Chemistry for arundhati roy

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